Q
stringlengths 4
3.96k
| A
stringlengths 1
3k
| Result
stringclasses 4
values |
|---|---|---|
Example 4.3.4 Let \( K = \mathbb{Q}\left( \sqrt{D}\right) \) with \( D \) a squarefree integer. Find an integral basis for \( {\mathcal{O}}_{K} \) .
|
Solution. An arbitrary element \( \alpha \) of \( K \) is of the form \( \alpha = {r}_{1} + {r}_{2}\sqrt{D} \) with \( {r}_{1},{r}_{2} \in \mathbb{Q} \) . Since \( \left\lbrack {K : \widehat{\mathbb{Q}}}\right\rbrack = 2,\alpha \) has only one conjugate: \( {r}_{1} - {r}_{2}\sqrt{D} \) . From Lemma 4.1.1 we know that if \( \alpha \) is an algebraic integer, then \( {\operatorname{Tr}}_{K}\left( \alpha \right) = 2{r}_{1} \) and\n\n\[ \n{\mathrm{N}}_{K}\left( \alpha \right) = \left( {{r}_{1} + {r}_{2}\sqrt{D}}\right) \left( {{r}_{1} - {r}_{2}\sqrt{D}}\right)\n\]\n\n\[ \n= {r}_{1}^{2} - D{r}_{2}^{2}\n\]\n\nare both integers. We note also that since \( \alpha \) satisfies the monic polynomial \( {x}^{2} - 2{r}_{1}x + {r}_{1}^{2} - D{r}_{2}^{2} \), if \( {\operatorname{Tr}}_{K}\left( \alpha \right) \) and \( {\mathrm{N}}_{K}\left( \alpha \right) \) are integers, then \( \alpha \) is an algebraic integer. If \( 2{r}_{1} \in \mathbb{Z} \) where \( {r}_{1} \in \mathbb{Q} \), then the denominator of \( {r}_{1} \) can be at most 2 . We also need \( {r}_{1}^{2} - D{r}_{2}^{2} \) to be an integer, so the denominator of \( {r}_{2} \) can be no more than 2 . Then let \( {r}_{1} = {g}_{1}/2,{r}_{2} = {g}_{2}/2 \), where \( {g}_{1},{g}_{2} \in \mathbb{Z} \) . The second condition amounts to\n\n\[ \n\frac{{g}_{1}^{2} - D{g}_{2}^{2}}{4} \in \mathbb{Z}\n\]\n\nwhich means that \( {g}_{1}^{2} - D{g}_{2}^{2} \equiv 0\left( {\;\operatorname{mod}\;4}\right) \), or \( {g}_{1}^{2} \equiv D{g}_{2}^{2}\left( {\;\operatorname{mod}\;4}\right) \) .\n\nWe will discuss two cases:\n\nCase 1. \( D \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) .\n\nIf \( D \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), and \( {g}_{1}^{2} \equiv D{g}_{2}^{2}\left( {\;\operatorname{mod}\;4}\right) \), then \( {g}_{1} \) and \( {g}_{2} \) are either both even or both odd. So if \( \alpha = {r}_{1} + {r}_{2}\sqrt{D} \) is an algebraic integer of \( \mathbb{Q}\left( \sqrt{D}\right) \), then either \( {r}_{1} \) and \( {r}_{2} \) are both integers, or they are both fractions with denominator 2.\n\nWe recall from Chapter 3 that if \( 4 \mid \left( {-D + 1}\right) \), then \( \left( {1 + \sqrt{D}}\right) /2 \) is an algebraic integer. This suggests that we use \( 1,\left( {1 + \sqrt{D}}\right) /2 \) as a basis; it is clear from the discussion above that this is in fact an integral basis.\n\nCase 2. \( D \equiv 2,3\left( {\;\operatorname{mod}\;4}\right) \) .\n\nIf \( {g}_{1}^{2} \equiv D{g}_{2}^{2}\left( {\;\operatorname{mod}\;4}\right) \), then both \( {g}_{1} \) and \( {g}_{2} \) must be even. Then a basis for \( {\mathcal{O}}_{K} \) is \( 1,\sqrt{D} \) ; again it is clear that this is an integral basis.
|
Yes
|
Let \( K = \mathbb{Q}\left( \alpha \right) \) where \( \alpha = {r}^{1/3}, r = a{b}^{2} \in \mathbb{Z} \) where \( {ab} \) is squarefree. If \( 3 \mid r \), assume that \( 3 \mid a,3 \nmid b \) . Find an integral basis for \( K \) .
|
The minimal polynomial of \( \alpha \) is \( f\left( x\right) = {x}^{3} - r \), and \( \alpha \) ’s conjugates are \( \alpha ,{\omega \alpha } \), and \( {\omega }^{2}\alpha \) where \( \omega \) is a primitive cube root of unity. By Exercise 4.3.3,\n\n\[ \n{d}_{K/\mathbb{Q}}\left( \alpha \right) = - \mathop{\prod }\limits_{{i = 1}}^{3}{f}^{\prime }\left( {\alpha }^{\left( i\right) }\right) = - {3}^{3}{r}^{2}.\n\]\n\nSo \( - {3}^{3}{r}^{2} = {m}^{2}{d}_{K} \) where \( m = \left\lbrack {{\mathcal{O}}_{K} : \mathbb{Z} + \mathbb{Z}\alpha + \mathbb{Z}{\alpha }^{2}}\right\rbrack \) . We note that \( f\left( x\right) \) is Eisensteinian for every prime divisor of \( a \) so by Example 4.3.1 if \( p \mid a \) , \( p \nmid m \) . Thus if \( 3 \mid a,{27}{a}^{2} \mid {d}_{K} \), and if \( 3 \nmid a \), then \( 3{a}^{2} \mid {d}_{K} \) .\n\nWe now consider \( \beta = {\alpha }^{2}/b \), which is a root of the polynomial \( {x}^{3} - {a}^{2}b \) . This polynomial is Eisensteinian for any prime which divides \( b \) . Therefore \( {b}^{2} \mid {d}_{K} \) . We conclude that \( {d}_{K} = - {3}^{n}{\left( ab\right) }^{2} \) where \( n = 3 \) if \( 3 \mid r \) and \( n = 1 \) or 3 otherwise. We will consider three cases: \( r ≢ 1,8\left( {\;\operatorname{mod}\;9}\right), r \equiv 1\left( {\;\operatorname{mod}\;9}\right) \) and \( r \equiv 8\left( {\;\operatorname{mod}\;9}\right) \) .\n\nCase 1. If \( r ≢ 1,8\left( {\;\operatorname{mod}\;9}\right) \), then \( {r}^{3} ≢ r\left( {\;\operatorname{mod}\;9}\right) \).\n\nThen the polynomial \( {\left( x + r\right) }^{3} - r \) is Eisensteinian with respect to the prime 3. A root of this polynomial is \( \alpha - r \) and \( {d}_{K/\mathbb{Q}}\left( {\alpha - r}\right) = {d}_{K/\mathbb{Q}}\left( \alpha \right) = \) \( - {27}{r}^{2} \) . This implies that \( 3 \nmid m \) and so \( m = b \) .\n\nWe can choose as our integral basis \( 1,\alpha ,{\alpha }^{2}/b \), all of which are algebraic integers. We verify that this is an integral basis by checking the index of \( \mathbb{Z} + \mathbb{Z}\alpha + \mathbb{Z}{\alpha }^{2} \) in \( \mathbb{Z} + \mathbb{Z}\alpha + \mathbb{Z}{\alpha }^{2}/b \), which is clearly \( b \) . Thus\n\n\[ \n{\mathcal{O}}_{K} = \mathbb{Z} + \mathbb{Z}\alpha + \mathbb{Z}\frac{{\alpha }^{2}}{b}\n\]
|
No
|
Theorem 5.1.2 Let \( R \) be a commutative ring with identity. Then:\n\n(a) \( \mathfrak{m} \) is a maximal ideal if and only if \( R/\mathfrak{m} \) is a field.
|
Proof. (a) By the correspondence between ideals of \( R \) containing \( \mathfrak{m} \) and ideals of \( R/\mathfrak{m}, R/\mathfrak{m} \) has a nontrivial ideal if and only if there is an ideal \( \mathfrak{a} \) of \( R \) strictly between \( \mathfrak{m} \) and \( R \) . Thus,\n\n\( \mathfrak{m} \) is maximal,\n\n\( \Leftrightarrow R/\mathfrak{m} \) has no nontrivial ideals,\n\n\( \Leftrightarrow R/\mathfrak{m} \) is a field.
|
Yes
|
Theorem 5.1.6 For \( \alpha \in \mathbb{C} \), the following are equivalent:\n\n(1) \( \alpha \) is integral over \( {\mathcal{O}}_{K} \) ;\n\n(2) \( {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \) is a finitely generated \( {\mathcal{O}}_{K} \) -module;\n\n(3) There is a finitely generated \( {\mathcal{O}}_{K} \) -module \( M \subseteq \mathbb{C} \) such that \( {\alpha M} \subseteq M \) .
|
Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) Let \( \alpha \in \mathbb{C} \) be integral over \( {\mathcal{O}}_{K} \) . Say \( \alpha \) satisfies a monic polynomial of degree \( n \) over \( {\mathcal{O}}_{K} \) . Then\n\n\[{\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack = {\mathcal{O}}_{K} + {\mathcal{O}}_{K}\alpha + {\mathcal{O}}_{K}{\alpha }^{2} + \cdots + {\mathcal{O}}_{K}{\alpha }^{n - 1}\]\n\nand so is a finitely generated \( {\mathcal{O}}_{K} \) -module.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) Certainly, \( \alpha {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \subseteq {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \), so if \( {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \) is a finitely generated \( {\mathcal{O}}_{K} \) -module, then (3) is satisfied with \( M = {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \) .\n\n\( \left( 3\right) \Rightarrow \left( 1\right) \) Let \( {u}_{1},{u}_{2},\ldots ,{u}_{n} \) generate \( M \) as an \( {\mathcal{O}}_{K} \) -module. Then \( \alpha {u}_{i} \in \) \( M \) for all \( i = 1,2,\ldots, n \) since \( {\alpha M} \subseteq M \) . Let\n\n\[ \alpha {u}_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{u}_{j} \]\n\nfor \( i \in \{ 1,2,\ldots, n\} ,{a}_{ij} \in {\mathcal{O}}_{K} \) . Let \( A = \left( {a}_{ij}\right), B = \alpha {I}_{n} - A = \left( {b}_{ij}\right) \) . Then\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}{b}_{ij}{u}_{j} = \mathop{\sum }\limits_{{j = 1}}^{n}\left( {\alpha {\delta }_{ij} - {a}_{ij}}\right) {u}_{ij} \]\n\n\[ = \alpha {u}_{i} - \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{u}_{j} \]\n\n\[ = 0\;\text{ for all }i\text{. } \]\n\nThus, \( B{\left( {u}_{1},{u}_{2},\ldots ,{u}_{n}\right) }^{T} = {\left( 0,0,\ldots ,0\right) }^{T} \) . But\n\n\[ {\left( 0,0,\ldots ,0\right) }^{T} \neq {\left( {u}_{1},{u}_{2},\ldots ,{u}_{n}\right) }^{T} \in {\mathbb{C}}^{n}. \]\n\nThus, \( \det \left( B\right) = 0 \) . But, the determinant of \( B \) is a monic polynomial in \( {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \), so \( \alpha \) is integral over \( {\mathcal{O}}_{K} \) .
|
Yes
|
Theorem 5.1.7 \( {\mathcal{O}}_{K} \) is integrally closed.
|
Proof. If \( \alpha \in K \) is integral over \( {\mathcal{O}}_{K} \), then let\n\n\[ M = {\mathcal{O}}_{K}{u}_{1} + \cdots + {\mathcal{O}}_{K}{u}_{n},\;{\alpha M} \subseteq M.\]\n\nLet \( {\mathcal{O}}_{K} = \mathbb{Z}{v}_{1} + \cdots + \mathbb{Z}{v}_{m} \), where \( \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) is a basis for \( K \) over \( \mathbb{Q} \) . Then \( M = \mathop{\sum }\limits_{{i = 1}}^{m}\mathop{\sum }\limits_{{j = 1}}^{n}\mathbb{Z}{v}_{i}{u}_{j} \) is a finitely generated \( \mathbb{Z} \) -module with \( {\alpha M} \subseteq M \), so \( \alpha \) is integral over \( \mathbb{Z} \) . By definition, \( \alpha \in {\mathcal{O}}_{K} \) .
|
Yes
|
Theorem 5.2.3 For any commutative ring \( R \) with identity, the following are equivalent:\n\n(1) \( R \) is Noetherian;\n\n(2) every nonempty set of ideals contains a maximal element; and\n\n(3) every ideal of \( R \) is finitely generated.
|
Proof. (1) \( \Rightarrow \) (2) Suppose that \( S \) is a nonempty set of ideals of \( R \) that does not contain a maximal element. Let \( {\mathfrak{a}}_{1} \in S.{\mathfrak{a}}_{1} \) is not maximal in \( S \), so there is an \( {\mathfrak{a}}_{2} \in S \) with \( {\mathfrak{a}}_{1} \varsubsetneq {\mathfrak{a}}_{2} \) . \( {\mathfrak{a}}_{2} \) is not a maximal element of \( S \) , so there exists an \( {\mathfrak{a}}_{3} \in S \) with \( {\mathfrak{a}}_{1} \varsubsetneq {\mathfrak{a}}_{2} \varsubsetneq {\mathfrak{a}}_{3} \) . Continuing in this way, we find an infinite ascending chain of ideals of \( R \) . This contradicts \( R \) being Noetherian, so every nonempty set of ideals contains a maximal element.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) Let \( \mathfrak{b} \) be an ideal of \( R \) . Let \( A \) be the set of ideals contained in \( \mathfrak{b} \) which are finitely generated. \( A \) is nonempty, since \( \left( 0\right) \in A \) . Thus, \( A \) has a maximal element, say \( \mathfrak{a} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) . If \( \mathfrak{a} \neq \mathfrak{b} \), then \( \exists x \in \mathfrak{b} \smallsetminus \mathfrak{a} \) . But then \( \mathfrak{a} + \left( x\right) = \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}, x}\right) \) is a larger finitely generated ideal contained in \( \mathfrak{b} \), contradicting the maximality of \( \mathfrak{b} \) . Thus, \( \mathfrak{b} = \mathfrak{a} \), so \( \mathfrak{b} \) is finitely generated. Thus, every ideal of \( R \) is finitely generated.\n\n\( \left( 3\right) \Rightarrow \left( 1\right) \) Let \( {\mathfrak{a}}_{1} \subseteq {\mathfrak{a}}_{2} \subseteq {\mathfrak{a}}_{3} \subseteq \cdots \) be an ascending chain of ideals of \( R \) . Then \( \mathfrak{a} = \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{\mathfrak{a}}_{i} \) is also an ideal of \( R \), and so is finitely generated, say \( \mathfrak{a} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) . Then \( {x}_{1} \in {\mathfrak{a}}_{{i}_{1}},\ldots ,{x}_{n} \in {\mathfrak{a}}_{{i}_{n}} \) . Let \( m = \max \left( {{i}_{1},\ldots ,{i}_{n}}\right) \) . Then \( \mathfrak{a} \subseteq {\mathfrak{a}}_{m} \), so \( \mathfrak{a} = {\mathfrak{a}}_{m} \) . Thus, \( {\mathfrak{a}}_{m} = {\mathfrak{a}}_{m + 1} = \cdots \), and the chain does terminate. Thus, \( R \) is Noetherian.
|
Yes
|
Lemma 5.3.3 Any proper ideal of \( {\mathcal{O}}_{K} \) contains a product of nonzero prime ideals.
|
Proof. Let \( S \) be the set of all proper ideals of \( {\mathcal{O}}_{K} \) that do not contain a product of prime ideals. We need to show that \( S \) is empty. If not, then since \( {\mathcal{O}}_{K} \) is Noetherian, \( S \) has a maximal element, say \( \mathfrak{a} \) . Then, \( \mathfrak{a} \) is not prime since \( \mathfrak{a} \in S \), so there exist \( a, b \in {\mathcal{O}}_{K} \), with \( {ab} \in \mathfrak{a}, a \notin \mathfrak{a}, b \notin \mathfrak{a} \) . Then, \( \left( {\mathfrak{a}, a}\right) \supsetneq \mathfrak{a},\left( {\mathfrak{a}, b}\right) \supsetneq \mathfrak{a} \) . Thus, \( \left( {\mathfrak{a}, a}\right) \notin S,\left( {\mathfrak{a}, b}\right) \notin S \), by the maximality of \( \mathfrak{a} \) . Thus, \( \left( {\mathfrak{a}, a}\right) \supseteq {\wp }_{1}\cdots {\wp }_{r} \) and \( \left( {\mathfrak{a}, b}\right) \supseteq {\wp }_{1}^{\prime }\cdots {\wp }_{s}^{\prime } \), with the \( {\wp }_{i} \) and \( {\wp }_{j}^{\prime } \) nonzero prime ideals. But \( {ab} \in \mathfrak{a} \), so \( \left( {\mathfrak{a},{ab}}\right) = \mathfrak{a} \) . Thus, \( \mathfrak{a} = \left( {\mathfrak{a},{ab}}\right) \supseteq \left( {\mathfrak{a}, a}\right) \left( {\mathfrak{a}, b}\right) \supseteq {\wp }_{1}\cdots {\wp }_{r}{\wp }_{1}^{\prime }\cdots {\wp }_{s}^{\prime } \) . Therefore \( \mathfrak{a} \) contains a product of prime ideals. This contradicts \( \mathfrak{a} \) being in \( S \), so \( S \) must actually be empty. Thus, any proper ideal of \( {\mathcal{O}}_{K} \) contains a product of nonzero prime ideals.
|
Yes
|
Lemma 5.3.4 Let \( \wp \) be a prime ideal of \( {\mathcal{O}}_{K} \). There exists \( z \in K, z \notin {\mathcal{O}}_{K} \), such that \( {z}_{\wp } \subseteq {\mathcal{O}}_{K} \).
|
Proof. Take \( x \in \wp \). From the previous lemma, \( \left( x\right) = x{\mathcal{O}}_{K} \) contains a product of prime ideals. Let \( r \) be the least integer such that \( \left( x\right) \) contains a product of \( r \) prime ideals, and say \( \left( x\right) \supseteq {\wp }_{1}\cdots {\wp }_{r} \), with the \( {\wp }_{i} \) nonzero prime ideals.\n\nSince \( \wp \supseteq {\wp }_{1}\cdots {\wp }_{r},\wp \supseteq {\wp }_{i} \), for some \( i \), from Theorem 5.1.2 (d). Without loss of generality, we can assume that \( i = 1 \), so \( \wp \supseteq {\wp }_{1} \). But \( {\wp }_{1} \) is a nonzero prime ideal of \( {\mathcal{O}}_{K} \), and so is maximal. Thus, \( \wp = {\wp }_{1} \).\n\nNow, \( {\wp }_{2}\cdots {\wp }_{r} \nsubseteq \left( x\right) \), since \( r \) was chosen to be minimal. Choose an element \( b \in {\wp }_{2}\cdots {\wp }_{r}, b \notin x{\mathcal{O}}_{K} \). Then\n\n\[ b{x}^{-1}\wp = b{x}^{-1}{\wp }_{1}\; \subseteq \;\left( {{\wp }_{2}\cdots {\wp }_{r}}\right) \left( {{x}^{-1}{\wp }_{1}}\right) \]\n\n\[ = {x}^{-1}\left( {{\wp }_{1}\cdots {\wp }_{r}}\right) \]\n\n\[ \subseteq \;{x}^{-1}x{\mathcal{O}}_{K} \]\n\n\[ = {\mathcal{O}}_{K} \]\n\nPut \( z = b{x}^{-1} \). Then \( z\wp \subseteq {\mathcal{O}}_{K} \). Now, if \( z \) were in \( {\mathcal{O}}_{K} \), we would have \( b{x}^{-1} \in {\mathcal{O}}_{K} \), and so \( b \in x{\mathcal{O}}_{K} \). But this is not the case, so \( z \notin {\mathcal{O}}_{K} \). Thus, we have found \( z \in K, z \notin {\mathcal{O}}_{K} \) with \( z\wp \subseteq {\mathcal{O}}_{K} \).
|
Yes
|
Theorem 5.3.5 Let \( \wp \) be a prime ideal of \( {\mathcal{O}}_{K} \) . Then \( {\wp }^{-1} \) is a fractional ideal and \( \wp {\wp }^{-1} = {\mathcal{O}}_{K} \) .
|
Proof. It is easily seen that \( {\wp }^{-1} \) is an \( {\mathcal{O}}_{K} \) -module. Now, \( \wp \cap \mathbb{Z} \neq \left( 0\right) \), from Exercise 4.4.1, so let \( n \in \wp \cap \mathbb{Z}, n \neq 0 \) . Then, \( n{\wp }^{-1} \subseteq \wp {\wp }^{-1} \subseteq {\mathcal{O}}_{K} \), by definition. Thus, \( {\wp }^{-1} \) is a fractional ideal.\n\nIt remains to show that \( \wp {\wp }^{-1} = {\mathcal{O}}_{K} \) . Since \( 1 \in {\wp }^{-1},\wp \subseteq \wp {\wp }^{-1} \subseteq {\mathcal{O}}_{K} \) . \( \wp {\wp }^{-1} \) is an ideal of \( {\mathcal{O}}_{K} \), since it is an \( {\mathcal{O}}_{K} \) -module contained in \( {\mathcal{O}}_{K} \) . But \( \wp \) is maximal, so either \( \wp {\wp }^{-1} = {\mathcal{O}}_{K} \), in which case we are done, or \( \wp {\wp }^{-1} = \wp \) .\n\nSuppose that \( \wp {\wp }^{-1} = \wp \) . Then \( x\wp \subseteq \wp \forall x \in {\wp }^{-1} \) . Since \( \wp \) is a finitely generated \( \mathbb{Z} \) -module (from Exercise 4.4.2), \( x \in {\mathcal{O}}_{K} \) for all \( x \in {\wp }^{-1} \), by Theorem 5.1.6. Thus, \( {\wp }^{-1} \subseteq {\mathcal{O}}_{K} \) . But \( 1 \in {\wp }^{-1} \), so \( {\wp }^{-1} = {\mathcal{O}}_{K} \) . From the comments above, and by the previous lemma, we know this is not true, so \( \wp {\wp }^{-1} \neq \wp \) . Thus, \( \wp {\wp }^{-1} = {\mathcal{O}}_{K} \) .
|
Yes
|
Theorem 5.3.6 Any ideal of \( {\mathcal{O}}_{K} \) can be written as a product of prime ideals uniquely.
