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Theorem 9.1. Let \( {X}_{2} \) be the cyclic double cover of \( {S}^{3} \) branched over a link \( L \) and suppose that \( A \) is a Seifert matrix for \( L \) with respect to some orientation and some Seifert surface. Then \( {H}_{1}\left( {X}_{2}\right) \) is presented, as an abelian group, by the matrix \( \left( {A + {A}^{\tau }}\right) \) .
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Proof. In the above notation, \( {\widehat{X}}_{2} = {Y}_{0} \cup {Y}_{1} \), where \( {Y}_{0} \cap {Y}_{1} \) is two disjoint copies of \( F \) . A presentation of \( {H}_{1}\left( {\widehat{X}}_{2}\right) \) can be obtained from the following exact Mayer-Vietoris sequence:\n\n\[ \rightarrow {H}_{1}\left( {{Y}_{0} \cap {Y}_{1}}\right) \overset{{\alpha }_{ \star }}{ \rightarrow }{H}_{1}\left( {Y}_{0}\right) \oplus {H}_{1}\left( {Y}_{1}\right) \overset{{\beta }_{ \star }}{ \rightarrow }{H}_{1}\left( {\widehat{X}}_{2}\right) \rightarrow \]\n\n\[ \rightarrow {H}_{0}\left( {{Y}_{0} \cap {Y}_{1}}\right) \overset{{\alpha }_{ \star }}{ \rightarrow }{H}_{0}\left( {Y}_{0}\right) \oplus {H}_{0}\left( {Y}_{1}\right) . \]\n\nThe situation is here very similar to that of Theorem 6.5, and the same sign conventions will be used. There is now a homeomorphism \( t : {\widehat{X}}_{2} \rightarrow {\widehat{X}}_{2} \) with \( {t}^{2} = 1 \) which interchanges \( {Y}_{0} \) and \( {Y}_{1} \) . As in Theorem 6.5, one can take a base \( \left\{ \left\lbrack {f}_{i}\right\rbrack \right\} \) for \( {H}_{1}\left( F\right) \), with corresponding Seifert matrix \( A \) and dual base \( \left\{ \left\lbrack {e}_{i}\right\rbrack \right\} \) for \( {H}_{1}\left( Y\right) \) . Transferring to \( {\widehat{X}}_{2} \), this gives a base \( \left\{ \left\lbrack {f}_{i}\right\rbrack \right\} \cup \left\{ \left\lbrack {t{f}_{i}}\right\rbrack \right\} \) for \( {H}_{1}\left( {{Y}_{0} \cap {Y}_{1}}\right) \) (since \( {Y}_{0} \cap {Y}_{1} \) is two copies of \( F \) ), a base \( \left\{ \left\lbrack {e}_{i}\right\rbrack \right\} \) for \( {H}_{1}\left( {Y}_{0}\right) \) and a base \( \left\{ \left\lbrack {t{e}_{i}}\right\rbrack \right\} \) for \( {H}_{1}\left( {Y}_{1}\right) \) . Then, with respect to these bases, \( {\alpha }_{ \star } \) is represented by the matrix\n\n\[ \left( \begin{array}{rr} - A & {A}^{\tau } \\ {A}^{\tau } & - A \end{array}\right) \]\n\nSimilarly, using bases represented by single points, the map \( {H}_{0}\left( {{Y}_{0} \cap {Y}_{1}}\right) \rightarrow \) \( {H}_{0}\left( {Y}_{0}\right) \oplus {H}_{0}\left( {Y}_{1}\right) \) is represented by \( \left( \begin{array}{rr} - 1 & 1 \\ 1 & - 1 \end{array}\right) \) . Thus the kernel of this last map is a copy of \( \mathbb{Z} \), and (recalling the definition of the maps in the Mayer-Vietoris sequence) any loop in \( {\widehat{X}}_{2} \) that cuts each of the two components of \( {Y}_{0} \cap {Y}_{1} \) at one point maps to a generator of this copy of \( \mathbb{Z} \) .
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Yes
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Let \( {X}_{2} \) be the double cover of \( {S}^{3} \) branched over a link \( L \) . The order of the group \( {H}_{1}\left( {X}_{2}\right) \) is the modulus of the determinant of \( \left( {A + {A}^{\tau }}\right) \), that is\n\n\[ \left| {{H}_{1}\left( {X}_{2}\right) }\right| = \left| {\det \left( {A + {A}^{\tau }}\right) }\right| = \left| {{\Delta }_{L}\left( {-1}\right) }\right| . \]
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Proof. Any finitely generated abelian group can be expressed as a direct sum of cyclic groups. Thus it has as a presentation matrix a diagonal matrix, the entries on the diagonal being the orders of the summands, with the convention that an infinite group has order zero. By Theorem 6.1, the determinant of a square presentation matrix is unique up to multiplication by a unit (that is, by \( \pm 1 \) ), so the result follows at once. The statement about the Alexander polynomial then follows from Theorem 6.5.
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Yes
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Any Goeritz matrix for a link \( L \), associated with the white regions of a diagram of \( L \), represents, with respect to some base, the Gordon-Litherland form \[ {\mathcal{G}}_{F} : {H}_{1}\left( F\right) \times {H}_{1}\left( F\right) \rightarrow \mathbb{Z}, \] where \( F \) is the spanning surface for \( L \) given by the black regions of the diagram.
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Let the white regions, \( {R}_{0},{R}_{1},\ldots ,{R}_{n} \), of the diagram inherit an orientation from the sphere \( {S}^{2} \) in which they are assumed to lie; thus each \( \partial {R}_{i} \) has an orientation. Let \( {f}_{i} \) be the oriented simple closed curve in \( F \) that consists of \( \partial {R}_{i} \) pushed into the union of the black regions. Then \( \left\{ {\left\lbrack {f}_{i}\right\rbrack : 0 \leq i \leq n}\right\} \) forms a set of generators for \( {H}_{1}\left( F\right) \) ; any subset of \( n \) of the \( \left\{ \left\lbrack {f}_{i}\right\rbrack \right\} \) forms a base for \( {H}_{1}\left( F\right) \) . Suppose that the white regions \( {R}_{i} \) and \( {R}_{j} \) are both incident at a crossing \( c \) where \( \zeta \left( c\right) = + 1 \) . Then in the above notation, the curve or curves \( {p}^{-1}{f}_{j} \), namely the push-off of \( {f}_{j} \) from \( F \) locally to both sides of \( F \), meet \( {R}_{i} \) in a positive point of intersection and meet \( {R}_{j} \) in a negative point of intersection near to \( c \) . See Figure 9.4. The sign is positive if the orientation of the region is in the sense of a right-hand screw with respect to the orientation of \( {p}^{-1}{f}_{j} \) . The signs are reversed if \( \zeta \left( c\right) = - 1 \) . Thus for \( i \neq j,\operatorname{lk}\left( {{p}^{-1}{f}_{j},{f}_{i}}\right) = \sum \zeta \left( c\right) \), where the sum is over all crossings at which \( {R}_{i} \) and \( {R}_{j} \) come together, and \( \operatorname{lk}\left( {{p}^{-1}{f}_{j},{f}_{j}}\right) = - \sum \zeta \left( c\right) \) , the sum being over all \( c \) at which \( {R}_{j} \) is incident with other regions. Note the two points of \( {p}^{-1}{f}_{j} \cap {R}_{j} \) near a crossing at which \( {R}_{j} \) is incident with itself cancel each other. Hence the quadratic form \( {\mathcal{G}}_{F} \) is represented with respect to the base \( \left\lbrack {f}_{1}\right\rbrack ,\left\lbrack {f}_{2}\right\rbrack ,\ldots ,\left\lbrack {f}_{n}\right\rbrack \) by the Goeritz matrix of the diagram with the above labelling of the white regions.
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Yes
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Corollary 9.5. The determinant of \( L,\left| {{\Delta }_{L}\left( {-1}\right) }\right| \), is equal to \( \left| {\det G}\right| \), where \( G \) is any Goertiz matrix for \( L \) .
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The proof of this is immediate from the last three theorems. It follows that \( \left| {\det G}\right| \) is an invariant of \( L \), and, as a Goeritz matrix is often easy to write down, it can be a useful invariant.
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Yes
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Theorem 9.6. Suppose that \( {L}_{ + },{L}_{ - },{L}_{0} \) and \( {L}_{\infty } \) are four links that have identical diagrams except near a point where they are as shown in Figure 9.5. Then\n\n\[ \n{\left( \det {L}_{ + }\right) }^{2} + {\left( \det {L}_{ - }\right) }^{2} = 2\left( {{\left( \det {L}_{0}\right) }^{2} + {\left( \det {L}_{\infty }\right) }^{2}}\right) .\n\]
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Proof. The diagram shows the four links together with connected shaded spanning surfaces \( {F}_{i} \) for \( i = + , - ,0,\infty \) . These can always be constructed by using Seifert’s method (see Chapter 2) for \( {F}_{0} \) and adding bands to get the other three surfaces. The four surfaces are taken to be identical outside the areas shown. Take closed curves in \( {F}_{0} \) representing a base of \( {H}_{1}\left( {F}_{0}\right) \) and, for bases of \( {H}_{1}\left( {F}_{i}\right) \) for \( i = + , - ,\infty \), take the classes of the extra curves shown in the diagrams (the ends of them are joined up outside the diagrams) together with the set of curves already chosen for \( {F}_{0} \) . Matrices \( {M}_{i} \) for the Gordon-Litherland forms \( {\mathcal{G}}_{{F}_{i}} \) with respect to these bases are of the following form:\n\n\[ \n{M}_{\infty } = \left( \begin{array}{rr} n & \rho \\ {\rho }^{\tau } & {M}_{0} \end{array}\right) ,\;{M}_{ \pm } = \left( \begin{matrix} n \mp 1 & \rho \\ {\rho }^{\tau } & {M}_{0} \end{matrix}\right) .\n\]\n\nThus\n\n\[ \n\det {M}_{ \pm } = \det {M}_{\infty } \mp \det {M}_{0} \n\]\n\nSquaring and adding give the required result.
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Yes
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Theorem 9.7. Let \( {X}_{r} \) be the cyclic \( r \) -fold cover of \( {S}^{3} \) branched over an \( n \) - component oriented link \( L \), and suppose that \( A \) is a Seifert matrix for \( L \) coming from a genus \( g \) Seifert surface. Then \( {H}_{1}\left( {X}_{r}\right) \) is presented, as an abelian group, by the \( r \times \left( {r + 1}\right) \) matrix of blocks\n\n\[ \left( \begin{matrix} - {A}^{\tau } & & & & A & B \\ A & - {A}^{\tau } & & & & B \\ & A & - {A}^{\tau } & & & B \\ & & \ddots & \ddots & & \vdots \\ & & & A & - {A}^{\tau } & B \end{matrix}\right) ,\]\n\nwhere \( B \) is the \( \left( {{2g} + n - 1}\right) \times \left( {n - 1}\right) \) matrix \( \left( \begin{array}{l} 0 \\ I \end{array}\right) \) .
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Assuming that a \
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No
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The order of the first homology group of \( {X}_{r} \), the cyclic \( r \) -fold cover of \( {S}^{3} \) branched over \( L \), is given by\n\n\[ \left| {{H}_{1}\left( {X}_{r}\right) }\right| = \left| {\mathop{\prod }\limits_{{v = 1}}^{{r - 1}}{\Delta }_{L}\left( {e}^{{2\pi }\imath \frac{v}{r}}\right) }\right| . \]
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Assuming that a \
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No
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Lemma 10.5. Suppose that \( L \) and \( {L}^{\prime } \) are oriented links having property \( \left( \star \right) \) which are the same except near one point, where they are as shown in Figure 10.1; then \( \mathcal{A}\left( L\right) = \mathcal{A}\left( {L}^{\prime }\right) \)
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Proof. The two segments shown on one of the two sides of Figure 10.1 must belong to the same component of the link. Suppose, without loss of generality, it is the two segments on the left side. Then using the Seifert circuit method of Theorem 2.2, a Seifert surface can be constructed for the left link that meets the neighbourhood of the point in question in the way indicated by the shading. Adding a band to that produces a Seifert surface for the right link as indicated. Now, as these two surfaces just differ by a band added to the boundary, the \( \mathbb{Z}/2\mathbb{Z} \) -homology of the second surface is just that of the first surface with an extra \( \mathbb{Z}/2\mathbb{Z} \) summand. However, that summand is in the image of the homology of the boundary of the surface; this image is disregarded (by means of the quotienting) in construction of the quadratic form that gives the Arf invariant.
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Yes
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Theorem 10.7. Let \( K \) be a knot. Then \( \mathcal{A}\left( K\right) \equiv {a}_{2}\left( K\right) \) modulo 2, where \( {a}_{2}\left( K\right) \) is the coefficient of \( {z}^{2} \) in the Conway polynomial \( {\nabla }_{K}\left( z\right) \) . The Arfinvariant of \( K \) is related to the Alexander polynomial by\n\n\[ \mathcal{A}\left( K\right) = \left\{ \begin{array}{ll} 0 & \text{ if }{\Delta }_{K}\left( {-1}\right) \equiv \pm 1\text{ modulo }8, \\ 1 & \text{ if }{\Delta }_{K}\left( {-1}\right) \equiv \pm 3\text{ modulo }8. \end{array}\right. \]\n\nIf \( K \) is a slice knot, then \( \mathcal{A}\left( K\right) = 0 \) .
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Proof. The formula \( \mathcal{A}\left( {L}_{ + }\right) - \mathcal{A}\left( {L}_{ - }\right) \equiv \operatorname{lk}\left( {L}_{0}\right) \) modulo 2, valid when \( {L}_{ + } \) has one component, allows calculation of \( \mathcal{A}\left( K\right) \) from \( \mathcal{A} \) (unknot) \( = 0 \) . However, this gives the same answer as the calculation, modulo 2, of \( {a}_{2}\left( K\right) \) using Proposition 8.7 (v). With the Conway normalisation, \( {\Delta }_{K}\left( {-1}\right) = {\nabla }_{K}\left( {-{2i}}\right) \) . However, \( {\nabla }_{K}\left( z\right) = \) \( 1 + {a}_{2}\left( K\right) {z}^{2} + {a}_{4}\left( K\right) {z}^{4} + \cdots \), and so \( {\nabla }_{K}\left( {-{2i}}\right) \equiv 1 - 4{a}_{2}\left( K\right) \) modulo 8 . Thus, modulo 8,\n\n\[ {\nabla }_{K}\left( {-{2i}}\right) \equiv \left\{ \begin{array}{ll} 1 & \text{ if }{a}_{2}\left( K\right) \equiv 0{\;\operatorname{modulo}\;2}, \\ - 3 & \text{ if }{a}_{2}\left( K\right) \equiv 1{\;\operatorname{modulo}\;2}. \end{array}\right. \]\n\nThis gives the required result. As remarked after Theorem 8.19, if \( K \) is a slice knot then \( {\Delta }_{K}\left( {-1}\right) \equiv \pm 1 \) modulo 8, and so, from the above discussion, \( \mathcal{A}\left( K\right) = 0 \) .
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Yes
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Theorem 11.1 (The Loop Theorem). Let \( M \) be a (possibly non-compact) 3- manifold with boundary \( \partial M \) such that the inclusion-induced homomorphism \( {\Pi }_{1}\left( {\partial M}\right) \rightarrow {\Pi }_{1}\left( M\right) \) is not injective. Then there exists a (piecewise linear) embedding of the disc \( e : {D}^{2} \rightarrow M \), with \( {e}^{-1}\left( {\partial M}\right) = \partial {D}^{2} \) such that the restriction \( e : \partial {D}^{2} \rightarrow \partial M \) is not homotopic to a constant map.
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Null
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No
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Theorem 11.2. Let \( X \) be the exterior of a knot \( K \) in \( {S}^{3} \) . If \( K \) is not the unknot, then the inclusion map induces an injection \( {\Pi }_{1}\left( {\partial X}\right) \rightarrow {\Pi }_{1}\left( X\right) \) .
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Proof. Suppose \( {\Pi }_{1}\left( {\partial X}\right) \rightarrow {\Pi }_{1}\left( X\right) \) is not injective. Then, by the loop theorem, there is an embedding \( e : {D}^{2} \rightarrow X \) sending \( \partial {D}^{2} \) into the torus \( \partial X \), to a simple closed curve not homotopically trivial in the torus. Now \( e\left( {\partial {D}^{2}}\right) \) is certainly the boundary of the disc \( e\left( {D}^{2}\right) \) and so represents a non-trivial element of the kernel of the map \( {H}_{1}\left( {\partial X}\right) \rightarrow {H}_{1}\left( X\right) \) ; the longitude of the knot \( K \) (with either orientation) is the only simple closed curve representing an element in this kernel (see Definition 1.6). The longitude is parallel to \( K \) in a small solid torus neighbourhood of \( K \), so expanding the disc \( e\left( {D}^{2}\right) \) by an annulus gives a disc embedded in \( {S}^{3} \) with \( K \) as its boundary. Thus \( e\left( {D}^{2}\right) \) when so expanded is a Seifert surface for \( K \) . This shows that \( K \) is unknotted.
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Yes
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Corollary 11.3. A knot \( K \) is the unknot if and only if \( {\Pi }_{1}\left( {{S}^{3} - K}\right) \) is infinite cyclic.
