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Verify the claims above that\n\n(a) Roots of unity are units, and\n\n(b) Powers of units are units. | One can either proceed from the definition or use the characterization \( {\mathrm{N}}_{K/\mathbb{Q}}\left( \alpha \right) = \pm 1 \) . If one definition seems more natural to you, use the other. | No |
Example 51.2.2 (Examples of \( \mu \left( {\mathcal{O}}_{K}\right) \) ) (a) If \( K = \mathbb{Q}\left( i\right) \), then \( {\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack i\right\rbrack \) . | \[ \mu \left( {\mathcal{O}}_{K}\right) = \{ \pm 1, \pm i\} \;\text{ where }K = \mathbb{Q}\left( i\right) . \] | Yes |
Example 51.4.1 \( \left( {{x}^{2} - 5{y}^{2} = \pm 1}}\right) \) | Set \( K = \mathbb{Q}\left( \sqrt{5}\right) \), so \( {\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack {\frac{1}{2}\left( {1 + \sqrt{5}}\right) }\right\rbrack \) . By Dirichlet’s unit theorem, \( {\mathcal{O}}_{K}^{ \times } \) is generated by a single element \( u \) . The choice\n\n\[ u = \frac{1}{2} + \frac{1}{2}\sqrt{5} ... | Yes |
Theorem 52.2.3 (Field extensions have multiplicative degree)\n\nLet \( F \subseteq K \subseteq L \) be fields with \( L/K, K/F \) finite. Then\n\n\[ \left\lbrack {L : K}\right\rbrack \left\lbrack {K : F}\right\rbrack = \left\lbrack {L : F}\right\rbrack . \] | Proof. Basis bash: you can find a basis of \( L \) over \( K \), and then expand that into a basis \( L \) over \( F \) . (Diligent readers can fill in details.) | No |
Theorem 52.3.1 (The \( n \) embeddings of a number field)\n\nLet \( K \) be a number field of degree \( n \) . Then there are exactly \( n \) field homomorphisms \( K \hookrightarrow \mathbb{C} \), say \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) which fix \( \mathbb{Q} \) . | Proof. This is actually kind of fun! Recall that any irreducible polynomial over \( \mathbb{Q} \) has distinct roots (Lemma 47.1.2). We’ll adjoin elements \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{m} \) one at a time to \( \mathbb{Q} \) , until we eventually get all of \( K \), that is,\n\n\[ K = \mathbb{Q}\lef... | Yes |
(a) \( \operatorname{Aut}\left( {\mathbb{Q}\left( i\right) /\mathbb{Q}}\right) \cong \mathbb{Z}/2\mathbb{Z} \) | with elements \( z \mapsto z \) and \( z \mapsto \bar{z} \) | Yes |
Lemma 52.5.4 (Root shuffling in \( \operatorname{Aut}\left( {K/F}\right) \) )\n\nLet \( f \in F\left\lbrack x\right\rbrack \), suppose \( K/F \) is a finite extension, and assume \( \alpha \in K \) is a root of \( f \) . Then for any \( \sigma \in \operatorname{Aut}\left( {K/F}\right) ,\sigma \left( \alpha \right) \) i... | Proof. Let \( f\left( x\right) = {c}_{n}{x}^{n} + {c}_{n - 1}{x}^{n - 1} + \cdots + {c}_{0} \), where \( {c}_{i} \in F \) . Thus,\n\n\[ 0 = \sigma \left( {f\left( \alpha \right) }\right) = \sigma \left( {{c}_{n}{\alpha }^{n} + \cdots + {c}_{0}}\right) = {c}_{n}\sigma {\left( \alpha \right) }^{n} + \cdots + {c}_{0} = f\... | Yes |
The splitting field of \( {x}^{3} - 2 \) over \( \mathbb{Q} \) is in fact \[ \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \] and not just \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) ! One must really adjoin all the roots, and it’s not necessarily the case that these roots will generate each other. | To be clear:\n- For \( {x}^{2} - 5 \), we adjoin \( \sqrt{5} \) and this will automatically include \( - \sqrt{5} \) .\n- For \( {x}^{2} + x + 1 \), we adjoin \( \omega \) and get the other root \( {\omega }^{2} \) for free.\n- But for \( {x}^{3} - 2 \), if we adjoin \( \sqrt[3]{2} \), we do NOT get \( \omega \sqrt[3]{... | Yes |
Example 52.5.9 (Examples and non-examples of Galois extensions)\n\n(a) The extension \( \mathbb{Q}\left( \sqrt{2}\right) /\mathbb{Q} \) is Galois, since it’s the splitting field of \( {x}^{2} - 2 \) over \( \mathbb{Q} \) . The Galois group has order two, \( \sqrt{2} \mapsto \pm \sqrt{2} \) . | The extension \( \mathbb{Q}\left( \sqrt{2}\right) /\mathbb{Q} \) is Galois, since it’s the splitting field of \( {x}^{2} - 2 \) over \( \mathbb{Q} \) . The Galois group has order two, \( \sqrt{2} \mapsto \pm \sqrt{2} \) . | Yes |
Example 52.5.10 (Galois closures, and the automorphism group of \( \\mathbb{Q}\\left( {\\sqrt[3]{2},\\omega }\\right) \) ) Let’s return to the field \( K = \\mathbb{Q}\\left( {\\sqrt[3]{2},\\omega }\\right) \), which is a field with \( \\left\\lbrack {K : \\mathbb{Q}}\\right\\rbrack = 6 \) . Consider the two automorphi... | Actually one can check explicitly that \[ \\operatorname{Gal}\\left( {K/\\mathbb{Q}}\\right) \\cong {S}_{3} \] is the symmetric group on 3 elements, with order \( 3! = 6 \) . | Yes |
Theorem 52.8.1 (Fixed field theorem)\n\nLet \( K \) be a field and \( G \) a subgroup of \( \operatorname{Aut}\left( K\right) \) . Then \( \left\lbrack {K : {K}^{G}}\right\rbrack = \left| G\right| \) . | The inequality itself is not difficult:\n\nExercise 52.8.2. Show that \( \left\lbrack {K : F}\right\rbrack \geq \left| {\operatorname{Aut}\left( {K/F}\right) }\right| \), and that equality holds if and only if the set of elements fixed by all \( \sigma \in \operatorname{Aut}\left( {K/F}\right) \) is exactly \( F \) . (... | No |
Lemma 53.2.2 (Finite fields have prime power orders)\n\nLet \( F \) be a finite field. Then\n\n(a) Its characteristic is nonzero, and hence some prime \( p \) .\n\n(b) The field \( F \) is a finite extension of \( {\\mathbb{F}}_{p} \), and in particular it is an \( {\\mathbb{F}}_{p} \) -vector space.\n\n(c) We have \( ... | Proof. Very briefly, since this is easy:\n\n(a) Apply Lagrange’s theorem (or pigeonhole principle!) to \( \\left( {F, + }\\right) \) to get the characteristic isn't zero.\n\n(b) The additive subgroup of \( \\left( {F, + }\\right) \) generated by \( {1}_{F} \) is an isomorphic copy of \( {\\mathbb{F}}_{p} \) .\n\n(c) Si... | Yes |
Theorem 53.2.6 (Freshman's dream)\n\nFor any \( a, b \in F \) we have\n\n\[{\left( a + b\right) }^{p} = {a}^{p} + {b}^{p}\] | Proof. Use the Binomial theorem, and the fact that \( \left( \begin{array}{l} p \\ i \end{array}\right) \) is divisible by \( p \) for \( 0 < i < p \) . | No |
