# Tuesday, November 09 :::{.remark} Today: proving/checking the axioms for Chern classes. Recall that given $E \mapsvia{\pi} X$, we can form a pullback: \begin{tikzcd} & {\pi^* E \cong \gamma \oplus E'} && E \\ {\CP^{n-1}} \\ & {\PP(E)} && X \arrow["\pi", from=1-4, to=3-4] \arrow["\pi"', from=3-2, to=3-4] \arrow[from=1-2, to=3-2] \arrow[from=1-2, to=1-4] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-2, to=3-4] \arrow[hook, from=2-1, to=3-2] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMSwwLCJcXHBpXiogRSBcXGNvbmcgXFxnYW1tYSBcXG9wbHVzIEUnIl0sWzMsMCwiRSJdLFszLDIsIlgiXSxbMSwyLCJcXFBQKEUpIl0sWzAsMSwiXFxDUF57bi0xfSJdLFsxLDIsIlxccGkiXSxbMywyLCJcXHBpIiwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbNCwzLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==) Here \( \gamma\dual \) is the dual of \( \gamma \) and \( c_1( \gamma\dual) = -_1( \gamma) \). By Leray-Hirsch, we have \[ 1 + c_1( \gamma\dual) + c_1(\gamma\dual)^2 + \cdots + c_1( \gamma\dual)^{n-1} \divides (c_1(\gamma\dual))^n \] and for $c_i(E)$ the $i$th Chern class of $E$, \[ c_1(\gamma\dual)^n + c_1(E) c_1(\gamma\dual)^{n-1} + \cdots + c_n(E) = 0 .\] ::: :::{.proposition title="?"} These satisfy some axioms: 1. Naturality: $c_1(f^* E) = f^* c_1(E)$ 2. Homomorphism: $c_(E_1 \oplus E_2) = c(E_1) \cupprod c(E_2)$. 3. $c_{>\dim E}(E) = 0$. 4. For $\gamma\to \CP^\infty$ the canonical, $c_1(\gamma)\in H^2(\CP^\infty)$ is the positive generator. ::: :::{.proof title="?"} Here (3) is clear by definition, since we don't even define $c_{n+1}$ or higher if $\dim E = n$. Number (4) isn't bad either, since $c_1(\gamma) = e(\gamma)$, using that $c(\mcl) = c_1(\mcl)$ for line bundles. For (1), consider a pullback: \begin{tikzcd} \textcolor{rgb,255:red,92;green,92;blue,214}{\PP(f)^* \gamma \oplus \PP(f)^* E' } &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\gamma \oplus E'} \\ & {f^* E} &&&& E \\ \\ \textcolor{rgb,255:red,92;green,92;blue,214}{\PP(f^* E)} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\PP(E)} \\ & {X'} &&&& X \arrow["f", from=5-2, to=5-6] \arrow[from=2-6, to=5-6] \arrow[from=2-2, to=2-6] \arrow[from=2-2, to=5-2] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=5-6] \arrow[color={rgb,255:red,214;green,153;blue,92}, from=4-5, to=5-6] \arrow[color={rgb,255:red,214;green,153;blue,92}, from=4-1, to=5-2] \arrow["{\exists \PP(f)}", color={rgb,255:red,92;green,92;blue,214}, from=4-1, to=4-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=1-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-5, to=4-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-1, to=4-1] \arrow["\lrcorner"{anchor=center, pos=0.125}, color={rgb,255:red,92;green,92;blue,214}, draw=none, from=1-1, to=4-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) We then use that on fibers, $\gamma_{\PP(f)^* E} \cong \PP(f)^* \gamma_{\PP(E)}$, so the canonical pulls back to something isomorphic to the canonical. Then $c_1(\gamma\dual_{\PP(f)^* E}) = \PP(f)^* c_1(\gamma\dual_{\PP(E)})$, so \[ (c_1 \gamma\dual _{\PP(E)} )^{n} + c_1(E) \cdot (c_1 \gamma\dual _{\PP(E)} )^{n-1} + c_2(E) \cdot (c_1 \gamma\dual _{\PP(E)} )^{n-2} + \cdots + c_n(E) ,\] and applying $\PP(f)^*$ to this entire expression yields zero. Thus \[ (c_1 \gamma\dual _{\PP(f)^* E } )^{n} \PP(f)^* c_1(E) \cdot (c_1 \gamma\dual _{\PP(f)^* E } )^{n-1} \PP(f)^* c_2(E) \cdot (c_1 \gamma\dual _{\PP(f)^* E } )^{n-2} + \cdots + + \PP(f)^* c_n(E) = 0 ,\] and we can note that by definition this equals \[ (c_1 \gamma\dual _{\PP(f)^* E } )^{n} c_1 f^* E \cdot (c_1 \gamma\dual _{\PP(f)^* E } )^{n-1} c_2 f^* E \cdot c_1 \gamma\dual _{\PP(f)^* E } )^{n-2} + \cdots + c_n f^* E = 0 .