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Proof. If \( {\Pi }_{1}\left( {{S}^{3} - K}\right) \) is isomorphic to \( \mathbb{Z} \), there can be no injection \( {\Pi }_{1}\left( {\partial X}\right) \rightarrow \) \( {\Pi }_{1}\left( X\right) \) (as \( {\Pi }_{1}\left( {\partial X}\right) \) is isomorphic to \( \mathbb{Z} \oplus \mathbb{Z} \) ).
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No
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Corollary 11.4. Let \( {X}_{1} \) and \( {X}_{2} \) be the exteriors of two non-trivial knots and let M be a 3-manifold formed by identifying their boundaries together using any homeomorphism. Then the inclusion into \( M \) of the torus \( T \) that comes from the identified boundaries induces an injection \( {\Pi }_{1}\left( T\right) \rightarrow {\Pi }_{1}\left( M\right) \) .
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Proof. This follows at once from the above theorem and from the Van Kam-pen theorem, which describes how fundamental groups behave when a space is described as a union of subspaces.
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Yes
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Theorem 11.5 (The Sphere Theorem). Suppose that \( M \) is an orientable 3-manifold and that there exists a map \( {S}^{2} \rightarrow M \) that is not homotopic to a constant map (that is, \( {\Pi }_{2}\left( M\right) \neq 0 \) ). Then there exists a (piecewise linear) embedding \( {S}^{2} \rightarrow M \) that is not homotopic to a constant map.
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Null
|
No
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Theorem 11.6. If \( K \) is a knot in \( {S}^{3} \) any map \( {S}^{2} \rightarrow {S}^{3} - K \) is homotopic to a constant map (that is, \( {\Pi }_{2}\left( {{S}^{3} - K}\right) = 0 \) ).
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Proof. If the statement is false then, by the sphere theorem, there exists a piecewise linear embedding \( e : {S}^{2} \rightarrow {S}^{3} - K \) that is not homotopic to a constant in \( \left( {{S}^{3} - K}\right) \) . Then, by the Schönflies theorem, \( e\left( {S}^{2}\right) \) separates \( {S}^{3} \) into two components, the closure of each of which is a ball with boundary \( e\left( {S}^{2}\right) \) . The knot \( K \), being connected and disjoint from \( e\left( {S}^{2}\right) \), lies in one of these balls, so \( e \) is homotopic to a constant using the other ball.
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Yes
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Theorem 11.7. If \( K \) is a knot in \( {S}^{3} \), any map \( {S}^{r} \rightarrow {S}^{3} - K \) is homotopic to a constant map (that is, \( {\Pi }_{r}\left( {{S}^{3} - K}\right) = 0 \) ) for all \( r \geq 2 \) .
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Proof. Let \( X \) be the exterior of \( K \) and let \( \widetilde{X} \) be the universal cover of \( X \) . Thus \( \widetilde{X} \) is the simply connected cover of \( X \), it is acted upon by \( {\Pi }_{1}\left( X\right) \), and the quotient of \( \widetilde{X} \) by this action is \( X \) . The operation of lifting maps and homotopies from \( X \) to \( \widetilde{X} \) shows that, for \( r \geq 2,{\Pi }_{r}\left( X\right) = 0 \) if and only if \( {\Pi }_{r}\left( \widetilde{X}\right) = 0 \) (or equivalently just use the homotopy long exact sequence of the covering). So certainly \( {\Pi }_{2}\left( \widetilde{X}\right) = 0 \) . Now the third homology of any non-compact connected 3-manifold is zero. A simplicial argument for this uses the fact that any 3-cycle would be a finite sum of oriented 3-simplexes, a neighbourhood of the union of those 3-simplexes is a compact 3-manifold \( N \) with non-empty boundary which can be taken to be connected; any such \( N \) deformation retracts to a 2-dimensional complex (by collapsing 3- simplexes from the boundary), and so \( {H}_{3}\left( N\right) = 0 \) . Of course, \( \widetilde{X} \) is non-compact because \( {\Pi }_{1}\left( X\right) \) is infinite (as \( {H}_{1}\left( X\right) \) is infinite), and so each simplex has infinitely many different lifts in \( \widetilde{X} \) . Thus \( {H}_{3}\left( \widetilde{X}\right) = 0 \) and \( {H}_{r}\left( \widetilde{X}\right) = 0 \) for \( r > 3 \), as then \( X \) has no \( r \) -simplex and so its \( {r}^{\text{th }} \) chain group is zero. Now, for a simply connected cell complex, the Hurewicz isomorphism theorem asserts that the first non-vanishing homology group and the first non-vanishing homotopy group occur in the same dimension and are isomorphic. Thus \( {\Pi }_{r}\left( \widetilde{X}\right) = 0 \) for all \( r \), and so \( \widetilde{X} \) is a contractible space. The above remark about lifting ensures that \( {\Pi }_{r}\left( X\right) = 0 \) for \( r \geq 2 \) .
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Yes
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Theorem 11.8. If there exists an isomorphism from \( {\Pi }_{1}\left( {{S}^{3} - {K}_{1}}\right) \) to \( {\Pi }_{1}\left( {{S}^{3} - {K}_{2}}\right) \) which sends \( \left\lbrack {\lambda }_{1}\right\rbrack \) to \( \left\lbrack {\lambda }_{2}\right\rbrack \) and \( \left\lbrack {\mu }_{1}\right\rbrack \) and \( \left\lbrack {\mu }_{2}\right\rbrack \), then \( {K}_{1} \) and \( {K}_{2} \) are equivalent knots.
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Null
|
No
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Theorem 11.9. If \( {K}_{1} \) and \( {K}_{2} \) are prime knots in \( {S}^{3} \) and \( {\Pi }_{1}\left( {{S}^{3} - {K}_{1}}\right) \) and \( {\Pi }_{1}\left( {{S}^{3} - }\right. \) \( \left. {K}_{2}\right) \) are isomorphic groups, then \( \left( {{S}^{3} - {K}_{1}}\right) \) and \( \left( {{S}^{3} - {K}_{2}}\right) \) are homeomorphic spaces.
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Thus, for prime knots, the knot group determines the complement of the knot. It is by no means obvious that this means that the knots are the same. Perhaps the homeomorphism might send a meridian to a non-meridian. That this is not so is the substance of one of the most impressive results in knot theory of the 1980's. It is due to Gordon and J. Luecke [37] and the proof is lengthy and intricate:
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No
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Theorem 11.10. If \( {K}_{1} \) and \( {K}_{2} \) are unoriented knots in \( {S}^{3} \) and there is an orientation preserving homeomorphism between their complements, then \( {K}_{1} \) and \( {K}_{2} \) are equivalent (as unoriented knots).
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Null
|
No
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Lemma 12.2. Suppose that \( U \) and \( V \) are 3-manifolds with homeomorphic boundaries, and that \( {h}_{0} : \partial U \rightarrow \partial V \) and \( {h}_{1} : \partial U \rightarrow \partial V \) are isotopic homeomorphisms. Then \( U{ \cup }_{{h}_{0}}V \) and \( U{ \cup }_{{h}_{1}}V \) are homeomorphic.
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Proof. Choose ([113],[47]) a collar neighbourhood \( C \) of \( \partial U \) in \( U \) ; \( C \) is a neighbourhood of \( \partial U \) homeomorphic to \( \partial U \times \left\lbrack {0,1}\right\rbrack \), with \( \partial U \) identified with \( \partial U \times 0 \) . A homeomorphism \( f : U{ \cup }_{{h}_{0}}V \rightarrow U{ \cup }_{{h}_{1}}V \) can be constructed by defining \( f \) to be the identity on \( \left( {U - C}\right) \cup V \) and on \( C \) defining \( f\left( {x, t}\right) = \) \( \left( {{h}_{1}^{-1}{h}_{t}x, t}\right) \) .
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Yes
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Lemma 12.5. Suppose oriented simple closed curves \( p \) and \( q \), contained in the interior of the surface \( F \), intersect transversely at precisely one point. Then \( p{ \sim }_{\tau }q \) .
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Proof. The first diagram of Figures 12.3 shows the intersection point of \( p \) and \( q \) and also a simple closed curve \( {C}_{1} \) that runs parallel to, and is slightly displaced from, \( q \) . Similarly, \( {C}_{2} \) is a slightly displaced copy of \( p \) . The second diagram shows \( {\tau }_{1}p \), where \( {\tau }_{1} \) is a twist about \( {C}_{1} \) . The third diagram shows \( {\tau }_{2}{\tau }_{1}p \), where \( {\tau }_{2} \) is a twist about \( {C}_{2} \) . In this diagram \( {\tau }_{2}{\tau }_{1}p \) has a doubled-back portion that can easily be moved by a homeomorphism isotopic to the identity (that is, a slide in \( F \) ) to change \( {\tau }_{2}{\tau }_{1}p \) to \( q \) .
|
Yes
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Lemma 12.6. Suppose that oriented simple closed curves \( p \) and \( q \) contained in the interior of the surface \( F \) are disjoint and that neither separates \( F \) (that is, \( \left\lbrack p\right\rbrack \neq 0 \neq \left\lbrack q\right\rbrack \) in \( \left. {{H}_{1}\left( {F,\partial F}\right) }\right) \) . Then \( p{ \sim }_{\tau }q \) .
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Proof. Consideration of the surface obtained by cutting \( F \) along \( p \cup q \) shows at once that there is a simple closed curve \( r \) in \( F \) that intersects each of \( p \) and \( q \) transversely at one point. Then, by Lemma 12.5, \( p{ \sim }_{\tau }r{ \sim }_{\tau }q \) .
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Yes
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Proposition 12.7. Suppose that oriented simple closed curves \( p \) and \( q \) are contained in the interior of the surface \( F \) and that neither separates \( F \) . Then \( p{ \sim }_{\tau }q \) .
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Proof. Changing \( q \) by means of a homeomorphism of \( F \) that is (close to and) isotopic to the identity, it can be assumed that \( p \) and \( q \) intersect transversely at \( n \) points. The proof is by induction on \( n \) ; Lemmas 12.5 and 12.6 start the induction, so assume that \( n \geq 2 \) and that the result is true for less that \( n \) points of intersection.\n\nLet \( A \) and \( B \) be consecutive points along \( p \) of \( p \cap q \) . Suppose firstly that \( p \) leaves \( A \) on one side of \( q \) and returns to \( B \) from the other side of \( q \) . Let \( r \) be a simple closed curve in \( F \) that starts near \( A \), follows close to \( p \) until near \( B \) and then returns to its start in a neighbourhood of \( q \) . As shown in the first diagram of Figure 12.4, \( r \) can be chosen so that \( p \cap r \) contains less than \( n \) points and \( q \cap r \) is one point. Hence \( p{ \sim }_{\tau }r \) by the induction hypothesis, and \( r{ \sim }_{\tau }q \) by Lemma 12.5.\n\nSuppose now that \( p \) leaves \( A \) on one side of \( q \) and returns to \( B \) from the same side of \( q \) . Let \( {r}_{1} \) and \( {r}_{2} \) be the two simple closed curves shown in the second\n\n \n\nFigure 12.4\n\ndiagram of Figure 12.4. Each starts near \( A \), proceeds near \( p \) until close to \( B \) and then back to its start following near to \( q \) . However, \( {r}_{1} \) starts on the right of \( p \) and \( {r}_{2} \) starts on the left. Now in \( {H}_{1}\left( {F,\partial F}\right) ,\left\lbrack {r}_{1}\right\rbrack - \left\lbrack {r}_{2}\right\rbrack = \left\lbrack q\right\rbrack \), and hence at least one of \( {r}_{1} \) and \( {r}_{2} \) does not separate (as \( \left\lbrack q\right\rbrack \neq 0 \) ). Let that curve be defined to be \( r \) . Then \( r \) is disjoint from \( q \), so \( r{ \sim }_{\tau }q \) by Lemma 12.6 and, as \( r \cap p \) has at most \( n - 2 \) points, \( p{ \sim }_{\tau }r \) by the induction hypothesis.
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Yes
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Corollary 12.8. Let \( {p}_{1},{p}_{2},\ldots ,{p}_{n} \) be disjoint simple closed curves in the interior of \( F \) the union of which does not separate \( F \) . Let \( {q}_{1},{q}_{2},\ldots ,{q}_{n} \) be another set of curves with the same properties. Then there is a homeomorphism \( h \) of \( F \) that is in the group generated by twists, so that \( h{p}_{i} = {q}_{i} \) for each \( i = 1,2,\ldots, n \) .
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Proof. Suppose inductively that such an \( h \) can be found so that \( h{p}_{i} = {q}_{i} \) for each \( i = 1,2,\ldots, n - 1 \) . Apply Proposition 12.7 to \( h{p}_{n} \) and \( {q}_{n} \) in \( F \) cut along \( {q}_{1} \cup {q}_{2} \cup \ldots \cup {q}_{n - 1} \)
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Yes
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Lemma 12.12. Any closed connected orientable 3-manifold has a Heegaard splitting.
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Proof. This is similar to the first part of the proof of Theorem 8.2. Take a triangulation of \( M \) as a simplicial complex \( K \) . The vertices of the first derived subdivision \( {K}^{\left( 1\right) } \) of \( K \) are the barycentres \( \widehat{A} \) of the simplexes \( A \) of \( K \) . The second derived subdivision \( {K}^{\left( 2\right) } \) of \( K \) is, of course, just \( {\left( {K}^{\left( 1\right) }\right) }^{\left( 1\right) } \) . The 1-skeleton of \( K \) (that is, the sub-complex consisting of the 0 -simplexes and 1 -simplexes of \( K \) ), being a graph, has, as intimated above, for its simplicial neighbourhood in \( {K}^{\left( 2\right) } \), a handlebody. The closure of the complement of this is the simplicial neighbourhood in \( {K}^{\left( 2\right) } \) of another graph. That graph, called the dual 1-skeleton of \( K \), is the sub-complex \( \mathop{\bigcup }\limits_{A}{C}_{A} \) of \( {K}^{\left( 1\right) } \), where the union is over all 3-simplexes \( A \), and \( {C}_{A} \) is the cone with vertex \( \widehat{A} \) on the barycentres of the 2-dimensional faces of \( A \) . Thus \( {K}^{\left( 2\right) } \) is expressed as the union of two handlebodies that intersect in their common boundary, and this is the required Heegaard splitting.
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Yes
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Theorem 12.13. Let \( M \) be a closed connected orientable 3-manifold. There exists finite sets of disjoint solid tori \( {T}_{1}^{\prime },{T}_{2}^{\prime },\ldots ,{T}_{N}^{\prime } \) in \( M \) and \( {T}_{1},{T}_{2},\ldots ,{T}_{N} \) in \( {S}^{3} \) such that \( M - { \cup }_{1}^{N}\operatorname{Int}\left( {T}_{i}^{\prime }\right) \) and \( {S}^{3} - { \cup }_{1}^{N}\operatorname{Int}\left( {T}_{i}\right) \) are homeomorphic.
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Proof. By Lemma 12.12, \( M \) has a Heegaard splitting, so for handlebodies \( U \) and \( V \) of some genus \( g \), and some homeomorphism \( h : \partial U \rightarrow \partial V, M = \) \( U{ \cup }_{h}V \) . Let \( {p}_{1}^{\prime },{p}_{2}^{\prime },\ldots ,{p}_{g}^{\prime } \) be disjoint simple closed curves in \( \partial U \), that bound disjoint discs in \( U \) and let \( {q}_{1},{q}_{2},\ldots {q}_{g} \) be disjoint simple closed curves in \( \partial V \) (one around each \
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No
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Theorem 12.14. Any closed connected orientable 3-manifold \( M \) can be obtained from \( {S}^{3} \) by a collection of 1 -surgeries, that is, by removing disjoint copies of \( {S}^{1} \times {D}^{2} \) and replacing them with copies of \( {D}^{2} \times {S}^{1} \) in the canonical way. Thus \( M \) bounds a 4-manifold that is a 4-ball to which a collection of 2-handles has been added.
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Null
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No
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Theorem 12.15. Two framed links in \( {S}^{3} \) give, by surgery, the same oriented 3- manifold if and only if they are related by a sequence of moves of two types. In a move of type 1, an extra unknotted component, unlinked from all other components, with framing 1 or -1 is added to or removed from the link. In a move of type 2, any two components that are, together with their framing curves, contained in a doubly punctured disc (itself possibly knotted up and linked with other components) in \( {S}^{3} \) , as on the left of Figure 12.6, can be changed to the two curves on the right, the new framing curves again being on the punctured disc.
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For the proof, which uses 4-dimensional Cerf theory, refer to [65]. If one considers the surgery information as a recipe for adding 2-handles on to a 4-ball to create a 4-manifold with the 3-manifold as its boundary, a move of type 2 corresponds to sliding one 2-handle over another. A type 1 move changes the 4-manifold by taking the connected sum with a complex projective plane (oriented in either way), or by removing such a summand. Neither manoeuvre changes the boundary of the 4-manifold. The two moves of Theorem 12.15 can be, and indeed have been, explored at length to give many examples of different framed links representing the same manifold [65].