Theorem 53.3.1 (Fermat's little theorem in finite fields)\n\nLet \( F \) be a finite field of order \( {p}^{n} \) . Then every element \( x \in F \) satisfies\n\n\[ {x}^{{p}^{n}} - x = 0. \] | Proof. If \( x = 0 \) it’s true; otherwise, use Lagrange’s theorem on the abelian group \( \left( {F, \times }\right) \) to get \( {x}^{{p}^{n} - 1} = {1}_{F} \) . | Yes |
A field \( F \) is a finite field with \( {p}^{n} \) elements if and only if it is a splitting field of \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p}. \) | Proof. By \ | No |
The polynomial \( {x}^{9} - x \) is separable modulo 3 and has factorization | \[ x\left( {x + 1}\right) \left( {x + 2}\right) \left( {{x}^{2} + 1}\right) \left( {{x}^{2} + x + 2}\right) \left( {{x}^{2} + {2x} + 2}\right) \;\left( {\;\operatorname{mod}\;3}\right) . \] So if \( F \) has order 9, then we intuitively expect it to be the field generated by adjoining all the roots: \( 0,1,2 \), as wel... | Yes |
Theorem 53.4.1 (The \( p \) th power automorphism)\n\nThe map \( {\sigma }_{p} : {\mathbb{F}}_{{p}^{n}} \rightarrow {\mathbb{F}}_{{p}^{n}} \) defined by\n\n\[ \n{\sigma }_{p}\left( x\right) = {x}^{p} \n\]\n\nis an automorphism, and moreover fixes \( {\mathbb{F}}_{p} \) . | Proof. It's a homomorphism since it fixes 1, respects multiplication, and respects addition.\n\nNext, we claim that it is injective. To see this, note that\n\n\[ \n{x}^{p} = {y}^{p} \Leftrightarrow {x}^{p} - {y}^{p} = 0 \Leftrightarrow {\left( x - y\right) }^{p} = 0 \Leftrightarrow x = y. \n\]\n\nHere we have again use... | Yes |
Theorem 53.4.3 (Galois group of the extension \( {\mathbb{F}}_{{p}^{n}}/{\mathbb{F}}_{p} \) ) We have \( \operatorname{Gal}\left( {{\mathbb{F}}_{{p}^{n}}/{\mathbb{F}}_{p}}\right) \cong \mathbb{Z}/n\mathbb{Z} \) with generator \( {\sigma }_{p} \). | Proof. Since \( \left\lbrack {{\mathbb{F}}_{{p}^{n}} : {\mathbb{F}}_{p}}\right\rbrack = n \), the Galois group \( G \) has order \( n \). So we just need to show \( {\sigma }_{p} \in G \) has order \( n \). Note that \( {\sigma }_{p} \) applied \( k \) times gives \( x \mapsto {x}^{{p}^{k}} \). Hence, \( {\sigma }_{p} ... | Yes |
Theorem 54.4.2 (Galois group acts transitively)\n\nLet \( K/\mathbb{Q} \) be Galois with \( G = \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) . Let \( \left\{ {\mathfrak{p}}_{i}\right\} \) be the set of distinct prime ideals in the factorization of \( p \cdot {\mathcal{O}}_{K} \) (in \( {\mathcal{O}}_{K} \) ).\n\nT... | Proof. Fairly slick. Suppose for contradiction that no \( \sigma \in G \) sends \( {\mathfrak{p}}_{1} \) to \( {\mathfrak{p}}_{2} \), say. By the Chinese remainder theorem, we can find an \( x \in {\mathcal{O}}_{K} \) such that\n\n\[ x \equiv 0\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}_{1}}\right) \]\n\n\[ x \equiv... | Yes |
Theorem 54.4.3 (Inertial degree and ramification indices are all equal)\n\nAssume \( K/\mathbb{Q} \) is Galois. Then for any rational prime \( p \) we have\n\n\[ p \cdot {\mathcal{O}}_{K} = {\left( {\mathfrak{p}}_{1}{\mathfrak{p}}_{2}\ldots {\mathfrak{p}}_{g}\right) }^{e} \]\n\nfor some \( e \), where the \( {\mathfrak... | Proof. To see that the inertial degrees are equal, note that each \( \sigma \) induces an isomorphism\n\n\[ {\mathcal{O}}_{K}/\mathfrak{p} \cong {\mathcal{O}}_{K}/\sigma \left( \mathfrak{p}\right) \]\n\nBecause the action is transitive, all \( {f}_{i} \) are equal. | No |
Example 54.4.5 (Factoring 5 in a Galois/non-Galois extension)\n\nLet \( p = 5 \) be a prime.\n\n(a) Let \( E = \\mathbb{Q}\\left( \\sqrt[3]{2}\\right) \) . One can show that \( {\\mathcal{O}}_{E} = \\mathbb{Z}\\left\\lbrack \\sqrt[3]{2}\\right\\rbrack \), so we use the Factoring Algorithm on the minimal polynomial \( {... | \[ \n\\left( 5\\right) = \\left( {5,\\sqrt[3]{2} - 3}\\right) \\left( {5,\\sqrt[3]{4} + 3\\sqrt[3]{2} + 9}\\right) \n\] \nwhich have inertial degrees 1 and 2, respectively. The fact that this is not uniform reflects that \( E \) is not Galois. | Yes |
Theorem 54.5.2 (Decomposition group and Galois group)\n\nDefine \( \theta \) as above. Then\n\n- \( \theta \) is surjective, and\n\n- its kernel is a group of order \( e \), the ramification index.\n\nIn particular, if \( p \) is unramified then \( {D}_{\mathfrak{p}} \cong \operatorname{Gal}\left( {\left( {{\mathcal{O}... | If \( p \) is unramified, then taking modulo \( \mathfrak{p} \) gives \( {D}_{\mathfrak{p}} \cong \operatorname{Gal}\left( {\left( {{\mathcal{O}}_{K}/\mathfrak{p}}\right) /{\mathbb{F}}_{p}}\right) \).\n\nBut we know exactly what \( \operatorname{Gal}\left( {\left( {{\mathcal{O}}_{K}/\mathfrak{p}}\right) /{\mathbb{F}}_{... | Yes |
Lemma 55.1.3 (Order of the Frobenius element)\n\nLet \( {\mathrm{{Frob}}}_{\mathfrak{p}} \) be a Frobenius element from an extension \( K/\mathbb{Q} \) . Then the order of \( \mathfrak{p} \) is equal to the inertial degree \( {f}_{\mathfrak{p}} \) . In particular, \( \left( p\right) \) splits completely in \( {\mathcal... | Exercise 55.1.4. Prove this lemma as by using the fact that \( {\mathcal{O}}_{K}/\mathfrak{p} \) is the finite field of order \( {f}_{\mathfrak{p}} \), and the Frobenius element is just \( x \mapsto {x}^{p} \) on this field. | No |
Theorem 55.2.2 (Conjugacy classes in Galois groups)\n\nThe set\n\n\[ \n\\left\\{ {{\\operatorname{Frob}}_{\\mathfrak{p}} \\mid \\mathfrak{p}\\text{ above }p}\\right\\} \n\]\n\nis one of the conjugacy classes of \( G \) . | Proof. We’ve used the fact that \( G = \\operatorname{Gal}\\left( {K/\\mathbb{Q}}\\right) \) is transitive to show that \( {\\operatorname{Frob}}_{{\\mathfrak{p}}_{1}} \) and \( {\\mathrm{{Frob}}}_{{\\mathfrak{p}}_{2}} \) are conjugate if they both lie above \( p \) ; hence it’s contained in some conjugacy class. So it... | Yes |
Theorem 55.3.1 (Chebotarev density theorem over \( \mathbb{Q} \) ) Let \( C \) be a conjugacy class of \( G = \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) . The density of (unramified) primes \( p \) such that \( \left\{ {{\operatorname{Frob}}_{\mathfrak{p}} \mid \mathfrak{p}\text{above}p}\right\} = C \) is exactl... | By density, I mean that the proportion of primes \( p \leq x \) that work approaches \( \frac{\left| C\right| }{\left| G\right| } \) as \( x \rightarrow \infty \) . Note that I’m throwing out the primes that ramify in \( K \) . This is no issue, since the only primes that ramify are those dividing \( {\Delta }_{K} \), ... | Yes |