|
## Proof.\n\nExistence. Let \( S \) be the set of ideals of \( {\mathcal{O}}_{K} \) that cannot be written as a product of prime ideals. If \( S \) is nonempty, then \( S \) has a maximal element, since \( {\mathcal{O}}_{K} \) is Noetherian. Let \( \mathfrak{a} \) be a maximal element of \( S \) . Then \( \mathfrak{a} \subseteq \wp \) for some maximal ideal \( \wp \), since \( {\mathcal{O}}_{K} \) is Noetherian. Recall that every maximal ideal of \( {\mathcal{O}}_{K} \) is prime. Since \( \mathfrak{a} \in S,\mathfrak{a} \neq \wp \) and therefore \( \mathfrak{a} \) is not prime.\n\nConsider \( {\wp }^{-1}\mathfrak{a}.{\wp }^{-1}\mathfrak{a} \subset {\wp }^{-1}\wp = {\mathcal{O}}_{K} \) . Since \( \mathfrak{a} \varsubsetneq \wp \) ,\n\n\[ \n{\wp }^{-1}\mathfrak{a} \varsubsetneq {\wp }^{-1}\wp = {\mathcal{O}}_{K} \n\] \nsince for any \( x \in \wp \smallsetminus \mathfrak{a} \) ,\n\n\[ \n{\wp }^{-1}x \subseteq {\wp }^{-1}\mathfrak{a}\; \Rightarrow \;x \in \wp {\wp }^{-1}\mathfrak{a} = {\mathcal{O}}_{K}\mathfrak{a} = \mathfrak{a}, \n\] \n\nwhich is not true. Thus, \( {\wp }^{-1}\mathfrak{a} \) is a proper ideal of \( {\mathcal{O}}_{K} \), and contains \( \mathfrak{a} \) properly since \( {\wp }^{-1} \) contains \( {\mathcal{O}}_{K} \) properly. Thus, \( {\wp }^{-1}\mathfrak{a} \notin S \), since \( \mathfrak{a} \) is a maximal element of \( S \) . Thus, \( {\wp }^{-1}\mathfrak{a} = {\wp }_{1}\cdots {\wp }_{r} \), for some prime ideals \( {\wp }_{i} \) . Then, \( \wp {\wp }^{-1}\mathfrak{a} = \wp {\wp }_{1}\cdots {\wp }_{r} \), so \( \mathfrak{a} = \wp {\wp }^{-1}\mathfrak{a} = \wp {\wp }_{1}\cdots {\wp }_{r} \) . But then \( \mathfrak{a} \notin S \), a contradiction.\n\nThus, \( S \) is empty, so every ideal of \( {\mathcal{O}}_{K} \) can be written as a product of prime ideals.\n\nUniqueness. Suppose that \( \mathfrak{a} = {\wp }_{1}\cdots {\wp }_{r} = {\wp }_{1}^{\prime }\cdots {\wp }_{s}^{\prime } \) are two factorizations of \( \mathfrak{a} \) as a product of prime ideals.\n\nThen, \( {\wp }_{1}^{\prime } \supseteq {\wp }_{1}^{\prime }\cdots {\wp }_{s}^{\prime } = {\wp }_{1}\cdots {\wp }_{r} \), so \( {\wp }_{1}^{\prime } \supseteq {\wp }_{i} \), for some \( i \), say \( {\wp }_{1}^{\prime } \supseteq {\wp }_{1} \) . But \( {\wp }_{1} \) is maximal, so \( {\wp }_{1}^{\prime } = {\wp }_{1} \) . Thus, multiplying both sides by \( {\left( {\wp }_{1}^{\prime }\right) }^{-1} \) and cancelling using \( {\left( {\wp }_{1}^{\prime }\right) }^{-1}{\wp }_{1}^{\prime } = {\mathcal{O}}_{K} \), we obtain\n\n\[ \n{\wp }_{2}^{\prime }\cdots {\wp }_{s}^{\prime } = {\wp }_{2}\cdots {\wp }_{r} \n\] \n\nThus, continuing in this way, we see that \( r = s \) and the primes are unique up to reordering.
|
Yes
|
Theorem 5.3.13 (Chinese Remainder Theorem) (a) Let \( \mathfrak{a},\mathfrak{b} \) be ideals so that \( \gcd \left( {\mathfrak{a},\mathfrak{b}}\right) = 1 \), i.e., \( \mathfrak{a} + \mathfrak{b} = {\mathcal{O}}_{K} \) . Given \( a, b \in {\mathcal{O}}_{K} \), we can solve\n\n\[ x \equiv a\;\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \]\n\n\[ x \equiv b\;\left( {\;\operatorname{mod}\;\mathfrak{b}}\right) . \]
|
Proof. (a) Since \( \mathfrak{a} + \mathfrak{b} = {\mathcal{O}}_{K},\exists {x}_{1} \in \mathfrak{a},{x}_{2} \in \mathfrak{b} \) with \( {x}_{1} + {x}_{2} = 1 \) . Let \( x = b{x}_{1} + a{x}_{2} \equiv a{x}_{2}\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \) . But, \( {x}_{2} = 1 - {x}_{1} \equiv 1\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \) . Thus, we have found an \( x \) such that \( x \equiv a\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \) . Similarly, \( x \equiv b\left( {\;\operatorname{mod}\;\mathfrak{b}}\right) \) .
|
Yes
|
Theorem 5.3.16 (a) If \( \mathfrak{a} = \mathop{\prod }\limits_{{i = 1}}^{r}{\wp }_{i}^{{e}_{i}} \), then\n\n\[ N\left( \mathfrak{a}\right) = \mathop{\prod }\limits_{{i = 1}}^{r}N\left( {\wp }_{i}^{{e}_{i}}\right) . \]
|
Proof. (a) Consider the map\n\n\[ \phi : {\mathcal{O}}_{K} \rightarrow {\bigoplus }_{i = 1}^{r}\left( {{\mathcal{O}}_{K}/{\wp }_{i}^{{e}_{i}}}\right) \]\n\n\[ x \rightarrow \left( {{x}_{1},\ldots ,{x}_{r}}\right) \]\n\nwhere \( {x}_{i} \equiv x\left( {\;\operatorname{mod}\;{\wp }_{i}^{{e}_{i}}}\right) \). \n\nThe function \( \phi \) is surjective by the Chinese Remainder Theorem, and \( \phi \) is a homomorphism since each of the \( r \) components \( x \rightarrow {x}_{i}\left( \right. \) mod \( \left. {\wp }_{i}^{{e}_{i}}\right) \) is a homomorphism.\n\nNext, we show by induction that \( \mathop{\bigcap }\limits_{{i = 1}}^{r}{\wp }_{i}^{{e}_{i}} = \mathop{\prod }\limits_{{i = 1}}^{r}{\wp }_{i}^{{e}_{i}} \) . The base case \( r = 1 \) is trivial. Suppose \( r > 1 \), and that the result is true for numbers smaller than \( r \). \n\n\[ \mathop{\bigcap }\limits_{{i = 1}}^{r}{\wp }_{i}^{{e}_{i}} = \operatorname{lcm}\left( {\mathop{\bigcap }\limits_{{i = 1}}^{{r - 1}}{\wp }_{i}^{{e}_{i}},{\wp }_{r}^{{e}_{r}}}\right) \]\n\n\[ = \operatorname{lcm}\left( {\mathop{\prod }\limits_{{i = 1}}^{{r - 1}}{\wp }_{i}^{{e}_{i}},{\wp }_{r}^{{e}_{r}}}\right) \]\n\n\[ = \mathop{\prod }\limits_{{i = 1}}^{r}{\wp }_{i}^{{e}_{i}} \]\n\nThus, \( \ker \left( \phi \right) = \mathop{\bigcap }\limits_{{i = 1}}^{r}{\wp }_{i}^{{e}_{i}} = \mathop{\prod }\limits_{{i = 1}}^{r}{\wp }_{i}^{{e}_{i}} \), which implies that\n\n\[ {\mathcal{O}}_{K}/\mathfrak{a} \simeq \oplus \left( {{\mathcal{O}}_{K}/{\wp }_{i}^{{e}_{i}}}\right) \]\n\nHence, \( N\left( \mathfrak{a}\right) = \mathop{\prod }\limits_{{i = 1}}^{r}N\left( {\wp }_{i}^{{e}_{i}}\right) \).
|
Yes
|
Theorem 5.4.3 Let \( \mathcal{D} \) be the different of an algebraic number field \( K \) . Then \( N\left( \mathcal{D}\right) = \left| {d}_{K}\right| \) .
|
Proof. For some \( m > 0, m{\mathcal{D}}^{-1} \) is an ideal of \( {\mathcal{O}}_{K} \) . Now,\n\n\[ m{\mathcal{D}}^{-1} = \mathbb{Z}m{\omega }_{1}^{ * } + \cdots + \mathbb{Z}m{\omega }_{n}^{ * } \]\n\nLet\n\n\[ m{\omega }_{i}^{ * } = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{\omega }_{j} \]\n\nso\n\n\[ {\omega }_{i}^{ * } = \mathop{\sum }\limits_{{j = 1}}^{n}\frac{{a}_{ij}}{m}{\omega }_{j} \]\n\nand\n\n\[ {\omega }_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}{b}_{ij}{\omega }_{j}^{ * } \]\n\nThus, \( \left( {b}_{ij}\right) \) is the matrix inverse of \( \left( {{a}_{ij}/m}\right) \) . But\n\n\[ \operatorname{Tr}\left( {{\omega }_{i}{\omega }_{j}}\right) = \operatorname{Tr}\left( {\mathop{\sum }\limits_{{k = 1}}^{n}{b}_{ik}{\omega }_{k}^{ * }{\omega }_{j}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{k = 1}}^{n}{b}_{ik}\operatorname{Tr}\left( {{\omega }_{k}^{ * }{\omega }_{j}}\right) \]\n\n\[ = {b}_{ij}\text{.} \]\n\nThus, \( \det \left( {b}_{ij}\right) = {d}_{K} \) .\n\nHowever, by Exercise 4.2.8, since \( m{\mathcal{D}}^{-1} \) is an ideal of \( {\mathcal{O}}_{K} \) with integral basis \( m{\omega }_{1}^{ * },\ldots, m{\omega }_{n}^{ * } \), we know that\n\n\[ {d}_{K/\mathbb{Q}}\left( {m{\omega }_{1}^{ * },\ldots, m{\omega }_{n}^{ * }}\right) = N{\left( m{\mathcal{D}}^{-1}\right) }^{2}{d}_{K} \]\n\nand from Exercise 4.2.7 we have\n\n\[ {d}_{K/\mathbb{Q}}\left( {m{\omega }_{1}^{ * },\ldots, m{\omega }_{n}^{ * }}\right) = {\left( \det \left( {a}_{ij}\right) \right) }^{2}{d}_{K}, \]\n\nwhich shows that\n\n\[ \left| {\det \left( {a}_{ij}\right) }\right| = N\left( {m{\mathcal{D}}^{-1}}\right) = {m}^{n}N\left( {\mathcal{D}}^{-1}\right) , \]\n\nand thus\n\n\[ \left| {\det \left( \frac{{a}_{ij}}{m}\right) }\right| = N\left( {\mathcal{D}}^{-1}\right) = N{\left( \mathcal{D}\right) }^{-1}. \]\n\nHence, \( \left| {d}_{K}\right| = \left| {\det \left( {b}_{ij}\right) }\right| = {\left| \det \left( {a}_{ij}/m\right) \right| }^{-1} = N\left( \mathcal{D}\right) \).
|
Yes
|
Theorem 5.4.4 Let \( p \in \mathbb{Z} \) be prime, \( \wp \subseteq {\mathcal{O}}_{K} \), a prime ideal and \( \mathcal{D} \) the different of \( K \) . If \( {\wp }^{e} \mid \left( p\right) \), then \( {\wp }^{e - 1} \mid \mathcal{D} \).
|
Proof. We may assume that \( e \) is the highest power of \( \wp \) dividing \( \left( p\right) \). So let \( \left( p\right) = {\wp }^{e}\mathfrak{a},\gcd \left( {\mathfrak{a},\wp }\right) = 1 \). Let \( x \in \wp \mathfrak{a} \). Then \( x = \mathop{\sum }\limits_{{i = 1}}^{n}{p}_{i}{a}_{i},{p}_{i} \in \wp ,{a}_{i} \in \mathfrak{a} \). Hence,\n\n\[ \n{x}^{p} \equiv \mathop{\sum }\limits_{{i = 1}}^{n}{p}_{i}^{p}{a}_{i}^{p}\;\left( {\;\operatorname{mod}\;p}\right) \n\]\n\nand\n\n\[ \n{x}^{{p}^{m}} \equiv \mathop{\sum }\limits_{{i = 1}}^{n}{p}_{i}^{{p}^{m}}{a}_{i}^{{p}^{m}}\;\left( {\;\operatorname{mod}\;p}\right) . \n\]\n\nFor sufficiently large \( m,{p}_{i}^{{p}^{m}} \in {\wp }^{e} \), so \( {x}^{{p}^{m}} \in {\wp }^{e} \) and thus, \( {x}^{{p}^{m}} \in {\wp }^{e}\mathfrak{a} = \left( p\right) \).\n\nTherefore, \( \operatorname{Tr}\left( {x}^{{p}^{m}}\right) \in p\mathbb{Z} \), which implies that\n\n\[ \n{\left( \operatorname{Tr}\left( x\right) \right) }^{{p}^{m}} \in p\mathbb{Z} \n\]\n\n\[ \n\Rightarrow \operatorname{Tr}\left( x\right) \in p\mathbb{Z} \n\]\n\n\[ \n\Rightarrow \operatorname{Tr}\left( {{p}^{-1}\wp \mathfrak{a}}\right) \subseteq \mathbb{Z} \n\]\n\n\[ \n\Rightarrow \;{p}^{-1}\wp \mathfrak{a} \subseteq {\mathcal{D}}^{-1} \n\]\n\n\[ \n\Rightarrow \;{\mathcal{D}p}^{-1}\wp \mathfrak{a} \subseteq {\mathcal{D}\mathcal{D}}^{-1} = {\mathcal{O}}_{K} \n\]\n\n\[ \n\Rightarrow \mathcal{D} \subseteq p{\wp }^{-1}{\mathfrak{a}}^{-1} = {\wp }^{e}\mathfrak{a}{\wp }^{-1}{\mathfrak{a}}^{-1} = {\wp }^{e - 1} \n\]\n\n\[ \n\Rightarrow \;{\wp }^{e - 1} \mid \mathcal{D}\text{. } \n\]
|
Yes
|
Theorem 6.2.2 There exists a constant \( {C}_{K} \) such that every ideal \( \mathfrak{a} \subseteq {\mathcal{O}}_{K} \) is equivalent to an ideal \( \mathfrak{b} \subseteq {\mathcal{O}}_{K} \) with \( N\left( \mathfrak{b}\right) \leq {C}_{K} \) .
|
Proof. Suppose \( \mathfrak{a} \) is an ideal of \( {\mathcal{O}}_{K} \) . Let \( \beta \in \mathfrak{a} \) be a non-zero element such that \( \left| {N\left( \beta \right) }\right| \) is minimal.\n\nFor each \( \alpha \in \mathfrak{a} \), by Exercise 6.1.2, we can find \( t \in \mathbb{Z},\left| t\right| \leq {H}_{K} \), and \( w \in {\mathcal{O}}_{K} \) such that\n\n\[ \left| {N\left( {{t\alpha } - {w\beta }}\right) }\right| < \left| {N\left( \beta \right) }\right| \text{.} \]\n\nMoreover, since \( \alpha ,\beta \in \mathfrak{a} \), so \( {t\alpha } - {w\beta } \in \mathfrak{a} \) ; and therefore, by the minimality of \( \left| {N\left( \beta \right) }\right| \), we must have \( {t\alpha } = {w\beta } \) . Thus, we have shown that for any \( \alpha \in \mathfrak{a} \) , there exist \( t \in \mathbb{Z},\left| t\right| \leq {H}_{K} \), and \( w \in {\mathcal{O}}_{K} \) such that \( {t\alpha } = {w\beta } .\n\nLet\n\n\[ M = \mathop{\prod }\limits_{{\left| t\right| \leq {H}_{K}}}t \]\n\nand we have \( M\mathfrak{a} \subseteq \left( \beta \right) \) . This means that \( \left( \beta \right) \) divides \( \left( M\right) \mathfrak{a} \), and so\n\n\[ \left( M\right) \mathfrak{a} = \left( \beta \right) \mathfrak{b} \]\n\nfor some ideal \( \mathfrak{b} \subseteq {\mathcal{O}}_{K} .\n\nObserve that \( \beta \in \mathfrak{a} \), so \( {M\beta } \in \left( \beta \right) \mathfrak{b} \), and hence \( \left( M\right) \subseteq \mathfrak{b} \) . This implies that \( \left| {N\left( \mathfrak{b}\right) }\right| \leq N\left( \left( M\right) \right) = {C}_{K} \) . Hence, \( \mathfrak{a} \sim \mathfrak{b} \), and \( {C}_{K} = N\left( \left( M\right) \right) \) satisfies the requirements.
|
No
|
Theorem 6.2.5 The number of equivalence classes of ideals is finite.
|
Proof. By Exercise 6.2.3, each equivalence class of ideals can be represented by an integral ideal. This integral ideal, by Theorem 6.2.2, is equivalent to another integral ideal with norm less than or equal to a given constant \( {C}_{K} \) . Apply Exercise 6.2.4, and we are done.
|
No
|
Show that the equation \( {x}^{2} + 5 = {y}^{3} \) has no integral solution.
|
Observe that if \( y \) is even, then \( x \) is odd, and \( {x}^{2} + 5 \equiv 0\left( {\;\operatorname{mod}\;4}\right) \), and hence \( {x}^{2} \equiv 3\left( {\;\operatorname{mod}\;4}\right) \), which is a contradiction. Therefore, \( y \) is odd. Also, if a prime \( p \mid \left( {x, y}\right) \), then \( p \mid 5 \), so \( p = 5 \); and hence, by dividing both sides of the equation by 5, we end up with \( 1 \equiv 0\left( {\;\operatorname{mod}\;5}\right) \), which is absurd. Thus, \( x \) and \( y \) are coprime.\n\nSuppose now that \( \left( {x, y}\right) \) is an integral solution to the given equation. We consider the factorization\n\n\[ \left( {x + \sqrt{-5}}\right) \left( {x - \sqrt{-5}}\right) = {y}^{3}, \]\n\nin the ring of integers \( \mathbb{Z}\left\lbrack \sqrt{-5}\right\rbrack \) .\n\nSuppose a prime \( \wp \) divides the gcd of \( \left( {x + \sqrt{-5}}\right) \) and \( \left( {x - \sqrt{-5}}\right) \) (which implies \( \wp \) divides \( \left( y\right) ) \). Then \( \wp \) divides \( \left( {2x}\right) \). Also, since \( y \) is odd, \( \wp \) does not divide (2). Thus, \( \wp \) divides \( \left( x\right) \). This is a contradiction to the fact that \( x \) and \( y \) are coprime. Hence, \( \left( {x + \sqrt{-5}}\right) \) and \( \left( {x - \sqrt{-5}}\right) \) are coprime ideals. This and equation (6.3) ensure (by Exercise 5.3.12) that\n\n\[ \left( {x + \sqrt{-5}}\right) = {\mathfrak{a}}^{3}\;\text{ and }\;\left( {x - \sqrt{-5}}\right) = {\mathfrak{b}}^{3}, \]\n\nfor some ideals \( \mathfrak{a} \) and \( \mathfrak{b} \).\n\nSince the class number of \( \mathbb{Q}\left( \sqrt{-5}\right) \) was found in Example 6.2.8 to be 2, \( {\mathfrak{c}}^{2} \) is principal for any ideal \( \mathfrak{c} \). Thus, since \( {\mathfrak{a}}^{3} \) and \( {\mathfrak{b}}^{3} \) are principal, we deduce that \( \mathfrak{a} \) and \( \mathfrak{b} \) are also principal. Moreover, since the units of \( \mathbb{Q}\left( \sqrt{-5}\right) \) are 1 and -1, which are both cubes, we conclude that\n\n\[ x + \sqrt{-5} = {\left( a + b\sqrt{-5}\right) }^{3}, \]\n\nfor some integers \( a \) and \( b \). This implies that\n\n\[ 1 = b\left( {3{a}^{2} - 5{b}^{2}}\right) \]\n\nIt is easy to see that \( b \mid 1 \), so \( b = \pm 1 \). Therefore, \( 3{a}^{2} - 5 = \pm 1 \). Both cases lead to contradiction with the fact that \( a \in \mathbb{Z} \).\n\nHence, the given equation does not have an integral solution.
|
Yes
|
Theorem 7.3.1 (Law of Quadratic Reciprocity) Let \( p \) and \( q \) be odd primes. Then\n\n\[ \left( \frac{p}{q}\right) = \left( \frac{q}{p}\right) {\left( -1\right) }^{\frac{p - 1}{2} \cdot \frac{q - 1}{2}} \]
|
Proof. From Exercise 7.2.2, we have\n\n\[ {S}^{q} \equiv \left( \frac{q}{p}\right) S\;\left( {\;\operatorname{mod}\;q}\right) \]\n\nThus, cancelling out an \( S \) from both sides will give us\n\n\[ {S}^{q - 1} \equiv \left( \frac{q}{p}\right) \;\left( {\;\operatorname{mod}\;q}\right) \]\n\nSince \( q \) is odd, \( q - 1 \) must be divisible by 2 . So\n\n\[ {S}^{q - 1} = {\left( {S}^{2}\right) }^{\left( {q - 1}\right) /2} = {\left\lbrack p\left( \frac{-1}{p}\right) \right\rbrack }^{\left( {q - 1}\right) /2}. \]\n\nThe last equality follows from Theorem 7.2.1. Thus,\n\n\[ \left( \frac{q}{p}\right) \equiv {\left\lbrack p\left( \frac{-1}{p}\right) \right\rbrack }^{\left( {q - 1}\right) /2}\;\left( {\;\operatorname{mod}\;q}\right) \]\n\nFrom Exercise 7.1.2, \( \left( {-1/p}\right) = {\left( -1\right) }^{\left( {p - 1}\right) /2} \) . We substitute this into the above equation to get\n\n\[ \left( \frac{q}{p}\right) \equiv {p}^{\frac{q - 1}{2}}{\left( -1\right) }^{\frac{p - 1}{2} \cdot \frac{q - 1}{2}}\;\left( {\;\operatorname{mod}\;q}\right) . \]\n\nExercise 7.1.3 tells us that \( {p}^{\left( {q - 1}\right) /2} \equiv \left( {p/q}\right) \left( {\;\operatorname{mod}\;q}\right) \) . So,\n\n\[ \left( \frac{q}{p}\right) \equiv \left( \frac{p}{q}\right) {\left( -1\right) }^{\frac{p - 1}{2} \cdot \frac{q - 1}{2}}\;\left( {\;\operatorname{mod}\;q}\right) . \]\n\nBut both sides only take on the value \( \pm 1 \), and since \( q \geq 3 \), the congruence can be replaced by an equals sign. This gives us\n\n\[ \left( \frac{p}{q}\right) = \left( \frac{q}{p}\right) {\left( -1\right) }^{\frac{p - 1}{2} \cdot \frac{q - 1}{2}} \]
|
No
|
Theorem 7.4.5 Suppose \( p = 2 \) . Then:\n\n(a) \( 2{\mathcal{O}}_{K} = {\wp }^{2} \) , \( \wp \) prime if and only if \( 2 \mid {d}_{K} \) ;
|
Proof. (a) \( \Leftarrow \) If \( 2 \mid {d}_{K} \), then \( d \equiv 2,3\left( {\;\operatorname{mod}\;4}\right) \) . If \( d \equiv 2\left( {\;\operatorname{mod}\;4}\right) \), then we claim that \( \left( 2\right) = {\left( 2,\sqrt{d}\right) }^{2} \) . Note that\n\n\[{\left( 2,\sqrt{d}\right) }^{2} = \left( {4,2\sqrt{d}, d}\right) = \left( 2\right) \left( {2,\sqrt{d}, d/2}\right) .\]\n\nSince \( d \) is squarefree, then 2 and \( d/2 \) are relatively prime and thus the second ideal above is actually \( {\mathcal{O}}_{K} \) . So \( \left( 2\right) = {\left( 2,\sqrt{d}\right) }^{2} \) .\n\nIf \( d \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) we claim that \( \left( 2\right) = {\left( 2,1 + \sqrt{d}\right) }^{2} \), since\n\n\[{\left( 2,1 + \sqrt{d}\right) }^{2} = \left( {4,2 + 2\sqrt{d},1 + d + 2\sqrt{d}}\right) = \left( 2\right) \left( {2,1 + \sqrt{d},\frac{1 + d}{2} + \sqrt{d}}\right) .\]\n\nNow we note that \( 1 + \sqrt{d} \) and \( \left( {1 + d}\right) /2 + \sqrt{d} \) are relatively prime, and so the second ideal is \( {\mathcal{O}}_{K} \) .\n\n\( \Rightarrow \) We consider \( {d}_{K} \), which we know is congruent to either 0 or 1 mod 4. Suppose that \( {d}_{K} \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . Then \( {\mathcal{O}}_{K} \) is generated as a \( \mathbb{Z} \) -module by \( 1,\left( {1 + \sqrt{d}}\right) /2 \) . There must exist some element \( a \) in \( \wp \) which is not in \( {\wp }^{2} \) . So \( a = m + n\left( {1 + \sqrt{d}}\right) /2 \) where we can assume that \( m \) and \( n \) are either 0 or 1, since for any \( \alpha \in {\mathcal{O}}_{K}, a + {2\alpha } \) is in \( \wp \) but not in \( {\wp }^{2} \) .\n\nNow, if \( n = 0 \), then \( m \neq 0 \) because otherwise \( a = 0 \) and is obviously in (2). But if \( m = 1 \), then \( a = 1 \) and \( a \notin \wp \) . So, \( n = 1 \) and \( m = 0 \) or 1 . We know that \( {a}^{2} \in \left( 2\right) \), and\n\n\[{a}^{2} = {\left( m + \frac{1 + \sqrt{d}}{2}\right) }^{2}\]\n\n\[= {m}^{2} + \frac{d - 1}{4} + \left( {{2m} + 1}\right) \frac{1 + \sqrt{d}}{2} \in \left( 2\right) .\]\n\nBut \( {2m} + 1 \) is odd and so \( {a}^{2} \notin \left( 2\right) \), and we have arrived at a contradiction.\n\nWe conclude that \( {d}_{K} \equiv 0\left( {\;\operatorname{mod}\;4}\right) \), and so clearly \( 2 \mid {d}_{K} \) .