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No
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Lemma 13.2. Suppose that \( {A}^{4} \) is not a \( {k}^{\text{th }} \) root of unity for \( k \leq n \) . Then there is a unique element \( {f}^{\left( n\right) } \in T{L}_{n} \) such that\n\n(i) \( {f}^{\left( n\right) }{e}_{i} = 0 = {e}_{i}{f}^{\left( n\right) } \) for \( 1 \leq i \leq n - 1 \) ,\n\n(ii) \( \left( {{f}^{\left( n\right) } - 1}\right) \) belongs to the algebra generated by \( \left\{ {{e}_{1},{e}_{2},\ldots ,{e}_{n - 1}}\right\} \) ,\n\n(iii) \( {f}^{\left( n\right) }{f}^{\left( n\right) } = {f}^{\left( n\right) } \) and\n\n(iv) \( {\Delta }_{n} = {\left( -1\right) }^{n}\left( {{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) /\left( {{A}^{2} - {A}^{-2}}\right) \) .
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Proof. Note that if \( {f}^{\left( n\right) } \) exists, \( \mathbf{1} - {f}^{\left( n\right) } \) is the identity of the algebra generated by \( \left\{ {{e}_{1},{e}_{2},\ldots ,{e}_{n - 1}}\right\} \), and so \( {f}^{\left( n\right) } \) is then certainly unique. Let \( {f}^{\left( 0\right) } \) be the empty diagram (so that \( {\Delta }_{0} = 1 \) ), let \( {f}^{\left( 1\right) } = \mathbf{1} \), and inductively assume that \( {f}^{\left( 2\right) },{f}^{\left( 3\right) },\ldots ,{f}^{\left( n\right) } \) have been defined with the above properties (i),(ii),(iii) and (iv). Observe that (i) and (ii) immediately imply (iii) and that this generalises to the identity shown in Figure 13.4 provided that \( \left( {i + j}\right) \leq n \) .\n\n\n\nFigure 13.4\n\nNow consider the element \( x \), say, of \( T{L}_{n - 1} \) shown at the start of Figure 13.5. The identity of Figure 13.4 implies that \( {f}^{\left( n - 1\right) }x = x \) . But \( {f}^{\left( n - 1\right) }x \) is, by (i), just some scalar multiple \( \lambda \) of \( {f}^{\left( n - 1\right) } \) (because \( x \) is a linear sum of 1 ’s and products of \( {e}_{i} \) ’s); the trick of placing squares in the plane and joining points on the left to points on the right, in the standard way, implies that the scalar \( \lambda \) is \( {\Delta }_{n}/{\Delta }_{n - 1} \) .\n\n\n\nFigure 13.5\n\nSuppose now that \( {A}^{4k} \neq 1 \) for \( k \leq n + 1 \), so that \( {\Delta }_{k} \neq 0 \) for \( k \leq n \) . Define \( {f}^{\left( n + 1\right) } \in T{L}_{n + 1} \) inductively by the equation of Figure 13.6.\n\n\n\nFigure 13.6\n\nProperties (i) and (ii) (and hence (iii)) for \( {f}^{\left( n + 1\right) } \) follow immediately, except perhaps for the fact that \( {f}^{\left( n + 1\right) }{e}_{n} = 0 \) . However, Figure 13.7 shows, using the identities of Figure 13.5 and Figure 13.4, why that also is true.\n\n\n\nFigure 13.7
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Yes
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Lemma 13.4. In \( \mathcal{S}\left( {{S}^{1} \times I,2\text{points}}\right) ,{a\omega } - {b\omega } \) is a linear sum of two elements, each of which contains a copy of \( {f}^{\left( r - 1\right) } \) . (That is, each of the two elements is the image of \( {f}^{\left( r - 1\right) } \) under some map \( T{L}_{r - 1} \rightarrow \mathcal{S}\left( {{S}^{1} \times I,2\text{points}}\right) \) formed by including a square into an annulus and joining up boundary points in some way.)
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Proof. Consider the inclusion, shown in Figure 13.10, of the \( T{L}_{n + 1} \) recurrence relation of Figure 13.6 into the annulus.\n\n\n\nFigure 13.10\n\nThe top boundary points on either side of the square are joined to the two points on the annulus boundary, and the other \( n \) points on the left of the square are joined to the \( n \) on the right by parallel arcs encircling the annulus. As in the proof of Lemma 13.2, the two small squares in the final diagram of Figure 13.10 can be slid together (using \( {f}^{\left( n\right) }{f}^{\left( n\right) } = {f}^{\left( n\right) } \) ) to become one square, and the equality can then be rearranged to become that of Figure 13.11. Sum these equalities from \( n = 0 \) to \( n = r - 2 \) (here \( {\Delta }_{-1} = 0 \) ). The right-hand side is \( {a\omega } \) . Rotate each annulus of Figure 13.11 through \( \pi \) and sum again. The right-hand side is now \( {b\omega } \) . The left-hand sides of the formulae so obtained are almost the same; recalling that \( {\Delta }_{-1} = 0 \), the difference of these left-hand sides is the difference of the first term of Figure 13.11, when \( n = r - 2 \), and its rotation; in each is a copy of \( {f}^{\left( r - 1\right) } \) .\n\n\n\nFigure 13.11
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Yes
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Lemma 13.5. Suppose that \( A \) is chosen so that \( {A}^{4} \) is a primitive \( {r}^{\text{th }} \) root of unity, \( r \geq 3 \) . Suppose that \( D \) is a planar diagram of a link of \( n \) (ordered) components. Suppose that \( {D}^{\prime } \) is another such diagram, obtained from \( D \) by a Kirby type 2 move, in which a parallel of the first component of \( D \) is joined by some band to another component (or, equivalently, a segment of the second component is moved up to and over the first). Then\n\n\[< \omega ,\ldots ,\;{ > }_{D} = < \omega ,\ldots ,\;{ > }_{{D}^{\prime }}.\]
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Proof. It must be checked that the elements of \( \mathcal{S}\left( {\mathbb{R}}^{2}\right) \), produced as described above from \( D \) and from \( {D}^{\prime } \), with \( \omega \) as the \
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No
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Corollary 13.6. If \( {A}^{4} \) is a primitive \( {r}^{\text{th }} \) root of unity, \( r \geq 3 \), and planar diagrams \( D \) and \( {D}^{\prime } \) are related by a sequence of Kirby moves of type 2, then\n\n\[< \omega ,\omega ,\ldots ,\omega { > }_{D} = < \omega ,\omega \ldots ,\omega { > }_{{D}^{\prime }}.\]
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Null
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No
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Theorem 13.8. Suppose that a closed oriented 3-manifold \( M \) is obtained by surgery on a framed link that is represented by a planar diagram D. Let \( {b}_{ + } \) be the number of positive eigenvalues and \( {b}_{ - } \) be the number negative eigenvalues of the linking matrix of this link. Suppose \( r \geq 3 \) and that \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. Then\n\n\[ < \omega ,\omega ,\ldots ,\omega { > }_{D} < \omega { > }_{{U}_{ + }}^{-{b}_{ + }} < \omega { > }_{{U}_{ - }}^{-{b}_{ - }} \]\n\nis a well-defined invariant of \( M \) .
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Proof. Note that \( A \) is a primitive \( 4{r}^{th} \) root of unity, and so, by Lemma 13.7, \( < \omega { > }_{{U}_{ + }} \) and \( < \omega { > }_{{U}_{ - }} \) are non-zero. It follows from the Corollary 13.6 and the preceding remarks about the linking matrix that the given expression is invariant under Kirby type 2 moves. The last two factors make it invariant under Kirby type 1 moves, and regular isotopy of \( D \) just induces regular isotopies of all the diagrams used in defining the expression.
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Yes
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Theorem 13.8. Suppose that a closed oriented 3-manifold \( M \) is obtained by surgery on a framed link that is represented by a planar diagram D. Let \( {b}_{ + } \) be the number of positive eigenvalues and \( {b}_{ - } \) be the number negative eigenvalues of the linking matrix of this link. Suppose \( r \geq 3 \) and that \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. Then\n\n\[< \omega ,\omega ,\ldots ,\omega { > }_{D} < \omega { > }_{{U}_{ + }}^{-{b}_{ + }} < \omega { > }_{{U}_{ - }}^{-{b}_{ - }}\]\n\nis a well-defined invariant of \( M \) .
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Proof. Note that \( A \) is a primitive \( 4{r}^{th} \) root of unity, and so, by Lemma 13.7, \( < \omega { > }_{{U}_{ + }} \) and \( < \omega { > }_{{U}_{ - }} \) are non-zero. It follows from the Corollary 13.6 and the preceding remarks about the linking matrix that the given expression is invariant under Kirby type 2 moves. The last two factors make it invariant under Kirby type 1 moves, and regular isotopy of \( D \) just induces regular isotopies of all the diagrams used in defining the expression.
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Yes
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Lemma 13.9. Suppose \( r \geq 3 \) and \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. The element of \( T{L}_{n} \) shown in Figure 13.13 is the zero map of outsides if \( 1 \leq n \leq r - 2 \) . When \( n = 0 \), the element acts as multiplication by \( \langle \omega {\rangle }_{U} \) .
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Proof. Consider first the element of \( T{L}_{n} \) that consists of \( {f}^{\left( n\right) } \) encircled by one simple closed curve. This is shown in Figure 13.14 for \( n = 4 \) . Figure 13.14 shows a calculation for that element. Firstly one crossing is removed in the two standard ways, the results being multiplied by \( A \) and \( {A}^{-1} \) and added. The two elements obtained are then simplified by removing kinks and multiplying by \( - {A}^{\pm 3} \) . Now, in the two resulting elements, removal of any of the crossings depicted in one of the standard ways gives zero (as \( {f}^{\left( n\right) }{e}_{i} = 0 \) ), so only the other standard way need be considered. It follows that \( {f}^{\left( n\right) } \) encircled by one simple closed curve is equal, in \( T{L}_{n} \), to \( \left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) {f}^{\left( n\right) } \) . Now the element required in this Lemma is \( {f}^{\left( n\right) } \) encircled by an \( \omega \), regarded as a map of outsides. Let this be denoted \( x \) . A small single unknotted simple closed curve inserted into this changes \( x \), in the usual way, to \( \left( {-{A}^{-2} - {A}^{2}}\right) x \) . However, that small curve can be slid right over the \( \omega \) without (by Lemma 13.5) changing the map of outsides, and then removed altogether (by the preceding paragraph) at the cost of multiplying by \( \left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) \) . Thus \( \left( {-{A}^{-2} - {A}^{2}}\right) x = \) \( \left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) x \) . Hence either \( x = 0 \) or \( {A}^{2\left( {n + 1}\right) } = {A}^{2} \) or \( {A}^{2\left( {n + 1}\right) } = {A}^{-2} \) . The two latter possibilities do not occur for \( 1 \leq n \leq r - 2 \), as \( A \) is a primitive \( 4{r}^{th} \) root of unity, so then \( x = 0 \) . When \( n = 0 \), it is trivial that \( x \) acts as multiplication by \( \langle \omega {\rangle }_{U} \) because there is nothing but the curve labelled \( \omega \) to consider.
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Yes
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Lemma 14.2. The element of \( T{L}_{n} \) shown in Figure 14.2, which consists of the idempotent \( {f}^{\left( n\right) } \) with all its strands encircled by a parallel strands that join up the ends of an idempotent \( {f}^{\left( a\right) } \), is\n\n\[ \n{\left( -1\right) }^{a}\frac{{A}^{2\left( {n + 1}\right) \left( {a + 1}\right) } - {A}^{-2\left( {n + 1}\right) \left( {a + 1}\right) }}{{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}{f}^{\left( n\right) }.\n\]
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Proof. The \( a \) parallel strands and the idempotent \( {f}^{\left( a\right) } \) can, as explained in Chapter 13, be thought of as \( {S}_{a}\left( \alpha \right) \) contained in an annulus encircling the strands of \( {f}^{\left( n\right) } \), where \( {S}_{a} \) is the \( {a}^{\text{th }} \) Chebyshev polynomial. Now, as in the proof of Lemma 13.9, \( {f}^{\left( n\right) } \) with a single strand encircling it (to be thought of as \( \alpha \) in the annulus) is \( \left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) {f}^{\left( n\right) } \) . Hence the element required here is \( {S}_{a}\left( {-{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) {f}^{\left( n\right) } \) . This immediately gives the result using the remarks about Chebyshev polynomials after Lemma 13.2.
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Yes
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Lemma 14.3. Suppose \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. Then\n\n\[< \omega { > }_{{U}_{ + }} = \frac{G}{2{A}^{\left( 3 + {r}^{2}\right) }\left( {{A}^{2} - {A}^{-2}}\right) },\]
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Proof. Recall that \( {U}_{ + } \) is the diagram of the unknot with one positive crossing,\n\n\[ \omega = \mathop{\sum }\limits_{{n = 0}}^{{r - 2}}{\Delta }_{n}{S}_{n}\left( \alpha \right) \text{ and }{\Delta }_{n} = \frac{{\left( -1\right) }^{n}\left( {{A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }}\right) }{{A}^{2} - {A}^{-2}}.\]\n\nSo use of Lemma 14.1 to remove the kink in each \( {S}_{n}\left( \alpha \right) \) shows that \( \langle \omega {\rangle }_{{U}_{ + }} \) is\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{{r - 2}}{\Delta }_{n}^{2}{\left( -1\right) }^{n}{A}^{{n}^{2} + {2n}} = {\left( {A}^{2} - {A}^{-2}\right) }^{-2}\mathop{\sum }\limits_{{n = 0}}^{{r - 2}}{\left( -1\right) }^{n}{A}^{{n}^{2} + {2n}}{\left( {A}^{2\left( {n + 1}\right) } - {A}^{-2\left( {n + 1}\right) }\right) }^{2}.\]\n\nNow elementary manoeuvres of algebraic number theory (see [74] for example) show that the summation in the last term is \( \frac{1}{2}{A}^{-\left( {3 + {r}^{2}}\right) }\left( {{A}^{2} - {A}^{-2}}\right) \mathop{\sum }\limits_{{n = 1}}^{{4r}}{A}^{{n}^{2}} \) .
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Yes
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Lemma 14.5.\n\n\\[ \n\\Gamma \\left( {x, y, z}\\right) = \\frac{{\\Delta }_{x + y + z}!{\\Delta }_{x - 1}!{\\Delta }_{y - 1}!{\\Delta }_{z - 1}!}{{\\Delta }_{y + z - 1}!{\\Delta }_{z + x - 1}!{\\Delta }_{x + y - 1}!}. \n\\]
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Proof. Consider the equations depicted in Figure 14.4; as usual a symbol beside a line is a count of the number of parallel arcs that it represents. The first equality follows from the defining relation of Figure 13.6 for \\( {f}^{\\left( y + z - 1\\right) } \\) (together with \\( {f}^{\\left( z\\right) }{e}_{z - 1} = 0 \\) ), and the second line follows by iterating the first line. Next, the defining relation for \\( {f}^{\\left( y + z\\right) } \\) followed by a double application of Figure 14.4 produces the identity of Figure 14.5.\n\nNow apply this last identity to Figure 14.3, using the formulae of Figures 13.4 and 13.5. The following recurrence relation results:\n\n\\[ \n\\Gamma \\left( {x, y, z}\\right) = \n\\]\n\n\\[ \n\\Gamma \\left( {x, y, z - 1}\\right) {\\Delta }_{x + z}/{\\Delta }_{x + z - 1} - \\Gamma \\left( {x + 1, y - 1, z - 1}\\right) {\\left( {\\Delta }_{y - 1}\\right) }^{2}/\\left( {{\\Delta }_{y + z - 1}{\\Delta }_{y + z - 2}}\\right) . \n\\]\n\nThis is ready for a verification of the given formula by induction on \\( z \\). That formula is clearly true when \\( z = 0 \\), and inserting it into this recurrence relation reduces the proof to a demonstration of the equality\n\n\\[ \n{\\Delta }_{x + y + z}{\\Delta }_{z - 1} = {\\Delta }_{x + z}{\\Delta }_{y + z - 1} - {\\Delta }_{y - 1}{\\Delta }_{x} \n\\]\n\nThe truth of this can however easily be checked either directly from the formula for \\( {\\Delta }_{n} \\) or using a double induction on\n\n\\[ \n{\\Delta }_{x + y} = {\\Delta }_{x}{\\Delta }_{y} - {\\Delta }_{x - 1}{\\Delta }_{y - 1} \n\\]
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Yes
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Lemma 14.7. Let \( \left( {a, b, c}\right) \) be admissible and let \( A \) be a primitive \( 4{r}^{\text{th }} \) root of unity. Then \( {\tau }_{a, b, c}^{ * } \) is non-zero if and only if \( a + b + c \leq 2\left( {r - 2}\right) \) .
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Proof. \( \mathcal{S}{D}^{\prime } \) has a base consisting of all diagrams in \( {D}^{\prime } \) with no crossing. For all but one of these diagrams there is an arc from a point of one of the three specified subsets (for example, that with \( a \) points) to another point of the same subset. As usual (using \( {f}^{\left( a\right) }{e}_{i} = 0 \) ), \( {\tau }_{a, b, c}^{ * } \) annihilates such an element. There remains to consider the base element of \( \mathcal{S}{D}^{\prime } \) that consists of \( z \) arcs from the first boundary subset to the second such subset, \( x \) from the second to the third and \( y \) from the third to the first. Of course, \( {T}_{a, b, c}^{ * } \) maps this element to \( \Gamma \left( {x, y, z}\right) \) . It follows from Lemma 14.5 that as \( x + y + z \) increases, this is non-zero until \( {\Delta }_{x + y + z}! = 0 \) and that this occurs when \( x + y + z = r - 1 \) .