Lemma 55.4.1 (Cyclotomic Frobenius elements)\n\nIn the cyclotomic setting \( L = \mathbb{Q}\left( {\zeta }_{q}\right) \), let \( p \) be a rational unramified prime and \( \mathfrak{p} \) above it. Then\n\n\[{\operatorname{Frob}}_{\mathfrak{p}} = {\sigma }_{p}\] | Proof. Observe that \( {\sigma }_{p} \) satisfies the functional equation (check on generators). Done by uniqueness. | No |
Theorem 55.5.1 (Restrictions of Frobenius elements)\n\nAssume \( L/\mathbb{Q} \) and \( K/\mathbb{Q} \) are both Galois. Let \( \mathfrak{P} \) and \( \mathfrak{p} \) be unramified as above. Then \( {\operatorname{Frob}}_{\mathfrak{P}}{ \upharpoonright }_{K} = {\operatorname{Frob}}_{\mathfrak{p}} \), i.e. for every \( ... | Proof. We know\n\n\[ \operatorname{Frob}\mathfrak{P}\left( \alpha \right) \equiv {\alpha }^{p}\;\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \;\forall \alpha \in {\mathcal{O}}_{L} \]\n\nfrom the definition. | No |
Theorem 55.6.1 (Quadratic reciprocity)\n\nLet \( p \) and \( q \) be distinct odd primes. Then\n\n\[ \left( \frac{p}{q}\right) \left( \frac{q}{p}\right) = {\left( -1\right) }^{\frac{p - 1}{2} \cdot \frac{q - 1}{2}} \] | ## §55.6.i Step 1: Setup\n\nFor this proof, we first define\n\n\[ L = \mathbb{Q}\left( {\zeta }_{q}\right) \]\n\nwhere \( {\zeta }_{q} \) is a primitive \( q \) th root of unity. Then \( L/\mathbb{Q} \) is Galois, with Galois group \( G \) . | Yes |
Theorem 55.6.4 (Quadratic reciprocity, equivalent formulation)\n\nFor distinct odd primes \( p, q \) we have\n\n\[ \left( \frac{p}{q}\right) = \left( \frac{{q}^{ * }}{p}\right) \] | ## §55.6.iv Finishing up\n\nWe already know by Lemma 55.4.1 that Frob \( \mathfrak{p} = {\sigma }_{p} \in H \) if and only if \( p \) is a quadratic residue. On the other hand,\n\nExercise 55.6.7. Show that \( p \) splits in \( {\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack {\frac{1}{2}\left( {1 + \sqrt{{q}^{ * }}}\right) ... | No |
Example 55.7.3 (Factoring \( {x}^{3} - 2\left( {\;\operatorname{mod}\;5}\right) \) ) | Let \( \alpha = \sqrt[3]{2} \) and \( f = {x}^{3} - 2 \), so \( E = \mathbb{Q}\left( \sqrt[3]{2}\right) \) . Set \( p = 5 \) and let finally, let \( K = \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) be the splitting field. Setup:\n\n be a prime number. Prove that there exists a prime number \( q \) such that for every integer \( n \), the number \( {n}^{p} - p \) is not divisible by \( q \) . | We will show, much more strongly, that there exist infinitely many primes \( q \) such that \( {X}^{p} - p \) is irreducible modulo \( q \) . Solution. Okay! First, we draw the tower of fields \[ \mathbb{Q} \subseteq \mathbb{Q}\left( \sqrt[p]{p}\right) \subseteq K \] where \( K \) is the splitting field of \( f\left( x... | Yes |
Consider \( K = \mathbb{Q} \) with infinite prime \( \infty \). Then - If we take \( \mathfrak{m} = 8 \) . Thus \( {I}_{\mathbb{Q}}\left( 8\right) \cong \left\{ {\frac{a}{b}\mathbb{Z} \mid a/b \equiv 1,3,5,7\left( {\;\operatorname{mod}\;8}\right) }\right\} \) . Moreover \[ {P}_{\mathbb{Q}}\left( 8\right) \cong \left\{ ... | You might at first glance think that the quotient is thus \( {\left( \mathbb{Z}/8\mathbb{Z}\right) }^{ \times } \) . But the issue is that we are dealing with ideals: specifically, we have \[ 7\mathbb{Z} = - 7\mathbb{Z} \in {P}_{\mathbb{Q}}\left( 8\right) \] because \( - 7 \equiv 1\left( {\;\operatorname{mod}\;8}\right... | Yes |
Let \( L = \mathbb{Q}\left( i\right), K = \mathbb{Q} \). We have \( \operatorname{Gal}\left( {L/K}\right) \cong \mathbb{Z}/2\mathbb{Z} \). If \( p \) is an odd prime with \( \mathfrak{P} \) above it, then Frob \( \mathfrak{P} \) is the unique element such that \[ {\left( a + bi\right) }^{p} \equiv {\operatorname{Frob}}... | In particular, \[ {\operatorname{Frob}}_{\mathfrak{P}}\left( i\right) = {i}^{p} = \left\{ \begin{array}{lll} i & p \equiv 1 & \left( {\;\operatorname{mod}\;4}\right) \\ - i & p \equiv 3 & \left( {\;\operatorname{mod}\;4}\right) \end{array}\right. \] From this we see that Frob \( \mathfrak{P} \) is the identity when \( ... | Yes |
Lemma 56.4.4 (Order of the Frobenius element)\n\nThe Frobenius element \( {\operatorname{Frob}}_{\mathfrak{P}} \in \operatorname{Gal}\left( {L/K}\right) \) of an extension \( L/K \) has order equal to the inertial degree of \( \mathfrak{P} \), that is,\n\n\[ \operatorname{ord}{\operatorname{Frob}}_{\mathfrak{P}} = f\le... | Proof. We want to understand the order of the map \( T : x \mapsto {x}^{N\mathfrak{p}} \) on the field \( {\mathcal{O}}_{K}/\mathfrak{P} \) . But the latter is isomorphic to the splitting field of \( {X}^{N\mathfrak{P}} - X \) in \( {\mathbb{F}}_{p} \), by Galois theory of finite fields. Hence the order is \( {\log }_{... | Yes |
Suppose we want to understand \( \left( {2/p}\right) \equiv {2}^{\frac{p - 1}{2}} \) where \( p > 2 \) is prime. | Consider the element\n\[ \left( \frac{\mathbb{Q}\left( \sqrt{2}\right) /\mathbb{Q}}{p\mathbb{Z}}\right) \in \operatorname{Gal}\left( {\mathbb{Q}\left( \sqrt{2}\right) /\mathbb{Q}}\right) \]\nIt is uniquely determined by where it sends \( a \) . But in fact we have\n\[ \left( \frac{\mathbb{Q}\left( \sqrt{2}\right) /\mat... | Yes |
Similarly, it also generalizes the cubic Legendre symbol. To see this, assume \( \theta \) is primary in \( K = \mathbb{Q}\left( \sqrt{-3}\right) = \mathbb{Q}\left( \omega \right) \) (thus \( {\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack \omega \right\rbrack \) is Eisenstein integers). Then for example | \[ \left( \frac{K\left( \sqrt[3]{2}\right) /K}{\theta {\mathcal{O}}_{K}}\right) \left( \sqrt[3]{2}\right) \equiv {\left( \sqrt[3]{2}\right) }^{N\left( \theta \right) } \equiv {2}^{\frac{{N\theta } - 1}{3}} \cdot \sqrt{2} \equiv {\left( \frac{2}{\theta }\right) }_{3}\sqrt[3]{2}.\;\left( {\;\operatorname{mod}\;\mathfrak{... | Yes |
