# Chern characteristic class

1. 3周前

### Phantom_Ghost

1楼 1月20日 数学版主, 物理版主, 优秀回答者
2周前Phantom_Ghost 重新编辑

A short note recording the content of the talk from the seminar of ''topological K-theory'' during WS18/19.

Reference: Allen Hatcher, Vector Bundles and K-Theory.

2. ### Phantom_Ghost

2楼 1月20日 数学版主, 物理版主, 优秀回答者
3周前Phantom_Ghost 重新编辑

Theorem 1: There is a unique sequence of maps $c_i$ assigning to vector bundle $E\to B$ a class $c_i(E)\in H^{2i}(B;\mathbb{Z})$, which depends on isomorphism class of $E$ and the following properties:

$i)$ $c_{i}(f^*E)=f^*c_i(E)$ ($f^*E$ is a pullback bundle)

$ii)$ $c(E_1\oplus E_2)=c_1(E_1)\smile c(E_2)$ where $c=1+\sum_{i>0}c_i$ or equivalently $c_k(E_1\oplus E_2)=\sum_{i+j=k}c_i(E_1)\smile c_j(E_2)$ where $c_0=1$.

$iii)$ $c_i(E)=0$ $i>\text{rank}\,E$

$iv)$ For the canonical line bundle $E\to\mathbb{C}P^\infty$, $c_1(E)$ is a fixed generator pf $H^2(\mathbb{C}P^\infty;\mathbb{Z})$

The proof of the theorem is based on Leray-Hirsch theorem:
Theorem 2 (Leray-Hirsch): Let $\pi:E\to B$ be a fiber bundle, $H^\bullet(E;R)$ is a module over $H^\bullet(B;R)$ ($R$ is commutative ring); $\alpha\in H^\bullet(B;R)$ and $\beta\in H^\bullet(E;R)$, $\alpha\cdot\beta=\pi^*\alpha\smile\beta$; $H^\bullet(E;R)$ is free over $H^\bullet(B;R)$. Evidently, $\forall x\in B$, $F_x\overset{i_x}{\hookrightarrow}E$ induces a surjection on $H^\bullet(E;R)\to H^\bullet(B;R)$ and $H^\bullet(F;R)$ is a free $R$-modlue of finite rank. Moreover we can obtain a basis by choosing elements that map to a basis in $H^\bullet(F;R)$ under $i^*$.

Proof of thm1(existence): $P(\pi):P(E)\to B$ is universal bundle, remove zero section from $\mathbb{C}P^{n-1}$ out of $\mathbb{C}^\times$. Wee want to find $x_i\in H^{2i}(P(E);\mathbb{Z})$ s.t. they restrict to generators of $H^{2i}(\mathbb{C}P^{n-1};\mathbb{Z})$. From universal bundle, there is $g:E\to\mathbb{C}P^\infty$ with linear injections as fibers. Projectivizing $g$ by deleting zero section and then factoring out scalar multiplication induces $P(g):P(E)\to\mathbb{C}P^\infty$
\begin{array}[c]{cc}
P(E) & \\
\scriptstyle{i}\uparrow&\searrow{}^{P(g)}\\
\;\;\;\;\;\;\mathbb{C}P^{n-1}\hookrightarrow& \mathbb{C}P^\infty\\
\end{array}

\begin{array}[c]{cc}
& H^{2i}(P(E);\mathbb{Z})\\
{}^{{}^{P(g)^*}\nearrow}&\downarrow\scriptstyle{i^*}\\
\;\;\;\;\;\;H^{2i}(\mathbb{C}P^\infty;\mathbb{Z})\rightarrow& H^{2i}(\mathbb{C}P^{n-1};\mathbb{Z})
\end{array}
Let $\alpha$ be a generator of $H^2(\mathbb{C}P^\infty;R)$, then $x^i=P(g)^*\alpha^i$, $1,x,x^2,\dots, x^{n-1}$ form a basis of $H^\bullet(P(E);R)$ as a $H^\bullet(E;R)$-module. Their images under $i^*$ map form a basis for the cohomology of fiber $H^\bullet(\mathbb{C}P^{n-1})$. Therefore $x^n+c_1(E)x^{n-1}+\cdots+c_n(E)\cdot 1=0$.
This is how we obtain the Chern class.

Proof of uniqueness follows from the proposition below.

Proposition 3 (Splitting principle): For each vector bundle $E\to B$, there is a space $F(E)$ with a map $F(E)\xrightarrow{p}B$ s.t. the pullback bundle $p^*E\to F(E)$ splits as a direct sum of line bundles and $p^*:H^\bullet(B;\mathbb{Z})\to H^\bullet(P(E);\mathbb{Z})$ is injective.
(This is a similar phenomenon compare with Bott periodicity for K-theory)

Proof of prop.: We have the map $P(E)\xrightarrow{P(\pi)}B$ and pull back $E$ along this map. There is a subbundle in bundle on $P(E)$: $L:=\{(\ell,v)\in P(E)\times|v\in\ell\}$ , $P(\pi)^*E=L\oplus L^\perp$. $H^\bullet(P(E);\mathbb{Z})$ is a free module with basis $1,x,x^2,\dots,x^{n-1}$. Repeat the same construction for $L^\perp$ and we are done.