\] ::: :::{.remark} Recall the alternative description of Chern classes: \begin{tikzcd} {\pi^* E \cong \RR \oplus E'} && E \\ \\ {\SS E} && X \arrow[from=1-3, to=3-3] \arrow["\pi"', from=3-1, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=1-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJcXFNTIEUiXSxbMiwyLCJYIl0sWzIsMCwiRSJdLFswLDAsIlxccGleKiBFIFxcY29uZyBcXFJSIFxcb3BsdXMgRSciXSxbMiwxXSxbMCwxLCJcXHBpIiwyXSxbMywwXSxbMywyXV0=) By the Gysin sequence, we obtained $H^j(X) \cong H^j(\SS E)$ for $j\leq n-2$, and there are $c_1(E'), \cdots, c_{n-2}(E')$ living in $H^2(\SS E), H^4(\SS E), \cdots, H^{2n-2}(\SS E)$. So we defined $c_i(E)$ such that $\pi^* c_i(E) = c_i(E')$ for $i\leq n-1$, so $c_n(E) = e(E)$. ::: ## Splitting Principle :::{.proposition title="?"} Given an $E\in \VectBundlerk{n}(\RR)\slice X$, there exists $Y\in \Top$ with $f:Y\to X$ such that 1. $f^* E = L_1 \oplus \cdots \oplus L_n$ with $L_i$ line bundles. 2. $f^*: H^*(X) \injects H^*(Y)$ is injective, so classes in $H^*(Y)$ can be uniquely pulled back. ::: :::{.proof title="?"} By induction, it suffices to find $Y$ where $f^* E \cong E' \oplus E''$ splits nontrivially and $f^*$ is injective. Taking the projectivization does exactly this, so take $Y \da \PP(E)$ and $f: \PP(E)\to X$ to be the projection. Then $f^* E = \gamma \oplus E'$, so checking the 2nd condition in Leray-Hirsch, we get that $H^*(Y)$ is generated over $H^*(X)$ by $1, c_1(\gamma),c_1(\gamma)^2, \cdots, c_1(\gamma)^{n-1}$. Equivalently, the following map is injective: \[ f^* : H^*(X) &\to H^*(Y) \\ \alpha & \mapsto 1\cdot \alpha .\] Here being generated means that \[ \beta\in H^*(Y) \implies \beta = \sum_{k=0}^{n-1} c_1(\gamma)^k \alpha_k, \quad \alpha_k \in H^*(X) .\] ::: :::{.corollary title="?"} If a polynomial identity holds for Chern classes in $E \da \bigoplus_i L_i$ for $L_i$ line bundles, then it holds for any bundle. ::: :::{.lemma title="?"} For line bundles $L_1,\cdots, L_n$, \[ c(\bigoplus L_i) = \prod c(L_i) = \prod(1 + c_1(L_i)) = 1 + \qty{\sum c_1(L_i) } + \cdots + \qty{\prod c_1(L_i)} ,\] so for $E \da \bigoplus L_i$, $c_i(E)$ is the $i$th symmetric polynomial in the $c_1(L)$. In other words, writing $\sigma_i(x_1, \cdots, x_n)$ as the $i$th symmetric polynomial, $c_i(E) = \sigma_i(c_1(L_1), c_1(L_2), \cdots, c_1(L_n))$. ::: :::{.corollary title="?"} The direct sum formula $c(E \oplus E') = c(E) c(E')$. Take a pullback \begin{tikzcd} {\bigoplus_{i} L_i \oplus \bigoplus_j L_j} && {E \oplus 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=WzAsNCxbMiwwLCJFIFxcb3BsdXMgRSciXSxbMiwyLCJYIl0sWzAsMiwiWSJdLFswLDAsIlxcYmlnb3BsdXNfe2l9IExfaSBcXG9wbHVzIFxcYmlnb3BsdXNfaiBMX2oiXSxbMywwXSxbMCwxXSxbMiwxXSxbMywyXSxbMywxLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) Then \[ c( \bigoplus_i L_i \oplus \bigoplus_j L_j) = \qty{\prod_i c(L_i)} \cdot \qty{\prod_j