The big miracle of quadratic reciprocity states that: for a fixed (squarefree) \( a \), the Legendre symbol \( \left( \frac{a}{p}\right) \) should only depend the residue of \( p \) modulo something. | Let \( L = \mathbb{Q}\left( \sqrt{a}\right), K = \mathbb{Q} \). Then we’ve already seen that the Artin symbol\n\n\[ \left( \frac{\mathbb{Q}\left( \sqrt{a}\right) /\mathbb{Q}}{ \bullet }\right) \]\n\nis the correct generalization of the Legendre symbol. Thus, Artin reciprocity tells us that there is a conductor \( \math... | Yes |
We are going to prove\n\n\[ \nH\left( {\mathbb{Q}\left( \zeta \right) /\mathbb{Q}, m\infty }\right) = {P}_{\mathbb{Q}}\left( {m\infty }\right) = \left\{ {\frac{a}{b}\mathbb{Z} \mid a/b \equiv 1\;\left( {\;\operatorname{mod}\;m}\right) }\right\} .\n\] | It’s well-known \( \mathbb{Q}\left( \zeta \right) /\mathbb{Q} \) is unramified outside finite primes dividing \( m \), so that the Artin symbol is defined on \( {I}_{K}\left( \mathfrak{m}\right) \) . Now the Artin map is given by\n\n\[ \n{I}_{\mathbb{Q}}\left( \mathfrak{m}\right) \overset{\left( \frac{\mathbb{Q}\left( ... | Yes |
Lemma 56.5.6 (Inclusion-reversing congruence subgroups)\n\nFix a modulus \( \mathfrak{m} \) . Let \( L/K \) and \( M/K \) be abelian extensions and suppose \( \mathfrak{m} \) is divisible by the conductors of \( L/K \) and \( M/K \) . Then\n\n\( L \subseteq M\; \) if and only if \( \;H\left( {M/K,\mathfrak{m}}\right) \... | Sketch of proof. Let us first prove the equivalence with \( \mathfrak{m} \) fixed. In one direction, assume \( L \subseteq M \) ; one can check from the definitions that the diagram\n\n\n\ncommutes, because it suffic... | No |
Example 57.4.4 \( \left( {{S}^{1} \vee {S}^{1}}\right. \) is a figure eight) | Let \( X = {S}^{1} \) and \( Y = {S}^{1} \), and let \( {x}_{0} \in X \) and \( {y}_{0} \in Y \) be any points. Then \( X \vee Y \) is a \ | No |
Example 57.5.2 \( \left( {D}^{2}\right. \) with \( \left. {2 + 2 + 1\text{and}1 + 1 + 1\text{cells}}\right) \) | (a) First, we start with \( {X}^{0} \) having two points \( {e}_{a}^{0} \) and \( {e}_{b}^{0} \) . Then, we join them with two 1-cells \( {D}^{1} \) (green), call them \( {e}_{c}^{1} \) and \( {e}_{d}^{1} \) . The endpoints of each 1-cell (the copy of \( {S}^{0} \) ) get identified with distinct points of \( {X}^{0} \)... | Yes |
\( {\mathbb{{RP}}}^{1} \) can be thought of as \( {S}^{1} \) modulo the relation the antipodal points are identified. Projecting onto a tangent line, we see that we get a copy of \( \mathbb{R} \) plus a single point at infinity, corresponding to the parallel line (drawn in cyan below). | Thus, the points of \( {\mathbb{{RP}}}^{1} \) have two forms:\n\n- \( \left( {x : 1}\right) \), which we think of as \( x \in \mathbb{R} \) (in dark red above), and\n\n- \( \left( {1 : 0}\right) \), which we think of as \( 1/0 = \infty \), corresponding to the cyan line above.\n\nSo, we can literally write\n\n\[ {\math... | Yes |
Example 58.2.3 (Loops in \( {S}^{2} \) are nulhomotopic) | As the following picture should convince you, every loop in the simply connected space \( {S}^{2} \) is nulhomotopic.\n\n(Starting with the purple loop, we contract to the red-brown point.)\n\n | No |
The fundamental group of \( \mathbb{C} \) is the trivial group: in the plane, every loop is nulhomotopic. | Proof: imagine it's a piece of rope and reel it in. | No |
Lemma 58.3.2 ( \( {f}_{\sharp } \) is homotopy invariant)\n\nIf \( {\gamma }_{1} \simeq {\gamma }_{2} \) are path-homotopic, then in fact\n\n\[ \n{f}_{\sharp }{\gamma }_{1} \simeq {f}_{\sharp }{\gamma }_{2} \n\] | Proof. Just take the homotopy \( h \) taking \( {\gamma }_{1} \) to \( {\gamma }_{2} \) and consider \( f \circ h \) . | Yes |
Example 58.4.3 \( \left( {{\pi }_{n}\left( {S}^{m}\right) \cong \{ 1\} \text{when}n < m}\right) \) | We saw that \( {\pi }_{1}\left( {S}^{2}\right) \cong \{ 1\} \), because a circle in \( {S}^{2} \) can just be reeled in to a point. It turns out that similarly, any smaller \( n \) -dimensional sphere can be reeled in on the surface of a bigger \( m \) -dimensional sphere. So in general, \( {\pi }_{n}\left( {S}^{m}\rig... | No |
Example 58.5.7 \( \left( {\mathbb{C}\smallsetminus \{ 0\} \text{is homotopy equivalent to}{S}^{1}}\right) \) | Consider the topological spaces \( \mathbb{C} \smallsetminus \{ 0\} \), the punctured plane, and the circle \( {S}^{1} \) viewed as a subset of \( {S}^{1} \) . We claim these spaces are actually homotopy equivalent! The necessary functions are the inclusion\n\n\[ \n{S}^{1} \hookrightarrow \mathbb{C} \smallsetminus \{ 0... | No |
Example 58.5.9 (Disk \( = \) Point, Annulus \( = \) Circle.) | By the same token, a disk is homotopic to a point; an annulus is homotopic to a circle. (This might be a little easier to visualize, since it's finite.) | No |
Theorem 58.5.10 (Homotopy equivalent spaces have isomorphic fundamental groups) Let \( X \) and \( Y \) be path-connected, homotopy-equivalent spaces. Then \( {\pi }_{n}\left( X\right) \cong {\pi }_{n}\left( Y\right) \) for every positive integer \( n \) . | Proof. Let \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow X \) be a loop. Let \( f : X \rightarrow Y \) and \( g : Y \rightarrow X \) be maps witnessing that \( X \) and \( Y \) are homotopy equivalent (meaning \( f \circ g \) and \( g \circ f \) are each homotopic to the identity). Then the composition\n\n\[ ... | Yes |
Theorem 58.6.2 (Functorial interpretation of fundamental groups) There is a functor\n\n\[ \n{\pi }_{1} : {\mathrm{{hTop}}}_{ * } \rightarrow \mathrm{{Grp}} \n\]\n\nsending\n\n\[ \n\left( {Y,{y}_{0}}\right) \rightarrow {\pi }_{1}\left( {Y,{y}_{0}}\right) \n\] | This implies several things, like\n\n- The functor bundles the information of \( {f}_{\sharp } \), including the fact that it respects composition. In the categorical language, \( {f}_{\sharp } \) is \( {\pi }_{1}\left( f\right) \).\n\n- Homotopic spaces have isomorphic fundamental group (since the spaces are isomorphi... | Yes |
Let’s take \( n \) disconnected copies of any space \( B \) : formally, \( E = B \times \{ 1,\ldots, n\} \) with the discrete topology on \( \{ 1,\ldots, n\} \) . Then there exists a tautological covering projection \( E \rightarrow B \) by \( \left( {x, m}\right) \mapsto x \) ; we just project all \( n \) copies. | This is a covering projection because every open set in \( B \) is evenly covered. | No |