Remark
1) $E=B\times\mathbb{C}$, it is a pullback of a bundle over one point, since $H^{2i}(\{\text{pt.}\};\mathbb{Z})=0$ for $i>0$, hence $c_i(E)=0$ for $i>0$.

2) $E=E_1\oplus\mathbb{n}$, $c(E)=c(E_1\oplus\mathbb{n})=c(E_1)\smile \underset{=1}{\underbrace{c(\mathbb{n})}}=c(E_1)$.

3) $E\cong L_1\oplus L_2\oplus\cdots\oplus L_n$, uniqueness comes from splitting principle and 4)

4) There is no bundle over $E\to\mathbb{C}P^\infty$ whose sum with the canonical line bundle over $\mathbb{C}P^\infty$ is trivial. i.e assume $E\oplus L\cong\mathbb{n}$, then $1=c(\mathbb{n})=c(E\oplus L)=c(E)\smile c(L)=(1+c_1(E)+\cdots)(1+c_1(L))$ by 1); hence $c(E)=(1+c_1(L))^{-1}=1-c_1(L)+c_1(L)^2+\cdots$ contradicts to 3).

3. ### Phantom_Ghost

3楼 1月21日 数学版主, 物理版主, 优秀回答者
2周前Phantom_Ghost 重新编辑

Chern character
\begin{align*}
\text{ch}(E_1\oplus E_2)&=\text{ch}(E_2\oplus E_1)=\text{ch}(E_1)+\text{ch}(E_2)\\
\text{ch}(E_1\otimes E_2)&=\text{ch}(E_1)\smile \text{ch}(E_2)\\
\text{ch}(L)&=e^{c_1(L)}=1+c_1(L)+\frac{1}{2!}(c_1(L))^2+\cdots\\
\text{ch}(L_1\otimes L_2)&=e^{c_1(L_1\otimes L_2)}=e^{c_1(L_1)+c_1(L_2)}=e^{c_1(L_1)}e^{c_1(L_1)}
\end{align*}

Proposition 4: $c_1:\text{Vect}_\mathbb{C}^1(B)\to H^2(B;\mathbb{Z})$ is a homomorphism of rings.
(Here the vector bundles equipped with operation $\oplus,\otimes$ become a ring, while the integral cohomology equipped with $+,\smile$ become a graded ring)

Proof: First show for line bundles: $c_1(L_1\otimes L_2)=c_1(L_1)+c_1(L_2)$ where $L_1\otimes L_2\to\mathbb{C}P^\infty\times\mathbb{C}P^\infty$ , $L_{1,2}\hookrightarrow\mathbb{C}P^\infty$ are the pullbacks of canonical line bundle. Since $c_1(L_i)$ is a generator of $H^2(\mathbb{C}P^\infty)$, we get $H^\bullet(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)\cong\mathbb{Z}[\alpha_1,\alpha_2]$ by Künneth formula where $\alpha_i=\text{pr}_i^*(c_1(L_i))$, $\text{pr}_{1,2}:\mathbb{C}P^\infty_1\times\mathbb{C}P^\infty_2\to \mathbb{C}P^\infty_{1,2}$.
\begin{align*}
H^2(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)&\cong[H^2(\mathbb{C}P^\infty)\otimes H^0(\mathbb{C}P^\infty)]\oplus[H^0(\mathbb{C}P^\infty)\otimes H^2(\mathbb{C}P^\infty)]\\
&\bigoplus_{i+j=2+1}\underset{=0\;\text{(torsion free)}}{\underbrace{\text{Tor}(H^i(\mathbb{C}P^\infty),H^j(\mathbb{C}P^\infty))}}
\end{align*}
$\Rightarrow$ $\mathbb{C}P^\infty\vee\mathbb{C}P^\infty\hookrightarrow\mathbb{C}P^\infty\times\mathbb{C}P^\infty$ induces isomorphism on cohomology $H^2(\mathbb{C}P^\infty\vee\mathbb{C}P^\infty)\cong H^2(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)$.
Therefore to compute $c_1(L_1\otimes L_2)$ it is sufficient to compute $c_1(L_1\otimes L_2|_{\mathbb{C}P^\infty\vee\mathbb{C}P^\infty})$:
$L_2$ is trivial on the first factor of $\mathbb{C}P^\infty\vee\mathbb{C}P^\infty$ while $L_1$ is trivial on the second factor: $c_1(L_1\otimes L_2|_{\mathbb{C}P^\infty\times\{\text{pt.}\}})=c_1(L_1)=\alpha_1$ , $c_1(L_1\otimes L_2|_{\{\text{pt.}\}\times\mathbb{C}P^\infty})=c_1(L_2)=\alpha_2$ $\Rightarrow$ $c_1(L_1\otimes L_2)=c_1(L_1)+c_1(L_2)$.