|
Yes
|
Lemma 8.1.4 (a) Let \( m, n \in \mathbb{Z} \) with \( 0 < m \leq n \) and let \( \Delta = \left( {d}_{ij}\right) \in \) \( {M}_{n \times m}\left( \mathbb{R}\right) \) . For any integer \( t > 1 \), there is a nonzero \( \mathbf{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \in \) \( {\mathbb{Z}}^{n} \) with each \( \left| {x}_{i}\right| \leq t \) such that, if \( \mathbf{y} = \mathbf{x}\Delta = \left( {{y}_{1},\ldots ,{y}_{m}}\right) \in {\mathbb{R}}^{m} \), then each \( \left| {y}_{i}\right| \leq c{t}^{1 - n/m} \), where \( c \) is a constant depending only on the matrix \( \Delta \) .
|
Proof. (a) Let \( \delta = \mathop{\max }\limits_{{1 \leq j \leq m}}\mathop{\sum }\limits_{{i = 1}}^{n}\left| {d}_{ij}\right| \) . Then, for\n\n\[ \mathbf{0} \neq \mathbf{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \in {\mathbb{Z}}_{ \geq 0}^{n} \]\n\nwith each \( \left| {x}_{i}\right| \leq t \) ,\n\n\[ \left| {y}_{j}\right| = \left| {\mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}{d}_{ij}}\right| \leq {\delta t} \]\n\nConsider the cube \( {\left\lbrack -\delta t,\delta t\right\rbrack }^{m} \in {\mathbb{R}}^{m} \) . Let \( h \) be an integer \( \geq 1 \) and divide the given cube into \( {h}^{m} \) equal subcubes so that each will have side length \( {2\delta t}/h \) . Now, for each \( \mathbf{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \in {\left\lbrack 0, t\right\rbrack }^{n} \cap {\mathbb{Z}}^{n},\mathbf{y} = \left( {{y}_{1},\ldots ,{y}_{m}}\right) \in \) \( {\left\lbrack -\delta t,\delta t\right\rbrack }^{m} \) which means that there are \( {\left( t + 1\right) }^{n} \) such points \( \mathbf{y} \in {\left\lbrack -\delta t,\delta t\right\rbrack }^{m} \) . Thus, if \( {h}^{m} < {\left( t + 1\right) }^{n} \), then two of the points must lie in the same subcube. That is, for some \( {\mathbf{x}}^{\prime } \neq {\mathbf{x}}^{\prime \prime } \), we have that, if \( \mathbf{y} = \left( {{\mathbf{x}}^{\prime } - {\mathbf{x}}^{\prime \prime }}\right) \Delta = \left( {{y}_{1},\ldots ,{y}_{m}}\right) \) , then each \( \left| {y}_{i}\right| \leq {2\delta t}/h \) .\n\nSince \( t > 1 \) and \( n/m \geq 1,{\left( t + 1\right) }^{n/m} > {t}^{n/m} + 1 \) so there exists an integer \( h \) with \( {t}^{n/m} < h < {\left( t + 1\right) }^{n/m} \) (in particular, \( {h}^{m} < {\left( t + 1\right) }^{n} \) ). Then \( \left| {y}_{i}\right| \leq {2\delta t}/h < {2\delta }{t}^{1 - n/m} \) for each \( i \) .
|
Yes
|
Lemma 8.1.5 Let \( E = A \cup B \) be a proper partition of \( E \). (a) There exists a sequence of nonzero integers \( \\left\\{ {\\alpha }_{v}\\right\\} \\subseteq {\\mathcal{O}}_{K} \) such that \[ \\left| {\\alpha }_{v}^{\\left( k\\right) }\\right| > \\left| {\\alpha }_{v + 1}^{\\left( k\\right) }\\right| \\;\\text{ for }k \\in A, \] \[ \\left| {\\alpha }_{v}^{\\left( k\\right) }\\right| < \\left| {\\alpha }_{v + 1}^{\\left( k\\right) }\\right| \\;\\text{ for }k \\in B, \] and \( \\left| {{N}_{K}\\left( {\\alpha }_{v}\\right) }\\right| \\leq {c}^{m} \), where \( c \) is a positive constant depending only on \( K \) and \( m = \\left| {A \\cup \\bar{A}}\\right| \).
|
Proof. (a) Let \( {t}_{1} \) be an integer greater than 1 and let \( \\left\\{ {t}_{v}\\right\\} \) be the sequence defined recursively by the relation \( {t}_{v + 1} = M{t}_{v} \) for all \( v \\geq 1 \), where \( M \) is a positive constant that will be suitably chosen. By Lemma 8.1.4, for each \( v,\\exists {\\alpha }_{v} \\in {\\mathcal{O}}_{K} \) such that \[ {c}^{1 - m}{t}_{v}^{1 - n/m} \\leq \\left| {\\alpha }_{v}^{\\left( k\\right) }\\right| \\leq c{t}_{v}^{1 - n/m}\\;\\text{ for }\\;k \\in A, \] \[ {c}^{-m}{t}_{v} \\leq \\left| {\\alpha }_{v}^{\\left( k\\right) }\\right| \\leq {t}_{v}\\;\\text{ for }k \\in B. \] Now, let \( \\kappa = \\min \\{ 1, n/m - 1\\} \) and choose \( M \) such that \( {M}^{\\kappa } > {c}^{m} \) so that both \( M > {c}^{m} \) and \( {M}^{n/m - 1} > {c}^{m} \). Then, if \( k \\in A \), \[ \\left| {\\alpha }_{v}^{\\left( k\\right) }\\right| \\geq {c}^{-m + 1}{t}_{v}^{1 - n/m} = {c}^{-m + 1}{\\left( \\frac{{t}_{v + 1}}{M}\\right) }^{1 - n/m} > c{t}_{v + 1}^{1 - n/m} \\geq \\left| {\\alpha }_{v + 1}^{\\left( k\\right) }\\right| \] and if \( k \\in B \), then \[ \\left| {\\alpha }_{v}^{\\left( k\\right) }\\right| \\leq {t}_{v} = \\frac{{t}_{v + 1}}{M} < {c}^{-m}{t}_{v + 1} \\leq \\left| {\\alpha }_{v + 1}^{\\left( k\\right) }\\right| . \] Also, \[ \\left| {{N}_{K}\\left( {\\alpha }_{v}\\right) }\\right| = \\mathop{\\prod }\\limits_{{i \\in A \\cup \\bar{A}}}\\left| {\\alpha }_{v}^{\\left( i\\right) }\\right| \\mathop{\\prod }\\limits_{{j \\in B \\cup \\bar{B}}}\\left| {\\alpha }_{v}^{\\left( j\\right) }\\right| \\leq {\\left( c{t}_{v}^{1 - n/m}\\right) }^{m}{t}_{v}^{n - m} = {c}^{m}. \]
|
Yes
|
Theorem 8.1.10 (a) Let \( {\zeta }_{m} = {e}^{{2\pi i}/m}, K = \mathbb{Q}\left( {\zeta }_{m}\right) \) . If \( m \) is even, the only roots of unity in \( K \) are the mth roots of unity, so that \( {W}_{K} \simeq \mathbb{Z}/m\mathbb{Z} \) . If \( m \) is odd, the only ones are the \( {2m} \) th roots of unity, so that \( {W}_{K} \simeq \) \( \mathbb{Z}/{2m}\mathbb{Z} \) .
|
Proof. (a) If \( m \) is odd, then \( {\zeta }_{2m} = - {\zeta }_{2m}^{m + 1} = - {\zeta }_{m}^{\left( {m + 1}\right) /2} \) which implies that \( \mathbb{Q}\left( {\zeta }_{m}\right) = \mathbb{Q}\left( {\zeta }_{2m}\right) \) . It will, therefore, suffice to establish the statement for \( m \) even. Suppose that \( \theta \in \mathbb{Q}\left( {\zeta }_{m}\right) \) is a primitive \( k \) th root of unity, \( k \nmid m \) . Then\n\n\( \mathbb{Q}\left( {\zeta }_{m}\right) \) contains a primitive \( r \) th root of unity, where \( r = \operatorname{lcm}\left( {k, m}\right) > m \) . Then \( \mathbb{Q}\left( {\zeta }_{r}\right) \subseteq \mathbb{Q}\left( {\zeta }_{m}\right) \)\n\n\[ \Rightarrow \;\varphi \left( r\right) = \left\lbrack {\mathbb{Q}\left( {\zeta }_{r}\right) : \mathbb{Q}}\right\rbrack \leq \left\lbrack {\mathbb{Q}\left( {\zeta }_{m}\right) : \mathbb{Q}}\right\rbrack = \varphi \left( m\right) \]\n\n(where \( \varphi \) denotes the Euler phi-function). But \( m \) is even and \( m \) properly divides \( r \) implies that \( \varphi \left( m\right) \) properly divides \( \varphi \left( r\right) \), so that, in particular, \( \varphi \left( m\right) < \varphi \left( r\right) \), a contradiction. Thus, the \( m \) th roots of unity are the only roots of unity in \( \mathbb{Q}\left( {\zeta }_{m}\right) \) .
|
Yes
|
Theorem 8.2.4 (a) Let \( \alpha \) be an irrational number and let \( {C}_{j} = {p}_{j}/{q}_{j} \) , for \( j \in \mathbb{N} \), be the convergents of the simple continued fraction of \( \alpha \) . If \( r, s \in \mathbb{Z} \) with \( s > 0 \) and \( k \) is a positive integer such that\n\n\[ \left| {{s\alpha } - r}\right| < \left| {{q}_{k}\alpha - {p}_{k}}\right| \]\n\nthen \( s \geq {q}_{k + 1} \) .
|
Proof. (a) Suppose, on the contrary, that \( 1 \leq s < {q}_{k + 1} \) . For each \( k \geq 0 \) , consider the system of linear equations\n\n\[ {p}_{k}x + {p}_{k + 1}y = r \]\n\n\[ {q}_{k}x + {q}_{k + 1}y = s. \]\n\nUsing Gaussian elimination, we easily find that\n\n\[ \left( {{p}_{k + 1}{q}_{k} - {p}_{k}{q}_{k + 1}}\right) y = r{q}_{k} - s{p}_{k}, \]\n\n\[ \left( {{p}_{k}{q}_{k + 1} - {p}_{k + 1}{q}_{k}}\right) x = r{q}_{k + 1} - s{p}_{k + 1}. \]\n\nBy Exercise 8.2.1 (b), \( {p}_{k + 1}{q}_{k} - {p}_{k}{q}_{k + 1} = {\left( -1\right) }^{k} \) so \( {p}_{k}{q}_{k + 1} - {p}_{k + 1}{q}_{k} = \) \( {\left( -1\right) }^{k + 1} \) . Thus, the unique solution to this system is given by\n\n\[ x = {\left( -1\right) }^{k}\left( {s{p}_{k + 1} - r{q}_{k + 1}}\right) \]\n\n\[ y = {\left( -1\right) }^{k}\left( {r{q}_{k} - s{p}_{k}}\right) \]\n\nWe will show that \( x \) and \( y \) are nonzero and have opposite signs.\n\nIf \( x = 0 \), then\n\n\[ \frac{r}{s} = \frac{{p}_{k + 1}}{{q}_{k + 1}} \]\n\nand since \( \left( {{p}_{k + 1},{q}_{k + 1}}\right) = 1 \), this implies that \( {q}_{k + 1} \mid s \), and so \( {q}_{k + 1} \leq s \) , contradicting our hypothesis.\n\nIf \( y = 0 \), then \( r = {p}_{k}x, s = {q}_{k}x \), so that\n\n\[ \left| {{s\alpha } - r}\right| = \left| x\right| \cdot \left| {{q}_{k}\alpha - {p}_{k}}\right| \geq \left| {{q}_{k}\alpha - {p}_{k}}\right| \]\n\nagain contradicting our hypothesis.\n\nSuppose now that \( y < 0 \) . Then since \( {q}_{k}x = s - {q}_{k + 1}y \) and each \( {q}_{j} \geq \) \( 0, x > 0 \) . On the other hand, if \( y > 0 \), then \( {q}_{k + 1}y \geq {q}_{k + 1} > s \) so \( {q}_{k}x = \) \( s - {q}_{k + 1}y < 0 \) and \( x < 0. \)\n\nBy Exercise 8.2.1 (d), if \( k \) is even,\n\n\[ \frac{{p}_{k}}{{q}_{k}} < \alpha < \frac{{p}_{k + 1}}{{q}_{k + 1}} \]\n\nwhile, if \( k \) is odd,\n\n\[ \frac{{p}_{k + 1}}{{q}_{k + 1}} < \alpha < \frac{{p}_{k}}{{q}_{k}} \]\n\nThus, in either case, \( {q}_{k}\alpha - {p}_{k} \) and \( {q}_{k + 1}\alpha - {p}_{k + 1} \) have opposite signs so that \( x\left( {{q}_{k}\alpha - {p}_{k}}\right) \) and \( y\left( {{q}_{k + 1}\alpha - {p}_{k + 1}}\right) \) have the same sign.\n\n\[ \Rightarrow \;\left| {{s\alpha } - r}\right| = \left| {\left( {{q}_{k}x + {q}_{k + 1}y}\right) \alpha - \left( {{p}_{k}x + {p}_{k + 1}y}\right) }\right| \]\n\n\[ = \left| {x\left( {{q}_{k}\alpha - {p}_{k}}\right) + y\left( {{q}_{k + 1}\alpha - {p}_{k + 1}}\right) }\right| \]\n\n\[ \geq \left| x\right| \cdot \left| {{q}_{k}\alpha - {p}_{k}}\right| + \left| y\right| \cdot \left| {{q}_{k + 1}\alpha - {p}_{k + 1}}\right| \]\n\n\[ \geq \left| x\right| \cdot \left| {{q}_{k}\alpha - {p}_{k}}\right| \geq \left| {{q}_{k}\alpha - {p}_{k}}\right| \]\n\na contradiction, thus establishing our assertion.
|
Yes
|
Theorem 9.1.10 If \( \chi \neq {\chi }_{0} \), then \( \left| {g\left( \chi \right) }\right| = \sqrt{p} \).
|
Proof. We first observe that \( a \neq 0 \) and \( \chi \neq {\chi }_{0} \) together imply that \( {g}_{a}\left( \chi \right) = \chi \left( {a}^{-1}\right) g\left( \chi \right) \) because\n\n\[ \chi \left( a\right) {g}_{a}\left( \chi \right) = \chi \left( a\right) \mathop{\sum }\limits_{{t \in {\mathbb{F}}_{p}}}\chi \left( t\right) {\zeta }^{at} \]\n\n\[ = \mathop{\sum }\limits_{{t \in {\mathbb{F}}_{p}}}\chi \left( {at}\right) {\zeta }^{at} \]\n\n\[ = g\left( \chi \right) \text{.} \]\n\nWith our conventions that \( {\chi }_{0}\left( 0\right) = 1 \), we see for \( a \neq 0 \) ,\n\n\[ {g}_{a}\left( {\chi }_{0}\right) = \mathop{\sum }\limits_{{t \in {\mathbb{F}}_{p}}}{\zeta }^{at} = 0 \]\n\nsince this is just the sum of the \( p \) th roots of unity. Finally, \( {g}_{0}\left( \chi \right) = 0 \) if \( \chi \neq {\chi }_{0} \) and \( {g}_{0}\left( {\chi }_{0}\right) = p \).\n\nNow, by our first observation,\n\n\[ \mathop{\sum }\limits_{{a \in {\mathbb{F}}_{p}}}{g}_{a}\left( \chi \right) \overline{{g}_{a}\left( \chi \right) } = {\left| g\left( \chi \right) \right| }^{2}\left( {p - 1}\right) .\n\nOn the other hand,\n\n\[ \mathop{\sum }\limits_{{a \in {\mathbb{F}}_{p}}}{g}_{a}\left( \chi \right) \overline{{g}_{a}\left( \chi \right) } = \mathop{\sum }\limits_{{s \in {\mathbb{F}}_{p}}}\mathop{\sum }\limits_{{t \in {\mathbb{F}}_{p}}}\chi \left( s\right) \chi \left( t\right) \mathop{\sum }\limits_{{a \in {\mathbb{F}}_{p}}}{\zeta }^{{as} - {at}}. \]\n\nIf \( s \neq t \), the innermost sum is zero, being the sum of all the \( p \) th roots of unity. If \( s = t \), the sum is \( p \). Hence \( {\left| g\left( \chi \right) \right| }^{2} = p \).
|
Yes
|
Lemma 9.1.13 Let \( \pi \) be a prime of \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) such that \( N\left( \pi \right) = p \equiv 1\left( {\;\operatorname{mod}\;3}\right) \) . The character \( {\chi }_{\pi } \) introduced above can be viewed as a character of the finite field \( \mathbb{Z}\left\lbrack \rho \right\rbrack /\left( \pi \right) \) of \( p \) elements. \( J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = \pi \) .
|
Proof. If \( \chi \) is any cubic character, Exercise 9.1.12 shows that \( g{\left( \chi \right) }^{3} = \) \( {pJ}\left( {\chi ,\chi }\right) \) since \( \chi \left( {-1}\right) = 1 \) . We can write \( J\left( {\chi ,\chi }\right) = a + {b\rho } \) for some \( a, b \in \mathbb{Z} \) . But\n\n\[ g{\left( \chi \right) }^{3} = {\left( \mathop{\sum }\limits_{t}\chi \left( t\right) {\zeta }^{t}\right) }^{3} \]\n\n\[ \equiv \mathop{\sum }\limits_{t}{\chi }^{3}\left( t\right) {\zeta }^{3t}\;\left( {{\;\operatorname{mod}\;3}\Omega }\right) \]\n\n\[ \equiv \mathop{\sum }\limits_{{t \neq 0}}{\zeta }^{3t}\;\left( {{\;\operatorname{mod}\;3}\Omega }\right) \]\n\n\[ \equiv - 1\;\left( {{\;\operatorname{mod}\;3}\Omega }\right) \text{.} \]\n\nTherefore, \( a + {b\rho } \equiv - 1\left( {{\;\operatorname{mod}\;3}\Omega }\right) \) . In a similar way,\n\n\[ g{\left( \bar{\chi }\right) }^{3} \equiv a + b\bar{\rho } \equiv - 1\;\left( {{\;\operatorname{mod}\;3}\Omega }\right) . \]\n\nThus, \( b\sqrt{-3} \equiv 0\left( {{\;\operatorname{mod}\;3}\Omega }\right) \) which means \( - 3{b}^{2}/9 \) is an algebraic integer and by Exercise 3.1.2, it is an ordinary integer. Thus, \( b \equiv 0\left( {\;\operatorname{mod}\;3}\right) \) and \( a \equiv - 1\left( {\;\operatorname{mod}\;3}\right) \) . Also, from Exercise 9.1.12 and Theorem 9.1.10, it is clear that \( {\left| J\left( \chi ,\chi \right) \right| }^{2} = p = J\left( {\chi ,\chi }\right) \overline{J\left( {\chi ,\chi }\right) } \) . Therefore, \( J\left( {\chi ,\chi }\right) \) is a primary prime of norm \( p \) . Set \( J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = {\pi }^{\prime } \) . Since \( \pi \bar{\pi } = p = {\pi }^{\prime }\overline{{\pi }^{\prime }} \), we have \( \pi \mid {\pi }^{\prime } \) or \( \pi \mid \overline{{\pi }^{\prime }} \) . We want to eliminate the latter possibility.\n\nBy definition,\n\n\[ J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = \mathop{\sum }\limits_{t}{\chi }_{\pi }\left( t\right) {\chi }_{\pi }\left( {1 - t}\right) \]\n\n\[ \equiv \mathop{\sum }\limits_{t}{t}^{\left( {p - 1}\right) /3}{\left( 1 - t\right) }^{\left( {p - 1}\right) /3}\;\left( {\;\operatorname{mod}\;\pi }\right) . \]\n\nThe polynomial \( {x}^{\left( {p - 1}\right) /3}{\left( 1 - x\right) }^{\left( {p - 1}\right) /3} \) has degree \( \frac{2}{3}\left( {p - 1}\right) < p - 1 \) . Let \( g \) be a primitive root \( \left( {\;\operatorname{mod}\;\pi }\right) \) . Then\n\n\[ \mathop{\sum }\limits_{t}{t}^{j} = \mathop{\sum }\limits_{{a = 0}}^{{p - 2}}{g}^{aj} \equiv 0\;\left( {\;\operatorname{mod}\;\pi }\right) \]\n\nif \( {g}^{j} ≢ 1\left( {\;\operatorname{mod}\;\pi }\right) \), which is the case since \( j < p - 1 \) . Thus, \( J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) \equiv 0 \) \( \left( {\;\operatorname{mod}\;\pi }\right) \) . Therefore \( \pi \mid {\pi }^{\prime } \), as desired.
