# Tuesday, October 05 :::{.remark} Recall that we were proving the Thom isomorphism theorem. Some motivation: ::: :::{.corollary title="The Gysin Sequence"} Consider an oriented sphere bundle: \begin{tikzcd} S^{n-1} \ar[r] & \SS E \ar[d] \\ & X \end{tikzcd} Then there is a LES induced by the Euler class $e\in H^n(X)$ \begin{tikzcd} \cdots \\ \\ {H^{j-n}(X)} && {H^j(X)} && {H^{j}(\SS E)} \\ &&&& {} \\ && \cdots && {H^{j-1}(\SS E)} \arrow[from=5-5, to=3-1] \arrow["{e\cupprod(\wait)}", from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=1-1] \arrow[from=5-3, to=5-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbNCwzXSxbMCwyLCJIXntqLW59KFgpIl0sWzQsMiwiSF57an0oXFxTUyBFKSJdLFsyLDIsIkheaihYKSJdLFs0LDQsIkhee2otMX0oXFxTUyBFKSJdLFswLDAsIlxcY2RvdHMiXSxbMiw0LCJcXGNkb3RzIl0sWzQsMV0sWzEsMywiZVxcY3VwcHJvZChcXHdhaXQpIl0sWzMsMl0sWzIsNV0sWzYsNF1d) ::: :::{.proof title="of corollary"} Corresponding to $\SS E \mapsvia{\pi} X$, the mapping cone of $\pi$ is $\DD E$. So consider the LES in relative homology: \begin{tikzcd} \cdots \\ \textcolor{rgb,255:red,214;green,92;blue,92}{u_E} && \textcolor{rgb,255:red,214;green,92;blue,92}{e} \\ {H^j(\DD E, \SS E)} && {H^j(\DD E)} && {H^j(\SS E)} \\ &&&& {} \\ &&&& {H^{j-1}(\SS E)} \\ \\ \textcolor{rgb,255:red,92;green,92;blue,214}{H^{j-n}(\DD E)} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^j(X)} \\ {} \\ \textcolor{rgb,255:red,92;green,92;blue,214}{H^{j-n}(X)} \arrow[curve={height=6pt}, from=5-5, to=3-1] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow["{u_E \cupprod(\wait), \cong}", color={rgb,255:red,92;green,92;blue,214}, from=7-1, to=3-1] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=9-1, to=7-1] \arrow["\cong", color={rgb,255:red,92;green,92;blue,214}, from=7-3, to=3-3] \arrow["{e\cupprod (\wait)}"', color={rgb,255:red,92;green,92;blue,214}, dashed, from=9-1, to=7-3] \arrow[color={rgb,255:red,214;green,92;blue,92}, maps to, from=2-1, to=2-3] \arrow[curve={height=6pt}, from=3-5, to=1-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) After identifying terms, we see that $u_E \in H^n(\DD E,\SS E)$ maps to the Euler class $e\in H^n(X)$, which carries interesting geometric information. ::: :::{.proof title="of theorem, for fields"} Suppose the claim holds for compact $X$, and write $X = \Union_i C_i$ for $C_i$ compact CW skeleta. Then $H_i(X) = \colim_j H_i(C_j)$ since simplices are also compact. Note that \[ H^i(X; k) &\cong H_i(X; k)\dual \\ &\da \Hom( H_i(X; k), k) \\ &= \Hom( \colim_j H_i(C_j;k); k) \\ &= \cocolim_j \Hom( H_i(C_j; k), k) \\ &= \cocolim_j H^i(C_j; k) .\] Similarly, \[ H^i(\DD E, \SS E; k) \iso \cocolim_j H^i\qty{ \ro {\DD E}{ C_j}, \ro{\SS E}{C_j}; k } = \begin{cases} 0 & i [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJIXmkoXFxybyB7XFxERCBFfSB7Q19qfTsgaykiXSxbMywyLCJIXmkoXFxybyB7XFxERCBFfSB7Q19qfSwgXFxyb3tcXFNTIEV9eyBDX2p9IDsgaykiXSxbMCwwLCJIXmkoXFxERCBFOyBrKSJdLFszLDAsIkhebShcXEREIEUsIFxcU1MgRTsgaykiXSxbMCwxLCJcXHJve3VfRX17XFxERCBFfVxcY3VwcHJvZChcXHdhaXQpLCBcXGNvbmciXSxbMiwzLCJ1X0UgXFxjdXBwcm9kKFxcd2FpdCkiXSxbMiwwXSxbMywxXV0=) Now use that the vertical maps become isomorphisms after a colimit, so the top map must become an isomorphism as well. ::: :::{.proof title="of theorem, for arbitrary rings"} Consider $H_i(\DD E, \SS E; \ZZ) \to \cocolim_j H_i( \ro{\DD E}{C_j}, \ro{\SS E}{C_j}; \ZZ)$ Using universal coefficients, we