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Yes
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Lemma 14.9. Suppose \( A \) is not a root of unity. \( A \) base for \( {Q}_{a, b, c, d} \) is the set of elements as in Figure 14.10 (the boundary of the disc is not shown), where \( j \) takes all values such that \( \left( {a, b, j}\right) \) and \( \left( {c, d, j}\right) \) are both admissible.
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Proof. Note that the proposed base elements each consist of two triads glued together; there is an \( {f}^{\left( j\right) } \) on the central line. Certainly \( {Q}_{a, b, c, d} \) is spanned by all elements of the form shown in Figure 14.11, where the lines all represent multiple parallel arcs, for, as usual, any other diagrams interact with the idempotents to give zero. Without loss of generality, it is assumed that \( b + d \geq a + c \), and it is clear that the diagonal line represents \( \frac{1}{2}\{ b + d - a - c\} \) parallel arcs. The number of arcs represented by the other lines can vary. Suppose there are \( j \) arcs crossing the vertical dotted line. In the Temperley-Lieb algebra \( T{L}_{j} \), recall that \( \mathbf{1} - {f}^{\left( j\right) } \) is in the ideal generated by the \( {e}_{i} \) . Thus a diagram with \( j \) arcs crossing the dotted line can be replaced with a linear sum of diagrams, one with \( j \) arcs containing an \( {f}^{\left( j\right) } \) and others that cross the vertical line fewer than \( j \) times (coming from the \( {e}_{i} \) ). Thus, by induction on the number of arcs crossing the vertical line, it is seen that the given elements span the space.
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Yes
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Lemma 14.10. Suppose \( A \) is a primitive \( 4{r}^{\text{th }} \) root of unity. A base for \( {Q}_{a, b, c, d}^{ * } \) (this being \( {Q}_{a, b, c, d} \) regarded as maps of diagrams outside the disc) is the set of elements as in Figure 14.10 where \( j \) takes all values such that \( \left( {a, b, j}\right) \) and \( \left( {c, d, j}\right) \) are both \( r \) -admissible.
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Proof. The proof that the given elements span is the same as in Lemma 14.9 with a small modification. Now, \( {f}^{\left( n\right) } \) does not exist for \( n \geq r \) . However, \( {f}^{\left( r - 1\right) } \) is the zero map of outsides. Thus working in this dual context, any diagram as in the above proof, with at least \( \left( {r - 1}\right) \) arcs crossing the dotted vertical line, can be replaced by a sum of diagrams with fewer such arcs. Further, any triad encountered that is not \( r \) -admissible may be discarded, since it represents the zero map. The proof of independence is essentially the same as before (though the map used now goes from and to the dual spaces).
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Yes
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Lemma 14.11. In \( \mathcal{S}\left( {{S}^{1} \times I}\right) ,{S}_{a}\left( \alpha \right) {S}_{b}\left( \alpha \right) = \mathop{\sum }\limits_{c}{S}_{c}\left( \alpha \right) \) where the summation is over all \( c \) such that \( \left( {a, b, c}\right) \) is admissible. If \( \bar{A} \) is a primitive \( 4{r}^{\text{th }} \) root of unity regarding both sides of the equation as maps of outsides (of immersed annuli as in Chapter 13), \( {S}_{a}\left( \alpha \right) {S}_{b}\left( \alpha \right) = \mathop{\sum }\limits_{c}{S}_{c}\left( \alpha \right) \), where now the sum is over all \( c \) such that \( \left( {a, b, c}\right) \) is \( r \) -admissible.
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Proof. In fact the first part of this lemma is almost immediate. This is because it is a result on Chebyshev polynomials that \( {S}_{a}\left( x\right) {S}_{b}\left( x\right) = \mathop{\sum }\limits_{c}{S}_{c}\left( x\right) \), the sum being over all \( c \) such that \( \left( {a, b, c}\right) \) is admissible. This follows by induction on \( b \) . However, another proof is shown in Figure 14.16, where the result of Figure 14.15 is first applied at the top of the diagram and then the result of Figure 14.8 is applied at the bottom. The advantage of this alternative proof is that it also works in the \( r \) -admissible case as well.
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Yes
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Theorem 14.12. Let \( {F}_{g} \) be the closed orientable surface of genus \( g \) and let \( A \) be a primitive \( 4{r}^{\text{th }} \) root of unity, \( r \geq 3 \) . Then \( {\mathcal{I}}_{A}\left( {{S}^{1} \times {F}_{g}}\right) \) is an integer. It is \( r - 1 \) when \( g = 1 \) . Otherwise it is the number of ways of labelling the \( 3\left( {g - 1}\right) \) edges of the graph of Figure 14.18 with integers \( {a}_{i},0 \leq {a}_{i} \leq r - 2 \), so that the three labels at any node form an \( r \) -admissible triple.
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Proof. The 3-manifold \( {S}^{1} \times {F}_{g} \) is obtained by surgery on a link that consists of \( g \) copies of the Borromean rings summed together on one component, each component having the zero framing. (Proving this is an interesting exercise.) A diagram \( D \) for such a link is obtained by taking \( g \) annuli, each containing a link as on the left of Figure 14.17, threading an unknotted closed curve through these annuli and then taking the resultant diagram of \( {2g} + 1 \) components. Then \( < \omega ,\omega ,\ldots ,\omega { > }_{D} = < \omega ,{\beta }^{g}{ > }_{H} \), where \( H \) is just the two-crossing diagram of the simple Hopf link of two curves. Thus, as the signature of the linking matrix of\n\n\n\nFigure 14.18\n\nthis link is zero (and using \( \left\langle {{\mu \omega }{ > }_{U} = {\mu }^{-1}}\right\rangle \) ,\n\n\[{\mathcal{I}}_{A}\left( {{S}^{1} \times {F}_{g}}\right) = {\mu }^{{2g} + 2} < \omega ,{\beta }^{g}{ > }_{H}.\n\]\n\nNow \( \beta = \mathop{\sum }\limits_{{a = 0}}^{{r - 2}}{\mu }^{-2}{\left( {S}_{a}\left( \alpha \right) \right) }^{2} \), so \( {\beta }^{g} \) can be expressed as a sum of the \( {S}_{n}\left( \alpha \right) \) ’s by Lemma 14.11. Then, by Lemma 13.9, \( \langle \omega ,{\beta }^{g}{\rangle }_{H} \) is \( {\mu }^{-2 - {2g}}N \), where \( N \) is the number of times \( {S}_{0}\left( \alpha \right) \) appears in the expansion (by Lemma 14.11) for \( {\left( \mathop{\sum }\limits_{{a = 0}}^{{r - 2}}{\left( {S}_{a}\left( \alpha \right) \right) }^{2}\right) }^{g} \) as a sum of the \( {S}_{n}\left( \alpha \right) \) ’s. This \( N \) is the number of \( r \) -admissible labellings of the edges of Figure 14.18.
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Yes
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Theorem 14.13. If \( p \) and \( q \) are coprime positive integers, then the Jones polynomial of the \( \left( {p, q}\right) \) -torus knot is\n\n\[ \n{t}^{\left( {p - 1}\right) \left( {q - 1}\right) /2}{\left( 1 - {t}^{2}\right) }^{-1}\left( {1 - {t}^{p + 1} - {t}^{q + 1} + {t}^{p + q}}\right) .\n\]
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Proof. Consider the diagram of Figure 14.23, which shows \( p \) arcs traversing a rectangle. Suppose \( q \) copies of this are placed side by side and the result is closed up by joining the \( p \) points on the left to those on the right, using \( p \) crossing-free arcs encircling an annulus \( {S}^{1} \times I \) to form a diagram \( T\left( {p, q}\right) \) in that annulus. It is desired to evaluate this diagram in the skein of the annulus in terms of the base elements \( \left\{ {{S}_{n}\left( \alpha \right) }\right\} \) . Then, placing the annulus in the plane will at once give a value for the Jones polynomial of the \( \left( {p, q}\right) \) -torus knot. For some fixed \( k \) (which will here later be taken to be 1), consider \( p \) arcs side by side in a diagram, each labelled with an \( {f}^{\left( k\right) } \) so that as usual each arc represents \( k \) parallel arcs with the idempotent inserted. Applying the identity of Figure 14.15 ( \( p - 1 \) ) times shows that this is of the form of Figure 14.24, where the coefficient \( \Lambda \left( {{i}_{1},{i}_{2},\ldots ,{i}_{p - 2}, a}\right) \) is the quotient of a product of \( \Delta \) ’s by a product of \( \theta \) ’s and the summation is over all \( \left( {{i}_{1},{i}_{2},\ldots ,{i}_{p - 2}, a}\right) \) that produce an admissible triple at each vertex of the diagram.\n\n\n\nFigure 14.23\n\n\n\n\n\nFigure 14.24\n\nThe diagram of Figure 14.25 is, of course, a multiple of \( {f}^{\left( a\right) } \) ; let it be denoted, without any summation convention, by\n\n\[ \n{\left\{ \Lambda \left( {i}_{1},{i}_{2},\ldots ,{i}_{p - 2}, a\right) \Lambda \left( {j}_{1},{j}_{2},\ldots ,{j}_{p - 2}, a\right) \right\} }^{-1/2}M{\left( a\right) }_{\mathbf{j}}^{\mathbf{i}}{f}^{\left( a\right) }.\n\]\n\nThe \( \left\{ {M{\left( a\right) }_{\mathrm{i}}^{\mathrm{i}}}\right\} \) will be regarded as a matrix \( M\left( a\right) \) with rows and columns indexed by i and \( \mathbf{j} \), each representing a multi-suffix \( \left( {{i}_{1},{i}_{2},\ldots ,{i}_{p - 2}}\right) \) or \( \left( {{j}_{1},{j}_{2},\ldots ,{j}_{p - 2}}\right) \) .\n\n\n\nFigure 14.25\n\nSuppose \( T{\left( p, q\right) }^{\left( k\right) } \) is the diagram \( T\left( {p, q}\right) \) decorated by \( {S}_{k}\left( \alpha \right) \) . Suppose that in \( T{\left( p, q\right) }^{\left( k\right) } \), between consecutive occurrences of (the \( k \) -weighted) Figure 14.23, the \( p \) parallel strings are \
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Yes
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Lemma 15.1. Suppose that \( p \) and \( q \) are two arcs in \( {\mathbb{R}}^{2} \) meeting only at their end points \( A \) and \( B \), and let \( R \) be the compact region bounded by \( p \cup q \) . Suppose that \( {t}_{1},{t}_{2},\ldots {t}_{n} \) are arcs in \( R \), each meeting \( p \cup q \) at just its end points, one in \( p \) and one in \( q \) . Suppose that every \( {t}_{i} \cap {t}_{j} \) is at most one point, that intersections of arcs are transverse and that there are no triple points. The graph, with vertices all intersections of these arcs and edges comprising \( p \cup q \cup \mathop{\bigcup }\limits_{i}{t}_{i} \), separates \( R \) into a collection of \( v \) -gons; amongst these \( v \) -gons there is a 3 -gon with an edge in \( p \) and a 3-gon with an edge in \( q \) .
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Proof. Proceed by induction on the number \( n \) of arcs. The result is trivial if \( n = 1 \), so assume \( n > 1 \) . Amongst the end points of the \( {t}_{i} \) that lie on \( p \), let \( X \) be the nearest to \( A \) . If then \( X \) is an end of \( {t}_{j} \), let \( {B}^{\prime } \) be the other end of \( {t}_{j} \) on \( q \) . If possible, from \( \left\{ {{t}_{i} : i \neq j}\right\} \) select a \( {t}_{k} \) with \( {t}_{k} \cap {t}_{j} = {X}^{\prime },{t}_{k} \cap q = {A}^{\prime } \), such that \( {t}_{k} \) has no point of intersection with a \( {t}_{i} \) between \( {A}^{\prime } \) and \( {X}^{\prime } \) . Select such a \( {t}_{k} \) with \( {X}^{\prime } \) as near as possible to \( {B}^{\prime } \), see Figure 15.1. If there is no such \( {t}_{k} \), select \( p \) instead, taking \( {A}^{\prime } = A \) and \( {X}^{\prime } = X \) . Now let \( {p}^{\prime } \) be an arc starting at \( {A}^{\prime } \), proceeding along\n\n\n\nFigure 15.1\n\n\( {t}_{k} \) (or \( p \) if there is no \( {t}_{k} \) ) to \( {X}^{\prime } \) and then along \( {t}_{j} \) to \( {B}^{\prime } \) (it helps not to think of a corner at \( {X}^{\prime } \) ). Let \( {q}^{\prime } \) be the sub-arc of \( q \) from \( {A}^{\prime } \) to \( {B}^{\prime } \) and let \( {R}^{\prime } \) be the region bounded by \( {p}^{\prime } \cup {q}^{\prime } \) . If no \( {t}_{i} \) meets the interior of \( {R}^{\prime } \), then \( {R}^{\prime } \) is a 3 -gon with an edge in \( q \) . Otherwise \( {R}^{\prime } \) meets \( \left\{ {{t}_{i} : i \neq j, i \neq k}\right\} \) in fewer than \( n \) arcs and so, by induction, there is a 3-gon in \( {R}^{\prime } \) with an edge in \( {q}^{\prime } \) . The choice made for \( {t}_{k} \) ensures that \( {A}^{\prime } \) is not a vertex of this 3-gon (which is important, as \( {X}^{\prime } \) is not a vertex of a \( v \) -gon of \( {R}^{\prime } \) ), and so it is one of the original 3-gons having an edge contained in \( q \) . Similarly, there is a 3-gon with an edge in \( p \) .
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Yes
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Theorem 15.5. There exists a function \[ \Lambda : \left\{ {\text{ Unoriented links diagrams in }{S}^{2}}\right\} \rightarrow \mathbb{Z}\left\lbrack {{a}^{\pm 1},{z}^{\pm 1}}\right\rbrack \] that is defined uniquely by the following: (i) \( \Lambda \left( U\right) = 1 \), where \( U \) is the zero-crossing diagram of the unknot; (ii) \( \Lambda \left( D\right) \) is unchanged by Reidemeister moves of Types II and III on the diagram \( D \) ; (iii) \( \Lambda \left( { \curvearrowright \sim }\right) = {a\Lambda }\left( \curvearrowright \right) \) ; (iv) If \( {D}_{ + },{D}_{ - },{D}_{0} \) and \( {D}_{\infty } \) are four diagrams exactly the same except near a point where they are as shown in Figure 15.6, then \[ \left( {\star \star }\right) \;\Lambda \left( {D}_{ + }\right) + \Lambda \left( {D}_{ - }\right) = z\left( {\Lambda \left( {D}_{0}\right) + \Lambda \left( {D}_{\infty }\right) }\right) . \]
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Proof. Note that, when considering a crossing in an unoriented diagram, it has no claim to be termed \( {D}_{ + } \) rather than \( {D}_{ - } \) in the above notation. However, this never matters, since \( {D}_{ + } \) and \( {D}_{ - } \) feature symmetrically in the formula \( \left( {\star \star }\right) \) ; the treatment of \( {D}_{0} \) and \( {D}_{\infty } \) is likewise symmetric. Observe that the equation \( \left( {\star \star }\right) \) determines uniquely any one of \( \Lambda \left( {D}_{ + }\right) ,\Lambda \left( {D}_{ - }\right) ,\Lambda \left( {D}_{0}\right) \) and \( \Lambda \left( {D}_{\infty }\right) \) from knowledge of the other three. Observe also that a solution to \( \left( {\star \star }\right) \) is \( \left( {\Lambda \left( {D}_{ + }\right) ,\Lambda \left( {D}_{ - }\right) ,\Lambda \left( {D}_{0}\right) ,\Lambda \left( {D}_{\infty }\right) }\right) = \left( {{ax},{a}^{-1}x, x,{\delta x}}\right) \), where \( x \) is arbitrary and \( \delta = \left( {a + {a}^{-1}}\right) {z}^{-1} - 1 \)
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Yes
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Proposition 16.2. If \( {L}_{1} \) and \( {L}_{2} \) are oriented links, then\n\n(i) \( P\left( {{L}_{1} + {L}_{2}}\right) = P\left( {L}_{1}\right) P\left( {L}_{2}\right) \) ;\n\n(ii) \( F\left( {{L}_{1} + {L}_{2}}\right) = F\left( {L}_{1}\right) F\left( {L}_{2}\right) \) ;\n\n(iii) \( P\left( {{L}_{1} \sqcup {L}_{2}}\right) = - \left( {l + {l}^{-1}}\right) {m}^{-1}P\left( {L}_{1}\right) P\left( {L}_{2}\right) \) ;\n\n(iv) \( F\left( {{L}_{1} \sqcup {L}_{2}}\right) = \left( {\left( {a + {a}^{-1}}\right) {z}^{-1} - 1}\right) F\left( {L}_{1}\right) F\left( {L}_{2}\right) \) .
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Null
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No
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Proposition 16.3. Both \( P\left( L\right) \) and \( F\left( L\right) \) are unchanged by mutation of \( L \) .
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Null
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No
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Proposition 16.4. If the oriented link \( {L}^{ * } \) is obtained from the oriented link \( L \) by reversing the orientation of one component \( K \), then\n\n\[ F\left( {L}^{ * }\right) = {a}^{4\operatorname{lk}\left( {K, L - K}\right) }F\left( L\right) . \]\n\nChanging the orientation of all components of \( L \) leaves both \( P\left( L\right) \) and \( F\left( L\right) \) unchanged.