|
Yes
|
Lemma 9.1.15 Let \( {\pi }_{1} = q \equiv 2\\left( {\\;\\operatorname{mod}\\;3}\\right) \) and \( {\pi }_{2} = \\pi \) be a prime of \( \\mathbb{Z}\\left\\lbrack \\rho \\right\\rbrack \) of norm \( p \) . Then \( {\\chi }_{\\pi }\\left( q\\right) = {\\chi }_{q}\\left( \\pi \\right) \) . In other words,\n\n\\[ \n{\\left( \\frac{q}{\\pi }\\right) }_{3} = {\\left( \\frac{\\pi }{q}\\right) }_{3}.\n\\]
|
Proof. Let \( {\\chi }_{\\pi } = \\chi \), and consider the Jacobi sum \( J\\left( {\\chi ,\\ldots ,\\chi }\\right) \) with \( q \) terms. Since \( 3 \\mid q + 1 \), we have, by Exercise 9.1.12, \( g{\\left( \\chi \\right) }^{q + 1} = {pJ}\\left( {\\chi ,\\ldots ,\\chi }\\right) \) . By Exercise 9.1.14, \( g{\\left( \\chi \\right) }^{3} = {p\\pi } \) so that\n\n\\[ \ng{\\left( \\chi \\right) }^{q + 1} = {\\left( p\\pi \\right) }^{\\left( {q + 1}\\right) /3}.\n\\]\n\nRecall that\n\n\\[ \nJ\\left( {\\chi ,\\ldots ,\\chi }\\right) = \\sum \\chi \\left( {t}_{1}\\right) \\cdots \\chi \\left( {t}_{q}\\right)\n\\]\n\nwhere the sum is over all \( {t}_{1},\\ldots ,{t}_{q} \\in \\mathbb{Z}/p\\mathbb{Z} \) such that \( {t}_{1} + \\cdots + {t}_{q} = 1 \) . The term in which \( {t}_{1} = \\cdots = {t}_{q} \) satisfies \( q{t}_{1} = 1 \) and so \( \\chi \\left( q\\right) \\chi \\left( {t}_{1}\\right) = 1 \) . Raising both sides to the \( q \) th power and noting that \( q \\equiv 2\\left( {\\;\\operatorname{mod}\\;3}\\right) \) gives\n\n\\[ \n\\chi {\\left( q\\right) }^{2}\\chi {\\left( {t}_{1}\\right) }^{q} = 1\n\\]\n\nand so \( \\chi {\\left( {t}_{1}\\right) }^{q} = \\chi \\left( q\\right) \) . Therefore, the \
|
No
|
Theorem 9.1.16 Let \( {\pi }_{1} \) and \( {\pi }_{2} \) be two primary primes of \( \mathbb{Z}\left\lbrack \rho \right\rbrack \), of norms \( {p}_{1},{p}_{2} \), respectively. Then \( {\chi }_{{\pi }_{1}}\left( {\pi }_{2}\right) = {\chi }_{{\pi }_{2}}\left( {\pi }_{1}\right) \). In other words,\n\n\[ \n{\left( \frac{{\pi }_{2}}{{\pi }_{1}}\right) }_{3} = {\left( \frac{{\pi }_{1}}{{\pi }_{2}}\right) }_{3} \n\]
|
Proof. To begin, let \( {\gamma }_{1} = \overline{{\pi }_{1}},{\gamma }_{2} = \overline{{\pi }_{2}} \). Then \( {p}_{1} = {\pi }_{1}{\gamma }_{1},{p}_{2} = {\pi }_{2}{\gamma }_{2} \), and \( {p}_{1},{p}_{2} \equiv 1\left( {\;\operatorname{mod}\;3}\right) \). Now,\n\n\[ \ng{\left( {\chi }_{{\gamma }_{1}}\right) }^{{p}_{2}} = J\left( {{\chi }_{{\gamma }_{1}},\ldots ,{\chi }_{{\gamma }_{1}}}\right) g\left( {\chi }_{{\gamma }_{1}}^{{p}_{2}}\right) \n\]\n\nby Exercise 9.1.11. (There are \( {p}_{2} \) characters in the Jacobi sum.) Since \( {p}_{2} \equiv 1\left( {\;\operatorname{mod}\;3}\right) ,{\chi }_{{\gamma }_{1}}^{{p}_{2}} = {\chi }_{{\gamma }_{1}} \). Thus,\n\n\[ \n{\left\lbrack g{\left( {\chi }_{{\gamma }_{1}}\right) }^{3}\right\rbrack }^{\left( {{p}_{2} - 1}\right) /3} = J\left( {{\chi }_{{\gamma }_{1}},\ldots ,{\chi }_{{\gamma }_{1}}}\right) . \n\]\n\nAs before, isolating the \
|
No
|
Theorem 10.1.4 Let \( {\left\{ {a}_{m}\right\} }_{m = 1}^{\infty } \) be a sequence of complex numbers, and let \( A\left( x\right) = \mathop{\sum }\limits_{{m \leq x}}{a}_{m} = O\left( {x}^{\delta }\right) \), for some \( \delta \geq 0 \) . Then\n\n\[ \mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{m}}{{m}^{s}} \]\n\nconverges for \( \operatorname{Re}\left( s\right) > \delta \) and in this half-plane we have\n\n\[ \mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{m}}{{m}^{s}} = s{\int }_{1}^{\infty }\frac{A\left( x\right) {dx}}{{x}^{s + 1}} \]\n\nfor \( \operatorname{Re}\left( s\right) > 1 \) .
|
Proof. We write\n\n\[ \mathop{\sum }\limits_{{m = 1}}^{M}\frac{{a}_{m}}{{m}^{s}} = \mathop{\sum }\limits_{{m = 1}}^{M}\left( {A\left( m\right) - A\left( {m - 1}\right) }\right) {m}^{-s} \]\n\n\[ = A\left( M\right) {M}^{-s} + \mathop{\sum }\limits_{{m = 1}}^{{M - 1}}A\left( m\right) \left\{ {{m}^{-s} - {\left( m + 1\right) }^{-s}}\right\} . \]\n\nSince\n\n\[ {m}^{-s} - {\left( m + 1\right) }^{-s} = s{\int }_{m}^{m + 1}\frac{dx}{{x}^{s + 1}} \]\n\nwe get\n\n\[ \mathop{\sum }\limits_{{m = 1}}^{M}\frac{{a}_{m}}{{m}^{s}} = \frac{A\left( M\right) }{{M}^{s}} + s{\int }_{1}^{M}\frac{A\left( x\right) {dx}}{{x}^{s + 1}}. \]\n\nFor \( \operatorname{Re}\left( s\right) > \delta \), we find\n\n\[ \mathop{\lim }\limits_{{M \rightarrow \infty }}\frac{A\left( M\right) }{{M}^{s}} = 0 \]\n\nsince \( A\left( x\right) = O\left( {x}^{\delta }\right) \) . Hence, the partial sums converge for \( \operatorname{Re}\left( s\right) > \delta \) and\n\nwe have\n\n\[ \mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{m}}{{m}^{s}} = s{\int }_{1}^{\infty }\frac{A\left( x\right) {dx}}{{x}^{s + 1}} \]\n\nin this half-plane.
|
Yes
|
Let \( K = \mathbb{Q}\left( i\right) \). Show that \( \left( {s - 1}\right) {\zeta }_{K}\left( s\right) \) extends to an analytic function for \( \operatorname{Re}\left( s\right) > \frac{1}{2} \).
|
Solution. Since every ideal \( \mathfrak{a} \) of \( {\mathcal{O}}_{K} \) is principal, we can write \( \mathfrak{a} = \left( {a + {ib}}\right) \) for some integers \( a, b \). Moreover, since\n\n\[ \mathfrak{a} = \left( {a + {ib}}\right) = \left( {-a - {ib}}\right) = \left( {-a + {ib}}\right) = \left( {a - {ib}}\right) \]\n\nwe can choose \( a, b \) to be both positive. In this way, we can associate with each ideal \( \mathfrak{a} \) a unique lattice point \( \left( {a, b}\right), a \geq 0, b \geq 0 \). Conversely, to each such lattice point \( \left( {a, b}\right) \) we can associate the ideal \( \mathfrak{a} = \left( {a + {ib}}\right) \). Moreover, \( N\mathfrak{a} = {a}^{2} + {b}^{2} \). Thus, if we write\n\n\[ {\zeta }_{K}\left( s\right) = \mathop{\sum }\limits_{\mathfrak{a}}\frac{1}{N{\mathfrak{a}}^{s}} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{n}^{s}} \]\n\nwe find that\n\n\[ A\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{a}_{n} \]\n\nis equal to the number of lattice points lying in the positive quadrant defined by the circle \( {a}^{2} + {b}^{2} \leq x \). We will call such a lattice point \( \left( {a, b}\right) \) internal if \( {\left( a + 1\right) }^{2} + {\left( b + 1\right) }^{2} \leq x \). Otherwise, we will call it a boundary lattice point. Let \( I \) be the number of internal lattice points, and \( B \) the number of boundary lattice points. Then\n\n\[ I \leq \frac{\pi }{4}x \leq I + B \]\n\nAny boundary point \( \left( {a, b}\right) \) is contained in the annulus\n\n\[ {\left( \sqrt{x} - \sqrt{2}\right) }^{2} \leq {a}^{2} + {b}^{2} \leq {\left( \sqrt{x} + \sqrt{2}\right) }^{2} \]\n\nand an upper bound for \( B \) is provided by the area of the annulus. This is easily seen to be\n\n\[ \pi {\left( \sqrt{x} + \sqrt{2}\right) }^{2} - \pi {\left( \sqrt{x} - \sqrt{2}\right) }^{2} = O\left( \sqrt{x}\right) . \]\n\nThus \( A\left( x\right) = {\pi x}/4 + O\left( \sqrt{x}\right) \). By Theorem 10.1.4, we deduce that\n\n\[ {\zeta }_{K}\left( s\right) = \frac{\pi }{4}s{\int }_{1}^{\infty }\frac{dx}{{x}^{s}} + s{\int }_{1}^{\infty }\frac{E\left( x\right) }{{x}^{s + 1}}{dx} \]\n\nwhere \( E\left( x\right) = O\left( \sqrt{x}\right) \), so that the latter integral converges for \( \operatorname{Re}\left( s\right) > \frac{1}{2} \).\n\nThus\n\n\[ \left( {s - 1}\right) {\zeta }_{K}\left( s\right) = \frac{\pi }{4}s + s\left( {s - 1}\right) {\int }_{1}^{\infty }\frac{E\left( x\right) }{{x}^{s + 1}}{dx} \]\n\nis analytic for \( \operatorname{Re}\left( s\right) > \frac{1}{2} \).
|
Yes
|
Theorem 10.2.8 (Dirichlet's Hyperbola Method) Let\n\n\\[ \nf\left( n\right) = \mathop{\sum }\limits_{{\delta \mid n}}g\left( \delta \right) h\left( \frac{n}{\delta }\right) \n\\]\n\nand define\n\n\\[ \nG\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}g\left( n\right) \n\\]\n\n\\[ \nH\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}h\left( n\right) \n\\]\n\nThen for any \\( y > 0 \\) ,\n\n\\[ \n\mathop{\sum }\limits_{{n \leq x}}f\left( n\right) = \mathop{\sum }\limits_{{\delta \leq y}}g\left( \delta \right) H\left( \frac{x}{\delta }\right) + \mathop{\sum }\limits_{{\delta < \frac{x}{y}}}h\left( \delta \right) G\left( \frac{x}{\delta }\right) - G\left( y\right) H\left( \frac{x}{y}\right) .\n\\]
|
Proof. We have\n\n\\[ \n\mathop{\sum }\limits_{{n \leq x}}f\left( n\right) = \mathop{\sum }\limits_{{{\delta e} \leq x}}g\left( \delta \right) h\left( e\right) \n\\]\n\n\\[ \n= \mathop{\sum }\limits_{\substack{{{\delta e} \leq x} \\ {\delta \leq y} }}g\left( \delta \right) h\left( e\right) + \mathop{\sum }\limits_{\substack{{{\delta e} \leq x} \\ {\delta > y} }}g\left( \delta \right) h\left( e\right) \n\\]\n\n\\[ \n= \mathop{\sum }\limits_{{\delta \leq y}}g\left( \delta \right) H\left( \frac{x}{\delta }\right) + \mathop{\sum }\limits_{{e \leq \frac{x}{y}}}h\left( e\right) \\{ {G\left( \frac{x}{e}\right) - G\left( y\right) }\\} \n\\]\n\n\\[ \n= \mathop{\sum }\limits_{{\delta \leq y}}g\left( \delta \right) H\left( \frac{x}{\delta }\right) + \mathop{\sum }\limits_{{e \leq \frac{x}{y}}}h\left( e\right) G\left( \frac{x}{e}\right) - G\left( y\right) H\left( \frac{x}{y}\right) \n\\]\n\nas desired.
|
Yes
|
Let \( K \) be a quadratic field, and \( {a}_{n} \) the number of ideals of norm \( n \) in \( {\mathcal{O}}_{K} \) . Show that\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{a}_{n} = {cx} + O\left( \sqrt{x}\right) \]\n\nwhere\n\n\[ c = \mathop{\sum }\limits_{{\delta = 1}}^{\infty }\left( \frac{{d}_{K}}{\delta }\right) \frac{1}{\delta } \]
|
Solution. By Exercise 10.2.5,\n\n\[ {a}_{n} = \mathop{\sum }\limits_{{\delta \mid n}}\left( \frac{{d}_{K}}{\delta }\right) \]\n\nso that we can apply Theorem 10.2.8 with \( g\left( \delta \right) = \left( \frac{{d}_{K}}{\delta }\right) \) and \( h\left( \delta \right) = 1 \) , \( y = \sqrt{x} \) . We get\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{a}_{n} = \mathop{\sum }\limits_{{\delta \leq \sqrt{x}}}\left( \frac{{d}_{K}}{\delta }\right) \left\lbrack \frac{x}{\delta }\right\rbrack + \mathop{\sum }\limits_{{\delta < \sqrt{x}}}G\left( \frac{x}{\delta }\right) - G\left( \sqrt{x}\right) \left\lbrack \sqrt{x}\right\rbrack . \]\n\nBy Exercise 10.2.7, \( \left| {G\left( x\right) }\right| \leq \left| {d}_{K}\right| \) . Hence\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{a}_{n} = \mathop{\sum }\limits_{{\delta \leq \sqrt{x}}}\left( \frac{{d}_{K}}{\delta }\right) \left\lbrack \frac{x}{\delta }\right\rbrack + O\left( \sqrt{x}\right) \]\n\nNow \( \left\lbrack {x/\delta }\right\rbrack = x/\delta + O\left( 1\right) \) so that\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{a}_{n} = \mathop{\sum }\limits_{{\delta \leq \sqrt{x}}}\left( \frac{{d}_{K}}{\delta }\right) \frac{x}{\delta } + O\left( \sqrt{x}\right) \]\n\nFinally,\n\n\[ \mathop{\sum }\limits_{{\delta \leq \sqrt{x}}}\left( \frac{{d}_{K}}{\delta }\right) \frac{1}{\delta } = \mathop{\sum }\limits_{{\delta = 1}}^{\infty }\left( \frac{{d}_{K}}{\delta }\right) \frac{1}{\delta } - \mathop{\sum }\limits_{{\delta > \sqrt{x}}}\left( \frac{{d}_{K}}{\delta }\right) \frac{1}{\delta } \]\n\nand by Theorem 10.1.4 we see that\n\n\[ c = \mathop{\sum }\limits_{{\delta = 1}}^{\infty }\left( \frac{{d}_{K}}{\delta }\right) \frac{1}{\delta } \]\n\nconverges and\n\n\[ \mathop{\sum }\limits_{{\delta > \sqrt{x}}}\left( \frac{{d}_{K}}{\delta }\right) \frac{1}{\delta } = O\left( \frac{1}{\sqrt{x}}\right) . \]\n\nTherefore\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{a}_{n} = {cx} + O\left( \sqrt{x}\right) \]
|
No
|
Lemma 10.4.1 Let \( \\left\\{ {a}_{n}\\right\\} \) be a sequence of nonnegative numbers. There exists a \( {\\sigma }_{0} \\in \\mathbb{R} \) (possibly infinite) such that\n\n\[ f\\left( s\\right) = \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }\\frac{{a}_{n}}{{n}^{s}} \]\n\nconverges for \( \\sigma > {\\sigma }_{0} \) and diverges for \( \\sigma < {\\sigma }_{0} \) . Moreover, if \( s \\in \\mathbb{C} \), with \( \\operatorname{Re}\\left( s\\right) > {\\sigma }_{0} \), then the series converges uniformly in \( \\operatorname{Re}\\left( s\\right) \\geq {\\sigma }_{0} + \\delta \) for any \( \\delta > 0 \) and\n\n\[ {f}^{\\left( k\\right) }\\left( s\\right) = {\\left( -1\\right) }^{k}\\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }\\frac{{a}_{n}{\\left( \\log n\\right) }^{k}}{{n}^{s}} \]\n\nfor \( \\operatorname{Re}\\left( s\\right) > {\\sigma }_{0} \) . ( \( {\\sigma }_{0} \) is called the abscissa of convergence of the (Dirichlet) series \( \\left. {\\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }{a}_{n}{n}^{-s}\\text{.}}\\right) \)
|
Proof. If there is no real value of \( s \) for which the series converges, we take \( {\\sigma }_{0} = \\infty \) . Therefore, suppose there is some real \( {s}_{0} \) for which the series converges. Clearly by the comparison test, the series converges for \( \\operatorname{Re}\\left( s\\right) > {s}_{0} \) since the coefficients are nonnegative. Now let \( {\\sigma }_{0} \) be the infimum of all real \( {s}_{0} \) for which the series converges. The uniform convergence in \( \\operatorname{Re}\\left( s\\right) \\geq {\\sigma }_{0} + \\delta \) for any \( \\delta > 0 \) is now immediate. Because of this, we can differentiate term by term to calculate \( {f}^{\\left( k\\right) }\\left( s\\right) \) for \( \\operatorname{Re}\\left( s\\right) > {\\sigma }_{0} \) .
|
Yes
|
Theorem 10.4.2 Let \( {a}_{n} \geq 0 \) be a sequence of nonnegative numbers. Then\n\n\[ f\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{n}^{s}} \]\n\ndefines a holomorphic function in \( \operatorname{Re}\left( s\right) > {\sigma }_{0} \) and \( s = {\sigma }_{0} \) is a singular point of \( f\left( s\right) \) . (Here \( {\sigma }_{0} \) is the abscissa of convergence of the Dirichlet series.)
|
Proof. By the previous lemma, it is clear that \( f\left( s\right) \) is holomorphic in \( \operatorname{Re}\left( s\right) > {\sigma }_{0} \) . If \( f \) is not singular at \( s = {\sigma }_{0} \), then there is a disk\n\n\[ D = \left\{ {s : \left| {s - {\sigma }_{1}}\right| < \delta }\right\} \]\n\nwhere \( {\sigma }_{1} > {\sigma }_{0} \) such that \( \left| {{\sigma }_{0} - {\sigma }_{1}}\right| < \delta \) and a holomorphic function \( g \) in \( D \) such that \( g\left( s\right) = f\left( s\right) \) for \( \operatorname{Re}\left( s\right) > {\sigma }_{0}, s \in D \) . By Taylor’s formula\n\n\[ g\left( s\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{g}^{\left( k\right) }\left( {\sigma }_{1}\right) }{k!}{\left( s - {\sigma }_{1}\right) }^{k} = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{f}^{\left( k\right) }\left( {\sigma }_{1}\right) }{k!}{\left( s - {\sigma }_{1}\right) }^{k} \]\nsince \( g\left( s\right) = f\left( s\right) \) for \( s \) in a neighborhood of \( {\sigma }_{1} \) . Thus, the series\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{\left( -1\right) }^{k}{f}^{\left( k\right) }\left( {\sigma }_{1}\right) }{k!}{\left( {\sigma }_{1} - s\right) }^{k} \]\n\nconverges absolutely for any \( s \in D \) . By the lemma, we can write this series as the double series\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{\left( {\sigma }_{1} - s\right) }^{k}}{k!}\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}{\left( \log n\right) }^{k}}{{n}^{{\sigma }_{1}}}. \]\n\nIf \( {\sigma }_{1} - \delta < s < {\sigma }_{1} \), this convergent double series consists of nonnegative terms and we may interchange the summations to find\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{n}^{{\sigma }_{1}}}\mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{\left( {\sigma }_{1} - s\right) }^{k}{\left( \log n\right) }^{k}}{k!} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{n}^{s}} < \infty . \]\n\nSince \( {\sigma }_{1} - \delta < {\sigma }_{0} < {\sigma }_{1} \), this is a contradiction for \( s = {\sigma }_{0} \) . Therefore, the abscissa of convergence is a singular point of \( f\left( s\right) \) .
|
Yes
|
Theorem 10.4.5 Let \( L\\left( {s,\\chi }\\right) \) be defined as above. Then \( L\\left( {1,\\chi }\\right) \\neq 0 \) for \( \\chi \\neq {\\chi }_{0} \) .
|
Proof. By the previous exercise, the abscissa of convergence of\n\n\\[ \nf\\left( s\\right) = \\mathop{\\prod }\\limits_{\\chi }L\\left( {s,\\chi }\\right) = \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }\\frac{{c}_{n}}{{n}^{s}} \\]\n\nis greater than or equal to \( 1/\\varphi \\left( m\\right) \) . If for some \( \\chi \\neq {\\chi }_{0} \) we have \( L\\left( {1,\\chi }\\right) = 0 \) , then \( f\\left( s\\right) \) is holomorphic at \( s = 1 \) since the zero of \( L\\left( {s,\\chi }\\right) \) cancels the simple pole at \( s = 1 \) of\n\n\\[ \nL\\left( {s,{\\chi }_{0}}\\right) = \\zeta \\left( s\\right) \\mathop{\\prod }\\limits_{{p \\mid m}}\\left( {1 - \\frac{1}{{p}^{s}}}\\right) .\n\\]\n\nBy Exercise 10.3.3, each \( L\\left( {s,\\chi }\\right) \) extends to an analytic function for \( \\operatorname{Re}\\left( s\\right) > \) 0 . By Exercise 10.1.5, \( \\zeta \\left( s\\right) \) (and hence \( L\\left( {s,{\\chi }_{0}}\\right) \) ) is analytic for \( \\operatorname{Re}\\left( s\\right) > 0 \) , \( s \\neq 1 \) . Thus, \( f\\left( s\\right) \) is analytic for \( \\operatorname{Re}\\left( s\\right) > 0 \) . By Theorem 10.4.2, the abscissa of convergence of the Dirichlet series\n\n\\[ \n\\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }\\frac{{c}_{n}}{{n}^{s}}\n\\]\n\nis not in \( \\operatorname{Re}\\left( s\\right) > 0 \) which contradicts the divergence of the series at \( s = \) \( 1/\\varphi \\left( m\\right) \) .
|
Yes
|
Theorem 11.2.4 Let \( n \) be the degree of \( K/\mathbb{Q} \) . If \( \chi \neq {\chi }_{0} \), then \( L\left( {s,\chi }\right) \) extends analytically to \( \Re \left( s\right) > 1 - \frac{1}{n} \) .
|
Proof. By Theorem 11.1.5, we have\n\n\[ \mathop{\sum }\limits_{C}\mathop{\sum }\limits_{{\mathfrak{a} \in C, N\left( \mathfrak{a}\right) \leq x}}\chi \left( \mathfrak{a}\right) = \mathop{\sum }\limits_{C}\chi \left( C\right) N\left( {x, C}\right) = O\left( {x}^{1 - \frac{1}{n}}\right) ,\]\n\nsince (by Exercise 11.2.2)\n\n\[ \mathop{\sum }\limits_{C}\chi \left( C\right) = 0 \]
|
No
|
Proposition 2.3. Let \( \mathbf{E} \) be a Banach space and \( x \neq 0 \) an element of \( \mathbf{E} \) . Then there exists a continuous linear map \( \lambda \) of \( \mathbf{E} \) into \( \mathbf{R} \) such that \( \lambda \left( x\right) \neq 0 \) .
|
One constructs \( \lambda \) by Zorn’s lemma, supposing that \( \lambda \) is defined on some subspace, and having a bounded norm. One then extends \( \lambda \) to the subspace generated by one additional element, without increasing the norm.
|
No
|
Proposition 2.5. Let \( \\mathbf{E},\\mathbf{F} \) be two Banach spaces. Then the set of toplinear isomorphisms \( \\operatorname{Lis}\\left( {\\mathbf{E},\\mathbf{F}}\\right) \) is open in \( L\\left( {\\mathbf{E},\\mathbf{F}}\\right) \) .
|
The proof is in fact quite simple. If \( \\operatorname{Lis}\\left( {\\mathbf{E},\\mathbf{F}}\\right) \) is not empty, one is immediately reduced to proving that \( \\operatorname{Laut}\\left( \\mathbf{E}\\right) \) is open in \( L\\left( {\\mathbf{E},\\mathbf{E}}\\right) \). We then remark that if \( u \\in L\\left( {\\mathbf{E},\\mathbf{E}}\\right) \), and \( \\left| u\\right| < 1 \), then the series\n\n\[ 1 + u + {u}^{2} + \\cdots \]\n\nconverges. Given any toplinear automorphism \( w \) of \( \\mathbf{E} \), we can find an open neighborhood by translating the open unit ball multiplicatively from 1 to \( w \) .
|
No
|
Proposition 3.1. If \( f : U \rightarrow V \) is differentiable at \( {x}_{0} \), if \( g : V \rightarrow W \) is differentiable at \( f\left( {x}_{0}\right) \), then \( g \circ f \) is differentiable at \( {x}_{0} \), and\n\n\[{\left( g \circ f\right) }^{\prime }\left( {x}_{0}\right) = {g}^{\prime }\left( {f\left( {x}_{0}\right) }\right) \circ {f}^{\prime }\left( {x}_{0}\right) .