have \[ H^i(\DD E, \SS E; \ZZ) \cong \Hom( H_i(\DD E, \SS E; \ZZ), \ZZ ) \oplus \Ext( H_{i-1}(\DD E, \SS E; \ZZ), \ZZ) = 0 && i < n ,\] since each summand will be zero. For $i=n$, the Ext term vanishes, and we have \[ H^n(\DD E, \SS E; \ZZ) &\cong \Hom( H_n(\DD E, \SS E; \ZZ), \ZZ ) \\ &\cong \Hom( \colim_j H_n( \ro{ \DD E}{C_j} , \ro{ \SS E}{C_j}; \ZZ), \ZZ ) \\ &\cong \cocolim_j \Hom( H_n( \ro{ \DD E}{C_j} , \ro{ \SS E}{C_j}; \ZZ), \ZZ ) \\ &\cong \cocolim_j H_n( \ro{ \DD E}{C_j} , \ro{ \SS E}{C_j}; \ZZ) \\ &\cong \gens{u_E} \cong \ZZ ,\] using the distinguished generator $u_E \in H^n(\DD E, \SS E)$. So we can define a chain map \[ u_E \capprod(\wait): C_{j+n}(\DD E, \SS E; \ZZ) \to C_j(\DD E) ,\] which shifts degree by $-n$ by capping against $u_E$. This induces the cup product $u_E \cupprod(\wait):H^j(\DD E; \ZZ) \to H^{j+n}(\DD E, \SS E; \ZZ)$. :::{.lemma title="Milnor-Stasheff 10.6"} Given chain complexes of \(\ZZ\dash\)modules $C_*$ and $D_*$ and a chain map $f:C_* \to D_*$, if $f^*: H^*(D_*; k) \to H_*(C_*; k)$ are isomorphisms for every $k\in \Field$, then $f_*, f^*$ are isomorphisms over any $R\in \Ring$. ::: We showed that $u_E \cupprod(\wait)$ was an isomorphism for all $k$, so now we get an isomorphism over every $R$ and this completes the proof. ::: :::{.remark} Without the oriented assumption, this theorem still holds with $C_2$ coefficients. Note also that we have naturality for characteristic classes: given a pullback \begin{tikzcd} {f^* E} && E \\ \\ Y && X \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=3-3] \arrow[from=3-1, to=3-3] \arrow[from=1-1, to=3-1] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJFIl0sWzIsMiwiWCJdLFswLDIsIlkiXSxbMCwwLCJmXiogRSJdLFszLDBdLFswLDFdLFsyLDFdLFszLDJdLFszLDEsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) Then \[ c(f^* E) = f^*c(E) .\] Note that we can always pull back the canonical: \begin{tikzcd} {f^* \gamma_n} && {\gamma_n} \\ \\ X && {\Gr_n(\RR^\infty)} \arrow[from=1-3, to=3-3] \arrow["f"', from=3-1, to=3-3] \arrow[from=1-1, to=1-3] \arrow[from=1-1, to=3-1] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJmXiogXFxnYW1tYV9uIl0sWzIsMCwiXFxnYW1tYV9uIl0sWzIsMiwiXFxHcl9uKFxcUlJeXFxpbmZ0eSkiXSxbMCwyLCJYIl0sWzEsMl0sWzMsMiwiZiIsMl0sWzAsMV0sWzAsM10sWzAsMiwiIiwxLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d) If the Euler class $e$ is natural, then $e(f^* \gamma_n) = f^* e(\gamma_n)$ where $e(\gamma_n) \in H^n(\Gr_n(\RR^\infty))$, so $e$ is the characteristic class defined by $e(\gamma_n)$. ::: :::{.lemma title="?"} The Euler class $e$ is natural with respect to pullback and thus a characteristic class. ::: :::{.proof title="?"} We'll check naturality for the Thom class $u_E$. Consider pulling back a disc bundle: \begin{tikzcd} {D'\da f^* \DD E} && {D \da \DD E} \\ \\ Y && X \arrow["f", from=3-1, to=3-3] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=1-1, to=1-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJEJ1xcZGEgZl4qIFxcREQgRSJdLFsyLDAsIkQgXFxkYSBcXEREIEUiXSxbMiwyLCJYIl0sWzAsMiwiWSJdLFszLDIsImYiXSxbMSwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) Recall that $u_E \in H^n(\DD E, \SS E; \ZZ)$ such that $\ro{u_E}{x}: H^n( \ro{\DD E}{x}, \ro{\SS E}{x}; \ZZ)$ is the generator. We get an element in the fibers of the pullback in the following way: \begin{tikzcd} {u_E} &&& {u'} \\ {H^n(\DD E, \SS E; \ZZ)} &&& {H^n(f^* \DD