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Null
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No
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Proposition 16.5. For an oriented link \( L \), the Conway-normalised Alexander polynomial \( {\Delta }_{L}\left( t\right) \) and the Jones polynomial \( V\left( L\right) \) are related to the HOMFLY polynomial \( P\left( L\right) \) by \[ {\Delta }_{L}\left( t\right) = P{\left( L\right) }_{\left( i, i\left( {t}^{1/2} - {t}^{-1/2}\right) \right) }\;\text{ and }\;V\left( L\right) = P{\left( L\right) }_{\left( i{t}^{-1}, i\left( {t}^{-1/2} - {t}^{1/2}\right) \right) }, \] where \( {i}^{2} = - 1 \) .
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Null
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No
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Proposition 16.6. For an oriented link \( L \) , \[ V\left( L\right) = F\left( L\right) \text{ when }\left( {a, z}\right) = \left( {-{t}^{-3/4},\;\left( {{t}^{-1/4} + {t}^{1/4}}\right) }\right) , \] \[ {\left( V\left( L\right) \right) }^{2} = {\left( -1\right) }^{\# L - 1}F\left( L\right) \text{ when }t = - {q}^{-2},\;\left( {a, z}\right) = \left( {{q}^{3},{q}^{-1} + q}\right) . \]
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Proof. Underlying the Jones polynomial is the Kauffman bracket. Reverting to the notation for that, given in Definition 3.1, \[ \langle > < \rangle = A\langle > < \rangle + {A}^{-1}\langle < \rangle \] \[ \langle > < \rangle = {A}^{-1}\langle > < \rangle + A\langle > < \rangle . \] Adding these equations gives \[ \langle > < \rangle + \langle > < \rangle = \left( {A + {A}^{-1}}\right) \left( {\langle > < \rangle +\langle < \rangle }\right) . \] However, \( \langle \sim \rangle = - {A}^{3}\langle \sim \rangle \), and the Kauffman bracket is invariant under regular isotopy (Reidemeister Type II and III moves). However, these are the defining rules for the polynomial \( \Lambda \left( D\right) \) of Theorem 15.5 with the variables changed to \( \left( {a, z}\right) = \left( {-{A}^{3},\left( {A + {A}^{-1}}\right) }\right) \) . The substitution \( t = {A}^{-4} \) gives the first result. Subtracting the square of one of the above two equations from the square of the other gives \[ {\left( > < \right) }^{2} - \langle > < {\rangle }^{2} = \left( {{A}^{2} - {A}^{-2}}\right) \left( {\langle > < {\rangle }^{2}-\langle < {\rangle }^{2}}\right) . \] Of course \( \langle \sim \sim {\rangle }^{2} = {A}^{6}\langle \sim {\rangle }^{2} \), and so the square of the Kauffman bracket is an instance of the \( {\Lambda }^{ \star }\left( D\right) \) polynomial, defined at the very end of Chapter 15, that satisfies \[ {\Lambda }^{ \star }\left( {D}_{ + }\right) - {\Lambda }^{ \star }\left( {D}_{ - }\right) = \omega \left( {{\Lambda }^{ \star }\left( {D}_{0}\right) - {\Lambda }^{ \star }\left( {D}_{\infty }\right) }\right) . \] Now, translating the notation by \( \omega = \left( {{A}^{2} - {A}^{-2}}\right) ,\alpha = {A}^{6}, t = - {q}^{-2} = {A}^{-4} \) and (from Chapter 15) \( \left( {a, z}\right) = \left( {{i\alpha }, - {i\omega }}\right) \) the second result follows.
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Yes
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The \( l \) -breadth of \( P\left( L\right) \) satisfies \( {E}_{l}\left( {P\left( L\right) }\right) - {e}_{l}\left( {P\left( L\right) }\right) \leq \) \( 2\left( {s\left( D\right) - 1}\right) \) .
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Null
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No
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Proposition 16.9. Suppose \( L \) is an oriented link with \( \# L \) components. Then \( (1 - \) \( \# L) \) is the lowest power both of \( m \) in \( P\left( L\right) \) and of \( z \) in \( F\left( L\right) \), and
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\[ {\left\lbrack {z}^{\# L - 1}F\left( L\right) \right\rbrack }_{\left( {a, z}\right) = \left( {l,0}\right) } = {\left\lbrack {\left( -m\right) }^{\# L - 1}P\left( L\right) \right\rbrack }_{m = 0}. \]
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Yes
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Proposition 16.10. Any oriented link in \( {S}^{3} \) is the closure \( \widehat{\xi } \) of some \( \xi \) belonging to the braid group \( {B}_{n} \), for some \( n \) . Oriented links \( \widehat{\xi } \) and \( \widehat{\eta } \) are equivalent if \( \xi \) and \( \eta \) differ by a sequence of (Markov) moves of the following two types and inverses of such moves:\n\n(i) Change an element of \( {B}_{n} \) to a conjugate element in that group;\n\n(ii) Change \( \xi \in {B}_{n} \) to \( {i}_{n}\left( \xi \right) {\sigma }_{n}^{\pm 1} \in {B}_{n + 1} \), where \( {i}_{n} : {B}_{n} \rightarrow {B}_{n + 1} \) is the inclusion (that disregards the \( \left( {n + 1}\right) \) th string).
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Null
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No
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Theorem 16.11. If an oriented link \( L \) is the closure of \( \xi \in {B}_{n} \), let \( T\left( L\right) \) be defined to be \( T\left( \xi \right) \) . This is a well-defined link invariant.
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Proof. Because \( T \) is essentially a trace function, if \( \xi ,\eta \in {B}_{n} \) then \( T\left( {{\eta }^{-1}{\xi \eta }}\right) = \n\n\( T\left( \xi \right) \) . Using the properties of \( \mu \), it is easy to show that \( T\left( {\xi {\sigma }_{n}}\right) = T\left( {\xi {\sigma }_{n}^{-1}}\right) = \n\n\( T\left( \xi \right) \) . The result then follows from Proposition 16.10.
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No
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Theorem 2.1 Suppose that \( f \) is an integrable function on the circle with \( \widehat{f}\left( n\right) = 0 \) for all \( n \in \mathbb{Z} \) . Then \( f\left( {\theta }_{0}\right) = 0 \) whenever \( f \) is continuous at the point \( {\theta }_{0} \) .
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Proof. We suppose first that \( f \) is real-valued, and argue by contradiction. Assume, without loss of generality, that \( f \) is defined on \( \left\lbrack {-\pi ,\pi }\right\rbrack \), that \( {\theta }_{0} = 0 \), and \( f\left( 0\right) > 0 \) . The idea now is to construct a family of trigonometric polynomials \( \left\{ {p}_{k}\right\} \) that \
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No
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Corollary 2.2 If \( f \) is continuous on the circle and \( \widehat{f}\left( n\right) = 0 \) for all \( n \in \mathbb{Z} \), then \( f = 0 \) .
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Null
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No
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Corollary 2.3 Suppose that \( f \) is a continuous function on the circle and that the Fourier series of \( f \) is absolutely convergent, \( \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\left| {\widehat{f}\left( n\right) }\right| < \infty \) . Then, the Fourier series converges uniformly to \( f \), that is,\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}{S}_{N}\left( f\right) \left( \theta \right) = f\left( \theta \right) \;\text{ uniformly in }\theta . \]
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Proof. Recall that if a sequence of continuous functions converges uniformly, then the limit is also continuous. Now observe that the assumption \( \sum \left| {\widehat{f}\left( n\right) }\right| < \infty \) implies that the partial sums of the Fourier\n\nseries of \( f \) converge absolutely and uniformly, and therefore the function \( g \) defined by\n\n\[ g\left( \theta \right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\widehat{f}\left( n\right) {e}^{in\theta } = \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\sum }\limits_{{n = - N}}^{N}\widehat{f}\left( n\right) {e}^{in\theta } \]\n\nis continuous on the circle. Moreover, the Fourier coefficients of \( g \) are precisely \( \widehat{f}\left( n\right) \) since we can interchange the infinite sum with the integral (a consequence of the uniform convergence of the series). Therefore, the previous corollary applied to the function \( f - g \) yields \( f = g \), as desired.
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Yes
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Corollary 2.4 Suppose that \( f \) is a twice continuously differentiable function on the circle. Then\n\n\[ \widehat{f}\left( n\right) = O\left( {1/{\left| n\right| }^{2}}\right) \;\text{ as }\left| n\right| \rightarrow \infty ,\] \n\nso that the Fourier series of \( f \) converges absolutely and uniformly to \( f \) .
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Proof. The estimate on the Fourier coefficients is proved by integrating by parts twice for \( n \neq 0 \) . We obtain\n\n\[ {2\pi }\widehat{f}\left( n\right) = {\int }_{0}^{2\pi }f\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = {\left\lbrack f\left( \theta \right) \cdot \frac{-{e}^{-{in\theta }}}{in}\right\rbrack }_{0}^{2\pi } + \frac{1}{in}{\int }_{0}^{2\pi }{f}^{\prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = \frac{1}{in}{\int }_{0}^{2\pi }{f}^{\prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = \frac{1}{in}{\left\lbrack {f}^{\prime }\left( \theta \right) \cdot \frac{-{e}^{-{in\theta }}}{in}\right\rbrack }_{0}^{2\pi } + \frac{1}{{\left( in\right) }^{2}}{\int }_{0}^{2\pi }{f}^{\prime \prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\n\[ = \frac{-1}{{n}^{2}}{\int }_{0}^{2\pi }{f}^{\prime \prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta } \]\n\nThe quantities in brackets vanish since \( f \) and \( {f}^{\prime } \) are periodic. Therefore\n\n\[ {2\pi }{\left| n\right| }^{2}\left| {\widehat{f}\left( n\right) }\right| \leq \left| {{\int }_{0}^{2\pi }{f}^{\prime \prime }\left( \theta \right) {e}^{-{in\theta }}{d\theta }}\right| \leq {\int }_{0}^{2\pi }\left| {{f}^{\prime \prime }\left( \theta \right) }\right| {d\theta } \leq C, \]\n\nwhere the constant \( C \) is independent of \( n \) . (We can take \( C = {2\pi B} \) where \( B \) is a bound for \( {f}^{\prime \prime } \) .) Since \( \sum 1/{n}^{2} \) converges, the proof of the corollary is complete.
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Yes
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Proposition 3.1 Suppose that \( f, g \), and \( h \) are \( {2\pi } \) -periodic integrable functions. Then:\n\n(i) \( f * \left( {g + h}\right) = \left( {f * g}\right) + \left( {f * h}\right) \) .
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Proof. Properties (i) and (ii) follow at once from the linearity of the integral.
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Yes
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Lemma 3.2 Suppose \( f \) is integrable on the circle and bounded by \( B \) . Then there exists a sequence \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \) of continuous functions on the circle so that\n\n\[ \mathop{\sup }\limits_{{x \in \left\lbrack {-\pi ,\pi }\right\rbrack }}\left| {{f}_{k}\left( x\right) }\right| \leq B\;\text{ for all }k = 1,2,\ldots ,\]\n\nand\n\n\[ {\int }_{-\pi }^{\pi }\left| {f\left( x\right) - {f}_{k}\left( x\right) }\right| {dx} \rightarrow 0\;\text{ as }k \rightarrow \infty . \]
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Using this result, we may complete the proof of the proposition as follows. Apply Lemma 3.2 to \( f \) and \( g \) to obtain sequences \( \left\{ {f}_{k}\right\} \) and \( \left\{ {g}_{k}\right\} \) of approximating continuous functions. Then\n\n\[ f * g - {f}_{k} * {g}_{k} = \left( {f - {f}_{k}}\right) * g + {f}_{k} * \left( {g - {g}_{k}}\right) .\n\nBy the properties of the sequence \( \left\{ {f}_{k}\right\} \) ,\n\n\[ \left| {\left( {f - {f}_{k}}\right) * g\left( x\right) }\right| \leq \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left| {f\left( {x - y}\right) - {f}_{k}\left( {x - y}\right) }\right| \left| {g\left( y\right) }\right| {dy}\n\n\[ \leq \frac{1}{2\pi }\mathop{\sup }\limits_{y}\left| {g\left( y\right) }\right| {\int }_{-\pi }^{\pi }\left| {f\left( y\right) - {f}_{k}\left( y\right) }\right| {dy}\n\n\[ \rightarrow 0\;\text{ as }k \rightarrow \infty \text{. }\n\nHence \( \left( {f - {f}_{k}}\right) * g \rightarrow 0 \) uniformly in \( x \) . Similarly, \( {f}_{k} * \left( {g - {g}_{k}}\right) \rightarrow 0 \) uniformly, and therefore \( {f}_{k} * {g}_{k} \) tends uniformly to \( f * g \) . Since each \( {f}_{k} * {g}_{k} \) is continuous, it follows that \( f * g \) is also continuous, and we have (v).
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Yes
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Theorem 4.1 Let \( {\left\{ {K}_{n}\right\} }_{n = 1}^{\infty } \) be a family of good kernels, and \( f \) an integrable function on the circle. Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {f * {K}_{n}}\right) \left( x\right) = f\left( x\right) \]\n\nwhenever \( f \) is continuous at \( x \) . If \( f \) is continuous everywhere, then the above limit is uniform.
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Proof of Theorem 4.1. If \( \epsilon > 0 \) and \( f \) is continuous at \( x \), choose \( \delta \) so that \( \left| y\right| < \delta \) implies \( \left| {f\left( {x - y}\right) - f\left( x\right) }\right| < \epsilon \) . Then, by the first property of good kernels, we can write\n\n\[ \left( {f * {K}_{n}}\right) \left( x\right) - f\left( x\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_{n}\left( y\right) f\left( {x - y}\right) {dy} - f\left( x\right) \]\n\n\[ = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_{n}\left( y\right) \left\lbrack {f\left( {x - y}\right) - f\left( x\right) }\right\rbrack {dy}. \]\n\nHence,\n\n\[ \left| {\left( {f * {K}_{n}}\right) \left( x\right) - f\left( x\right) }\right| = \left| {\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_{n}\left( y\right) \left\lbrack {f\left( {x - y}\right) - f\left( x\right) }\right\rbrack {dy}}\right| \]\n\n\[ \leq \frac{1}{2\pi }{\int }_{\left| y\right| < \delta }\left| {{K}_{n}\left( y\right) }\right| \left| {f\left( {x - y}\right) - f\left( x\right) }\right| {dy} \]\n\n\[ + \frac{1}{2\pi }{\int }_{\delta \leq \left| y\right| \leq \pi }\left| {{K}_{n}\left( y\right) }\right| \left| {f\left( {x - y}\right) - f\left( x\right) }\right| {dy} \]\n\n\[ \leq \frac{\epsilon }{2\pi }{\int }_{-\pi }^{\pi }\left| {{K}_{n}\left( y\right) }\right| {dy} + \frac{2B}{2\pi }{\int }_{\delta \leq \left| y\right| \leq \pi }\left| {{K}_{n}\left( y\right) }\right| {dy} \]\n\nwhere \( B \) is a bound for \( f \) . The first term is bounded by \( {\epsilon M}/{2\pi } \) because of the second property of good kernels. By the third property we see that for all large \( n \), the second term will be less than \( \epsilon \) . Therefore, for some constant \( C > 0 \) and all large \( n \) we have\n\n\[ \left| {\left( {f * {K}_{n}}\right) \left( x\right) - f\left( x\right) }\right| \leq {C\epsilon } \]\n\nthereby proving the first assertion in the theorem. If \( f \) is continuous everywhere, then it is uniformly continuous, and \( \delta \) can be chosen independent of \( x \) . This provides the desired conclusion that \( f * {K}_{n} \rightarrow f \) uniformly.
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Yes
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Lemma 5.1 We have\n\n\[ \n{F}_{N}\left( x\right) = \frac{1}{N}\frac{{\sin }^{2}\left( {{Nx}/2}\right) }{{\sin }^{2}\left( {x/2}\right) }\n\]\n\nand the Fejér kernel is a good kernel.
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The proof of the formula for \( {F}_{N} \) (a simple application of trigonometric identities) is outlined in Exercise 15. To prove the rest of the lemma, note that \( {F}_{N} \) is positive and \( \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{F}_{N}\left( x\right) {dx} = 1 \), in view of the fact that a similar identity holds for the Dirichlet kernels \( {D}_{n} \) . However, \( {\sin }^{2}\left( {x/2}\right) \geq \) \( {c}_{\delta } > 0 \), if \( \delta \leq \left| x\right| \leq \pi \), hence \( {F}_{N}\left( x\right) \leq 1/\left( {N{c}_{\delta }}\right) \), from which it follows that\n\n\[ \n{\int }_{\delta \leq \left| x\right| \leq \pi }\left| {{F}_{N}\left( x\right) }\right| {dx} \rightarrow 0\;\text{ as }N \rightarrow \infty .\n\]
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No
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Theorem 5.2 If \( f \) is integrable on the circle, then the Fourier series of \( f \) is Cesàro summable to \( f \) at every point of continuity of \( f \) .
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Null
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No
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Corollary 5.3 If \( f \) is integrable on the circle and \( \widehat{f}\left( n\right) = 0 \) for all \( n \) , then \( f = 0 \) at all points of continuity of \( f \) .
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The proof is immediate since all the partial sums are 0 , hence all the Cesàro means are 0 .
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Yes
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Corollary 5.4 Continuous functions on the circle can be uniformly approximated by trigonometric polynomials.