|
Proof. We leave it as a simple (and classical) exercise.
|
No
|
Proposition 3.4. Let \( U \) be open in \( \mathbf{E} \), and let \( {f}_{i} : U \rightarrow {\mathbf{F}}_{i}\left( {i = 1,\ldots, n}\right) \) be continuous maps into spaces \( {\mathbf{F}}_{i} \) . Let \( f = \left( {{f}_{1},\ldots ,{f}_{n}}\right) \) be the map of \( U \) into the product of the \( {\mathbf{F}}_{i} \) . Then \( f \) is of class \( {C}^{p} \) if and only if each \( {f}_{i} \) is of class \( {C}^{p} \), and in that case
|
\[ {D}^{p}f = \left( {{D}^{p}{f}_{1},\ldots ,{D}^{p}{f}_{n}}\right) . \]
|
Yes
|
Proposition 3.5. Let \( {U}_{1},\ldots ,{U}_{n} \) be open in the spaces \( {\mathbf{E}}_{1},\ldots ,{\mathbf{E}}_{n} \) and let \( f : {U}_{1} \times \cdots \times {U}_{n} \rightarrow \mathbf{F} \) be a continuous map. Then \( f \) is of class \( {C}^{p} \) if and only if each partial derivative \( {D}_{i}f : {U}_{1} \times \cdots {U}_{n} \rightarrow L\left( {{\mathbf{E}}_{i},\mathbf{F}}\right) \) exists and is of class \( {C}^{p - 1} \) . If that is the case, then for \( x = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) and
|
\[ v = \left( {{v}_{1},\ldots ,{v}_{n}}\right) \in {\mathbf{E}}_{1} \times \cdots \times {\mathbf{E}}_{n} \] we have \[ {Df}\left( x\right) \cdot \left( {{v}_{1},\ldots ,{v}_{n}}\right) = \sum {D}_{i}f\left( x\right) \cdot {v}_{i}. \]
|
Yes
|
Proposition 3.7. Let \( \\mathbf{E},\\mathbf{F},\\mathbf{G} \) be Banach spaces, and \( U \) open in \( \\mathbf{E} \) . Let \( f : U \\rightarrow \\mathbf{F} \) be of class \( {C}^{p} \) and \( g : \\mathbf{F} \\rightarrow \\mathbf{G} \) continuous and linear. Then \( g \\circ f \) is of class \( {C}^{p} \) and
|
\[ {D}^{p}\\left( {g \\circ f}\\right) = g \\circ {D}^{p}f. \]
|
Yes
|
Proposition 3.9. Let \( \\mathbf{E},\\mathbf{F} \) be Banach spaces which are toplinearly isomorphic. If \( u : \\mathbf{E} \\rightarrow \\mathbf{F} \) is a toplinear isomorphism, we denote its inverse by \( {u}^{-1} \). Then the map
|
\[ u \\mapsto {u}^{-1} \] from \( \\operatorname{Lis}\\left( {\\mathbf{E},\\mathbf{F}}\\right) \) to \( \\operatorname{Lis}\\left( {\\mathbf{F},\\mathbf{E}}\\right) \) is a \( {C}^{\\infty } \)-isomorphism. Its derivative at a point \( {u}_{0} \) is the linear map of \( L\\left( {\\mathbf{E},\\mathbf{F}}\\right) \) into \( L\\left( {\\mathbf{F},\\mathbf{E}}\\right) \) given by the formula \[ v \\mapsto {u}_{0}^{-1}v{u}_{0}^{-1} \]
|
Yes
|
Proposition 3.10. Let \( U \) be open in the Banach space \( \mathbf{E} \) and let \( \mathbf{F},\mathbf{G} \) be Banach spaces.\n\n(i) If \( f : U \rightarrow L\left( {\mathbf{E},\mathbf{F}}\right) \) is a \( {C}^{p} \) -morphism, then the map of \( U \times \mathbf{E} \) into F given by\n\n\[ \left( {x, v}\right) \mapsto f\left( x\right) v \]\n\nis a morphism.\n\n(ii) If \( f : U \rightarrow L\left( {\mathbf{E},\mathbf{F}}\right) \) and \( g : U \rightarrow L\left( {\mathbf{F},\mathbf{G}}\right) \) are morphisms, then so is \( \gamma \left( {f, g}\right) \) ( \( \gamma \) being the composition).\n\n(iii) If \( f : U \rightarrow \mathbf{R} \) and \( g : U \rightarrow L\left( {\mathbf{E},\mathbf{F}}\right) \) are morphisms, so is \( {fg} \) (the value of \( {fg} \) at \( x \) is \( f\left( x\right) g\left( x\right) \), ordinary multiplication by scalars).\n\n(iv) If \( f, g : U \rightarrow L\left( {\mathbf{E},\mathbf{F}}\right) \) are morphisms, so is \( f + g \) .
|
This proposition concludes our summary of results assumed without proof.
|
No
|
Proposition 4.1. Let \( \lambda : \mathbf{E} \rightarrow \mathbf{R} \) be a continuous linear map and let \( f : I \rightarrow \mathbf{E} \) be ruled. Then \( {\lambda f} = \lambda \circ f \) is ruled, and\n\n\[ \lambda {\int }_{a}^{b}f\left( t\right) {dt} = {\int }_{a}^{b}{\lambda f}\left( t\right) {dt} \]
|
Proof. If \( {f}_{n} \) is a sequence of step functions converging uniformly to \( f \) , then \( \lambda {f}_{n} \) is ruled and converges uniformly to \( {\lambda f} \) . Our formula follows at once.
|
Yes
|
Corollary 4.2. Let \( \\mathbf{E} \) , \( \\mathbf{F} \) be two Banach spaces, \( U \) open in \( \\mathbf{E} \), and \( x, z \) two distinct points of \( U \) such that the segment \( x + t\\left( {z - x}\\right) \\left( {0 \\leqq t \\leqq 1}\\right) \) lies in U. Let \( f : U \\rightarrow \\mathbf{F} \) be continuous and of class \( {C}^{1} \). Then\n\n\[ \n\\left| {f\\left( z\\right) - f\\left( x\\right) }\\right| \\leqq \\left| {z - x}\\right| \\sup \\left| {{f}^{\\prime }\\left( \\xi \\right) }\\right|\n\]\n\nthe sup being taken over \( \\xi \) in the segment.
|
Proof. This comes from the usual estimations of the integral. Indeed, for any continuous map \( g : I \\rightarrow \\mathbf{F} \) we have the estimate\n\n\[ \n\\left| {{\\int }_{a}^{b}g\\left( t\\right) {dt}}\\right| \\leqq K\\left( {b - a}\\right)\n\]\n\nif \( K \) is a bound for \( g \) on \( I \), and \( a \\leqq b \). This estimate is obvious for step functions, and therefore follows at once for continuous functions.
|
Yes
|
Corollary 4.3. Let the hypotheses be as in Corollary 4.2. Let \( {x}_{0} \) be a point on the segment between \( x \) and \( z \) . Then\n\n\[ \left| {f\left( z\right) - f\left( x\right) - {f}^{\prime }\left( {x}_{0}\right) \left( {z - x}\right) }\right| \leqq \left| {z - x}\right| \sup \left| {{f}^{\prime }\left( \xi \right) - {f}^{\prime }\left( {x}_{0}\right) }\right| \] \nthe sup taken over all \( \xi \) on the segment.
|
Proof. We apply Corollary 4.2 to the map\n\n\[ g\left( x\right) = f\left( x\right) - {f}^{\prime }\left( {x}_{0}\right) x. \]
|
Yes
|
Lemma 5.1 (Contraction Lemma or Shrinking Lemma). Let \( M \) be a complete metric space, with distance function \( d \), and let \( f : M \rightarrow M \) be a mapping of \( M \) into itself. Assume that there is a constant \( K,0 < K < 1 \) , such that, for any two points \( x, y \) in \( M \), we have\n\n\[ d\left( {f\left( x\right), f\left( y\right) }\right) \leqq {Kd}\left( {x, y}\right) . \]\n\nThen \( f \) has a unique fixed point (a point such that \( f\left( x\right) = x \) ). Given any point \( {x}_{0} \) in \( M \), the fixed point is equal to the limit of \( {f}^{n}\left( {x}_{0}\right) \) (iteration of \( f \) repeated \( n \) times) as \( n \) tends to infinity.
|
Proof. This is a trivial exercise in the convergence of the geometric series, which we leave to the reader.
|
No
|
Lemma 5.4. Let \( U \) be open in \( \mathbf{E} \), and let \( f : U \rightarrow \mathbf{E} \) be of class \( {C}^{1} \) . Assume that \( f\left( 0\right) = 0,{f}^{\prime }\left( 0\right) = I \) . Let \( r > 0 \) and assume that \( {\bar{B}}_{r}\left( 0\right) \subset U \) . Let \( 0 < s < 1 \), and assume that\n\n\[ \left| {{f}^{\prime }\left( z\right) - {f}^{\prime }\left( x\right) }\right| \leqq s \]\n\nfor all \( x, z \in {\bar{B}}_{r}\left( 0\right) \) . If \( y \in \mathbf{E} \) and \( \left| y\right| \leqq \left( {1 - s}\right) r \), then there exists a\n\nunique \( x \in {\bar{B}}_{r}\left( 0\right) \) such that \( f\left( x\right) = y \) .
|
Proof. The map \( {g}_{y} \) given by \( {g}_{y}\left( x\right) = x - f\left( x\right) + y \) is defined for \( \left| x\right| \leqq r \) and \( \left| y\right| \leqq \left( {1 - s}\right) r \), and maps \( {\bar{B}}_{r}\left( 0\right) \) into itself because, from the estimate\n\n\[ \left| {f\left( x\right) - x}\right| = \left| {f\left( x\right) - f\left( 0\right) - {f}^{\prime }\left( 0\right) x}\right| \leqq \left| x\right| \sup \left| {{f}^{\prime }\left( z\right) - {f}^{\prime }\left( 0\right) }\right| \leqq {sr} \]\n\nwe obtain\n\n\[ \left| {{g}_{y}\left( x\right) }\right| \leqq {sr} + \left( {1 - s}\right) r = r. \]\n\nFurthermore, \( {g}_{y} \) is a shrinking map because, from the mean value theorem,\n\nwe get\n\n\[ \left| {{g}_{y}\left( {x}_{1}\right) - {g}_{y}\left( {x}_{2}\right) }\right| = \left| {{x}_{1} - {x}_{2} - \left( {f\left( {x}_{1}\right) - f\left( {x}_{2}\right) }\right) }\right| \]\n\n\[ = \left| {{x}_{1} - {x}_{2} - {f}^{\prime }\left( 0\right) \left( {{x}_{1} - {x}_{2}}\right) + \delta \left( {{x}_{1},{x}_{2}}\right) }\right| \]\n\n\[ = \left| {\delta \left( {{x}_{1},{x}_{2}}\right) }\right| \]\n\nwhere\n\n\[ \left| {\delta \left( {{x}_{1},{x}_{2}}\right) }\right| \leqq \left| {{x}_{1} - {x}_{2}}\right| \sup \left| {{f}^{\prime }\left( z\right) - {f}^{\prime }\left( 0\right) }\right| \leqq s\left| {{x}_{1} - {x}_{2}}\right| . \]\n\nHence \( {g}_{y} \) has a unique fixed point \( x \in {\bar{B}}_{r}\left( 0\right) \) which is such that \( f\left( x\right) = y \) . This proves the lemma.
|
Yes
|
Corollary 5.5. Let \( U \) be an open subset of \( \mathbf{E} \), and \( f : U \rightarrow {\mathbf{F}}_{1} \times {\mathbf{F}}_{2}a \) morphism of \( U \) into a product of Banach spaces. Let \( {x}_{0} \in U \), suppose that \( f\left( {x}_{0}\right) = \left( {0,0}\right) \) and that \( {f}^{\prime }\left( {x}_{0}\right) \) induces a toplinear isomorphism of \( \mathbf{E} \) and \( {\mathbf{F}}_{1} = {\mathbf{F}}_{1} \times 0 \) . Then there exists a local isomorphism \( g \) of \( {\mathbf{F}}_{1} \times {\mathbf{F}}_{2} \) at \( \left( {0,0}\right) \) such that\n\n\[ g \circ f : U \rightarrow {\mathbf{F}}_{1} \times {\mathbf{F}}_{2} \]\n\nmaps an open subset \( {U}_{1} \) of \( U \) into \( {\mathbf{F}}_{1} \times 0 \) and induces a local isomorphism of \( {U}_{1} \) at \( {x}_{0} \) on an open neighborhood of 0 in \( {\mathbf{F}}_{1} \) .
|
Proof. We may assume without loss of generality that \( {\mathbf{F}}_{1} = \mathbf{E} \) (identify by means of \( {f}^{\prime }\left( {x}_{0}\right) \) ) and \( {x}_{0} = 0 \) . We define\n\n\[ \varphi : U \times {\mathbf{F}}_{2} \rightarrow {\mathbf{F}}_{1} \times {\mathbf{F}}_{2} \]\n\nby the formula\n\n\[ \varphi \left( {x,{y}_{2}}\right) = f\left( x\right) + \left( {0,{y}_{2}}\right) \]\n\nfor \( x \in U \) and \( {y}_{2} \in {\mathbf{F}}_{2} \) . Then \( \varphi \left( {x,0}\right) = f\left( x\right) \), and\n\n\[ {\varphi }^{\prime }\left( {0,0}\right) = {f}^{\prime }\left( 0\right) + \left( {0,{\mathrm{{id}}}_{2}}\right) \]\n\nSince \( {f}^{\prime }\left( 0\right) \) is assumed to be a toplinear isomorphism onto \( {\mathbf{F}}_{1} \times 0 \), it follows that \( {\varphi }^{\prime }\left( {0,0}\right) \) is also a toplinear isomorphism. Hence by the theorem, it has a local inverse, say \( g \), which obviously satisfies our requirements.
|
Yes
|
Corollary 5.7. Let \( U \) be an open subset of a product of Banach spaces \( {\mathbf{E}}_{1} \times {\mathbf{E}}_{2} \) and \( \left( {{a}_{1},{a}_{2}}\right) \) a point of \( U \) . Let \( f : U \rightarrow \mathbf{F} \) be a morphism into a Banach space, say \( f\left( {{a}_{1},{a}_{2}}\right) = 0 \), and assume that the partial derivative\n\n\[ \n{D}_{2}f\left( {{a}_{1},{a}_{2}}\right) : {\mathbf{E}}_{2} \rightarrow \mathbf{F} \n\]\n\nis a toplinear isomorphism. Then there exists a local isomorphism \( h \) of a product \( {V}_{1} \times {V}_{2} \) onto an open neighborhood of \( \left( {{a}_{1},{a}_{2}}\right) \) contained in \( U \) such that the composite map\n\n\[ \n{V}_{1} \times {V}_{2}\overset{h}{ \rightarrow }U\overset{f}{ \rightarrow }\mathbf{F} \n\]\n\nis a projection (on the second factor).
|
Proof. We may assume \( \left( {{a}_{1},{a}_{2}}\right) = \left( {0,0}\right) \) and \( {\mathbf{E}}_{2} = \mathbf{F} \) . We define\n\n\[ \n\varphi : {\mathbf{E}}_{1} \times {\mathbf{E}}_{2} \rightarrow {\mathbf{E}}_{1} \times {\mathbf{E}}_{2} \n\]\n\nby\n\n\[ \n\varphi \left( {{x}_{1},{x}_{2}}\right) = \left( {{x}_{1}, f\left( {{x}_{1},{x}_{2}}\right) }\right) \n\]\n\nlocally at \( \left( {{a}_{1},{a}_{2}}\right) \) . Then \( {\varphi }^{\prime } \) is represented by the matrix\n\n\[ \n\left( \begin{matrix} {\mathrm{{id}}}_{1} & O \\ {D}_{1}f & {D}_{2}f \end{matrix}\right) \n\]\n\nand is therefore a toplinear isomorphism at \( \left( {{a}_{1},{a}_{2}}\right) \) . By the theorem, it has a local inverse \( h \) which clearly satisfies our requirements.
|
Yes
|
Corollary 5.8. Let \( U \) be an open subset of a Banach space \( \mathbf{E} \) and \( f : U \rightarrow \mathbf{F} \) a morphism into a Banach space \( \mathbf{F} \). Let \( {x}_{0} \in U \) and assume that \( {f}^{\prime }\left( {x}_{0}\right) \) is surjective, and that its kernel splits. Then there exists an open subset \( {U}^{\prime } \) of \( U \) containing \( {x}_{0} \) and an isomorphism\n\n\[ h : {V}_{1} \times {V}_{2} \rightarrow {U}^{\prime } \]\n\nsuch that the composite map \( f \circ h \) is a projection\n\n\[ {V}_{1} \times {V}_{2} \rightarrow {V}_{1} \rightarrow \mathbf{F} \]
|
Proof. Again this is essentially a reformulation of the corollary, taking into account the splitting assumption.
|
No
|
Lemma 2.1. Let \( {U}_{1},{U}_{2},{V}_{1},{V}_{2} \) be open subsets of Banach spaces, and \( g : {U}_{1} \times {U}_{2} \rightarrow {V}_{1} \times {V}_{2} \) a \( {C}^{p} \) -morphism. Let \( {a}_{2} \in {U}_{2} \) and \( {b}_{2} \in {V}_{2} \) and assume that \( g \) maps \( {U}_{1} \times {a}_{2} \) into \( {V}_{1} \times {b}_{2} \) . Then the induced map\n\n\[ \n{g}_{1} : {U}_{1} \rightarrow {V}_{1} \n\]\n\nis also a morphism.
|
Indeed, it is obtained as a composite map\n\n\[ \n{U}_{1} \rightarrow {U}_{1} \times {U}_{2} \rightarrow {V}_{1} \times {V}_{2} \rightarrow {V}_{1} \n\]\n\nthe first map being an inclusion and the third a projection.
|
Yes
|
Proposition 2.2. Let \( X, Y \) be manifolds of class \( {C}^{p}\left( {p \geqq 1}\right) \) modeled on Banach spaces. Let \( f : X \rightarrow Y \) be a \( {C}^{p} \) -morphism. Let \( x \in X \) . Then:\n\n(i) \( f \) is an immersion at \( x \) if and only if there exists a chart \( \left( {U,\varphi }\right) \) at \( x \) and \( \left( {V,\psi }\right) \) at \( f\left( x\right) \) such that \( {f}_{V, U}^{\prime }\left( {\varphi x}\right) \) is injective and splits.\n\n(ii) \( f \) is a submersion at \( x \) if and only if there exists a chart \( \left( {U,\varphi }\right) \) at \( x \) and \( \left( {V,\psi }\right) \) at \( f\left( x\right) \) such that \( {f}_{V, U}^{\prime }\left( {\varphi x}\right) \) is surjective and its kernel splits.
|
Proof. This is an immediate consequence of Corollaries 5.4 and 5.6 of the inverse mapping theorem.
|
Yes
|
Proposition 2.4. Let \( X, Y \) be manifolds of class \( {C}^{p}\left( {p \geqq 1}\right) \) modeled on Banach spaces. Let \( f : X \rightarrow Y \) be a \( {C}^{p} \) -morphism, and \( W \) a submanifold of \( Y \) . The map \( f \) is transversal over \( W \) if and only if for each \( x \in X \) such that \( f\left( x\right) \) lies in \( W \), the composite map\n\n\[ \n{T}_{x}\left( X\right) \overset{{T}_{x}f}{ \rightarrow }{T}_{w}\left( Y\right) \rightarrow {T}_{w}\left( Y\right) /{T}_{w}\left( W\right) \n\]\n\nwith \( w = f\left( x\right) \) is surjective and its kernel splits.
|
Proof. If \( f \) is transversal over \( W \), then for each point \( x \in X \) such that \( f\left( x\right) \) lies in \( W \), we choose charts as in the definition, and reduce the question to one of maps of open subsets of Banach spaces. In that case, the conclusion concerning the tangent spaces follows at once from the assumed direct product decompositions. Conversely, assume our condition on the tangent map. The question being local, we can assume that \( Y = \) \( {V}_{1} \times {V}_{2} \) is a product of open sets in Banach spaces such that \( W = {V}_{1} \times 0 \) , and we can also assume that \( X = U \) is open in some Banach space, \( x = 0 \) . Then we let \( g : U \rightarrow {V}_{2} \) be the map \( \pi \circ f \) where \( \pi \) is the projection, and note that our assumption means that \( {g}^{\prime }\left( 0\right) \) is surjective and its kernel splits. Furthermore, \( {g}^{-1}\left( 0\right) = {f}^{-1}\left( W\right) \) . We can then use Corollary 5.7 of the inverse mapping theorem to conclude the proof.
|
Yes
|
Proposition 2.5. Let \( f : X \rightarrow Z \) and \( g : Y \rightarrow Z \) be two \( {C}^{p} \) -morphisms with \( p \geqq 1 \) . If they are transversal, then\n\n\[{\left( f \times g\right) }^{-1}\left( {\Delta }_{Z}\right)\]\n\ntogether with the natural morphisms into \( X \) and \( Y \) (obtained from the projections), is a fiber product of \( f \) and \( g \) over \( Z \) .
|
Proof. Obvious.
|
No
|
Proposition 2.6. Assume that each \( {P}_{i} \) admits a manifold structure (compatible with its topology) such that these maps are morphisms, making \( {P}_{i} \) into a fiber product of \( {f}_{i} \) and \( {g}_{i} \) . Then \( P \), with its natural projections, is a fiber product of \( f \) and \( g \) .
|
To prove the above assertion, we observe that the \( {P}_{i} \) form a covering of \( P \) . Furthermore, the manifold structure on \( {P}_{i} \cap {P}_{j} \) induced by that of \( {P}_{i} \) or \( {P}_{j} \) must be the same, because it is the unique fiber product structure over \( {V}_{i} \cap {V}_{j} \), for the maps \( {f}_{ij} \) and \( {g}_{ij} \) (defined on \( {f}^{-1}\left( {{V}_{i} \cap {V}_{j}}\right) \) and \( {g}^{-1}\left( {{V}_{i} \cap {V}_{j}}\right) \) respectively). Thus we can give \( P \) a manifold structure, in such a way that the two projections into \( X \) and \( Y \) are morphisms, and make \( P \) into a fiber product of \( f \) and \( g \) .
|
Yes
|
Proposition 3.1. If \( X \) is a paracompact space, and if \( \left\{ {U}_{i}\right\} \) is an open covering, then there exists a locally finite open covering \( \left\{ {V}_{i}\right\} \) such that \( {V}_{i} \subset {U}_{i} \) for each \( i \) .
|
Proof. Let \( \left\{ {V}_{k}\right\} \) be a locally finite open refinement of \( \left\{ {U}_{i}\right\} \) . For each \( k \) there is an index \( i\left( k\right) \) such that \( {V}_{k} \subset {U}_{i\left( k\right) } \) . We let \( {W}_{i} \) be the union of those \( {V}_{k} \) such that \( i\left( k\right) = i \) . Then the \( {W}_{i} \) form a locally finite open covering, because any neighborhood of a point which meets infinitely many \( {W}_{i} \) must also meet infinitely many \( {V}_{k} \) .
|
Yes
|
Proposition 3.2. If \( X \) is paracompact, then \( X \) is normal. If, furthermore, \( \left\{ {U}_{i}\right\} \) is a locally finite open covering of \( X \), then there exists a locally finite open covering \( \left\{ {V}_{i}\right\} \) such that \( {\bar{V}}_{i} \subset {U}_{i} \) .
|
Proof. We refer the reader to Bourbaki [Bou 68].
|
No
|
Theorem 3.3. Let \( X \) be a manifold which is locally compact, Hausdorff, and whose topology has a countable base. Given an open covering of \( X \) , then there exists an atlas \( \left\{ \left( {{V}_{k},{\varphi }_{k}}\right) \right\} \) such that the covering \( \left\{ {V}_{k}\right\} \) is locally finite and subordinated to the given covering, such that \( {\varphi }_{k}{V}_{k} \) is the open ball \( {B}_{3} \), and such that the open sets \( {W}_{k} = {\varphi }_{k}^{-1}\left( {B}_{1}\right) \) cover \( X \) .