For the general case $E\cong L_1\oplus\cdots\oplus L_n$, $t_i:=c_1(L_i)$, then $\text{ch}(E)=\sum_i e^{t_i}=n+(t_1+\cdots+t_n)+\cdots+\frac{1}{k!}(t_1^k+\cdots+t_n^k)+\cdots$
$t_1^k+\cdots+t_n^k=s_k(\sigma_1,\dots,\sigma_n)$ with $\sigma_i=c_i(E)$ where $s_k(\cdot)$ is Newton polynomial. Then $\text{ch}(E)=\text{dim}\,E+\sum_{k>0}s_k(c_1(E),\dots,c_n(E))$.

Proposition 5 $\text{ch}(E_1\oplus E_2)=\text{ch}(E_1)+\text{ch}(E_2)$ ; $\text{ch}(E_1\otimes E_2)=\text{ch}(E_1)\text{ch}(E_2)$.

Proof: Splitting principle works also for $\mathbb{Q}$. This time we can pull back $E_1$ to a sum of $L_1\oplus\cdots L_n$ over $F(E_1)$. Then pull back $E_2$ over $F(E_1)$ so as a result $F(E_1,E_2)\to B$, where we pull back both $E_1$ and $E_2$ to the line budle.
$\text{ch}(E_1\oplus E_2)=\text{ch}(\bigoplus_{i=1,2,j>0}L_{i,j})e^{c_1(L_{i,j})}=\text{ch}(E_1)+\text{ch}(E_2)$
$\text{ch}(E_1\otimes E_2)=\text{ch}(\bigoplus_{i,j>0}L_{1,i}\otimes L_{2,j})=\sum_{i,j}\text{ch}(L_{1,i}\otimes L_{2,j})=\text{ch}(E_1)\text{ch}(E_2)$

Fact: $K(X)\xrightarrow{\text{ch}}H^\bullet(X;\mathbb{Q})$ is homomorphism
\begin{array}[c]{ccc}
K(X)&\xrightarrow{\text{ch}}&H^\bullet(X;\mathbb{Q})\\
\downarrow&&\downarrow\\
K(\{\text{pt.}\})&\xrightarrow{\text{ch}}&H^\bullet(\{\text{pt.}\};\mathbb{Q})
\end{array}
This yields the functoriality and there is a natural transormation between the functors $K(X)\xrightarrow{\text{ch}}H^\bullet(X;\mathbb{Q})$ and $\widetilde{K}(X)\xrightarrow{\text{ch}}\widetilde{H}^\bullet(X;\mathbb{Q})$.

Proposition 6: $\widetilde{K}(S^{2n})\xrightarrow{\text{ch}}\widetilde{H}^{2n}(S^{2n};\mathbb{Q})$ injective, $\text{im}(\text{ch})\subset\widetilde{H}^{2n}(S^{2n};\mathbb{Z})\subset\widetilde{H}^{2n}(S^{2n};\mathbb{Q})$.

Proof: Since
\begin{array}[c]{ccc}
\widetilde{K}(X)&\xrightarrow{\sim}&\widetilde{K}(\Sigma^2X)\\
\scriptstyle{\text{ch}}\downarrow&&\downarrow\scriptstyle{\text{ch}}\\
\widetilde{H}^{2n}(X;\mathbb{Q})&\xrightarrow{\sim}&\widetilde{H}^{2n}(\Sigma^2X;\mathbb{Q})
\end{array}
where $\Sigma^2X$ is the double suspension of $X$. The upper row in the diagram is due to Bott periodicity while the lower row is due to Künneth formula.
$\text{ch}(H-1)=\text{ch}(H)-\text{ch}(1)=1+c_1(H)-1=c_1(H)$ ($H$ denotes the canonical line bundle over $S^2\approx \mathbb{C}P^1$).
Perform induction on $n$:
$X=S^0$, it holds.
$n\to n+1$ follows by the commutative diagram
\begin{array}[c]{ccc}
\widetilde{K}(X)&\xrightarrow{\sim}&\widetilde{K}(\Sigma^2X)&\to& \widetilde{K}(\Sigma^4X)&\to& \cdots\\
\scriptstyle{\text{ch}}\downarrow&&\downarrow\scriptstyle{\text{ch}}&&\downarrow\\
\widetilde{H}^{2n}(X;\mathbb{Q})&\xrightarrow{\sim}&\widetilde{H}^{2n}(\Sigma^2X;\mathbb{Q})&\to&\widetilde{H}^{2n+4}(\Sigma^4X;\mathbb{Q})&\to& \cdots
\end{array}

Corollary 7: A calss in $H^{2n}(S^{2n};\mathbb{Z})$ is a Chern characteristic $c_n(E)$ $\Rightarrow$ It is divisble by $(n-1)!$

Proof: $E\to S^{2n}$ we get $c_i(E)=0$ for $n>i>0$.
$\text{ch}(E)=\text{dim}\,E+\frac{1}{n!}s_n(\sigma_1,\dots,\sigma_n)=\text{dim}\,E+\frac{n}{(n+1)!}\sigma_n$ since $s_n(\sigma_1,\dots,\sigma_n)=\sigma_1 s_{n-1}-\sigma_2 s_{n-2}+\cdots+(-1)^{n-2}\sigma_{n-1}s_1+(-1)^{n-1}n\sigma_n$.