c(L_j)} = c\qty{ \bigoplus_i L_i} \cdot c\qty{\bigoplus L_j} = f^* c(E) \cdot f^* c(E') ,\] so $f^*c(E \oplus E') = f^* c(E) \cdot f^* c(E') = f^*(c(E) \cdot c(E'))$, and by injectivity, $c(E \oplus E') = c(E) c(E')$. ::: :::{.proof title="of lemma"} Write $E = L_1 \oplus \cdots \oplus L_n$ where $E\to X$. Then pull back along $\pi:\PP E\to X$ to get $\pi^* E = \pi^* L_1 \oplus \cdots \oplus \pi^* L_n$. Note that $\gamma \subseteq \pi^* E$. :::{.claim} \[ \prod_{i=1}^n \qty{ c_1(\gamma\dual) + c_1( \pi^* L_i) } = 0 .\] ::: Using the claim, we get \[ 0 = c_1( \gamma\dual)^n + \qty{ \sum c_1 \pi^* L_i }(c_1(\gamma\dual))^{n-1} + \cdots + \prod c_1(\pi^* L_i) \cdot c_1(\gamma\dual)^1 = c_1(\gamma\dual)^{n} + \pi^* c_1(E) \cdot (c_1(\gamma\dual ))^{n-1} + \cdots + \pi^* c_n(E) \cdot c_1(\gamma\dual)^1 ,\] which proves e.g. that $c_1(E) = \sigma_1(c_1L_1, \cdots, c_1 L_n)$. The idea is that e.g. for the first term, $c_1( \gamma\dual) = -c_1(\gamma)$ and $c_1\pi^* L_1 = c_1( \gamma)$, yielding zero. If we can cover $E$ by opens where this happens for at least one term, then the entire product must be zero on $E$. Take a SES and apply $\Hom(\gamma, \wait)$: \begin{tikzcd} 0 && \gamma && {\bigoplus_i \pi^* L_i} && {\gamma^\perp} && 0 \\ \\ && {\Hom(\gamma, \gamma)} && {\Hom\qty{\gamma, \bigoplus_i \pi^* L_i} \cong \bigoplus_i \Hom(\gamma, \pi^* L_i)} \arrow[from=1-1, to=1-3] \arrow[""{name=0, anchor=center, inner sep=0}, "\iota", from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[""{name=1, anchor=center, inner sep=0}, "{\iota_*}", from=3-3, to=3-5] \arrow["{\Hom(\gamma,\wait)}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwwLCIwIl0sWzIsMCwiXFxnYW1tYSJdLFs0LDAsIlxcYmlnb3BsdXNfaSBcXHBpXiogTF9pIl0sWzYsMCwiXFxnYW1tYV5cXHBlcnAiXSxbOCwwLCIwIl0sWzIsMiwiXFxIb20oXFxnYW1tYSwgXFxnYW1tYSkiXSxbNCwyLCJcXEhvbVxccXR5e1xcZ2FtbWEsIFxcYmlnb3BsdXNfaSBcXHBpXiogTF9pfSBcXGNvbmcgXFxiaWdvcGx1c19pIFxcSG9tKFxcZ2FtbWEsIFxccGleKiBMX2kpIl0sWzAsMV0sWzEsMiwiXFxpb3RhIl0sWzIsM10sWzMsNF0sWzUsNiwiXFxpb3RhXyoiXSxbOCwxMSwiXFxIb20oXFxnYW1tYSxcXHdhaXQpIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==) Now note that we can get a splitting: \begin{tikzcd} {\Hom(\gamma, \gamma)} && {\bigoplus_i \Hom(\gamma, \pi^* L_i)} \\ \\ & X \arrow[from=1-3, to=3-2] \arrow[from=1-1, to=3-2] \arrow[from=1-1, to=1-3] \arrow["s"{description}, curve={height=-24pt}, dashed, from=3-2, to=1-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXEhvbShcXGdhbW1hLCBcXGdhbW1hKSJdLFsyLDAsIlxcYmlnb3BsdXNfaSBcXEhvbShcXGdhbW1hLCBcXHBpXiogTF9pKSJdLFsxLDIsIlgiXSxbMSwyXSxbMCwyXSxbMCwxXSxbMiwwLCJzIiwxLHsiY3VydmUiOi00LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) So take $X = \PP E$ and pick a nonvanishing section $s$ such that $\iota_* \circ s = (s_1, s_2,\cdots s_n)$ is a sum of sections. Then let $U_i = \ts{x\in \PP E\st s_i(x) \neq 0} \subseteq \PP E$ by the complement of $Z(s_i)$, then $\Union_i U_i = \PP E$ since $s$ is nonvanishing. So $\ro{\Hom(\gamma, \pi^* L_i)}{U_i}$ is trivial, this yields $\ro{\gamma}{U_i} \cong \ro{\pi^* L_i}{U_i}$, making the restrictions of $c(\wait)$ equal, so $c_1(\pi^* L_i) - c_1(\gamma) = 0$ on $U_i$, and this is equal to $c_1(\pi^* L_i) + c_1( \gamma\dual)$. Since $\ts{U_i}\covers \PP E$, this concludes the proof. :::