|
Proof. Let \( {U}_{1},{U}_{2},\ldots \) be a basis for the open sets of \( X \) such that each \( {\bar{U}}_{i} \) is compact. We construct inductively a sequence \( {A}_{1},{A}_{2},\ldots \) of compact sets whose union is \( X \), such that \( {A}_{i} \) is contained in the interior of \( {A}_{i + 1} \) . We let \( {A}_{1} = {\bar{U}}_{1} \) . Suppose we have constructed \( {A}_{i} \) . We let \( j \) be the smallest integer such that \( {A}_{i} \) is contained in \( {U}_{1} \cup \cdots \cup {U}_{j} \) . We let \( {A}_{i + 1} \) be the closed and compact set\n\n\[ \n{\bar{U}}_{1} \cup \cdots \cup {\bar{U}}_{j} \cup {\bar{U}}_{i + 1} \n\]\n\nFor each point \( x \in X \) we can find an arbitrarily small chart \( \left( {{V}_{x},{\varphi }_{x}}\right) \) at \( x \) such that \( {\varphi }_{x}{V}_{x} \) is the ball of radius 3 (so that each \( {V}_{x} \) is contained in some element of \( U \) ). We let \( {W}_{x} = {\varphi }_{x}^{-1}\left( {B}_{1}\right) \) be the ball of radius 1 in this chart. We can cover the set\n\n\[ \n{A}_{i + 1} - \operatorname{Int}\left( {A}_{i}\right) \n\]\n\n(intuitively the closed annulus) by a finite number of these balls of radius 1, say \( {W}_{1},\ldots ,{W}_{n} \), such that, at the same time, each one of \( {V}_{1},\ldots ,{V}_{n} \) is contained in the open set \( \operatorname{Int}\left( {A}_{i + 2}\right) - {A}_{i - 1} \) (intuitively, the open annulus of the next bigger size). We let \( {\mathfrak{B}}_{i} \) denote the collection \( {V}_{1},\ldots ,{V}_{n} \) and let \( \mathfrak{B} \) be composed of the union of the \( {\mathfrak{B}}_{i} \) . Then \( \mathfrak{B} \) is locally finite, and we are done.
|
Yes
|
Corollary 3.4. Let \( X \) be a manifold which is locally compact Hausdorff, and whose topology has a countable base. Then \( X \) admits partitions of unity.
|
Proof. Let \( \left\{ \left( {{V}_{k},{\varphi }_{k}}\right) \right\} \) be as in the theorem, and \( {W}_{k} = {\varphi }_{k}^{-1}\left( {B}_{1}\right) \) . We can find a function \( {\psi }_{k} \) of class \( {C}^{p} \) such that \( 0 \leqq {\psi }_{k} \leqq 1 \), such that \( {\psi }_{k}\left( x\right) = 1 \) for \( x \in {W}_{k} \) and \( {\psi }_{k}\left( x\right) = 0 \) for \( x \notin {V}_{k} \) . (The proof is recalled below.) We now let\n\n\[ \psi = \sum {\psi }_{k} \]\n\n(a sum which is finite at each point), and we let \( {\gamma }_{k} = {\psi }_{k}/\psi \) . Then \( \left\{ \left( {{V}_{k},{\gamma }_{k}}\right) \right\} \) is the desired partition of unity.\n\nWe now recall the argument giving the function \( {\psi }_{k} \) . First, given two real numbers \( r, s \) with \( 0 \leqq r < s \), the function defined by\n\n\[ \exp \left( \frac{-1}{\left( {t - r}\right) \left( {s - t}\right) }\right) \]\n\nin the open interval \( r < t < s \) and 0 outside the interval determines a bell-shaped \( {C}^{\infty } \) -function from \( \mathbf{R} \) into \( \mathbf{R} \) . Its integral from minus infinity to \( t \) , divided by the area under the bell yields a function which lies strictly between 0 and 1 in the interval \( r < t < s \), is equal to 0 for \( t \leqq r \) and is equal to 1 for \( t \geqq s \) . (The function is even monotone increasing.)\n\nWe can therefore find a real valued function of a real variable, say \( \eta \left( t\right) \) , such that \( \eta \left( t\right) = 1 \) for \( \left| t\right| < 1 \) and \( \eta \left( t\right) = 0 \) for \( \left| t\right| \geqq 1 + \delta \) with small \( \delta \), and such that \( 0 \leqq \eta \leqq 1 \) . If \( \mathbf{E} \) is a Hilbert space, then \( \eta \left( {\left| x\right| }^{2}\right) = \psi \left( x\right) \) gives us a function which is equal to 1 on the ball of radius 1 and 0 outside the ball of radius \( 1 + \delta \) . This function can then be transported to the manifold by any given chart whose image is the ball of radius 3 . For convenience, we state separately what we have just proved.\n\nLemma 3.5. Let \( E \) be a Hilbert space. There exists a \( {C}^{\infty } \) real function \( \psi \) on \( E \) such that \( \psi \left( x\right) = 1 \) for \( \left| x\right| \leqq 1,\psi \left( x\right) > 0 \) for \( \left| x\right| < 1 + \delta \), and \( \psi \left( x\right) = 0 \) for \( \left| x\right| \geqq 1 + \delta \) . Alternatively, there exists a \( {C}^{\infty } \) function \( h \) such that\n\n\[ h\left( x\right) > 0\;\text{ for }\;\left| x\right| < 1\;\text{ and }\;h\left( x\right) = 0\;\text{ for }\;\left| x\right| \geqq 1. \]\n\nIn other words, one would construct a function which is \( > 0 \) on a given ball and \( = 0 \) outside this ball.
|
Yes
|
Lemma 3.6. Let \( E \) be a separable Hilbert space and \( A \subset E \) a closed nonempty subset. Then there exists a real \( {C}^{\infty } \) function \( \psi \) on \( E \) such that \( \psi \left( x\right) = 0 \) for \( x \in A \) and \( \psi \left( x\right) > 0 \) for \( x \notin A \) .
|
Proof. Let \( h \) be as in Lemma 3.5. Since \( E \) is separable, there exists a sequence \( \left\{ {x}_{n}\right\} \) in the complement \( {A}^{c} \), and dense in this complement. Then\n\n\[ \n{A}^{c} = \mathop{\bigcup }\limits_{n}{B}_{n} \n\]\n\nwhere \( {B}_{n} = B\left( {{x}_{n},{r}_{n}}\right) \) is the ball of radius \( {r}_{n} \) with \( {r}_{n} = d\left( {{x}_{n}, A}\right) \). For each positive integer \( p \), the \( p \) -th derivative \( {D}^{p}h : E \rightarrow {L}^{p}\left( {E,\mathbf{R}}\right) \) is continuous with compact support. Let\n\n\[ \n{M}_{n} = \mathop{\max }\limits_{{p \leqq n}}\frac{{\begin{Vmatrix}{D}^{p}h\end{Vmatrix}}_{\infty }}{{\left( {r}_{n}\right) }^{p}} < \infty .\n\]\n\nLet\n\n\[ \n{\psi }_{n}\left( x\right) = \frac{1}{{2}^{n}{M}_{n}}h\left( {{r}_{n}^{-1}\left( {x - {x}_{n}}\right) }\right) \;\text{ and }\;\psi = \mathop{\sum }\limits_{n}{\psi }_{n}.\n\]\n\nThen for all integers \( p \geqq 0 \) and \( n \geqq p \) we have \( {\begin{Vmatrix}{D}^{p}{\psi }_{n}\end{Vmatrix}}_{\infty } \leqq 1/{2}^{n} \). Hence the series \( \sum {D}^{p}{\psi }_{n} \) converges uniformly on \( E \), and so \( \psi \) is well-defined and of class \( {C}^{\infty } \). Finally we note that \( \psi \left( x\right) = 0 \) for \( x \in A \) and \( \psi \left( x\right) > 0 \) for \( x \notin A \). This concludes the proof of the lemma.
|
Yes
|
Theorem 3.7. Let \( {A}_{1},{A}_{2} \) be non-void, closed, disjoint subsets of a separable Hilbert space \( \mathbf{E} \) . Then there exists a \( {C}^{\infty } \) -function \( \psi : \mathbf{E} \rightarrow \mathbf{R} \) such that \( \psi \left( x\right) = 0 \) if \( x \in {A}_{1} \) and \( \psi \left( x\right) = 1 \) if \( x \in {A}_{2} \), and \( 0 \leqq \psi \left( x\right) \leqq 1 \) for all \( x \) .
|
Proof. In the previous lemma, we use functions \( {\psi }_{1} \) and \( {\psi }_{2} \) corresponding to the closed sets \( {A}_{1} \) and \( {A}_{2} \), and we let\n\n\[ \psi = \frac{{\psi }_{1}}{{\psi }_{1} + {\psi }_{2}} \]\n\nto conclude the proof.
|
No
|
Corollary 3.8. Let \( X \) be a paracompact manifold of class \( {C}^{p} \), modeled on a separable Hilbert space \( \mathbf{E} \) . Then \( X \) admits partitions of unity (of class \( {C}^{p} \) ).
|
Proof. It is trivially verified that an open ball of finite radius in \( \mathbf{E} \) is \( {C}^{\infty } \) -isomorphic to \( \mathbf{E} \) . (We reproduce the formula in Chapter VII.) Given any point \( x \in X \), and a neighborhood \( N \) of \( x \), we can therefore always find a chart \( \left( {G,\gamma }\right) \) at \( x \) such that \( {\gamma G} = \mathbf{E} \), and \( G \subset N \) . Hence, given an open covering of \( X \), we can find an atlas \( \left\{ \left( {{G}_{\alpha },{\gamma }_{\alpha }}\right) \right\} \) subordinated to the given covering, such that \( {\gamma }_{\alpha }{G}_{\alpha } = \mathbf{E} \) . By paracompactness, we can find a refinement \( \left\{ {U}_{i}\right\} \) of the covering \( \left\{ {G}_{\alpha }\right\} \) which is locally finite. Each \( {U}_{i} \) is contained in some \( {G}_{\alpha \left( i\right) } \) and we let \( {\varphi }_{i} \) be the restriction of \( {\gamma }_{\alpha \left( i\right) } \) to \( {U}_{i} \) . We now find open refinements \( \left\{ {V}_{i}\right\} \) and then \( \left\{ {W}_{i}\right\} \) such that\n\n\[ \n{\bar{W}}_{i} \subset {V}_{i} \subset {\bar{V}}_{i} \subset {U}_{i} \n\]\n\nthe bar denoting closure in \( X \) . Each \( {\bar{V}}_{i} \) being closed in \( X \), it follows from our construction that \( {\varphi }_{i}{\bar{V}}_{i} \) is closed in \( \mathbf{E} \), and so is \( {\varphi }_{i}{\bar{W}}_{i} \) . Using the theorem, and transporting functions on \( \mathbf{E} \) to functions on \( X \) by means of the \( {\varphi }_{i} \), we can find for each \( i \) a \( {C}^{p} \) -function \( {\psi }_{i} : X \rightarrow \mathbf{R} \) with is 1 on \( {\bar{W}}_{i} \) and 0 on \( X - {V}_{i} \) . We let \( \psi = \sum {\psi }_{i} \) and \( {\theta }_{i} = {\psi }_{i}/\psi \) . Then the collection \( \left\{ {\theta }_{i}\right\} \) is the desired partition of unity.
|
Yes
|
Proposition 4.1. Let \( f : U \rightarrow \mathbf{F} \) and \( g : U \rightarrow \mathbf{F} \) be two morphisms of class \( {C}^{p}\left( {p \geqq 1}\right) \) defined on an open subset \( U \) of \( \mathbf{E} \) . Assume that \( f \) and \( g \) have the same restriction to \( U \cap {\mathbf{E}}_{\lambda }^{ + } \) for some half plane \( {\mathbf{E}}_{\lambda }^{ + } \), and let\n\n\[ x \in U \cap {\mathbf{E}}_{\lambda }^{ + } \]\n\nThen \( {f}^{\prime }\left( x\right) = {g}^{\prime }\left( x\right) \) .
|
Proof. After considering the difference of \( f \) and \( g \), we may assume without loss of generality that the restriction of \( f \) to \( U \cap {\mathbf{E}}_{\lambda }^{ + } \) is 0 . It is then obvious that \( {f}^{\prime }\left( x\right) = 0 \) .
|
Yes
|
Proposition 4.2. Let \( U \) be open in \( \mathbf{E} \) . Let \( \mu \) be a non-zero functional on \( \mathbf{F} \) and let \( f : U \rightarrow {\mathbf{F}}_{\mu }^{ + } \) be a morphism of class \( {C}^{p} \) with \( p \geqq 1 \) . If \( x \) is a point of \( U \) such that \( f\left( x\right) \) lies in \( {\mathbf{F}}_{\mu }^{0} \) then \( {f}^{\prime }\left( x\right) \) maps \( \mathbf{E} \) into \( {\mathbf{F}}_{\mu }^{0} \) .
|
Proof. Without loss of generality, we may assume that \( x = 0 \) and \( f\left( x\right) = 0 \) . Let \( W \) be a given neighborhood of 0 in \( \mathbf{F} \) . Suppose that we can find a small element \( v \in \mathbf{E} \) such that \( \mu {f}^{\prime }\left( 0\right) v \neq 0 \) . We can write (for small \( t) \) :\n\n\[ f\left( {tv}\right) = t{f}^{\prime }\left( 0\right) v + o\left( t\right) {w}_{t} \]\n\nwith some element \( {w}_{t} \in W \) . By assumption, \( f\left( {tv}\right) \) lies in \( {\mathbf{F}}_{\mu }^{ + } \) . Applying \( \mu \) we get\n\n\[ {t\mu }{f}^{\prime }\left( 0\right) v + o\left( t\right) \mu \left( {w}_{t}\right) \geqq 0. \]\n\nDividing by \( t \), this yields\n\n\[ \mu {f}^{\prime }\left( 0\right) v \geqq \frac{o\left( t\right) }{t}\mu \left( {w}_{t}\right) \]\n\nReplacing \( t \) by \( - t \), we get a similar inequality on the other side. Letting \( t \) tend to 0 shows that \( \mu {f}^{\prime }\left( 0\right) v = 0 \), a contradiction.
|
Yes
|
Proposition 4.3. Let \( \lambda \) be a functional on \( \mathbf{E} \) and \( \mu \) a functional on \( \mathbf{F} \) . Let \( U \) be open in \( {\mathbf{E}}_{\lambda }^{ + } \) and \( V \) open in \( {\mathbf{F}}_{\mu }^{ + } \) and assume \( U \cap {\mathbf{E}}_{\lambda }^{0}, V \cap {\mathbf{F}}_{\mu }^{0} \) are not empty. Let \( f : U \rightarrow V \) be an isomorphism of class \( {C}^{p}\left( {p \geqq 1}\right) \) . Then \( \lambda \neq 0 \) if and only if \( \mu \neq 0 \) . If \( \lambda \neq 0 \), then \( f \) induces a \( {C}^{p} \) -isomorphism of \( \operatorname{Int}\left( U\right) \) on \( \operatorname{Int}\left( V\right) \) and of \( \partial U \) on \( \partial V \) .
|
Proof. By the functoriality of the derivative, we know that \( {f}^{\prime }\left( x\right) \) is a toplinear isomorphism for each \( x \in U \) . Our first assertion follows from the preceding proposition. We also see that no interior point of \( U \) maps on a boundary point of \( V \) and conversely. Thus \( f \) induces a bijection of \( \partial U \) on \( \partial V \) and a bijection of \( \operatorname{Int}\left( U\right) \) on \( \operatorname{Int}\left( V\right) \) . Since these interiors are open in their respective spaces, our definition of derivative shows that \( f \) induces an isomorphism between them. As for the boundary, it is a submanifold of the full space, and locally, our definition of derivative, together with the product structure, shows that the restriction of \( f \) to \( \partial U \) must be an isomorphism on \( \partial V \) .
|
Yes
|
Proposition 1.1. Let \( \mathbf{E},\mathbf{F} \) be finite dimensional vector spaces. Let \( U \) be open in some Banach space. Let\n\n\[ f : U \times \mathbf{E} \rightarrow \mathbf{F} \]\n\nbe a morphism such that for each \( x \in U \), the map\n\n\[ {f}_{x} : \mathbf{E} \rightarrow \mathbf{F} \]\n\ngiven by \( {f}_{x}\left( v\right) = f\left( {x, v}\right) \) is a linear map. Then the map of \( U \) into \( L\left( {\mathbf{E},\mathbf{F}}\right) \) given by \( x \mapsto {f}_{x} \) is a morphism.
|
Proof. We can write \( \mathbf{F} = {\mathbf{R}}_{1} \times \cdots \times {\mathbf{R}}_{n} \) ( \( n \) copies of \( \mathbf{R} \) ). Using the fact that \( L\left( {\mathbf{E},\mathbf{F}}\right) = L\left( {\mathbf{E},{\mathbf{R}}_{1}}\right) \times \cdots \times L\left( {\mathbf{E},{\mathbf{R}}_{n}}\right) \), it will suffice to prove our assertion when \( \mathbf{F} = \mathbf{R} \) . Similarly, we can assume that \( \mathbf{E} = \mathbf{R} \) also. But in that case, the function \( f\left( {x, v}\right) \) can be written \( g\left( x\right) v \) for some map \( g : U \rightarrow \mathbf{R} \) . Since \( f \) is a morphism, it follows that as a function of each argument \( x, v \) it is also a morphism. Putting \( v = 1 \) shows that \( g \) is a morphism and concludes the proof.
|
Yes
|
Proposition 1.2. Let \( X \) be a manifold, and \( \pi : E \rightarrow X \) a mapping from some set \( E \) into \( X \) . Let \( \left\{ {U}_{i}\right\} \) be an open covering of \( X \), and for each \( i \) suppose that we are given a Banach space \( \mathbf{E} \) and a bijection (commuting with the projection on \( {U}_{i} \) ),\n\n\[{\tau }_{i} : {\pi }^{-1}\left( {U}_{i}\right) \rightarrow {U}_{i} \times \mathbf{E}\]\n\nsuch that for each pair \( i, j \) and \( x \in {U}_{i} \cap {U}_{j} \), the map \( {\left( {\tau }_{j}{\tau }_{i}^{-1}\right) }_{x} \) is a toplinear isomorphism, and condition \( \mathbf{{VB}}\mathbf{3} \) is satisfied as well as the cocycle condition. Then there exists a unique structure of manifold on \( E \) such that \( \pi \) is a morphism, such that \( {\tau }_{i} \) is an isomorphism making \( \pi \) into a vector bundle, and \( \left\{ \left( {{U}_{i},{\tau }_{i}}\right) \right\} \) into a trivialising covering.
|
Proof. By Proposition 3.10 of Chapter I and our condition VB 3, we conclude that the map\n\n\[{\tau }_{j}{\tau }_{i}^{-1} : \left( {{U}_{i} \cap {U}_{j}}\right) \times \mathbf{E} \rightarrow \left( {{U}_{i} \cap {U}_{j}}\right) \times \mathbf{E}\]\n\nis a morphism, and in fact an isomorphism since it has an inverse. From the definition of atlases, we conclude that \( E \) has a unique manifold structure such that the \( {\tau }_{i} \) are isomorphisms. Since \( \pi \) is obtained locally as a composite of morphisms (namely \( {\tau }_{i} \) and the projections of \( {U}_{i} \times \mathbf{E} \) on the first factor), it becomes a morphism. On each fiber \( {\pi }^{-1}\left( x\right) \), we can transport the topological vector space structure of any \( \mathbf{E} \) such that \( x \) lies in \( {U}_{i} \), by means of \( {\tau }_{ix} \) . The result is independent of the choice of \( {U}_{i} \) since \( {\left( {\tau }_{j}{\tau }_{i}^{-1}\right) }_{x} \) is a toplinear isomorphism. Our proposition is proved.
|
Yes
|
Proposition 1.3. Let \( \pi ,{\pi }^{\prime } \) be two vector bundles over manifolds \( X,{X}^{\prime } \) respectively. Let \( {f}_{0} : X \rightarrow {X}^{\prime } \) be a morphism, and suppose that we are given for each \( x \in X \) a continuous linear map\n\n\[ \n{f}_{x} : {\pi }_{x} \rightarrow {\pi }_{{f}_{0}\left( x\right) }^{\prime }\n\]\n\nsuch that, for each \( {x}_{0} \), condition \( \mathbf{{VB}} \) Mor 2 is satisfied. Then the map \( f \)\n\nfrom \( \pi \) to \( {\pi }^{\prime } \) defined by \( {f}_{x} \) on each fiber is a VB-morphism.
|
Proof. One must first check that \( f \) is a morphism. This can be done under the assumption that \( \pi ,{\pi }^{\prime } \) are trivial, say equal to \( U \times \mathbf{E} \) and \( {U}^{\prime } \times {\mathbf{E}}^{\prime } \) (following the notation of VB Mor 2), with trivialising maps equal to the identity. Our map \( f \) is then given by\n\n\[ \n\left( {x, v}\right) \mapsto \left( {{f}_{0}x,{f}_{x}v}\right)\n\]\n\nUsing Proposition 3.10 of Chapter I, we conclude that \( f \) is a morphism, and hence that \( \left( {{f}_{0}, f}\right) \) is a VB-morphism.
|
Yes
|
Proposition 3.1. Let \( X \) be a manifold and let\n\n\[ f : {\pi }^{\prime } \rightarrow \pi \]\n\nbe a VB-morphism of vector bundles over \( X \) . Assume that, for each \( x \in X \), the continuous linear map\n\n\[ {f}_{x} : {E}_{x}^{\prime } \rightarrow {E}_{x} \]\n\nis injective and splits. Then the sequence\n\n\[ 0 \rightarrow {\pi }^{\prime }\overset{f}{ \rightarrow }\pi \]\n\nis exact.
|
Proof. We can assume that \( X \) is connected and that the fibers of \( {E}^{\prime } \) and \( E \) are constant, say equal to the Banach spaces \( {\mathbf{E}}^{\prime } \) and \( \mathbf{E} \) . Let \( a \in X \) . Corresponding to the splitting of \( {f}_{a} \) we know that we have a product decomposition \( \mathbf{E} = {\mathbf{E}}^{\prime } \times \mathbf{F} \) and that there exists an open set \( U \) of \( X \) containing \( a \), together with trivializing maps\n\n\[ \tau : {\pi }^{-1}\left( U\right) \rightarrow U \times \mathbf{E}\;\text{ and }\;{\tau }^{\prime } : {\pi }^{\prime - 1}\left( U\right) \rightarrow U \times {\mathbf{E}}^{\prime }\]\n\nsuch that the composite map\n\n\[ {\mathbf{E}}^{\prime }\overset{{\tau }_{a}^{\prime } - 1}{ \rightarrow }{E}_{a}^{\prime }\overset{{f}_{a}}{ \rightarrow }{E}_{a}\overset{{\tau }_{a}}{ \rightarrow }{\mathbf{E}}^{\prime } \times \mathbf{F} \]\n\nmaps \( {\mathbf{E}}^{\prime } \) on \( {\mathbf{E}}^{\prime } \times 0 \) .\n\nFor any point \( x \) in \( U \), we have a map\n\n\[ {\left( \tau f{\tau }^{\prime - 1}\right) }_{x} : {\mathbf{E}}^{\prime } \rightarrow {\mathbf{E}}^{\prime } \times \mathbf{F} \]\n\nwhich can be represented by a pair of continuous linear maps\n\n\[ \left( {{h}_{11}\left( x\right) ,{h}_{21}\left( x\right) }\right) \]\n\nWe define\n\n\[ h\left( x\right) : {\mathbf{E}}^{\prime } \times \mathbf{F} \rightarrow {\mathbf{E}}^{\prime } \times \mathbf{F} \]\n\nby the matrix\n\n\[ \left( \begin{matrix} {h}_{11}\left( x\right) & 0 \\ {h}_{21}\left( x\right) & \text{ id } \end{matrix}\right) \]\n\noperating on the right on a vector \( \left( {v, w}\right) \in {\mathbf{E}}^{\prime } \times \mathbf{F} \) . Then \( h\left( x\right) \) restricted to \( {\mathbf{E}}^{\prime } \times 0 \) has the same action as \( {\left( \tau f{\tau }^{\prime - 1}\right) }_{x} \).\n\nThe map \( x \mapsto h\left( x\right) \) is a morphism of \( U \) into \( L\left( {\mathbf{E},\mathbf{E}}\right) \) and since it is continuous, it follows that for \( U \) small enough around our fixed point \( a \), it maps \( U \) into the group of toplinear automorphisms of \( \mathbf{E} \) . This proves our proposition.