Example 59.1.5 (Covering projection of \( {S}^{1} \) )\n\nTake \( p : \mathbb{R} \rightarrow {S}^{1} \) by \( \theta \mapsto {e}^{2\pi i\theta } \) . This is essentially wrapping the real line into a single helix and projecting it down. | We claim this is a covering projection. Indeed, consider the point \( 1 \in {S}^{1} \) (where we view \( {S}^{1} \) as the unit circle in the complex plane). We can draw a small open neighborhood of it whose pre-image is a bunch of copies in \( \mathbb{R} \) . | No |
Theorem 59.2.4 (Lifting paths)\n\nSuppose \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow B \) is a path with \( \gamma \left( 0\right) = {b}_{0} \), and \( p : \left( {E,{e}_{0}}\right) \rightarrow \left( {B,{b}_{0}}\right) \) is a covering projection. Then there exists a unique lifting \( \widetilde{\gamma } ... | Proof. For every point \( b \in B \), consider an evenly covered open neighborhood \( {U}_{b} \) in \( B \) . Then the family of open sets\n\n\[ \left\{ {{\gamma }^{\text{pre }}\left( {U}_{b}\right) \mid b \in B}\right\} \]\n\nis an open cover of \( \left\lbrack {0,1}\right\rbrack \) . As \( \left\lbrack {0,1}\right\rb... | Yes |
Example 59.3.6 (Fundamental group of \( {S}^{1} \) ) | Let’s return to our standard \( p : \mathbb{R} \rightarrow {S}^{1} \) . Since \( \mathbb{R} \) is simply connected, this is a universal cover of \( {S}^{1} \) . And indeed, the fiber of any point in \( {S}^{1} \) is a copy of the integers: naturally in bijection with loops in \( {S}^{1} \) . You can show (and it's intu... | No |
Example 59.4.5 (Fundamental group of \( {\mathbb{{RP}}}^{2} \) ) | As above, we saw that there was a covering projection \( {S}^{2} \rightarrow {\mathbb{{RP}}}^{2} \) . Moreover the fiber of any point has size two. Since \( {S}^{2} \) is simply connected, we have a natural bijection \( {\pi }_{1}\left( {\mathbb{{RP}}}^{2}\right) \) to a set of size two; that is,\n\n\[ \left| {{\pi }_{... | Yes |
Theorem 59.5.8 (Group theory via covering spaces)\n\nSuppose \( B \) is a locally connected, semi-locally simply connected space. Then:\n\n- Every subgroup \( H \subseteq {\pi }_{1}\left( B\right) \) corresponds to exactly one covering projection \( p : E \rightarrow B \) with \( E \) path-connected (up to isomorphism)... | - Moreover, the normal subgroups of \( {\pi }_{1}\left( B\right) \) correspond exactly to the regular covering projections. Hence it’s possible to understand the group theory of \( {\pi }_{1}\left( B\right) \) completely in terms of the covering projections.\n\nMoreover, this is how the \ | No |
Example 60.2.8 (Posets are categories)\n\nLet \( \mathcal{P} \) be a partially ordered set. We can construct a category \( P \) for it as follows:\n\n- The objects of \( P \) are going to be the elements of \( \mathcal{P} \).\n\n- The arrows of \( P \) are defined as follows:\n\n- For every object \( p \in P \), we add... | For example, for the poset \( \mathcal{P} \) on four objects \( \{ a, b, c, d\} \) with \( a \leq b \) and \( a \leq c \leq d \), we get:\n\n\n\n\n\nThis illustrates the point that\n\nThe arrows of a category can be ... | No |
Example 60.3.1 (Initial object)\n\nAn initial object of \( \mathcal{A} \) is an object \( {A}_{\text{init }} \in \mathcal{A} \) such that for any \( A \in \mathcal{A} \) (possibly\n\n\( A = {A}_{\text{init }} \) ), there is exactly one arrow from \( {A}_{\text{init }} \) to \( A \) . | For example,\n\n(a) The initial object of Set is the empty set \( \varnothing \) .\n\n(b) The initial object of Grp is the trivial group \( \{ 1\} \) .\n\n(c) The initial object of CRing is the ring \( \mathbb{Z} \) (recall that ring homomorphisms \( R \rightarrow S \) map \( {1}_{R} \) to \( {1}_{S} \) ).\n\n(d) The i... | Yes |
Proposition 60.4.3 (Uniqueness of products)\n\nWhen they exist, products are unique up to isomorphism: given two products \( {P}_{1} \) and \( {P}_{2} \) of \( X \) and \( Y \) there is an isomorphism between the two objects. | Proof. This is very similar to the proof that initial objects are unique up to unique isomorphism. Consider two such objects \( {P}_{1} \) and \( {P}_{2} \), and the associated projection maps. So, we have a diagram\n\n\n\n(a) In CRing, for instance, the inclusion \( \\mathbb{Z} \\hookrightarrow \\mathbb{Q} \) is epic (and not surjective).. Indeed, if two homomorphisms \( \\mathbb{Q} \\rightarrow A \) agree on every integer then they agree everywhere (why?) | Indeed, if two homomorphisms \( \\mathbb{Q} \\rightarrow A \) agree on every integer then they agree everywhere (why?) | No |
Fix an \( A \in \mathcal{A} \) . For a category \( \mathcal{A} \), define the covariant Yoneda functor \( {H}^{A} : \mathcal{A} \rightarrow \) Set by defining\n\n\[ \n{H}^{A}\left( {A}_{1}\right) \mathrel{\text{:=}} {\operatorname{Hom}}_{\mathcal{A}}\left( {A,{A}_{1}}\right) \in \text{Set.} \n\]\n\nHence each \( {A}_{1... | There’s only one reasonable way to do this: take the composition\n\n\[ \nA\xrightarrow[]{p}{A}_{1}\xrightarrow[]{f}{A}_{2} \n\]\n\nIn other words, \( {H}_{A}\left( f\right) \) is \( p \mapsto f \circ p \) . In still other words, \( {H}_{A}\left( f\right) = f \circ - \) ; the - is a slot for the input to go into. | Yes |
Theorem 61.2.8 (Functors preserve isomorphism)\n\nIf \( {A}_{1} \cong {A}_{2} \) are isomorphic objects in \( \mathcal{A} \) and \( F : \mathcal{A} \rightarrow \mathcal{B} \) is a functor then \( F\left( {A}_{1}\right) \cong \) \( F\left( {A}_{2}\right) \) . | Proof. Try it yourself! The picture is: \n\nYou'll need to use both key properties of functors: they preserve composition and the identity map. | No |
Example 61.3.2 \( \left( {V \mapsto {V}^{ \vee }}\right. \) is contravariant \( ) \) | Consider the functor \( {\operatorname{Vect}}_{k} \rightarrow {\operatorname{Vect}}_{k} \) by \( V \mapsto {V}^{ \vee } \). If we were trying to specify a covariant functor, we would need, for every linear map \( T : {V}_{1} \rightarrow {V}_{2} \), a linear map \( {T}^{ \vee } : {V}_{1}^{ \vee } \rightarrow {V}_{2}^{ \... | Yes |