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This means that if \( f \) is continuous on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) with \( f\left( {-\pi }\right) = f\left( \pi \right) \) and \( \epsilon > 0 \), then there exists a trigonometric polynomial \( P \) such that\n\n\[ \left| {f\left( x\right) - P\left( x\right) }\right| < \epsilon \;\text{ for all } - \pi \leq x \leq \pi . \]\n\nThis follows immediately from the theorem since the partial sums, hence the Cesàro means, are trigonometric polynomials.
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Yes
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Lemma 5.5 If \( 0 \leq r < 1 \), then\n\n\[ \n{P}_{r}\left( \theta \right) = \frac{1 - {r}^{2}}{1 - {2r}\cos \theta + {r}^{2}}.\n\]
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Proof. The identity \( {P}_{r}\left( \theta \right) = \frac{1 - {r}^{2}}{1 - {2r}\cos \theta + {r}^{2}} \) has already been derived in Section 1.1. Note that\n\n\[ \n1 - {2r}\cos \theta + {r}^{2} = {\left( 1 - r\right) }^{2} + {2r}\left( {1 - \cos \theta }\right) .\n\]\n\nHence if \( 1/2 \leq r \leq 1 \) and \( \delta \leq \left| \theta \right| \leq \pi \), then\n\n\[ \n1 - {2r}\cos \theta + {r}^{2} \geq {c}_{\delta } > 0.\n\]\n\nThus \( {P}_{r}\left( \theta \right) \leq \left( {1 - {r}^{2}}\right) /{c}_{\delta } \) when \( \delta \leq \left| \theta \right| \leq \pi \), and the third property of good kernels is verified. Clearly \( {P}_{r}\left( \theta \right) \geq 0 \), and integrating the expression (4) term by term (which is justified by the absolute convergence of the series) yields\n\n\[ \n\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{P}_{r}\left( \theta \right) {d\theta } = 1\n\]\n\nthereby concluding the proof that \( {P}_{r} \) is a good kernel.
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Yes
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Theorem 5.6 The Fourier series of an integrable function on the circle is Abel summable to \( f \) at every point of continuity. Moreover, if \( f \) is continuous on the circle, then the Fourier series of \( f \) is uniformly Abel summable to \( f \) .
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Null
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No
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Theorem 1.1 Suppose \( f \) is integrable on the circle. Then\n\n\[ \frac{1}{2\pi }{\int }_{0}^{2\pi }{\left| f\left( \theta \right) - {S}_{N}\left( f\right) \left( \theta \right) \right| }^{2}{d\theta } \rightarrow 0\;\text{ as }N \rightarrow \infty . \]
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Null
|
No
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Lemma 1.2 (Best approximation) If \( f \) is integrable on the circle with Fourier coefficients \( {a}_{n} \), then\n\n\[ \begin{Vmatrix}{f - {S}_{N}\left( f\right) }\end{Vmatrix} \leq \begin{Vmatrix}{f - \mathop{\sum }\limits_{{\left| n\right| \leq N}}{c}_{n}{e}_{n}}\end{Vmatrix} \]\n\nfor any complex numbers \( {c}_{n} \) . Moreover, equality holds precisely when \( {c}_{n} = {a}_{n} \) for all \( \left| n\right| \leq N \) .
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Proof. This follows immediately by applying the Pythagorean theorem to\n\n\[ f - \mathop{\sum }\limits_{{\left| n\right| \leq N}}{c}_{n}{e}_{n} = f - {S}_{N}\left( f\right) + \mathop{\sum }\limits_{{\left| n\right| \leq N}}{b}_{n}{e}_{n} \]\n\nwhere \( {b}_{n} = {a}_{n} - {c}_{n} \) .
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Yes
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Theorem 1.3 Let \( f \) be an integrable function on the circle with \( f \sim \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}{e}^{in\theta } \) . Then we have:\n\n(i) Mean-square convergence of the Fourier series\n\n\[ \frac{1}{2\pi }{\int }_{0}^{2\pi }{\left| f\left( \theta \right) - {S}_{N}\left( f\right) \left( \theta \right) \right| }^{2}{d\theta } \rightarrow 0\;\text{ as }N \rightarrow \infty .\n\]\n\n(ii) Parseval's identity\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\left| {a}_{n}\right| }^{2} = \frac{1}{2\pi }{\int }_{0}^{2\pi }{\left| f\left( \theta \right) \right| }^{2}{d\theta }.\n\]
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Null
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No
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Theorem 1.4 (Riemann-Lebesgue lemma) If \( f \) is integrable on the circle, then \( \widehat{f}\left( n\right) \rightarrow 0 \) as \( \left| n\right| \rightarrow \infty \) .
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Null
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No
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Lemma 1.5 Suppose \( F \) and \( G \) are integrable on the circle with\n\n\[ F \sim \sum {a}_{n}{e}^{in\theta }\;\text{ and }\;G \sim \sum {b}_{n}{e}^{in\theta }.\]\n\nThen\n\n\[ \frac{1}{2\pi }{\int }_{0}^{2\pi }F\left( \theta \right) \overline{G\left( \theta \right) }{d\theta } = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{a}_{n}\overline{{b}_{n}}.\]
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Proof. The proof follows from Parseval's identity and the fact that\n\n\[ \left( {F, G}\right) = \frac{1}{4}\left\lbrack {\parallel F + G{\parallel }^{2} - \parallel F - G{\parallel }^{2} + i\left( {\parallel F + {iG}{\parallel }^{2} - \parallel F - {iG}{\parallel }^{2}}\right) }\right\rbrack \]\n\nwhich holds in every Hermitian inner product space. The verification of this fact is left to the reader.
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No
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Theorem 2.1 Let \( f \) be an integrable function on the circle which is differentiable at a point \( {\theta }_{0} \) . Then \( {S}_{N}\left( f\right) \left( {\theta }_{0}\right) \rightarrow f\left( {\theta }_{0}\right) \) as \( N \) tends to infinity.
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Proof. Define\n\n\[ F\left( t\right) = \left\{ \begin{array}{ll} \frac{f\left( {{\theta }_{0} - t}\right) - f\left( {\theta }_{0}\right) }{t} & \text{ if }t \neq 0\text{ and }\left| t\right| < \pi \\ - {f}^{\prime }\left( {\theta }_{0}\right) & \text{ if }t = 0. \end{array}\right. \]\n\nFirst, \( F \) is bounded near 0 since \( f \) is differentiable there. Second, for all small \( \delta \) the function \( F \) is integrable on \( \left\lbrack {-\pi , - \delta }\right\rbrack \cup \left\lbrack {\delta ,\pi }\right\rbrack \) because \( f \) has this property and \( \left| t\right| > \delta \) there. As a consequence of Proposition 1.4 in the appendix, the function \( F \) is integrable on all of \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . We know that \( {S}_{N}\left( f\right) \left( {\theta }_{0}\right) = \left( {f * {D}_{N}}\right) \left( {\theta }_{0}\right) \), where \( {D}_{N} \) is the Dirichlet kernel. Since \( \frac{1}{2\pi }\int {D}_{N} = 1 \), we find that\n\n\[ {S}_{N}\left( f\right) \left( {\theta }_{0}\right) - f\left( {\theta }_{0}\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( {{\theta }_{0} - t}\right) {D}_{N}\left( t\right) {dt} - f\left( {\theta }_{0}\right) \]\n\n\[ = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left\lbrack {f\left( {{\theta }_{0} - t}\right) - f\left( {\theta }_{0}\right) }\right\rbrack {D}_{N}\left( t\right) {dt} \]\n\n\[ = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }F\left( t\right) t{D}_{N}\left( t\right) {dt} \]\n\nWe recall that\n\n\[ t{D}_{N}\left( t\right) = \frac{t}{\sin \left( {t/2}\right) }\sin \left( {\left( {N + 1/2}\right) t}\right) ,\]\n\nwhere the quotient \( \frac{t}{\sin \left( {t/2}\right) } \) is continuous in the interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . Since we can write\n\n\[ \sin \left( {\left( {N + 1/2}\right) t}\right) = \sin \left( {Nt}\right) \cos \left( {t/2}\right) + \cos \left( {Nt}\right) \sin \left( {t/2}\right) ,\]\n\nwe can apply the Riemann-Lebesgue lemma to the Riemann integrable functions \( F\left( t\right) t\cos \left( {t/2}\right) /\sin \left( {t/2}\right) \) and \( F\left( t\right) t \) to finish the proof of the theorem.
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Yes
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Theorem 2.2 Suppose \( f \) and \( g \) are two integrable functions defined on the circle, and for some \( {\theta }_{0} \) there exists an open interval \( I \) containing \( {\theta }_{0} \) such that\n\n\[ f\left( \theta \right) = g\left( \theta \right) \;\text{ for all }\theta \in I. \]\n\nThen \( {S}_{N}\left( f\right) \left( {\theta }_{0}\right) - {S}_{N}\left( g\right) \left( {\theta }_{0}\right) \rightarrow 0 \) as \( N \) tends to infinity.
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Proof. The function \( f - g \) is 0 in \( I \), so it is differentiable at \( {\theta }_{0} \), and we may apply the previous theorem to conclude the proof.
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Yes
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Lemma 2.3 Suppose that the Abel means \( {A}_{r} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{r}^{n}{c}_{n} \) of the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n} \) are bounded as \( r \) tends to 1 (with \( r < 1 \) ). If \( {c}_{n} = O\left( {1/n}\right) \), then the partial sums \( {S}_{N} = \mathop{\sum }\limits_{{n = 1}}^{N}{c}_{n} \) are bounded.
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Proof. Let \( r = 1 - 1/N \) and choose \( M \) so that \( n\left| {c}_{n}\right| \leq M \) . We estimate the difference\n\n\[ \n{S}_{N} - {A}_{r} = \mathop{\sum }\limits_{{n = 1}}^{N}\left( {{c}_{n} - {r}^{n}{c}_{n}}\right) - \mathop{\sum }\limits_{{n = N + 1}}^{\infty }{r}^{n}{c}_{n} \n\]\n\nas follows:\n\n\[ \n\left| {{S}_{N} - {A}_{r}}\right| \leq \mathop{\sum }\limits_{{n = 1}}^{N}\left| {c}_{n}\right| \left( {1 - {r}^{n}}\right) + \mathop{\sum }\limits_{{n = N + 1}}^{\infty }{r}^{n}\left| {c}_{n}\right| \n\]\n\n\[ \n\leq M\mathop{\sum }\limits_{{n = 1}}^{N}\left( {1 - r}\right) + \frac{M}{N}\mathop{\sum }\limits_{{n = N + 1}}^{\infty }{r}^{n} \n\]\n\n\[ \n\leq {MN}\left( {1 - r}\right) + \frac{M}{N}\frac{1}{1 - r} \n\]\n\n\[ \n= {2M} \n\]\n\nwhere we have used the simple observation that\n\n\[ \n1 - {r}^{n} = \left( {1 - r}\right) \left( {1 + r + \cdots + {r}^{n - 1}}\right) \leq n\left( {1 - r}\right) . \n\]\n\nSo we see that if \( M \) satisfies both \( \left| {A}_{r}\right| \leq M \) and \( n\left| {c}_{n}\right| \leq M \), then \( \left| {S}_{N}\right| \leq \) \( {3M} \) .
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Yes
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Lemma 2.4\n\n\[ \n{S}_{M}\left( {P}_{N}\right) = \left\{ \begin{array}{ll} {P}_{N} & \text{ if }M \geq {3N} \\ {\widetilde{P}}_{N} & \text{ if }M = {2N} \\ 0 & \text{ if }M < N \end{array}\right. \n\]
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This is clear from what has been said above and from Figure 3.
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No
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Lemma 2.2 If \( f \) is continuous and periodic of period 1, and \( \gamma \) is irrational, then\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {n\gamma }\right) \rightarrow {\int }_{0}^{1}f\left( x\right) {dx}\;\text{ as }N \rightarrow \infty .\n\]
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The proof of the lemma is divided into three steps.\n\nStep 1. We first check the validity of the limit in the case when \( f \) is one of the exponentials \( 1,{e}^{2\pi ix},\ldots ,{e}^{2\pi ikx},\ldots \) . If \( f = 1 \), the limit\nsurely holds. If \( f = {e}^{2\pi ikx} \) with \( k \neq 0 \), then the integral is 0 . Since \( \gamma \) is irrational, we have \( {e}^{2\pi ik\gamma } \neq 1 \), therefore\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {n\gamma }\right) = \frac{{e}^{2\pi ik\gamma }}{N}\frac{1 - {e}^{2\pi ikN\gamma }}{1 - {e}^{2\pi ik\gamma }}\n\]\n\nwhich goes to 0 as \( N \rightarrow \infty \) .\n\nStep 2. It is clear that if \( f \) and \( g \) satisfy the lemma, then so does \( {Af} + {Bg} \) for any \( A, B \in \mathbb{C} \) . Therefore, the first step implies that the lemma is true for all trigonometric polynomials.\n\nStep 3. Let \( \epsilon > 0 \) . If \( f \) is any continuous periodic function of period 1, choose a trigonometric polynomial \( P \) so that \( \mathop{\sup }\limits_{{x \in \mathbb{R}}}\left| {f\left( x\right) - P\left( x\right) }\right| < \epsilon /3 \) (this is possible by Corollary 5.4 in Chapter 2). Then, by step 1, for all large \( N \) we have\n\n\[ \n\left| {\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}P\left( {n\gamma }\right) - {\int }_{0}^{1}P\left( x\right) {dx}}\right| < \epsilon /3\n\]\n\nTherefore\n\n\[ \n\left| {\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}f\left( {n\gamma }\right) - {\int }_{0}^{1}f\left( x\right) {dx}}\right| \leq \frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}\left| {f\left( {n\gamma }\right) - P\left( {n\gamma }\right) }\right| +\n\]\n\n\[ \n+ \left| {\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}P\left( {n\gamma }\right) - {\int }_{0}^{1}P\left( x\right) {dx}}\right| +\n\]\n\n\[ \n+ {\int }_{0}^{1}\left| {P\left( x\right) - f\left( x\right) }\right| {dx}\n\]\n\n\[ \n< \epsilon\n\]\n\nand the lemma is proved.
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Yes
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Corollary 2.3 The conclusion of Lemma 2.2 holds for every function \( f \) which is Riemann integrable in \( \left\lbrack {0,1}\right\rbrack \), and periodic of period 1 .
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Proof. Assume \( f \) is real-valued, and consider a partition of the interval \( \left\lbrack {0,1}\right\rbrack \), say \( 0 = {x}_{0} < {x}_{1} < \cdots < {x}_{N} = 1 \) . Next, define \( {f}_{U}\left( x\right) = \) \( \mathop{\sup }\limits_{{{x}_{j - 1} \leq y \leq {x}_{j}}}f\left( y\right) \) if \( x \in \left\lbrack {{x}_{j - 1},{x}_{j}}\right) \) and \( {f}_{L}\left( x\right) = \mathop{\inf }\limits_{{{x}_{j - 1} \leq y \leq {x}_{j}}}f\left( y\right) \) for \( x \in \) \( \left( {{x}_{j - 1},{x}_{j}}\right) \) . Then clearly \( {f}_{L} \leq f \leq {f}_{U} \) and\n\n\[ \n{\int }_{0}^{1}{f}_{L}\left( x\right) {dx} \leq {\int }_{0}^{1}f\left( x\right) {dx} \leq {\int }_{0}^{1}{f}_{U}\left( x\right) {dx}.\n\]\n\nMoreover, by making the partition sufficiently fine we can guarantee that for a given \( \epsilon > 0 \) ,\n\n\[ \n{\int }_{0}^{1}{f}_{U}\left( x\right) {dx} - {\int }_{0}^{1}{f}_{L}\left( x\right) {dx} \leq \epsilon \n\]\n\nHowever,\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}{f}_{L}\left( {n\gamma }\right) \rightarrow {\int }_{0}^{1}{f}_{L}\left( x\right) {dx} \n\]\n\nby the theorem, because each \( {f}_{L} \) is a finite linear combination of characteristic functions of intervals; similarly we have\n\n\[ \n\frac{1}{N}\mathop{\sum }\limits_{{n = 1}}^{N}{f}_{U}\left( {n\gamma }\right) \rightarrow {\int }_{0}^{1}{f}_{U}\left( x\right) {dx}.\n\]\n\nFrom these two assertions we can conclude the proof of the corollary by using the previous approximation argument.
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Yes
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Lemma 3.3 If \( {2N} = {2}^{n} \), then\n\n\[{\bigtriangleup }_{2N}\left( f\right) - {\bigtriangleup }_{N}\left( f\right) = {2}^{-{n\alpha }}{e}^{i{2}^{n}x}.\]
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This follows from our previous observation (6) because \( {\bigtriangleup }_{2N}\left( f\right) = \) \( {S}_{2N}\left( f\right) \) and \( {\bigtriangleup }_{N}\left( f\right) = {S}_{N}\left( f\right) \) .
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No
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Proposition 1.1 The integral of a function of moderate decrease defined by (5) satisfies the following properties:\n\n(i) Linearity: if \( f, g \in \mathcal{M}\left( \mathbb{R}\right) \) and \( a, b \in \mathbb{C} \), then\n\n\[ \n{\int }_{-\infty }^{\infty }\left( {{af}\left( x\right) + {bg}\left( x\right) }\right) {dx} = a{\int }_{-\infty }^{\infty }f\left( x\right) {dx} + b{\int }_{-\infty }^{\infty }g\left( x\right) {dx}. \n\]
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We say a few words about the proof. Property (i) is immediate.