|
Yes
|
Proposition 3.2. Let \( X \) be a manifold and let\n\n\[ g : \pi \rightarrow {\pi }^{\prime \prime } \]\n\nbe a VB-morphism of vector bundles over \( X \) . Assume that for each \( x \in X \), the continuous linear map\n\n\[ {g}_{x} : {E}_{x} \rightarrow {E}_{x}^{\prime \prime } \]\n\n is surjective and has a kernel that splits. Then the sequence\n\n\[ \pi \overset{g}{ \rightarrow }{\pi }^{\prime \prime } \rightarrow 0 \]\n\nis exact.
|
Proof. It is dual to the preceding one and we leave it to the reader.
|
No
|
Theorem 4.1. Let \( \lambda \) be a functor as above, of class \( {C}^{p}, p \geqq 0 \) . Then for each manifold \( X \), there exists a functor \( {\lambda }_{X} \), on vector bundles (of class \( {C}^{p} \) ) satisfying the following properties. For any bundles \( \alpha ,\beta \) in \( \operatorname{VB}\left( {X,\mathfrak{A}}\right) \) and \( \operatorname{VB}\left( {X,\mathfrak{B}}\right) \) respectively, and VB-morphisms \( f : {\alpha }^{\prime } \rightarrow \alpha \;\text{ and }\;g : \beta \rightarrow {\beta }^{\prime } \) in the respective categories, and for each \( x \in X \), we have: OP 1. \( {\lambda }_{X}{\left( \alpha ,\beta \right) }_{x} = \lambda \left( {{\alpha }_{x},{\beta }_{x}}\right) \). OP 2. \( {\lambda }_{X}{\left( f, g\right) }_{x} = \lambda \left( {{f}_{x},{g}_{x}}\right) \). OP 3. If \( \alpha \) is the trivial bundle \( X \times \mathbf{E} \) and \( \beta \) the trivial bundle \( X \times \mathbf{F} \) , then \( {\lambda }_{X}\left( {\alpha ,\beta }\right) \) is the trivial bundle \( X \times \lambda \left( {\mathbf{E},\mathbf{F}}\right) \) . OP 4. If \( h : Y \rightarrow X \) is a \( {C}^{p} \) -morphism, then \( {\lambda }_{Y}^{ * }\left( {{h}^{ * }\alpha ,{h}^{ * }\beta }\right) = {h}^{ * }{\lambda }_{X}\left( {\alpha ,\beta }\right) .
|
Proof. We may assume that \( X \) is connected, so that all the fibers are toplinearly isomorphic to a fixed space. For each open subset \( U \) of \( X \) we let the total space \( {\lambda }_{U}\left( {{E}_{\alpha },{E}_{\beta }}\right) \) of \( {\lambda }_{U}\left( {\alpha ,\beta }\right) \) be the union of the sets \( \{ x\} \times \lambda \left( {{\alpha }_{x},{\beta }_{x}}\right) \) (identified harmlessly throughout with \( \lambda \left( {{\alpha }_{x},{\beta }_{x}}\right) \) ), as \( x \) ranges over \( U \) . We can find a covering \( \left\{ {U}_{i}\right\} \) of \( X \) with trivializing maps \( \left\{ {\tau }_{i}\right\} \) for \( \alpha \), and \( \left\{ {\sigma }_{i}\right\} \) for \( \beta \), \( {\tau }_{i} : {\alpha }^{-1}\left( {U}_{i}\right) \rightarrow {U}_{i} \times \mathbf{E} \) \( {\sigma }_{i} : {\beta }^{-1}\left( {U}_{i}\right) \rightarrow {U}_{i} \times \mathbf{F}. \) We have a bijection \( \lambda \left( {{\tau }_{i}^{-1},{\sigma }_{i}}\right) : {\lambda }_{{U}_{i}}\left( {{E}_{\alpha },{E}_{\beta }}\right) \rightarrow {U}_{i} \times \lambda \left( {\mathbf{E},\mathbf{F}}\right) \) obtained by taking on each fiber the map \( \lambda \left( {{\tau }_{ix}^{-1},{\sigma }_{ix}}\right) : \lambda \left( {{\alpha }_{x},{\beta }_{x}}\right) \rightarrow \lambda \left( {\mathbf{E},\mathbf{F}}\right) . \) We must verify that VB 3 is satisfied. This means looking at the map \( x \rightarrow \lambda \left( {{\tau }_{jx}^{-1},{\sigma }_{jx}}\right) \circ \lambda {\left( {\tau }_{ix}^{-1},{\sigma }_{ix}\right) }^{-1}. \) The expression on the right is equal to \( \lambda \left( {{\tau }_{ix}{\tau }_{jx}^{-1},{\sigma }_{jx}{\sigma }_{ix}^{-1}}\right) \) Since \( \lambda \) is a functor of class \( {\mathbf{C}}^{p} \), we see that we get a map \( {U}_{i} \cap {U}_{j} \rightarrow L\left( {\lambda \left( {\mathbf{E},\mathbf{F}}\right) ,\lambda \left( {\mathbf{E},\mathbf{F}}\right) }\right) \) which is a \( {C}^{p} \) -morphism. Furthermore, since \( \lambda \) is a functor, the transition mappings are in fact toplinear isomorphism, and VB 2, VB 3 are proved. The proof of the analogous statement for \( {\lambda }_{X}\left( {f, g}\right) \), to the effect that it is a VB-morphism, proceeds in an analogous way, again using the hypothesis that \( \lambda \) is of class \( {C}^{p} \) . Condition OP 3 is obviously satisfied, and OP 4 follows by localizing. This proves our theorem.
|
Yes
|
If \( \mu \) is another functor of class \( {C}^{p} \) with the same variance as \( \lambda \), and if we have a natural transformation of functors \( t : \lambda \rightarrow \mu \), then for each \( X \), the mapping\n\n\[ \n{t}_{X} : {\lambda }_{X} \rightarrow {\mu }_{X} \n\]\n\ndefined on each fiber by the map\n\n\[ \nt\left( {{\alpha }_{x},{\beta }_{x}}\right) : \lambda \left( {{\alpha }_{x},{\beta }_{x}}\right) \rightarrow \mu \left( {{\alpha }_{x},{\beta }_{x}}\right) ,\n\]\n\nis a natural transformation of functors (in the VB-category).
|
Proof. For simplicity of notation, assume that \( \lambda \) and \( \mu \) are both functors of one variable, and both covariant. For each open set \( U = {U}_{i} \) of a trivializing covering for \( \beta \), we have a commutative diagram: \n\nThe vertical maps are trivializing VB-isomorphism, and the top horizontal map is a VB-morphism. Hence \( {t}_{U} \) is a VB-morphism, and our assertion is proved.
|
Yes
|
Proposition 5.1. Let \( X, Y \) be manifolds and \( {f}_{0} : X \rightarrow Y \) a morphism. Let \( \alpha ,\beta \) be vector bundles over \( X, Y \) respectively, and let \( f, g : \alpha \rightarrow \beta \) be two VB-morphisms over \( {f}_{0} \) . Then the map \( f + g \) defined by the formula\n\n\[ \n{\left( f + g\right) }_{x} = {f}_{x} + {g}_{x} \n\]\n\nis also a VB-morphism. Furthermore, if \( \psi : Y \rightarrow \mathbf{R} \) is a function on \( Y \) , then the map \( {\psi f} \) defined by\n\n\[ \n{\left( \psi f\right) }_{x} = \psi \left( {{f}_{0}\left( x\right) }\right) {f}_{x} \n\]\n\nis also a VB-morphism.
|
Proof. Both assertions are immediate consequences of Proposition 3.10 of Chapter I.
|
No
|
Proposition 5.2. Let \( X \) be a manifold admitting partitions of unity. Let \( 0 \rightarrow \alpha \overset{f}{ \rightarrow }\beta \) be an exact sequence of vector bundles over \( X \) . Then there exists a surjective VB-morphism \( g : \beta \rightarrow \alpha \) whose kernel splits at each point, such that \( g \circ f = \mathrm{{id}} \) .
|
Proof. By the definition of exact sequence, there exists a partition of unity \( \left\{ \left( {{U}_{i},{\psi }_{i}}\right) \right\} \) on \( X \) such that for each \( i \), we can split the sequence over \( {U}_{i} \) . In other words, there exists for each \( i \) a VB-morphism\n\n\[ \n{g}_{i} : \beta \left| {{U}_{i} \rightarrow \alpha }\right| {U}_{i} \n\] \n\nwhich is surjective, whose kernel splits, and such that \( {g}_{i} \circ {f}_{i} = {\mathrm{{id}}}_{i} \) . We let \( g = \sum {\psi }_{i}{g}_{i} \) . Then \( g \) is a VB-morphism of \( \beta \) into \( \alpha \) by what we have just \n\nseen, and \n\[ \ng \circ f = \sum {\psi }_{i}{g}_{i}{f}_{i} = \mathrm{{id}}. \n\] \n\nIt is trivial that \( g \) is surjective because \( g \circ f = \mathrm{{id}} \) . The kernel of \( {g}_{x} \) splits at each point \( x \) because it has a closed complement, namely \( {f}_{x}{\alpha }_{x} \) . This concludes the proof.
|
Yes
|
Proposition 5.3. Let \( X \) be a manifold admitting partitions of unity. Let \( \pi \) be a vector bundle of finite type in \( \operatorname{VB}\left( {X,\mathbf{E}}\right) \), where \( \mathbf{E} \) is a Banach space. Then there exists an integer \( k > 0 \) and a vector bundle \( \alpha \) in \( \operatorname{VB}\left( {X,{\mathbf{E}}^{k}}\right) \) such that \( \pi \oplus \alpha \) is trivializable.
|
Proof. We shall prove that there exists an exact sequence\n\n\[ 0 \rightarrow \pi \overset{f}{ \rightarrow }\beta \]\n\nwith \( {E}_{\beta } = X \times {\mathbf{E}}^{k} \) . Our theorem will follow from the preceding proposition.\n\nLet \( \left\{ {{U}_{i},{\tau }_{i})}\right\} \) be a finite trivialization of \( \pi \) with \( i = 1,\ldots, k \) . Let \( \left\{ \left( {{U}_{i},{\psi }_{i}}\right) \right\} \) be a partition of unity. We define\n\n\[ f : {E}_{\pi } \rightarrow X \times {\mathbf{E}}^{k} \]\n\nas follows. If \( x \in X \) and \( v \) is in the fiber of \( {E}_{\pi } \) at \( x \), then\n\n\[ {f}_{x}\left( v\right) = \left( {x,{\psi }_{1}\left( x\right) {\tau }_{1}\left( v\right) ,\ldots ,{\psi }_{k}\left( x\right) {\tau }_{k}\left( v\right) }\right) . \]\n\nThe expression on the right makes sense, because in case \( x \) does not lie in \( {U}_{i} \) then \( {\psi }_{i}\left( x\right) = 0 \) and we do not have to worry about the expression \( {\tau }_{i}\left( v\right) \) . If \( x \) lies in \( {U}_{i} \), then \( {\tau }_{i}\left( v\right) \) means \( {\tau }_{ix}\left( v\right) \) .\n\nGiven any point \( x \), there exists some index \( i \) such that \( {\psi }_{i}\left( x\right) > 0 \) and hence \( f \) is injective. Furthermore, for this \( x \) and this index \( i,{f}_{x} \) maps \( {E}_{x} \) onto a closed subspace of \( {\mathbf{E}}^{k} \), which admits a closed complement, namely\n\n\[ \mathbf{E} \times \cdots \times 0 \times \cdots \times \mathbf{E} \]\n\nwith 0 in the \( i \) -th place. This proves our proposition.
|
Yes
|
Proposition 1.1. Let \( J \) be an open interval of \( \mathbf{R} \) containing 0, and \( U \) open in the Banach space \( \mathbf{E} \) . Let \( {x}_{0} \) be a point of \( U \), and \( a > 0, a < 1 \) a real number such that the closed ball \( {\bar{B}}_{3a}\left( {x}_{0}\right) \) lies in \( U \) . Assume that we have a continuous map\n\n\[ \nf : J \times U \rightarrow \mathbf{E} \n\]\n\nwhich is bounded by a constant \( L \geqq 1 \) on \( J \times U \), and satisfies a Lipschitz condition on \( U \) uniformly with respect to \( J \), with constant \( K \geqq 1 \) . If \( b < a/{LK} \), then for each \( x \) in \( {\bar{B}}_{a}\left( {x}_{0}\right) \) there exists a unique flow\n\n\[ \n\alpha : {J}_{b} \times {B}_{a}\left( {x}_{0}\right) \rightarrow U.\n\]\n\nIf \( f \) is of class \( {C}^{p}\left( {p \geqq 1}\right) \), then so is each integral curve \( {\alpha }_{x} \) .
|
Proof. Let \( {I}_{b} \) be the closed interval \( - b \leqq t \leqq b \), and let \( x \) be a fixed point in \( {\bar{B}}_{a}\left( {x}_{0}\right) \) . Let \( M \) be the set of continuous maps\n\n\[ \n\alpha : {I}_{b} \rightarrow {\bar{B}}_{2a}\left( {x}_{0}\right) \n\]\n\nof the closed interval into the closed ball of center \( {x}_{0} \) and radius \( {2a} \), such that \( \alpha \left( 0\right) = x \) . Then \( M \) is a complete metric space if we define as usual the distance between maps \( \alpha ,\beta \) to be\n\n\[ \n\mathop{\sup }\limits_{{t \in {I}_{b}}}\left| {\alpha \left( t\right) - \beta \left( t\right) }\right| \n\]\n\n\nWe shall now define a mapping\n\n\[ \nS : M \rightarrow M \n\]\n\nof \( M \) into itself. For each \( \alpha \) in \( M \), we let \( {S\alpha } \) be defined by\n\n\[ \n\left( {S\alpha }\right) \left( t\right) = x + {\int }_{0}^{t}f\left( {u,\alpha \left( u\right) }\right) {du}. \n\]\n\nThen \( {S\alpha } \) is certainly continuous, we have \( {S\alpha }\left( 0\right) = x \), and the distance of any point on \( {S\alpha } \) from \( x \) is bounded by the norm of the integral, which is bounded by\n\n\[ \nb\sup \left| {f\left( {u, y}\right) }\right| \leqq {bL} < a. \n\]\n\nThus \( {S\alpha } \) lies in \( M \) .\n\nWe contend that our map \( S \) is a shrinking map. Indeed,\n\n\[ \n\left| {{S\alpha } - {S\beta }}\right| \leqq b\sup \left| {f\left( {u,\alpha \left( u\right) }\right) - f\left( {u,\beta \left( u\right) }\right) }\right| \n\]\n\n\[ \n\leqq {bK}\left| {\alpha - \beta }\right| \n\]\n\nthereby proving our contention.\n\nBy the shrinking lemma (Chapter I, Lemma 5.1) our map has a unique fixed point \( \alpha \), and by definition, \( \alpha \left( t\right) \) satisfies the desired integral relation. Our remark above concludes the proof.
|
Yes
|
Corollary 1.2. The local flow \( \alpha \) in Proposition 1.1 is continuous. Furthermore, the map \( x \mapsto {\alpha }_{x} \) of \( {\bar{B}}_{a}\left( {x}_{0}\right) \) into the space of curves is continuous, and in fact satisfies a Lipschitz condition.
|
Proof. The second statement obviously implies the first. So fix \( x \) in \( {\bar{B}}_{a}\left( {x}_{0}\right) \) and take \( y \) close to \( x \) in \( {\bar{B}}_{a}\left( {x}_{0}\right) \) . We let \( {S}_{x} \) be the shrinking map of the theorem, corresponding to the initial condition \( x \) . Then\n\n\[ \begin{Vmatrix}{{\alpha }_{x} - {S}_{y}{\alpha }_{x}}\end{Vmatrix} = \begin{Vmatrix}{{S}_{x}{\alpha }_{x} - {S}_{y}{\alpha }_{x}}\end{Vmatrix} \leqq \left| {x - y}\right| .\n\]\n\nLet \( C = {bK} \) so \( 0 < C < 1 \) . Then\n\n\[ \begin{Vmatrix}{{\alpha }_{x} - {S}_{y}^{n}{\alpha }_{x}}\end{Vmatrix} \leqq \begin{Vmatrix}{{\alpha }_{x} - {S}_{y}{\alpha }_{x}}\end{Vmatrix} + \begin{Vmatrix}{{S}_{y}{\alpha }_{x} - {S}_{y}^{2}{\alpha }_{x}}\end{Vmatrix} + \cdots + \begin{Vmatrix}{{S}_{y}^{n - 1}{\alpha }_{x} - {S}_{y}^{n}{\alpha }_{x}}\end{Vmatrix}\n\]\n\n\[ \leqq \left( {1 + C + \cdots + {C}^{n - 1}}\right) \left| {x - y}\right| \text{.} \]\n\nSince the limit of \( {S}_{v}^{n}{\alpha }_{x} \) is equal to \( {\alpha }_{y} \) as \( n \) goes to infinity, the continuity of the map \( x \mapsto {\alpha }_{x} \) follows at once. In fact, the map satisfies a Lipschitz condition as stated.
|
Yes
|
Theorem 1.3 (Uniqueness Theorem). Let \( U \) be open in \( \mathbf{E} \) and let \( f : U \rightarrow E \) be a vector field of class \( {C}^{p}, p \geqq 1 \) . Let\n\n\[ \n{\alpha }_{1} : {J}_{1} \rightarrow U\;\text{ and }\;{\alpha }_{2} : {J}_{2} \rightarrow U \]\n\nbe two integral curves for \( f \) with the same initial condition \( {x}_{0} \) . Then \( {\alpha }_{1} \) and \( {\alpha }_{2} \) are equal on \( {J}_{1} \cap {J}_{2} \) .
|
Proof. Let \( Q \) be the set of numbers \( b \) such that \( {\alpha }_{1}\left( t\right) = {\alpha }_{2}\left( t\right) \) for\n\n\[ \n0 \leqq t < b. \]\n\nThen \( Q \) contains some number \( b > 0 \) by the local uniqueness theorem. If \( Q \) is not bounded from above, the equality of \( {\alpha }_{1}\left( t\right) \) and \( {\alpha }_{2}\left( t\right) \) for all \( t > 0 \) follows at once. If \( Q \) is bounded from above, let \( b \) be its least upper bound. We must show that \( b \) is the right end point of \( {J}_{1} \cap {J}_{2} \) . Suppose that this is not the case. Define curves \( {\beta }_{1} \) and \( {\beta }_{2} \) near 0 by\n\n\[ \n{\beta }_{1}\left( t\right) = {\alpha }_{1}\left( {b + t}\right) \;\text{ and }\;{\beta }_{2}\left( t\right) = {\alpha }_{2}\left( {b + t}\right) . \]\n\nThen \( {\beta }_{1} \) and \( {\beta }_{2} \) are integral curves of \( f \) with the initial conditions \( {\alpha }_{1}\left( b\right) \) and \( {\alpha }_{2}\left( b\right) \) respectively. The values \( {\beta }_{1}\left( t\right) \) and \( {\beta }_{2}\left( t\right) \) are equal for small negative \( t \) because \( b \) is the least upper bound of \( Q \) . By continuity it follows that \( {\alpha }_{1}\left( b\right) = {\alpha }_{2}\left( b\right) \), and finally we see from the local uniqueness theorem that\n\n\[ \n{\beta }_{1}\left( t\right) = {\beta }_{2}\left( t\right) \]\n\nfor all \( t \) in some neighborhood of 0, whence \( {\alpha }_{1} \) and \( {\alpha }_{2} \) are equal in a neighborhood of \( b \), contradicting the fact that \( b \) is a least upper bound of \( Q \) . We can argue the same way towards the left end points, and thus prove our statement.
|
Yes
|
Proposition 1.4. Let \( {\varphi }_{1} \) and \( {\varphi }_{2} \) be two \( {\epsilon }_{1} \) - and \( {\epsilon }_{2} \) -approximate solutions of \( f \) on \( {J}_{0} \) respectively, and let \( \epsilon = {\epsilon }_{1} + {\epsilon }_{2} \) . Assume that \( f \) is Lipschitz with constant \( K \) on \( U \) uniformly in \( {J}_{0} \), or that \( {D}_{2}f \) exists and is bounded by \( K \) on \( J \times U \) . Let \( {t}_{0} \) be a point of \( {J}_{0} \) . Then for any \( t \) in \( {J}_{0} \), we have\n\n\[ \left| {{\varphi }_{1}\left( t\right) - {\varphi }_{2}\left( t\right) }\right| \leqq \left| {{\varphi }_{1}\left( {t}_{0}\right) - {\varphi }_{2}\left( {t}_{0}\right) }\right| {e}^{K\left| {t - {t}_{0}}\right| } + \frac{\epsilon }{K}{e}^{K\left| {t - {t}_{0}}\right| }.\]
|
Proof. By assumption, we have\n\n\[ \left| {{\varphi }_{1}^{\prime }\left( t\right) - f\left( {t,{\varphi }_{1}\left( t\right) }\right) }\right| \leqq {\epsilon }_{1} \]\n\n\[ \left| {{\varphi }_{2}^{\prime }\left( t\right) - f\left( {t,{\varphi }_{2}\left( t\right) }\right) }\right| \leqq {\epsilon }_{2} \]\n\nFrom this we get\n\n\[ \left| {{\varphi }_{1}^{\prime }\left( t\right) - {\varphi }_{2}^{\prime }\left( t\right) + f\left( {t,{\varphi }_{2}\left( t\right) }\right) - f\left( {t,{\varphi }_{1}\left( t\right) }\right) }\right| \leqq \epsilon . \]\n\nSay \( t \geqq {t}_{0} \) to avoid putting bars around \( t - {t}_{0} \) . Let\n\n\[ \psi \left( t\right) = \left| {{\varphi }_{1}\left( t\right) - {\varphi }_{2}\left( t\right) }\right| \]\n\n\[ \omega \left( t\right) = \left| {f\left( {t,{\varphi }_{1}\left( t\right) }\right) - f\left( {t,{\varphi }_{2}\left( t\right) }\right) }\right| . \]\n\nThen, after integrating from \( {t}_{0} \) to \( t \), and using triangle inequalities we obtain\n\n\[ \left| {\psi \left( t\right) - \psi \left( {t}_{0}\right) }\right| \leqq \epsilon \left( {t - {t}_{0}}\right) + {\int }_{{t}_{0}}^{t}\omega \left( u\right) {du} \]\n\n\[ \leqq \epsilon \left( {t - {t}_{0}}\right) + K{\int }_{{t}_{0}}^{t}\psi \left( u\right) {du} \]\n\n\[ \leqq K{\int }_{{t}_{0}}^{t}\left\lbrack {\psi \left( u\right) + \epsilon /K}\right\rbrack {du} \]\n\nand finally the recurrence relation\n\n\[ \psi \left( t\right) \leqq \psi \left( {t}_{0}\right) + K{\int }_{{t}_{0}}^{t}\left\lbrack {\psi \left( u\right) + \epsilon /K}\right\rbrack {du}. \]\n\nOn any closed subinterval of \( {J}_{0} \), our map \( \psi \) is bounded. If we add \( \epsilon /K \) to both sides of this last relation, then we see that our proposition will follow from the next lemma.