Theorem 61.6.5 (Yoneda lemma)\n\nLet \( \mathcal{A} \) be a category, pick \( A \in \mathcal{A} \), and let \( {H}_{A} \) be the contravariant Yoneda functor. Let \( X : {\mathcal{A}}^{\mathrm{{op}}} \rightarrow \) Set be a contravariant functor. Then the map\n\n\[ \n\\left\\{ {\\text{ Natural transformations }{\\mathc... | This might be startling at first sight. Here's an unsatisfying explanation why this might not be too crazy: in category theory, a rule of thumb is that \ | No |
Consider the map \( \mathbb{Z}/6\mathbb{Z} \rightarrow {D}_{12} = \left\langle {r, s \mid {r}^{6} = {s}^{2} = 1,{rs} = s{r}^{-1}}\right\rangle \) . Then the cokernel of this map in Grp is \( {D}_{12}/\langle r\rangle \cong \mathbb{Z}/2\mathbb{Z} \) . | This doesn't always work out quite the way we want since in general the image of a homomorphism need not be normal in the codomain. Nonetheless, we can use this to define:\n\nDefinition 63.1.5. The image of \( A\overset{f}{ \rightarrow }B \) is the kernel of \( \operatorname{coker}f \) . We denote \( \operatorname{Im}f... | No |
A map \( A\overset{f}{ \rightarrow }B \) is monic if and only if its kernel is \( 0 \rightarrow A \) . Dually, \( A\overset{f}{ \rightarrow }B \) is epic if and only if its cokernel is \( B \rightarrow 0 \) . | The easy direction is: Exercise 63.2.5. Show that if \( A\overset{f}{ \rightarrow }B \) is monic, then \( 0 \rightarrow A \) is a kernel. (This holds even in non-abelian categories.) Of course, since kernels are unique up to isomorphism, monic \( \Rightarrow 0 \) kernel. On the other hand, assume that \( 0 \rightarrow ... | No |
Proposition 63.2.6 (Isomorphism \( \Leftrightarrow \) monic and epic)\n\nIn an abelian category, a map is an isomorphism if and only if it is monic and epic. | Proof. Omitted. (The Mitchell embedding theorem presented later implies this anyways for most situations we care about, by looking at a small sub-category.) | No |
Lemma 63.4.4 (Short five lemma)\n\nIn an abelian category, consider the commutative diagram\n\n\n\nand assume the top and bottom rows are exact. If \( \alpha \) and \( \gamma \) are isomorphisms, then so is \( \beta ... | Proof. We prove that \( \beta \) is epic (with a similar proof to get monic). By the embedding theorem we can treat the category as \( R \) -modules over some \( R \) . This lets us do a so-called \ | No |
Theorem 64.1.5 \( \left( {{\partial }^{2} = 0}\right) \)\n\nFor any chain \( c,\partial \left( {\partial \left( c\right) }\right) = 0 \) . | Proof. Essentially identical to Problem 44B: this is just a matter of writing down a bunch of \( \sum \) signs. Diligent readers are welcome to try the computation. | No |
Let’s compute \( {H}_{0}\left( X\right) \) for a topological space \( X \) . We take \( {C}_{0}\left( X\right) \), which is just formal linear sums of points of \( X \) . | First, we consider the kernel of \( \partial : {C}_{0}\left( X\right) \rightarrow 0 \), so the kernel of \( \partial \) is the entire space \( {C}_{0}\left( X\right) \) : that is, every point is a \ | No |
Consider \( {S}^{2} \), the two-dimensional sphere. Since it’s path connected, we have \( {H}_{0}\left( {S}^{2}\right) = \mathbb{Z} \). We also have \( {H}_{1}\left( {S}^{2}\right) = 0 \), for the same reason that \( {\pi }_{1}\left( {S}^{2}\right) \) is trivial as well. On the other hand we claim that \[ {H}_{2}\left(... | The elements of \( {H}_{2}\left( {S}^{2}\right) \) correspond to wrapping \( {S}^{2} \) in a tetrahedral bag (or two bags, or three bags, etc.). Thus, the second homology group lets us detect the spherical cavity of \( {S}^{2} \). | No |
Theorem 64.3.2 (Homology is a functor hTop \( \rightarrow \) Grp)\n\nFor any particular \( n,{H}_{n} \) is a functor hTop \( \rightarrow \) Grp. In particular,\n\n- Given any map \( f : X \rightarrow Y \), we get an induced map \( {f}_{ * } : {H}_{n}\left( X\right) \rightarrow {H}_{n}\left( Y\right) \) .\n\n- For two h... | In order to do this, we have to describe how to take a map \( f : X \rightarrow Y \) and obtain a map \( {H}_{n}\left( f\right) : {H}_{n}\left( X\right) \rightarrow {H}_{n}\left( Y\right) \) . Then we have to show that this map doesn’t depend on the choice of homotopy. (This is the analog of the work we did with \( {f}... | No |
Proposition 64.3.8 (Chain homotopic maps induce the same map on homology groups) Let \( f, g : {A}_{ \bullet } \rightarrow {B}_{ \bullet } \) be chain homotopic maps \( {A}_{ \bullet } \rightarrow {B}_{ \bullet } \) . Then the induced maps \( {f}_{ * },{g}_{ * } : {H}_{n}\left( {A}_{ \bullet }\right) \rightarrow {H}_{n... | Proof. It’s equivalent to show \( g - f \) gives the zero map on homology groups, In other words, we need to check that every cycle of \( {A}_{n} \) becomes a boundary of \( {B}_{n} \) under \( g - f \) . | No |
Lemma 64.3.10 (Map of space \( \Rightarrow \) map of singular chain complexes)\n\nEach \( f : X \rightarrow Y \) induces a map \( {C}_{n}\left( X\right) \rightarrow {C}_{n}\left( Y\right) \) . | Proof. Take the composition\n\n\[ \n{\Delta }^{n}\overset{\sigma }{ \rightarrow }X\overset{f}{ \rightarrow }Y \n\] \n\nIn other words, a path in \( X \) becomes a path in \( Y \), et cetera. (It’s not hard to see that the squares involving \( \partial \) commute; check it if you like.) | No |
Question 64.4.2. What’s the homology of the above chain at \( \mathbb{Z} \) ? (Hint: you need \( X \) nonempty.) | Obviously \( {\widetilde{H}}_{n}\left( X\right) \cong {H}_{n}\left( X\right) \) for \( n > 0 \) . But when \( n = 0 \), the map \( {H}_{0}\left( X\right) \rightarrow \mathbb{Z} \) by \( \varepsilon \) has kernel \( {\widetilde{H}}_{0}\left( X\right) \), thus \( {H}_{0}\left( X\right) \cong {\widetilde{H}}_{0}\left( X\r... | No |
Example 65.1.2 (Mayer-Vietoris short exact sequence and its augmentation) Let \( X = U \cup V \) be an open cover. For each \( n \) consider\n\n\[ \n{C}_{n}\left( {U \cap V}\right) \hookrightarrow {C}_{n}\left( U\right) \oplus {C}_{n}\left( V\right) \rightarrow {C}_{n}\left( {U + V}\right) \n\]\n\n\[ \nc \vdash \; \rig... | One can easily see (by taking a suitable basis) that the kernel of the latter map is exactly the image of the first map. This generates a short exact sequence\n\n\[ \n0 \rightarrow {C}_{ \bullet }\left( {U \cap V}\right) \hookrightarrow {C}_{ \bullet }\left( U\right) \oplus {C}_{ \bullet }\left( V\right) \rightarrow {C... | No |