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No
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Proposition 1.2 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) then:\n\n(i) \( f\left( {x + h}\right) \rightarrow \widehat{f}\left( \xi \right) {e}^{2\pi ih\xi } \) whenever \( h \in \mathbb{R} \) .\n\n(ii) \( f\left( x\right) {e}^{-{2\pi ixh}} \rightarrow \widehat{f}\left( {\xi + h}\right) \) whenever \( h \in \mathbb{R} \) .\n\n(iii) \( f\left( {\delta x}\right) \rightarrow {\delta }^{-1}\widehat{f}\left( {{\delta }^{-1}\xi }\right) \) whenever \( \delta > 0 \) .\n\n(iv) \( {f}^{\prime }\left( x\right) \rightarrow {2\pi i\xi }\widehat{f}\left( \xi \right) \) .\n\n(v) \( - {2\pi ixf}\left( x\right) \rightarrow \frac{d}{d\xi }\widehat{f}\left( \xi \right) \) .
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Proof. Property (i) is an immediate consequence of the translation invariance of the integral, and property (ii) follows from the definition. Also, the third property of Proposition 1.1 establishes (iii).\n\nIntegrating by parts gives\n\n\[{\int }_{-N}^{N}{f}^{\prime }\left( x\right) {e}^{-{2\pi ix\xi }}{dx} = {\left\lbrack f\left( x\right) {e}^{-{2\pi ix\xi }}\right\rbrack }_{-N}^{N} + {2\pi i\xi }{\int }_{-N}^{N}f\left( x\right) {e}^{-{2\pi ix\xi }}{dx},\]\n\nso letting \( N \) tend to infinity gives (iv).\n\nFinally, to prove property (v), we must show that \( \widehat{f} \) is differentiable and find its derivative. Let \( \epsilon > 0 \) and consider\n\n\[ \frac{\widehat{f}\left( {\xi + h}\right) - \widehat{f}\left( \xi \right) }{h} - \left( {-\widehat{2\pi ix}f}\right) \left( \xi \right) = \]\n\n\[ {\int }_{-\infty }^{\infty }f\left( x\right) {e}^{-{2\pi ix\xi }}\left\lbrack {\frac{{e}^{-{2\pi ixh}} - 1}{h} + {2\pi ix}}\right\rbrack {dx}. \]\n\nSince \( f\left( x\right) \) and \( {xf}\left( x\right) \) are of rapid decrease, there exists an integer \( N \) so that \( {\int }_{\left| x\right| \geq N}\left| {f\left( x\right) }\right| {dx} \leq \epsilon \) and \( {\int }_{\left| x\right| \geq N}\left| x\right| \left| {f\left( x\right) }\right| {dx} \leq \epsilon \) . Moreover, for \( \left| x\right| \leq N \), there exists \( {h}_{0} \) so that \( \left| h\right| < {h}_{0} \) implies\n\n\[ \left| {\frac{{e}^{-{2\pi ixh}} - 1}{h} + {2\pi ix}}\right| \leq \frac{\epsilon }{N} \]\n\nHence for \( \left| h\right| < {h}_{0} \) we have\n\n\[ \left| {\frac{\widehat{f}\left( {\xi + h}\right) - \widehat{f}\left( \xi \right) }{h} - \left( {\widehat{-{2\pi ix}}f}\right) \left( \xi \right) }\right| \]\n\n\[ \leq {\int }_{-N}^{N}\left| {f\left( x\right) {e}^{-{2\pi ix\xi }}\left\lbrack {\frac{{e}^{-{2\pi ixh}} - 1}{h} + {2\pi ix}}\right\rbrack }\right| {dx} + {C\epsilon } \]\n\n\[ \leq {C}^{\prime }\epsilon \]
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Yes
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Theorem 1.3 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then \( \widehat{f} \in \mathcal{S}\left( \mathbb{R}\right) \) .
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The proof is an easy application of the fact that the Fourier transform interchanges differentiation and multiplication. In fact, note that if \( f \in \) \( \mathcal{S}\left( \mathbb{R}\right) \), its Fourier transform \( \widehat{f} \) is bounded; then also, for each pair of non-negative integers \( \ell \) and \( k \), the expression\n\n\[ \n{\xi }^{k}{\left( \frac{d}{d\xi }\right) }^{\ell }\widehat{f}\left( \xi \right)\n\]\n\nis bounded, since by the last proposition, it is the Fourier transform of\n\n\[ \n\frac{1}{{\left( 2\pi i\right) }^{k}}{\left( \frac{d}{dx}\right) }^{k}\left\lbrack {{\left( -2\pi ix\right) }^{\ell }f\left( x\right) }\right\rbrack .\n\]
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Yes
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Theorem 1.4 If \( f\left( x\right) = {e}^{-\pi {x}^{2}} \), then \( \widehat{f}\left( \xi \right) = f\left( \xi \right) \) .
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Proof. Define\n\n\[ F\left( \xi \right) = \widehat{f}\left( \xi \right) = {\int }_{-\infty }^{\infty }{e}^{-\pi {x}^{2}}{e}^{-{2\pi ix\xi }}{dx} \]\n\nand observe that \( F\left( 0\right) = 1 \), by our previous calculation. By property (v) in Proposition 1.2, and the fact that \( {f}^{\prime }\left( x\right) = - {2\pi xf}\left( x\right) \), we obtain\n\n\[ {F}^{\prime }\left( \xi \right) = {\int }_{-\infty }^{\infty }f\left( x\right) \left( {-{2\pi ix}}\right) {e}^{-{2\pi ix\xi }}{dx} = i{\int }_{-\infty }^{\infty }{f}^{\prime }\left( x\right) {e}^{-{2\pi ix\xi }}{dx}. \]\n\nBy (iv) of the same proposition, we find that\n\n\[ {F}^{\prime }\left( \xi \right) = i\left( {2\pi i\xi }\right) \widehat{f}\left( \xi \right) = - {2\pi \xi F}\left( \xi \right) . \]\n\nIf we define \( G\left( \xi \right) = F\left( \xi \right) {e}^{\pi {\xi }^{2}} \), then from what we have seen above, it follows that \( {G}^{\prime }\left( \xi \right) = 0 \), hence \( G \) is constant. Since \( F\left( 0\right) = 1 \), we conclude that \( G \) is identically equal to 1, therefore \( F\left( \xi \right) = {e}^{-\pi {\xi }^{2}} \), as was to be shown.
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Yes
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Corollary 1.5 If \( \delta > 0 \) and \( {K}_{\delta }\left( x\right) = {\delta }^{-1/2}{e}^{-\pi {x}^{2}/\delta } \), then \( \widehat{{K}_{\delta }}\left( \xi \right) = {e}^{-{\pi \delta }{\xi }^{2}} \) .
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Null
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No
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Theorem 1.6 The collection \( {\left\{ {K}_{\delta }\right\} }_{\delta > 0} \) is a family of good kernels as \( \delta \rightarrow 0 \) .
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Proof. First, we claim that \( f \) is uniformly continuous on \( \mathbb{R} \) . Indeed, given \( \epsilon > 0 \) there exists \( R > 0 \) so that \( \left| {f\left( x\right) }\right| < \epsilon /4 \) whenever \( \left| x\right| \geq R \) . Moreover, \( f \) is continuous, hence uniformly continuous on the compact interval \( \left\lbrack {-R, R}\right\rbrack \), and together with the previous observation, we can find \( \eta > 0 \) so that \( \left| {f\left( x\right) - f\left( y\right) }\right| < \epsilon \) whenever \( \left| {x - y}\right| < \eta \) . Now we argue as usual. Using the first property of good kernels, we can write\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) = {\int }_{-\infty }^{\infty }{K}_{\delta }\left( t\right) \left\lbrack {f\left( {x - t}\right) - f\left( x\right) }\right\rbrack {dt}, \]\n\nand since \( {K}_{\delta } \geq 0 \), we find\n\n\[ \left| {\left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) }\right| \leq {\int }_{\left| t\right| > \eta } + {\int }_{\left| t\right| \leq \eta }{K}_{\delta }\left( t\right) \left| {f\left( {x - t}\right) - f\left( x\right) }\right| {dt}. \]\n\nThe first integral is small by the third property of good kernels, and the fact that \( f \) is bounded, while the second integral is also small since \( f \) is uniformly continuous and \( \int {K}_{\delta } = 1 \) . This concludes the proof of the corollary.
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Yes
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Corollary 1.7 If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) \rightarrow f\left( x\right) \;\text{ uniformly in }x\text{ as }\delta \rightarrow 0. \]
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Proof. First, we claim that \( f \) is uniformly continuous on \( \mathbb{R} \) . Indeed, given \( \epsilon > 0 \) there exists \( R > 0 \) so that \( \left| {f\left( x\right) }\right| < \epsilon /4 \) whenever \( \left| x\right| \geq R \) . Moreover, \( f \) is continuous, hence uniformly continuous on the compact interval \( \left\lbrack {-R, R}\right\rbrack \), and together with the previous observation, we can find \( \eta > 0 \) so that \( \left| {f\left( x\right) - f\left( y\right) }\right| < \epsilon \) whenever \( \left| {x - y}\right| < \eta \) . Now we argue as usual. Using the first property of good kernels, we can write\n\n\[ \left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) = {\int }_{-\infty }^{\infty }{K}_{\delta }\left( t\right) \left\lbrack {f\left( {x - t}\right) - f\left( x\right) }\right\rbrack {dt}, \]\n\nand since \( {K}_{\delta } \geq 0 \), we find\n\n\[ \left| {\left( {f * {K}_{\delta }}\right) \left( x\right) - f\left( x\right) }\right| \leq {\int }_{\left| t\right| > \eta } + {\int }_{\left| t\right| \leq \eta }{K}_{\delta }\left( t\right) \left| {f\left( {x - t}\right) - f\left( x\right) }\right| {dt}. \]\n\nThe first integral is small by the third property of good kernels, and the fact that \( f \) is bounded, while the second integral is also small since \( f \) is uniformly continuous and \( \int {K}_{\delta } = 1 \) . This concludes the proof of the corollary.
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Yes
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Proposition 1.8 If \( f, g \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[{\int }_{-\infty }^{\infty }f\left( x\right) \widehat{g}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }\widehat{f}\left( y\right) g\left( y\right) {dy}.\]
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To prove the proposition, we need to digress briefly to discuss the interchange of the order of integration for double integrals. Suppose \( F\left( {x, y}\right) \) is a continuous function in the plane \( \left( {x, y}\right) \in {\mathbb{R}}^{2} \) . We will assume the following decay condition on \( F \) :\n\n\[ \left| {F\left( {x, y}\right) }\right| \leq A/\left( {1 + {x}^{2}}\right) \left( {1 + {y}^{2}}\right) . \]\n\nThen, we can state that for each \( x \) the function \( F\left( {x, y}\right) \) is of moderate decrease in \( y \), and similarly for each fixed \( y \) the function \( F\left( {x, y}\right) \) is of moderate decrease in \( x \) . Moreover, the function \( {F}_{1}\left( x\right) = {\int }_{-\infty }^{\infty }F\left( {x, y}\right) {dy} \) is continuous and of moderate decrease; similarly for the function \( {F}_{2}\left( y\right) = \) \( {\int }_{-\infty }^{\infty }F\left( {x, y}\right) {dx} \) . Finally\n\n\[ {\int }_{-\infty }^{\infty }{F}_{1}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }{F}_{2}\left( y\right) {dy}. \]\n\nThe proof of these facts may be found in the appendix.\n\nWe now apply this to \( F\left( {x, y}\right) = f\left( x\right) g\left( y\right) {e}^{-{2\pi ixy}} \) . Then \( {F}_{1}\left( x\right) = \) \( f\left( x\right) \widehat{g}\left( x\right) \), and \( {F}_{2}\left( y\right) = \widehat{f}\left( y\right) g\left( y\right) \) so\n\n\[ {\int }_{-\infty }^{\infty }f\left( x\right) \widehat{g}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }\widehat{f}\left( y\right) g\left( y\right) {dy} \]\n\nwhich is the assertion of the proposition.
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Yes
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Theorem 1.9 (Fourier inversion) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ f\left( x\right) = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {e}^{2\pi ix\xi }{d\xi } \]
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Proof. We first claim that\n\n\[ f\left( 0\right) = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {d\xi } \]\n\nLet \( {G}_{\delta }\left( x\right) = {e}^{-{\pi \delta }{x}^{2}} \) so that \( \widehat{{G}_{\delta }}\left( \xi \right) = {K}_{\delta }\left( \xi \right) \). By the multiplication formula we get\n\n\[ {\int }_{-\infty }^{\infty }f\left( x\right) {K}_{\delta }\left( x\right) {dx} = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {G}_{\delta }\left( \xi \right) {d\xi }. \]\n\nSince \( {K}_{\delta } \) is a good kernel, the first integral goes to \( f\left( 0\right) \) as \( \delta \) tends to 0. Since the second integral clearly converges to \( {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {d\xi } \) as \( \delta \) tends to 0, our claim is proved. In general, let \( F\left( y\right) = f\left( {y + x}\right) \) so that\n\n\[ f\left( x\right) = F\left( 0\right) = {\int }_{-\infty }^{\infty }\widehat{F}\left( \xi \right) {d\xi } = {\int }_{-\infty }^{\infty }\widehat{f}\left( \xi \right) {e}^{2\pi ix\xi }{d\xi }. \]
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Yes
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Corollary 1.10 The Fourier transform is a bijective mapping on the Schwartz space.
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Null
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No
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Proposition 1.11 If \( f, g \in \mathcal{S}\left( \mathbb{R}\right) \) then:\n\n(i) \( f * g \in \mathcal{S}\left( \mathbb{R}\right) \) .\n\n(ii) \( f * g = g * f \) .\n\n(iii) \( \widehat{\left( f * g\right) }\left( \xi \right) = \widehat{f}\left( \xi \right) \widehat{g}\left( \xi \right) \) .
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Proof. To prove that \( f * g \) is rapidly decreasing, observe first that for any \( \ell \geq 0 \) we have \( \mathop{\sup }\limits_{x}{\left| x\right| }^{\ell }\left| {g\left( {x - y}\right) }\right| \leq {A}_{\ell }{\left( 1 + \left| y\right| \right) }^{\ell } \), because \( g \) is rapidly decreasing (to check this assertion, consider separately the two cases \( \left. {\left| x\right| \leq 2\left| y\right| \text{and}\left| x\right| \geq 2\left| y\right| }\right) \) . From this, we see that\n\n\[ \mathop{\sup }\limits_{x}\left| {{x}^{\ell }\left( {f * g}\right) \left( x\right) }\right| \leq {A}_{\ell }{\int }_{-\infty }^{\infty }\left| {f\left( y\right) }\right| {\left( 1 + \left| y\right| \right) }^{\ell }{dy} \]\n\nso that \( {x}^{\ell }\left( {f * g}\right) \left( x\right) \) is a bounded function for every \( \ell \geq 0 \) . These estimates carry over to the derivatives of \( f * g \), thereby proving that \( f * g \in \mathcal{S}\left( \mathbb{R}\right) \) because\n\n\[ {\left( \frac{d}{dx}\right) }^{k}\left( {f * g}\right) \left( x\right) = \left( {f * {\left( \frac{d}{dx}\right) }^{k}g}\right) \left( x\right) \;\text{ for }k = 1,2,\ldots \]\n\nThis identity is proved first for \( k = 1 \) by differentiating under the integral defining \( f * g \) . The interchange of differentiation and integration is justified in this case by the rapid decrease of \( {dg}/{dx} \) . The identity then follows for every \( k \) by iteration.\n\nFor fixed \( x \), the change of variables \( x - y = u \) shows that\n\n\[ \left( {f * g}\right) \left( x\right) = {\int }_{-\infty }^{\infty }f\left( {x - u}\right) g\left( u\right) {du} = \left( {g * f}\right) \left( x\right) . \]\n\nThis change of variables is a composition of two changes, \( y \mapsto - y \) and \( y \mapsto y - h \) (with \( h = x \) ). For the first one we use the observation that \( {\int }_{-\infty }^{\infty }F\left( x\right) {dx} = {\int }_{-\infty }^{\infty }F\left( {-x}\right) {dx} \) for any Schwartz function \( F \), and for the second, we apply (ii) of Proposition 1.1\n\nFinally, consider \( F\left( {x, y}\right) = f\left( y\right) g\left( {x - y}\right) {e}^{-{2\pi ix\xi }} \) . Since \( f \) and \( g \) are rapidly decreasing, considering separately the two cases \( \left| x\right| \leq 2\left| y\right| \) and \( \left| x\right| \geq 2\left| y\right| \), we see that the discussion of the change of order of integration after Proposition 1.8 applies to \( F \) . In this case \( {F}_{1}\left( x\right) = \left( {f * g}\right) \left( x\right) {e}^{-{2\pi ix\xi }} \) , and \( {F}_{2}\left( y\right) = f\left( y\right) {e}^{-{2\pi iy\xi }}\widehat{g}\left( \xi \right) \) . Thus \( {\int }_{-\infty }^{\infty }{F}_{1}\left( x\right) {dx} = {\int }_{-\infty }^{\infty }{F}_{2}\left( y\right) {dy} \), which implies (iii). The proposition is therefore proved.