|
Yes
|
Lemma 1.5. Let \( g \) be a positive real valued function on an interval, bounded by a number \( L \) . Let \( {t}_{0} \) be in the interval, say \( {t}_{0} \leqq t \), and assume that there are numbers \( A, K \geqq 0 \) such that\n\n\[ g\left( t\right) \leqq A + K{\int }_{{t}_{0}}^{t}g\left( u\right) {du} \]\n\nThen for all integers \( n \geqq 1 \) we have\n\n\[ g\left( t\right) \leqq A\left\lbrack {1 + \frac{K\left( {t - {t}_{0}}\right) }{1!} + \cdots + \frac{{K}^{n - 1}{\left( t - {t}_{0}\right) }^{n - 1}}{\left( {n - 1}\right) !}}\right\rbrack + \frac{L{K}^{n}{\left( t - {t}_{0}\right) }^{n}}{n!}. \]
|
Proof. The statement is an assumption for \( n = 1 \) . We proceed by induction. We integrate from \( {t}_{0} \) to \( t \), multiply by \( K \), and use the recurrence relation. The statement with \( n + 1 \) then drops out of the statement with \( n \) .
|
No
|
Corollary 1.6. Let \( f : J \times U \rightarrow \mathbf{E} \) be continuous, and satisfy a Lipschitz condition on \( U \) uniformly with respect to \( J \) . Let \( {x}_{0} \) be a point of \( U \) . Then there exists an open subinterval \( {J}_{0} \) of \( J \) containing 0, and an open subset of \( U \) containing \( {x}_{0} \) such that \( f \) has a unique flow\n\n\[ \alpha : {J}_{0} \times {U}_{0} \rightarrow U. \]\n\nWe can select \( {J}_{0} \) and \( {U}_{0} \) such that \( \alpha \) is continuous and satisfies a Lipschitz condition on \( {J}_{0} \times {U}_{0} \) .
|
Proof. Given \( x, y \) in \( {U}_{0} \) we let \( {\varphi }_{1}\left( t\right) = \alpha \left( {t, x}\right) \) and \( {\varphi }_{2}\left( t\right) = \alpha \left( {t, y}\right) \), using Proposition 1.6 to get \( {J}_{0} \) and \( {U}_{0} \) . Then \( {\epsilon }_{1} = {\epsilon }_{2} = 0 \) . For \( s, t \) in \( {J}_{0} \) we obtain\n\n\[ \left| {\alpha \left( {t, x}\right) - \alpha \left( {s, y}\right) }\right| \leqq \left| {\alpha \left( {t, x}\right) - \alpha \left( {t, y}\right) }\right| + \left| {\alpha \left( {t, y}\right) - \alpha \left( {s, y}\right) }\right| \]\n\n\[ \leqq \left| {x - y}\right| {e}^{K} + \left| {t - s}\right| L \]\n\nif we take \( {J}_{0} \) of small length, and \( L \) is a bound for \( f \) . Indeed, the term containing \( \left| {x - y}\right| \) comes from Proposition 1.4, and the term containing \( \left| {t - s}\right| \) comes from the definition of the integral curve by means of an integral and the bound \( L \) for \( f \) . This proves our corollary.
|
Yes
|
Corollary 1.7. Let \( J \) be an open interval of \( \mathbf{R} \) containing 0 and let \( U \) be open in \( \mathbf{E} \). Let \( f : J \times U \rightarrow \mathbf{E} \) be a continuous map, which is Lipschitz on \( U \) uniformly for every compact subinterval of \( J \). Let \( {t}_{0} \in J \) and let \( {\varphi }_{1} \), \( {\varphi }_{2} \) be two morphisms of class \( {C}^{1} \) such that \( {\varphi }_{1}\left( {t}_{0}\right) = {\varphi }_{2}\left( {t}_{0}\right) \) and satisfying the relation\n\n\[ \n{\varphi }^{\prime }\left( t\right) = f\left( {t,\varphi \left( t\right) }\right) \n\]\n\nfor all \( t \) in \( J \). Then \( {\varphi }_{1}\left( t\right) = {\varphi }_{2}\left( t\right) \).
|
Proof. We can take \( \epsilon = 0 \) in the proposition.
|
Yes
|
Corollary 1.8. Let \( J \) be the open interval \( \left( {a, b}\right) \) and let \( U \) be open in \( \mathbf{E} \) . Let \( f : J \times U \rightarrow \mathbf{E} \) be a continuous map, which is Lipschitz on \( U \), uniformly for every compact subset of \( J \). Let \( \alpha \) be an integral curve of \( f \), defined on a maximal open subinterval \( \left( {{a}_{0},{b}_{0}}\right) \) of \( J \). Assume: (i) There exists \( \epsilon > 0 \) such that \( \overline{\alpha \left( \left( {{b}_{0} - \epsilon ,{b}_{0}}\right) \right) } \) is contained in \( U \). (ii) There exists a number \( B > 0 \) such that \( \left| {f\left( {t,\alpha \left( t\right) }\right) }\right| \leqq B \) for all \( t \) in \( \left( {{b}_{0} - \epsilon ,{b}_{0}}\right) \nThen \( {b}_{0} = b \) .
|
Proof. From the integral expression for \( \alpha \), namely \[ \alpha \left( t\right) = \alpha \left( {t}_{0}\right) + {\int }_{{t}_{0}}^{t}f\left( {u,\alpha \left( u\right) }\right) {du} \] we see that for \( {t}_{1},{t}_{2} \) in \( \left( {{b}_{0} - \epsilon ,{b}_{0}}\right) \) we have \[ \left| {\alpha \left( {t}_{1}\right) - \alpha \left( {t}_{2}\right) }\right| \leqq B\left| {{t}_{1} - {t}_{2}}\right| \] From this it follows that the limit \[ \mathop{\lim }\limits_{{t \rightarrow {b}_{0}}}\alpha \left( t\right) \] exists, and is equal to an element \( {x}_{0} \) of \( U \) (by hypothesis (i)). Assume that \( {b}_{0} \neq b \). By the local existence theorem, there exists an integral curve \( \beta \) of \( f \) defined on an open interval containing \( {b}_{0} \) such that \( \beta \left( {b}_{0}\right) = {x}_{0} \) and \( {\beta }^{\prime }\left( t\right) = f\left( {t,\beta \left( t\right) }\right) \). Then \( {\beta }^{\prime } = {\alpha }^{\prime } \) on an open interval to the left of \( {b}_{0} \), and hence \( \alpha ,\beta \) differ by a constant on this interval. Since their limit as \( t \rightarrow {b}_{0} \) are equal, this constant is 0. Thus we have extended the domain of definition of \( \alpha \) to a larger interval, as was to be shown.
|
Yes
|
Corollary 1.10. Let the notation be as in Proposition 1.9. For each \( x \in V \) and \( z \in E \) the curve\n\n\[ \beta \left( {t, x, z}\right) = \lambda \left( {t, x}\right) z \]\n\nwith initial condition \( \beta \left( {0, x, z}\right) = z \) is a solution of the differential\nequation\n\n\[ {D}_{1}\beta \left( {t, x, z}\right) = g\left( {t, x}\right) \beta \left( {t, x, z}\right) . \]\n\nFurthermore, \( \beta \) is continuous in its three variables.
|
Proof. Obvious.
|
No
|
Lemma 1.13. Let \( {x}_{0} \in U \) . Let \( a > 0 \) be such that \( {Df} \) is bounded, say by a number \( {C}_{1} > 0 \), on the ball \( {B}_{a}\left( {x}_{0}\right) \) (we can always find such a since \( {Df} \) is continuous at \( \left. {x}_{0}\right) \) . Let \( b < 1/{C}_{1} \) . Then \( {D}_{2}T\left( {x,\sigma }\right) \) is invertible for all \( \left( {x,\sigma }\right) \) in \( {B}_{a}\left( {x}_{0}\right) \times V \) .
|
Proof. We have an estimate\n\n\[ \left| {{\int }_{0}^{t}{Df}\left( {\sigma \left( u\right) }\right) h\left( u\right) {du}}\right| \leqq b{C}_{1}\parallel h\parallel .\n\]\n\nThis means that\n\n\[ \left| {{D}_{2}T\left( {x,\sigma }\right) + I}\right| < 1 \]\n\nand hence that \( {D}_{2}T\left( {x,\sigma }\right) \) is invertible, as a continuous linear map, thus proving Lemma 1.13.
|
Yes
|
Lemma 1.15. Let \( f : U \rightarrow E \) be a \( {C}^{p} \) vector field on the open set \( U \) of \( E \), and let \( \alpha \) be its flow. Abbreviate \( \alpha \left( {t, x}\right) \) by \( {tx} \), if \( \left( {t, x}\right) \) is in the domain of definition of the flow. Let \( x \in U \) . If \( {t}_{0} \) lies in \( J\left( x\right) \), then
|
Proof. The two curves defined by\n\n\[ t \mapsto \alpha \left( {t,\alpha \left( {{t}_{0}, x}\right) }\right) \;\text{ and }\;t \mapsto \alpha \left( {t + {t}_{0}, x}\right) \]\n\nare integral curves of the same vector field, with the same initial condition \( {t}_{0}x \) at \( t = 0 \) . Hence they have the same domain of definition \( J\left( {{t}_{0}x}\right) \) . Hence \( {t}_{1} \) lies in \( J\left( {{t}_{0}x}\right) \) if and only if \( {t}_{1} + {t}_{0} \) lies in \( J\left( x\right) \) . This proves the first assertion. The second assertion comes from the uniqueness of the integral curve having given initial condition, whence the theorem follows.
|
Yes
|
Theorem 2.1. Let \( {\alpha }_{1} : {J}_{1} \rightarrow X \) and \( {\alpha }_{2} : {J}_{2} \rightarrow X \) be two integral curves of the vector field \( \xi \) on \( X \), with the same initial condition \( {x}_{0} \) . Then \( {\alpha }_{1} \) and \( {\alpha }_{2} \) are equal on \( {J}_{1} \cap {J}_{2} \) .
|
Proof. Let \( {J}^{ * } \) be the set of points \( t \) such that \( {\alpha }_{1}\left( t\right) = {\alpha }_{2}\left( t\right) \) . Then \( {J}^{ * } \) certainly contains a neighborhood of 0 by the local uniqueness theorem. Furthermore, since \( X \) is Hausdorff, we see that \( {J}^{ * } \) is closed. We must show that it is open. Let \( {t}^{ * } \) be in \( {J}^{ * } \) and define \( {\beta }_{1},{\beta }_{2} \) near 0 by\n\n\[ \n{\beta }_{1}\left( t\right) = {\alpha }_{1}\left( {{t}^{ * } + t}\right)\n\]\n\n\[ \n{\beta }_{2}\left( t\right) = {\alpha }_{2}\left( {{t}^{ * } + t}\right)\n\]\n\nThen \( {\beta }_{1} \) and \( {\beta }_{2} \) are integral curves of \( \xi \) with initial condition \( {\alpha }_{1}\left( {t}^{ * }\right) \) and \( {\alpha }_{2}\left( {t}^{ * }\right) \) respectively, so by the local uniqueness theorem, \( {\beta }_{1} \) and \( {\beta }_{2} \) agree in a neighborhood of 0 and thus \( {\alpha }_{1},{\alpha }_{2} \) agree in a neighborhood of \( {t}^{ * } \) , thereby proving our theorem.
|
Yes
|
Theorem 2.2. Let \( \xi \) be a vector field on \( X \), and \( \alpha \) its flows. Let \( x \) be a point of \( X \) . If \( {t}_{0} \) lies in \( J\left( x\right) \), then\n\n\[ J\left( {{t}_{0}x}\right) = J\left( x\right) - {t}_{0} \]\n\n(translation of \( J\left( x\right) \) by \( - {t}_{0} \) ), and we have for all \( t \) in \( J\left( x\right) - {t}_{0} \) :\n\n\[ t\left( {{t}_{0}x}\right) = \left( {t + {t}_{0}}\right) x \]
|
Proof. Our first assertion follows immediately from the maximality assumption concerning the domains of the integral curves. The second is equivalent to saying that the two curves given by the left-hand side and right-hand side of the last equality are equal. They are both integral curves for the vector field, with initial condition \( {t}_{0}x \) and must therefore be equal.
|
Yes
|
Theorem 2.3. Let \( \xi \) be a vector field on \( X \), and \( x \) a point of \( X \) . Assume that \( {t}^{ + }\left( x\right) < \infty \) . Given a compact set \( A \subset X \), there exists \( \epsilon > 0 \) such that for all \( t > {t}^{ + }\left( x\right) - \epsilon \), the point \( {tx} \) does not lie in \( A \), and similarly for \( {t}^{ - } \) .
|
Proof. Suppose such \( \epsilon \) does not exist. Then we can find a sequence \( {t}_{n} \) of real numbers approaching \( {t}^{ + }\left( x\right) \) from below, such that \( {t}_{n}x \) lies in \( A \) . Since \( A \) is compact, taking a subsequence if necessary, we may assume that \( {t}_{n}x \) converges to a point in \( A \) . By the local existence theorem, there exists a neighborhood \( U \) of this point \( y \) and a number \( \delta > 0 \) such that \( {t}^{ + }\left( z\right) > \delta \) for all \( z \in U \) . Taking \( n \) large, we have\n\n\[ \n{t}^{ + }\left( x\right) < \delta + {t}_{n} \n\]\n\nand \( {t}_{n}x \) is in \( U \) . Then by Theorem 2.2,\n\n\[ \n{t}^{ + }\left( x\right) = {t}^{ + }\left( {{t}_{n}x}\right) + {t}_{n} > \delta + {t}_{n} > {t}^{ + }\left( x\right) \n\]\n\ncontradiction.
|
Yes
|
Proposition 2.5. Let \( \\mathbf{E} \) be a Banach space, and \( X \) an \( \\mathbf{E} \) -manifold. Let \( \\xi \) be a vector field on \( X \) . Assume that there exist numbers \( a > 0 \) and \( K > 0 \) such that every point \( x \) of \( X \) admits a chart \( \\left( {U,\\varphi }\\right) \) at \( x \) such that the local representation \( f \) of the vector field on this chart is bounded by \( K \) , and so is its derivative \( {f}^{\\prime } \) . Assume also that \( {\\varphi U} \) contains a ball of radius a around \( {\\varphi x} \) . Then \( \\mathfrak{D}\\left( \\xi \\right) = \\mathbf{R} \\times X \) .
|
Proof. This follows at once from the global continuation theorem, and the uniformity of Proposition 1.1.
|
No
|
Theorem 2.9. Let \( \xi \) be a vector field on \( X \) and \( \alpha \) its flow. Let \( {\mathfrak{D}}_{t}\left( \xi \right) \) be the set of points \( x \) of \( X \) such that \( \left( {t, x}\right) \) lies in \( \mathfrak{D}\left( \xi \right) \) . Then \( {\mathfrak{D}}_{t}\left( \xi \right) \) is open for each \( t \in \mathbf{R} \), and \( {\alpha }_{t} \) is an isomorphism of \( {\mathfrak{D}}_{t}\left( \xi \right) \) onto an open subset of \( X \) . In fact, \( {\alpha }_{t}\left( {\mathfrak{D}}_{t}\right) = {\mathfrak{D}}_{-t} \) and \( {\alpha }_{t}^{-1} = {\alpha }_{-t} \) .
|
Proof. Immediate from the preceding theorem.
|
No
|
Theorem 2.9. Let \( \xi \) be a vector field on \( X \) and \( \alpha \) its flow. Let \( {\mathfrak{D}}_{t}\left( \xi \right) \) be the set of points \( x \) of \( X \) such that \( \left( {t, x}\right) \) lies in \( \mathfrak{D}\left( \xi \right) \) . Then \( {\mathfrak{D}}_{t}\left( \xi \right) \) is open for each \( t \in \mathbf{R} \), and \( {\alpha }_{t} \) is an isomorphism of \( {\mathfrak{D}}_{t}\left( \xi \right) \) onto an open subset of \( X \) . In fact, \( {\alpha }_{t}\left( {\mathfrak{D}}_{t}\right) = {\mathfrak{D}}_{-t} \) and \( {\alpha }_{t}^{-1} = {\alpha }_{-t} \) .
|
Proof. Immediate from the preceding theorem.
|
No
|
Proposition 2.11. If \( \alpha \) is an integral curve of a \( {C}^{1} \) vector field, \( \xi \), and \( \alpha \) passes through a critical point, then \( \alpha \) is constant, that is \( \alpha \left( t\right) = {x}_{0} \) for all \( t \) .
|
Proof. The constant curve through \( {x}_{0} \) is an integral curve for the vector field, and the uniqueness theorem shows that it is the only one.
|
No
|
Proposition 2.12. Let \( \xi \) be a vector field and \( \alpha \) an integral curve for \( \xi \) . Assume that all \( t \geqq 0 \) are in the domain of \( \alpha \), and that\n\n\[ \mathop{\lim }\limits_{{t \rightarrow 0}}\alpha \left( t\right) = {x}_{1} \]\n\nexists. Then \( {x}_{1} \) is a critical point for \( \xi \), that is \( \xi \left( {x}_{1}\right) = 0 \) .
|
Proof. Selecting \( t \) large, we may assume that we are dealing with the local representation \( f \) of the vector field near \( {x}_{1} \) . Then for \( {t}^{\prime } > t \) large, we have\n\n\[ \alpha \left( {t}^{\prime }\right) - \alpha \left( t\right) = {\int }_{t}^{{t}^{\prime }}f\left( {\alpha \left( u\right) }\right) {du}. \]\n\nWrite \( f\left( {\alpha \left( u\right) }\right) = f\left( {x}_{1}\right) + g\left( u\right) \), where \( \lim g\left( u\right) = 0 \) . Then\n\n\[ \left| {f\left( {x}_{1}\right) }\right| \left| {{t}^{\prime } - t}\right| \leqq \left| {\alpha \left( {t}^{\prime }\right) - \alpha \left( t\right) }\right| + \left| {{t}^{\prime } - t}\right| \sup \left| {g\left( u\right) }\right| ,\]\n\nwhere the sup is taken for \( u \) large, and hence for small values of \( g\left( u\right) \) . Dividing by \( \left| {{t}^{\prime } - t}\right| \) shows that \( f\left( {x}_{1}\right) \) is arbitrarily small, hence equal to 0, as was to be shown.
|
Yes
|
Proposition 2.13. Suppose on the other hand that \( {x}_{0} \) is not a critical point of the vector field \( \xi \) . Then there exists a chart at \( {x}_{0} \) such that the local representation of the vector field on this chart is constant.
|
Proof. In an arbitrary chart the vector field has a representation as a morphism\n\n\[\n\xi : U \rightarrow E\n\]\n\nnear \( {x}_{0} \) . Let \( \alpha \) be its flow. We wish to \
|
No
|
Proposition 3.2. In a chart \( U \times \mathbf{E} \) for \( {TX} \), let \( f : U \times \mathbf{E} \rightarrow \mathbf{E} \times \mathbf{E} \) represent \( F \), with \( f = \left( {{f}_{1},{f}_{2}}\right) \) . Then \( f \) represents a spray if and only if, for all \( s \in \mathbf{R} \) we have\n\n\[ \n{f}_{2}\left( {x,{sv}}\right) = {s}^{2}{f}_{2}\left( {x, v}\right) \n\]
|
Proof. The proof follows at once from the definitions and the formula giving the chart representation of \( s{\left( {s}_{TX}\right) }_{ * } \) .\n\nThus we see that the condition SPR 1 (in addition to being a second-order vector field), simply means that \( {f}_{2} \) is homogeneous of degree 2 in the variable \( v \) . By the remark in Chapter I,§3, it follows that \( {f}_{2} \) is a quadratic map in its second variable, and specifically, this quadratic map is given by\n\n\[ \n{f}_{2}\left( {x, v}\right) = \frac{1}{2}{D}_{2}^{2}{f}_{2}\left( {x,0}\right) \left( {v, v}\right) .\n\]\n\nThus the spray is induced by a symmetric bilinear map given at each point \( x \) in a chart by\n\n(2)\n\n\[ \nB\left( x\right) = \frac{1}{2}{D}_{2}^{2}{f}_{2}\left( {x,0}\right) .\n\]
|
Yes
|
Proposition 3.4. Suppose we are given a covering of the manifold \( X \) by open sets corresponding to charts \( U, V,\ldots \), and for each \( U \) we are given a morphism \[ {B}_{U} : U \rightarrow {L}_{\mathrm{{sym}}}^{2}\left( {\mathbf{E},\mathbf{E}}\right) \] which transforms according to the formula of Proposition 3.3 under an isomorphism \( h : U \rightarrow V \) . Then there exists a unique spray whose associated bilinear map in the chart \( U \) is given by \( {B}_{U} \) .
|
Proof. We leave the verification to the reader.
|
No
|
Theorem 4.1. Let \( X \) be a manifold and \( F \) a spray on \( X \) . Then\n\n\[{\exp }_{x} : {T}_{x} \rightarrow X\]\n\ninduces a local isomorphism at \( {0}_{x} \), and in fact \( {\left( {\exp }_{x}\right) }_{ * } \) is the identity at \( {0}_{x} \) .
|
Proof. We prove the second assertion first because the main assertion follows from it by the inverse mapping theorem. Furthermore, since \( {T}_{x} \) is a vector space, it suffices to determine the derivative of \( {\exp }_{x} \) on rays, in other words, to determine the derivative with respect to \( t \) of a curve \( {\exp }_{x}\left( {tv}\right) \) . This is done by using SPR 3, and we find\n\n\[ \frac{d}{dt}\pi {\beta }_{tv} = {\beta }_{tv} \]\n\nEvaluating this at \( t = 0 \) and taking into account that \( {\beta }_{w} \) has \( w \) as initial condition for any \( w \) gives us\n\n\[ {\left( {\exp }_{x}\right) }_{ * }\left( {0}_{x}\right) = \mathrm{{id}}. \]\n\nThis concludes the proof of Theorem 4.1.
|
Yes
|
Proposition 4.2. The images of straight segments through the origin in \( {T}_{x} \), under the exponential map \( {\exp }_{x} \), are geodesics. In other words, if \( v \in {T}_{x} \) and we let\n\n\[ \n\alpha \left( {v, t}\right) = {\alpha }_{v}\left( t\right) = {\exp }_{x}\left( {tv}\right) \n\]\n\nthen \( {\alpha }_{v} \) is a geodesic. Conversely, let \( \alpha : J \rightarrow X \) be a \( {C}^{2} \) geodesic defined on an interval \( J \) containing 0, and such that \( \alpha \left( 0\right) = x \) . Let \( {\alpha }^{\prime }\left( 0\right) = v \) . Then \( \alpha \left( t\right) = {\exp }_{x}\left( {tv}\right) \) .
|
Proof. The first statement by definition means that \( {\alpha }_{v}^{\prime } \) is an integral curve of the spray \( F \) . Indeed, by the SPR conditions, we know that\n\n\[ \n\alpha \left( {v, t}\right) = {\alpha }_{v}\left( t\right) = \pi {\beta }_{tv}\left( 1\right) = \pi {\beta }_{v}\left( t\right) , \n\]\n\nand \( {\left( \pi {\beta }_{v}\right) }^{\prime } = {\beta }_{v} \) is indeed an integral curve of the spray. Thus our assertion that the curves \( t \mapsto \exp \left( {tv}\right) \) are geodesics is obvious from the definition of the exponential map and the SPR conditions.\n\nConversely, given a geodesic \( \alpha : J \rightarrow X \), by definition \( {\alpha }^{\prime } \) satisfies the differential equation\n\n\[ \n{\alpha }^{\prime \prime }\left( t\right) = F\left( {{\alpha }^{\prime }\left( t\right) }\right) \n\]\n\nThe two curves \( t \mapsto \alpha \left( t\right) \) and \( t \mapsto {\exp }_{x}\left( {tv}\right) \) satisfy the same differential equation and have the same initial conditions, so the two curves are equal. This proves the second statement and concludes the proof of the proposition.
|
Yes
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.