Example 65.1.4 (Relative chain short exact sequence)\n\nSince \( {C}_{n}\left( {X, A}\right) \mathrel{\text{:=}} {C}_{n}\left( X\right) /{C}_{n}\left( A\right) \), we have a short exact sequence\n\n\[ 0 \rightarrow {C}_{ \bullet }\left( A\right) \hookrightarrow {C}_{ \bullet }\left( X\right) \rightarrow {C}_{ \bullet }... | for every space \( X \) and subspace \( A \) . This can be augmented: we get\n\n\[ 0 \rightarrow {\widetilde{C}}_{ \bullet }\left( A\right) \hookrightarrow {\widetilde{C}}_{ \bullet }\left( X\right) \rightarrow {C}_{ \bullet }\left( {X, A}\right) \rightarrow 0 \] | Yes |
Theorem 65.2.1 (Short exact \( \Rightarrow \) long exact)\n\nLet \( 0 \rightarrow {A}_{ \bullet }\overset{f}{ \rightarrow }{B}_{ \bullet }\overset{g}{ \rightarrow }{C}_{ \bullet } \rightarrow 0 \) be any short exact sequence of chain complexes we like. Then there is an exact sequence ![1dfd4520-d1c6-4b94-9568-57d535640... | Proof. A very long diagram chase, valid over any abelian category. (Alternatively, it's actually possible to use the snake lemma twice.) | No |
Theorem 65.3.4 (The homology groups of \( {S}^{m} \) )\n\nFor integers \( m \) and \( n \) ,\n\n\[ \n{\widetilde{H}}_{n}\left( {S}^{m}\right) \cong \left\{ \begin{array}{ll} \mathbb{Z} & n = m \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nThe generator \( {\widetilde{H}}_{n}\left( {S}^{n}\right) \) is an \( n \)... | Proof. This one’s fun, so I’ll only spoil the case \( m = 1 \), and leave the rest to you. Decompose the circle \( {S}^{1} \) into two arcs \( U \) and \( V \), as shown:\n\n\n\nEach of \( U \) and \( V \) is contrac... | No |
Proposition 65.3.5 (The homology groups of the figure eight)\n\nLet \( X = {S}^{1} \land {S}^{1} \) be the figure eight. Then\n\n\[ \n{\widetilde{H}}_{n}\left( X\right) \cong \left\{ \begin{array}{ll} {\mathbb{Z}}^{\oplus 2} & n = 1 \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nThe generators for \( {\widetilde{... | Proof. Again, for simplicity we work with reduced homology groups. Let \( U \) be the \ | No |
Lemma 66.1.1 (Homology relative to contractible spaces)\n\nLet \( X \) be a topological space, and let \( A \subseteq X \) be contractible. For all \( n \) ,\n\n\[ \n{H}_{n}\left( {X, A}\right) \cong {\widetilde{H}}_{n}\left( X\right) \n\] | Proof. Since \( A \) is contractible, we have \( {\widetilde{H}}_{n}\left( A\right) = 0 \) for every \( n \) . For each \( n \) there’s a segment of the long exact sequence given by\n\n\[ \n\cdots \rightarrow \underset{ = 0}{\underbrace{{\widetilde{H}}_{n}\left( A\right) }} \rightarrow {\widetilde{H}}_{n}\left( X\right... | Yes |
Theorem 66.3.2 (Relative homology \( \Rightarrow \) quotient space)\n\nLet \( X \) be a space and \( A \) be a subspace such that \( A \) is a deformation retract of some open set \( V \subseteq X \) . Then the quotient map \( q : X \rightarrow X/A \) induces an isomorphism\n\n\[ \n{H}_{n}\left( {X, A}\right) \cong {H}... | Proof. By hypothesis, we can consider the following maps of pairs:\n\n\[ \nr : \left( {V, A}\right) \rightarrow \left( {A, A}\right) \n\]\n\n\[ \nq : \left( {X, A}\right) \rightarrow \left( {X/A, A/A}\right) \n\]\n\n\[ \n\widehat{q} : \left( {X - A, V - A}\right) \rightarrow \left( {X/A - A/A, V/A - A/A}\right) .\n\]\n... | Yes |
Theorem 66.4.1 (Homology of wedge sums)\n\nLet \( X \) and \( Y \) be spaces with basepoints \( {x}_{0} \in X \) and \( {y}_{0} \in Y \), and assuming each point is a deformation retract of some open neighborhood. Then for every \( n \) we have\n\n\[ \n{\widetilde{H}}_{n}\left( {X \vee Y}\right) = {\widetilde{H}}_{n}\l... | Proof. Apply Theorem 66.3.2 with the subset \( \left\{ {{x}_{0},{y}_{0}}\right\} \) of \( X \coprod Y \) ,\n\n\[ \n{\widetilde{H}}_{n}\left( {X \vee Y}\right) \cong {\widetilde{H}}_{n}\left( {\left( {X \coprod Y}\right) /\left\{ {{x}_{0},{y}_{0}}\right\} }\right) \cong {H}_{n}\left( {X \coprod Y,\left\{ {{x}_{0},{y}_{0... | Yes |
Example 66.4.2 (The long exact sequence for \( \\left( {X, A}\\right) = \\left( {{D}^{2},{S}^{1}}\\right) \) ) | Consider \( {D}^{2} \) (which is contractible) with boundary \( {S}^{1} \) . Clearly \( {S}^{1} \) is a deformation retraction of \( {D}^{2} \\smallsetminus \\{ 0\\} \), and if we fuse all points on the boundary together we get \( {D}^{2}/{S}^{1} \\cong {S}^{2} \) . So we have a long exact sequence\n\n![1dfd4520-d1c6-4... | Yes |
Theorem 66.5.2 (Invariance of dimension, Brouwer 1910)\n\nLet \( U \subseteq {\mathbb{R}}^{n} \) and \( V \subseteq {\mathbb{R}}^{m} \) be nonempty open sets. If \( U \) and \( V \) are homeomorphic, then \( m = n \) . | Proof. Consider a point \( x \in U \) and its local homology groups. By excision,\n\n\[ \n{H}_{k}\left( {{\mathbb{R}}^{n},{\mathbb{R}}^{n}\smallsetminus \{ x\} }\right) \cong {H}_{k}\left( {U, U\smallsetminus \{ x\} }\right) .\n\]\n\nBut since \( {\mathbb{R}}^{n} \smallsetminus \{ x\} \) is homotopic to \( {S}^{n - 1} ... | Yes |
Theorem 67.1.3 (Hairy ball theorem)\n\nIf \( n > 0 \) is even, then \( {S}^{n} \) doesn’t have a continuous field of nonzero tangent vectors. | Proof. If the vectors are nonzero then WLOG they have norm 1 ; that is for every \( x \) we have an orthogonal unit vector \( v\left( x\right) \) . Then we can construct a homotopy map \( F : {S}^{n} \times \left\lbrack {0,1}\right\rbrack \rightarrow {S}^{n} \) by\n\n\[ \left( {x, t}\right) \mapsto \left( {\cos {\pi t}... | Yes |
Lemma 67.2.1 (CW homology groups)\n\nLet \( X \) be a CW complex. Then\n\n\[ \n{H}_{k}\left( {{X}^{n},{X}^{n - 1}}\right) \cong \left\{ \begin{array}{ll} {\mathbb{Z}}^{\oplus \# n\text{-cells of }X} & k = n \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nand\n\n\[ \n{H}_{k}\left( {X}^{n}\right) \cong \left\{ \begi... | Proof. The first part is immediate by noting that \( \left( {{X}^{n},{X}^{n - 1}}\right) \) is a good pair and \( {X}^{n}/{X}^{n - 1} \) is a wedge sum of two spheres. For the second part, fix \( k \) and note that, as long as \( n \leq k - 1 \) or \( n \geq k + 2 \) ,\n\n\[ \n\underset{ = 0}{\underbrace{{H}_{k + 1}\le... | Yes |