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Yes
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Theorem 1.12 (Plancherel) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) then \( \parallel \widehat{f}\parallel = \parallel f\parallel \) .
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Proof. If \( f \in \mathcal{S}\left( \mathbb{R}\right) \) define \( {f}^{b}\left( x\right) = \overline{f\left( {-x}\right) } \) . Then \( \widehat{{f}^{b}}\left( \xi \right) = \overline{\widehat{f}\left( \xi \right) } \) . Now let \( h = f * {f}^{b} \) . Clearly, we have\n\n\[ \n\widehat{h}\left( \xi \right) = {\left| \widehat{f}\left( \xi \right) \right| }^{2}\;\text{ and }\;h\left( 0\right) = {\int }_{-\infty }^{\infty }{\left| f\left( x\right) \right| }^{2}{dx}.\n\]\n\nThe theorem now follows from the inversion formula applied with \( x = 0 \) , that is,\n\n\[ \n{\int }_{-\infty }^{\infty }\widehat{h}\left( \xi \right) {d\xi } = h\left( 0\right)\n\]
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Yes
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Corollary 2.2 \( u\left( {\cdot, t}\right) \) belongs to \( \mathcal{S}\left( \mathbb{R}\right) \) uniformly in \( t \), in the sense that for any \( T > 0 \)\n\n(9)\n\n\[ \mathop{\sup }\limits_{\substack{{x \in \mathbb{R}} \\ {0 < t < T} }}{\left| x\right| }^{k}\left| {\frac{{\partial }^{\ell }}{\partial {x}^{\ell }}u\left( {x, t}\right) }\right| < \infty \;\text{ for each }k,\ell \geq 0. \]
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Proof. This result is a consequence of the following estimate:\n\n\[ \left| {u\left( {x, t}\right) }\right| \leq {\int }_{\left| y\right| \leq \left| x\right| /2}\left| {f\left( {x - y}\right) }\right| {\mathcal{H}}_{t}\left( y\right) {dy} + {\int }_{\left| y\right| \geq \left| x\right| /2}\left| {f\left( {x - y}\right) }\right| {\mathcal{H}}_{t}\left( y\right) {dy} \]\n\n\[ \leq \frac{{C}_{N}}{{\left( 1 + \left| x\right| \right) }^{N}} + \frac{C}{\sqrt{t}}{e}^{-c{x}^{2}/t}. \]\n\nIndeed, since \( f \) is rapidly decreasing, we have \( \left| {f\left( {x - y}\right) }\right| \leq {C}_{N}/{\left( 1 + \left| x\right| \right) }^{N} \) when \( \left| y\right| \leq \left| x\right| /2 \) . Also, if \( \left| y\right| \geq \left| x\right| /2 \) then \( {\mathcal{H}}_{t}\left( y\right) \leq C{t}^{-1/2}{e}^{-c{x}^{2}/t} \), and we obtain the above inequality. Consequently, we see that \( u\left( {x, t}\right) \) is rapidly decreasing uniformly for \( 0 < t < T \) .\n\nThe same argument can be applied to the derivatives of \( u \) in the \( x \) variable since we may differentiate under the integral sign and apply the above estimate with \( f \) replaced by \( {f}^{\prime } \), and so on.
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Yes
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Theorem 2.3 Suppose \( u\left( {x, t}\right) \) satisfies the following conditions:\n\n(i) \( u \) is continuous on the closure of the upper half-plane.\n\n(ii) \( u \) satisfies the heat equation for \( t > 0 \) .\n\n(iii) \( u \) satisfies the boundary condition \( u\left( {x,0}\right) = 0 \) .\n\n(iv) \( u\left( {\cdot, t}\right) \in \mathcal{S}\left( \mathbb{R}\right) \) uniformly in \( t \), as in (9).\n\nThen, we conclude that \( u = 0 \) .
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Proof. We define the energy at time \( t \) of the solution \( u\left( {x, t}\right) \) by\n\n\[ E\left( t\right) = {\int }_{\mathbb{R}}{\left| u\left( x, t\right) \right| }^{2}{dx} \]\n\nClearly \( E\left( t\right) \geq 0 \) . Since \( E\left( 0\right) = 0 \) it suffices to show that \( E \) is a decreasing function, and this is achieved by proving that \( {dE}/{dt} \leq 0 \) . The assumptions on \( u \) allow us to differentiate \( E\left( t\right) \) under the integral sign\n\n\[ \frac{dE}{dt} = {\int }_{\mathbb{R}}\left\lbrack {{\partial }_{t}u\left( {x, t}\right) \bar{u}\left( {x, t}\right) + u\left( {x, t}\right) {\partial }_{t}\bar{u}\left( {x, t}\right) }\right\rbrack {dx}. \]\n\nBut \( u \) satisfies the heat equation, therefore \( {\partial }_{t}u = {\partial }_{x}^{2}u \) and \( {\partial }_{t}\bar{u} = {\partial }_{x}^{2}\bar{u} \), so that after an integration by parts, where we use the fact that \( u \) and its \( x \) derivatives decrease rapidly as \( \left| x\right| \rightarrow \infty \), we find\n\n\[ \frac{dE}{dt} = {\int }_{\mathbb{R}}\left\lbrack {{\partial }_{x}^{2}u\left( {x, t}\right) \bar{u}\left( {x, t}\right) + u\left( {x, t}\right) {\partial }_{x}^{2}\bar{u}\left( {x, t}\right) }\right\rbrack {dx} \]\n\n\[ = - {\int }_{\mathbb{R}}\left\lbrack {{\partial }_{x}u\left( {x, t}\right) {\partial }_{x}\bar{u}\left( {x, t}\right) + {\partial }_{x}u\left( {x, t}\right) {\partial }_{x}\bar{u}\left( {x, t}\right) }\right\rbrack {dx} \]\n\n\[ = - 2{\int }_{\mathbb{R}}{\left| {\partial }_{x}u\left( x, t\right) \right| }^{2}{dx} \]\n\n\[ \leq 0 \]\n\nas claimed. Thus \( E\left( t\right) = 0 \) for all \( t \), hence \( u = 0 \) .
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Yes
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Lemma 2.4 The following two identities hold:\n\n\[ \n{\int }_{-\infty }^{\infty }{e}^{-{2\pi }\left| \xi \right| y}{e}^{2\pi i\xi x}{d\xi } = {\mathcal{P}}_{y}\left( x\right) \n\]\n\n\[ \n{\int }_{-\infty }^{\infty }{\mathcal{P}}_{y}\left( x\right) {e}^{-{2\pi ix\xi }}{dx} = {e}^{-{2\pi }\left| \xi \right| y}. \n\]
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Proof. The first formula is fairly straightforward since we can split the integral from \( - \infty \) to 0 and 0 to \( \infty \) . Then, since \( y > 0 \) we have\n\n\[ \n{\int }_{0}^{\infty }{e}^{-{2\pi \xi y}}{e}^{2\pi i\xi x}{d\xi } = {\int }_{0}^{\infty }{e}^{{2\pi i}\left( {x + {iy}}\right) \xi }{d\xi } = {\left\lbrack \frac{{e}^{{2\pi i}\left( {x + {iy}}\right) \xi }}{{2\pi i}\left( {x + {iy}}\right) }\right\rbrack }_{0}^{\infty } = \n\]\n\n\[ \n- \frac{1}{{2\pi i}\left( {x + {iy}}\right) } \n\]\n\nand similarly,\n\n\[ \n{\int }_{-\infty }^{0}{e}^{2\pi \xi y}{e}^{2\pi i\xi x}{d\xi } = \frac{1}{{2\pi i}\left( {x - {iy}}\right) }. \n\]\n\nTherefore\n\n\[ \n{\int }_{-\infty }^{\infty }{e}^{-{2\pi }\left| \xi \right| y}{e}^{2\pi i\xi x}{d\xi } = \frac{1}{{2\pi i}\left( {x - {iy}}\right) } - \frac{1}{{2\pi i}\left( {x + {iy}}\right) } = \frac{y}{\pi \left( {{x}^{2} + {y}^{2}}\right) }. \n\]\n\nThe second formula is now a consequence of the Fourier inversion theorem applied in the case when \( f \) and \( \widehat{f} \) are of moderate decrease.
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Yes
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Lemma 2.5 The Poisson kernel is a good kernel on \( \mathbb{R} \) as \( y \rightarrow 0 \) .
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Proof. Setting \( \xi = 0 \) in the second formula of the lemma shows that \( {\int }_{-\infty }^{\infty }{\mathcal{P}}_{y}\left( x\right) {dx} = 1 \), and clearly \( {\mathcal{P}}_{y}\left( x\right) \geq 0 \), so it remains to check the last property of good kernels. Given a fixed \( \delta > 0 \), we may change variables \( u = x/y \) so that\n\n\[{\int }_{\delta }^{\infty }\frac{y}{{x}^{2} + {y}^{2}}{dx} = {\int }_{\delta /y}^{\infty }\frac{du}{1 + {u}^{2}} = {\left\lbrack \arctan u\right\rbrack }_{\delta /y}^{\infty } = \pi /2 - \arctan \left( {\delta /y}\right) ,\]\n\nand this quantity goes to 0 as \( y \rightarrow 0 \) . Since \( {\mathcal{P}}_{y}\left( x\right) \) is an even function, the proof is complete.
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Yes
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Theorem 2.6 Given \( f \in \mathcal{S}\left( \mathbb{R}\right) \), let \( u\left( {x, y}\right) = \left( {f * {\mathcal{P}}_{y}}\right) \left( x\right) \) . Then:\n\n(i) \( u\left( {x, y}\right) \) is \( {C}^{2} \) in \( {\mathbb{R}}_{ + }^{2} \) and \( \bigtriangleup u = 0 \) .\n\n(ii) \( u\left( {x, y}\right) \rightarrow f\left( x\right) \) uniformly as \( y \rightarrow 0 \) .\n\n(iii) \( {\int }_{-\infty }^{\infty }{\left| u\left( x, y\right) - f\left( x\right) \right| }^{2}{dx} \rightarrow 0 \) as \( y \rightarrow 0 \) .\n\n(iv) If \( u\left( {x,0}\right) = f\left( x\right) \), then \( u \) is continuous on the closure \( \overline{{\mathbb{R}}_{ + }^{2}} \) of the upper half-plane, and vanishes at infinity in the sense that\n\n\[ u\left( {x, y}\right) \rightarrow 0\;\text{ as }\left| x\right| + y \rightarrow \infty . \]
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Proof. The proofs of parts (i), (ii), and (iii) are similar to the case of the heat equation, and so are left to the reader. Part (iv) is a consequence of two easy estimates whenever \( f \) is of moderate decrease. First, we have\n\n\[ \left| {\left( {f * {\mathcal{P}}_{y}}\right) \left( x\right) }\right| \leq C\left( {\frac{1}{\left( 1 + {x}^{2}\right) } + \frac{y}{{x}^{2} + {y}^{2}}}\right) \]\n\nwhich is proved (as in the case of the heat equation) by splitting the integral \( {\int }_{-\infty }^{\infty }f\left( {x - t}\right) {\mathcal{P}}_{y}\left( t\right) {dt} \) into the part where \( \left| t\right| \leq \left| x\right| /2 \) and the part where \( \left| t\right| \geq \left| x\right| /2 \) . Also, we have \( \left| {\left( {f * {\mathcal{P}}_{y}}\right) \left( x\right) }\right| \leq C/y \), since \( \mathop{\sup }\limits_{x}{\mathcal{P}}_{y}\left( x\right) \leq \) \( c/y \) .\n\nUsing the first estimate when \( \left| x\right| \geq \left| y\right| \) and the second when \( \left| x\right| \leq \left| y\right| \) gives the desired decrease at infinity.
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No
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Theorem 2.7 Suppose \( u \) is continuous on the closure of the upper half-plane \( \overline{{\mathbb{R}}_{ + }^{2}} \), satisfies \( \bigtriangleup u = 0 \) for \( \left( {x, y}\right) \in {\mathbb{R}}_{ + }^{2}, u\left( {x,0}\right) = 0 \), and \( u\left( {x, y}\right) \) vanishes at infinity. Then \( u = 0 \) .
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To prove Theorem 2.7 we argue by contradiction. Considering separately the real and imaginary parts of \( u \), we may suppose that \( u \) itself is real-valued, and is somewhere strictly positive, say \( u\left( {{x}_{0},{y}_{0}}\right) > 0 \) for some \( {x}_{0} \in \mathbb{R} \) and \( {y}_{0} > 0 \) . We shall see that this leads to a contradiction. First, since \( u \) vanishes at infinity, we can find a large semi-disc of radius \( R,{D}_{R}^{ + } = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} \leq R,\;y \geq 0}\right\} \) outside of which \( u\left( {x, y}\right) \leq \) \( \frac{1}{2}u\left( {{x}_{0},{y}_{0}}\right) \) . Next, since \( u \) is continuous in \( {D}_{R}^{ + } \), it attains its maximum \( M \) there, so there exists a point \( \left( {{x}_{1},{y}_{1}}\right) \in {D}_{R}^{ + } \) with \( u\left( {{x}_{1},{y}_{1}}\right) = M \), while \( u\left( {x, y}\right) \leq M \) in the semi-disc; also, since \( u\left( {x, y}\right) \leq \frac{1}{2}u\left( {{x}_{0},{y}_{0}}\right) \leq M/2 \) outside of the semi-disc, we have \( u\left( {x, y}\right) \leq M \) throughout the entire upper half-plane. Now the mean-value property for harmonic functions implies\n\n\[ u\left( {{x}_{1},{y}_{1}}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {{x}_{1} + \rho \cos \theta ,{y}_{1}
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Yes
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Lemma 2.8 (Mean-value property) Suppose \( \Omega \) is an open set in \( {\mathbb{R}}^{2} \) and let \( u \) be a function of class \( {C}^{2} \) with \( \bigtriangleup u = 0 \) in \( \Omega \) . If the closure of the disc centered at \( \left( {x, y}\right) \) and of radius \( R \) is contained in \( \Omega \), then\n\n\[ u\left( {x, y}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }u\left( {x + r\cos \theta, y + r\sin \theta }\right) {d\theta } \]\n\nfor all \( 0 \leq r \leq R \) .
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Proof. Let \( U\left( {r,\theta }\right) = u\left( {x + r\cos \theta, y + r\sin \theta }\right) \) . Expressing the Laplacian in polar coordinates, the equation \( \bigtriangleup u = 0 \) then implies\n\n\[ 0 = \frac{{\partial }^{2}U}{\partial {\theta }^{2}} + r\frac{\partial }{\partial r}\left( {r\frac{\partial U}{\partial r}}\right) . \]\n\nIf we define \( F\left( r\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi }U\left( {r,\theta }\right) {d\theta } \), the above gives\n\n\[ r\frac{\partial }{\partial r}\left( {r\frac{\partial F}{\partial r}}\right) = \frac{1}{2\pi }{\int }_{0}^{2\pi } - \frac{{\partial }^{2}U}{\partial {\theta }^{2}}\left( {r,\theta }\right) {d\theta }. \]\n\nThe integral of \( {\partial }^{2}U/\partial {\theta }^{2} \) over the circle vanishes since \( \partial U/\partial \theta \) is periodic, hence \( r\frac{\partial }{\partial r}\left( {r\frac{\partial F}{\partial r}}\right) = 0 \), and consequently \( r\partial F/\partial r \) must be constant. Evaluating this expression at \( r = 0 \) we find that \( \partial F/\partial r = 0 \) . Thus \( F \) is constant, but since \( F\left( 0\right) = u\left( {x, y}\right) \), we finally find that \( F\left( r\right) = u\left( {x, y}\right) \) for all \( 0 \leq r \leq R \), which is the mean-value property.\n\nFinally, note that the argument above is implicit in the proof of Theorem 5.7, Chapter 2.
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Yes
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Theorem 3.1 (Poisson summation formula) If \( f \in \mathcal{S}\left( \mathbb{R}\right) \), then\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }f\left( {x + n}\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\widehat{f}\left( n\right) {e}^{2\pi inx}. \]\n\nIn particular, setting \( x = 0 \) we have\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }f\left( n\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\widehat{f}\left( n\right) \]\n\nIn other words, the Fourier coefficients of the periodization of \( f \) are given precisely by the values of the Fourier transform of \( f \) on the integers.
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Proof. To check the first formula it suffices, by Theorem 2.1 in Chapter 2, to show that both sides (which are continuous) have the same Fourier coefficients (viewed as functions on the circle). Clearly, the \( {m}^{\text{th }} \) Fourier coefficient of the right-hand side is \( \widehat{f}\left( m\right) \) . For the left-hand side we have\n\n\[ {\int }_{0}^{1}\left( {\mathop{\sum }\limits_{{n = - \infty }}^{\infty }f\left( {x + n}\right) }\right) {e}^{-{2\pi imx}}{dx} = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\int }_{0}^{1}f\left( {x + n}\right) {e}^{-{2\pi imx}}{dx} \]\n\n\[ = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{\int }_{n}^{n + 1}f\left( y\right) {e}^{-{2\pi imy}}{dy} \]\n\n\[ = {\int }_{-\infty }^{\infty }f\left( y\right) {e}^{-{2\pi imy}}{dy} \]\n\n\[ = \widehat{f}\left( m\right) \]\n\nwhere the interchange of the sum and integral is permissible since \( f \) is rapidly decreasing. This completes the proof of the theorem.
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Yes
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