E, f^* \SS E; \ZZ)} \\ \\ {H^n(\DD E\ro{}{f(x)}, \SS E\ro{}{f(x)}; \ZZ)} &&& {H^n(f^* \DD E\ro{}{f(x)}, f^* \SS E\ro{}{f(x)}; \ZZ)} \\ {\ro{u_E}{x}} &&& {\ro{u'}{x}} \arrow["\cong", from=4-1, to=4-4] \arrow[from=2-4, to=4-4] \arrow[from=2-1, to=4-1] \arrow[from=2-1, to=2-4] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwzLCJIXm4oXFxERCBFXFxyb3t9e2YoeCl9LCBcXFNTIEVcXHJve317Zih4KX07IFxcWlopIl0sWzMsMywiSF5uKGZeKiBcXEREIEVcXHJve317Zih4KX0sIGZeKiBcXFNTIEVcXHJve317Zih4KX07IFxcWlopIl0sWzAsMSwiSF5uKFxcREQgRSwgXFxTUyBFOyBcXFpaKSJdLFszLDEsIkhebihmXiogXFxERCBFLCBmXiogXFxTUyBFOyBcXFpaKSJdLFswLDAsInVfRSJdLFszLDAsInUnIl0sWzAsNCwiXFxyb3t1X0V9e3h9Il0sWzMsNCwiXFxyb3t1J317eH0iXSxbMCwxLCJcXGNvbmciXSxbMywxXSxbMiwwXSxbMiwzXV0=) We then get naturality of the Euler class from the following: \begin{tikzcd} {u_E} &&& e \\ {H^n(\DD E, \SS E; \ZZ)} &&& {H^n(\DD E; \ZZ) \cong H^n(X)} \\ \\ {H^n(f^* \DD E, f^* \SS E; \ZZ)} &&& {H^n(Y)} \\ {u'} &&& {e'} \arrow[from=2-4, to=4-4] \arrow[from=4-1, to=4-4] \arrow[from=2-1, to=2-4] \arrow[from=2-1, to=4-1] \arrow[from=1-1, to=1-4] \arrow[from=5-1, to=5-4] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwxLCJIXm4oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzMsMSwiSF5uKFxcREQgRTsgXFxaWikgXFxjb25nIEhebihYKSJdLFswLDMsIkhebihmXiogXFxERCBFLCBmXiogXFxTUyBFOyBcXFpaKSJdLFszLDMsIkhebihZKSJdLFswLDAsInVfRSJdLFszLDAsImUiXSxbMCw0LCJ1JyJdLFszLDQsImUnIl0sWzEsM10sWzIsM10sWzAsMV0sWzAsMl0sWzQsNV0sWzYsN11d) ::: :::{.example title="?"} $e(\CP^1)$ is a generator of $H^2(\CP^1)$ Apply the Gysin sequence, taking the canonical line bundle $\SS E$ over $\CP^1$: \begin{tikzcd} {H^{j-1}(\SS E)} && {H^{j-2}(\CP^1)} && {H^{j}(\CP^1)} && {H^j(\CP^1)} && {H^j(\SS E)} \\ &&&&&& {} \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbNiwxXSxbMiwwLCJIXntqLTJ9KFxcQ1BeMSkiXSxbNCwwLCJIXntqfShcXENQXjEpIl0sWzYsMCwiSF5qKFxcQ1BeMSkiXSxbMCwwLCJIXntqLTF9KFxcU1MgRSkiXSxbOCwwLCJIXmooXFxTUyBFKSJdLFs0LDFdLFsxLDJdLFsyLDNdLFszLDVdXQ==) The claim is that the total space here is the Hopf fibration: \begin{tikzcd} S_1\to \SS E \cong S^3 \ar[r, "\subseteq"] \ar[dr] & E \ar[d] \\ & \CP^1\cong S^2 \end{tikzcd} What is the Hopf fibration? Write $\CP^1 = \ts{\tv{z_0: z_1} \st z_0^2 + z_1^2 = 1}/\sim$, then realize $S^3 = \ts{z_0^2 + z_1^2 = 1} \subseteq \CC^2$. Then take a map $S^3\to \CP^1$, whose fibers are $\ts{\lambda \in \CC \st \abs \lambda = 1} = \CC\units \cong S^1$. Then identifying elements and maps in the Gysin sequence yields the following: \begin{tikzcd} & {H^j(\SS E)} && \cdots \\ {j=3:} & \textcolor{rgb,255:red,92;green,92;blue,214}{H^{j}(\SS^3)} \\ \\ & {H^{j-1}(\SS E)} && {H^{j-2}(\CP^1)} && {H^{j}(\CP^1)} \\ {j=2:} & \textcolor{rgb,255:red,92;green,92;blue,214}{H^{j-1}(S^3) = 0} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^0(\CP^1)} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^2(\CP^1)} && {} \arrow[from=4-2, to=4-4] \arrow[from=4-4, to=4-6] \arrow["{e\cupprod(\wait)}", color={rgb,255:red,92;green,92;blue,214}, from=5-4, to=5-6] \arrow[from=4-6, to=1-2] \arrow[from=1-2, to=1-4] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) So $e$ is a generator of $H^2(\CP^1)$ ::: :::{.exercise title="?"} Check that $\SS E\to \CP^1$ with $\SS E$ the canonical is the Hopf fibration. ::: :::{.exercise title="?"} Try to compute $e(\T S^2)$! You may need to add on a bundle to trivialize it. :::