Theorem 67.2.4 (Cellular chain complex gives \( {H}_{n}\left( X\right) \) ) The \( k \) th homology group of the cellular chain complex is isomorphic to \( {H}_{k}\left( X\right) \) . | Proof. Follows from the diagram; Problem 67D. | No |
Theorem 67.2.5 (Euler characteristic via Betti numbers)\n\nFor any finite CW complex \( X \) we have\n\n\[ \chi \left( X\right) = \mathop{\sum }\limits_{n}{\left( -1\right) }^{n}\operatorname{rank}{H}_{n}\left( X\right) \]\n\nThus \( \chi \left( X\right) \) does not depend on the choice of CW decomposition. The numbers... | Proof. We quote the fact that if \( 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow D \rightarrow 0 \) is exact then \( \operatorname{rank}B + \) \( \operatorname{rank}D = \operatorname{rank}A + \operatorname{rank}C \) . Then for example the row\n\n\[ \underset{ = 0}{\underbrace{{H}_{2}\left( {X}^{1}\right) }} ... | Yes |
Theorem 67.3.1 (Cellular boundary formula for \( k = 1 \) ) | For \( k = 1 \) , \[ {d}_{1} : {\operatorname{Cells}}_{1}\left( X\right) \rightarrow {\operatorname{Cells}}_{0}\left( X\right) \] is just the boundary map. | No |
Example 67.3.3 (Cellular homology of a torus)\n\nConsider the torus built from \( {e}^{0},{e}_{a}^{1},{e}_{b}^{1} \) and \( {e}^{2} \) as before, where \( {e}^{2} \) is attached via the word \( {ab}{a}^{-1}{b}^{-1} \) . For example, \( {X}^{1} \) is\n\n was the boundary formula. We have \( {d}_{1}\left( {e}_{a}^{1}\right) = {e}_{0} - {e}_{0} = 0 \) and similarly \( {d}_{1}\left( {e}_{b}^{1}\right) = 0 \) . So \( {d}_{1} = 0 \) .\n\n- For \( {d}_{2} \), consider the image of the boundary \( {e}^{2... | Yes |
Example 68.2.3 \( \left( {{C}^{0}\left( {X;G}\right) ,{C}^{1}\left( {X;G}\right) \text{, and}{H}^{0}\left( {X;G}\right) }\right) \) | Let \( X \) be a topological space and consider \( {C}^{ \bullet }\left( X\right) \). - \( {C}_{0}\left( X\right) \) is the free abelian group on \( X \), and \( {C}^{0}\left( X\right) = \operatorname{Hom}\left( {{C}_{0}\left( X\right), G}\right) \). So a 0 -cochain is a function that takes every point of \( X \) to an... | Yes |
Theorem 68.3.4 \( \left( {{H}^{n}\left( {-;G}\right) : {\operatorname{hTop}}^{\mathrm{{op}}} \rightarrow \mathsf{{Grp}}}\right) \)\n\nFor every \( n,{H}^{n}\left( {-;G}\right) \) is a contravariant functor from \( {\mathrm{{hTop}}}^{\mathrm{{op}}} \) to Grp. | Proof. The idea is to leverage the work we already did in constructing the prism operator earlier. First, we construct the entire sequence of functors from \( {\operatorname{Top}}^{\mathrm{{op}}} \rightarrow \) Grp:\n\n\[ \n{\operatorname{Top}}^{\text{op }}\xrightarrow[]{{C}_{ \bullet }}{\operatorname{Cmplx}}^{\text{op... | Yes |
Theorem 68.4.1 (Universal coefficient theorem)\n\nLet \( {A}_{ \bullet } \) be a chain complex of free abelian groups, and let \( G \) be another abelian group. Then there is a natural short exact sequence\n\n\[ 0 \rightarrow \operatorname{Ext}\left( {{H}_{n - 1}\left( {A}_{ \bullet }\right), G}\right) \rightarrow {H}^... | Fortunately, in our case of interest, \( {A}_{ \bullet } \) is \( {C}_{ \bullet }\left( X\right) \) which is by definition free.\n\nThere are two things we need to explain, what the map \( h \) is and the map Ext is.\n\nIt's not too hard to guess how\n\n\[ h : {H}^{n}\left( {{A}_{ \bullet };G}\right) \rightarrow \opera... | Yes |
Lemma 68.4.2 (Computing the Ext functor)\n\nFor any abelian groups \( G, H,{H}^{\prime } \) we have\n\n(a) \( \operatorname{Ext}\left( {H \oplus {H}^{\prime }, G}\right) = \operatorname{Ext}\left( {H, G}\right) \oplus \operatorname{Ext}\left( {{H}^{\prime }, G}\right) \) . | Proof. For (a), note that if \( \cdots \rightarrow {F}_{1} \rightarrow {F}_{0} \rightarrow H \rightarrow 0 \) and \( \cdots \rightarrow {F}_{1}^{\prime } \rightarrow {F}_{0}^{\prime } \rightarrow {F}_{0}^{\prime } \rightarrow {H}^{\prime } \rightarrow \n\n0 are free resolutions, then so is \( {F}_{1} \oplus {F}_{1}^{\p... | Yes |
Example 68.5.2 (Cohomolgy groups of torus) | This example has no nonzero Ext terms either, since this time \( {H}^{n}\left( {{S}^{1} \times {S}^{1}}\right) \) is always free. So we obtain\n\n\[ \n{H}^{n}\left( {{S}^{1} \times {S}^{1}}\right) \cong \operatorname{Hom}\left( {{H}_{n}\left( {{S}^{1} \times {S}^{1}}\right), G}\right) .\n\]\n\nSince \( {H}_{n}\left( {{... | Yes |
Lemma 68.5.3 (0th homology groups are just duals)\n\nFor \( n = 0 \) and \( n = 1 \), we have\n\n\[ \n{H}^{n}\left( {X;G}\right) \cong \operatorname{Hom}\left( {{H}_{n}\left( X\right), G}\right) .\n\] | Proof. It’s already been shown for \( n = 0 \) . For \( n = 1 \), notice that \( {H}_{0}\left( X\right) \) is free, so the Ext term vanishes. | No |
Example 68.5.4 (Cohomolgy groups of Klein bottle)\n\nThis example will actually have Ext term. Recall that if \( K \) is a Klein Bottle then its\n\nhierarchy groups are \( \mathbb{Z} \) in dimension \( n = 0 \) and \( \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \) in \( n = 1 \), and 0 elsewhere. | For \( n = 0 \), we again just have \( {H}^{0}\left( {K;G}\right) \cong \operatorname{Hom}\left( {\mathbb{Z}, G}\right) \cong G \) . For \( n = 1 \), the Ext term is \( \operatorname{Ext}\left( {{H}_{0}\left( K\right), G}\right) \cong \operatorname{Ext}\left( {\mathbb{Z}, G}\right) = 0 \) so\n\n\[
{H}^{1}\left( {K;G}\r... | Yes |
Corollary 69.1.2 (Symmetry of Betti numbers)\n\nLet \( M \) be a smooth oriented compact \( n \) -manifold, and let \( {b}_{k} \) denote its Betti number. Then\n\n\[ \n{b}_{k} = {b}_{n - k} \n\] | Proof. Problem \( {69}{\mathrm{\;A}}^{ \dagger } \) . | No |
Theorem 69.2.2 (de Rham's theorem)\n\nFor any smooth manifold \( M \), we have a natural isomorphism\n\n\[ \n{H}^{k}\left( {M;\mathbb{R}}\right) \cong {H}_{\mathrm{{dR}}}^{k}\left( M\right) \n\]\n\nSo the theorem is that the real cohomology groups of manifolds \( M \) are actually just given by the behavior of differen... | Why does this happen? In fact, we observed already behavior of differential forms which reflects holes in the space. For example, let \( M = {S}^{1} \) be a circle and consider the angle form \( \alpha \) (see Example 43.5.4). The from \( \alpha \) is closed, but not exact, because it is possible to run a full circle a... | No |
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