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I'm reading 'Lectures on Chern-Weil theory and Witten deformations' by Weiping Zhang. Now I have many questions:

  1. He defined $tr:\Omega^*(M,End(E))\to\Omega^*(M)$ by $\omega A\mapsto\omega tr[A]$ where $tr[A]$ as fiberwise to be a smooth function. Then he proved a lemma: Let $\nabla^E$ is a connection on $E$, then for any $A\in\Omega^*(M,End(E))$ one has $dtr[A]=tr[[\nabla^E,A]]$. However, I know $tr[[\nabla^E,A]]=tr[\nabla^EA]-tr[A\nabla^E]$. So I know what is $tr[\nabla^EA]$, but I don't know what is $tr[A\nabla^E]$? I can't make it clearly and percisely.

  2. As 1, we consider $d$ a trivial connection on $\mathbb{C}^N|_M$ and $g$ is a section of $Aut(\mathbb{C}^N|_M)$, then he said $g^{-1}dg\in\Omega^1(M,End(\mathbb{C}|_M))$. Why???

  3. Let $F\subset TM$ be a subbundle where $X,Y\in\Gamma(F)$, then $[X,Y]\in\Gamma(F)$. Take $p_{\lambda}(TM/F)$ be Pontrjagin classes of $TM/F$. We have the Bott vanishing theorem: If $i_1+...+i_k>(\dim M-\dim F)/2$, then $p_{i_1}(TM/F)\cdots p_{i_k}(TM/F)=0$ in $H^{4i_1+...+i_k)}_{dR}(M,\mathbb{R})$. When we prove this theorem, we take a Riemannian metric $g^{TM}$ over $TM$, then $TM=F\oplus(TM/F)$ as orthogonal decomposition. Take Levi-Cevita connection $\nabla^{TM}$. Then Take $g^F,g^{TM/F}$ induced by $g^{TM}$ and let $\nabla^F=p\nabla^{TM}p,\nabla^{TM/F}=p'\nabla^{TM}p'$ where $p,p'$ are projections. Then we define the Bott connection $\nabla^{TM/F,B}$ on $TM/F$ as: for any $X\in\Gamma(TM),U\in\Gamma(TM/F)$, if $X\in\Gamma(F)$, then $\nabla^{TM/F,B}_XU=p'[X,U]$; if $X\in\Gamma(TM/F)$, then $\nabla^{TM/F,B}_XU=\nabla^{TM/F}_XU$. Take $R^{TM/F,B}$ are curvature of $\nabla^{TM/F,B}$, one can easy to see that $R^{TM/F,B}(X,Y)=0$ for any $X,Y\in\Gamma(F)$. Then he said that from this we have $R^{TM/F,B}\in\Gamma((TM/F)^*)\wedge\Omega^*(M,End(TM/F))$ where $(TM/F)^*$ is the dual bundle. But I don't know why??? Actually, this is the most important part of the proof! Moreover, the definition of $\nabla^{TM/F,B}$ is also strange because the Bott connection $\nabla^{TM/F,B}$ on $TM/F$, why we consider $X\in\Gamma(F)$? So as $R^{TM/F,B}(X,Y)$.

Thank you for your help!

WakeUp-X.Liu
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  • Oh, I found that $[\nabla^E,A]$ is $C^{\infty}-$linear, then we can define $tr[\nabla^E,A]$. But how about the second and the third? – WakeUp-X.Liu Aug 07 '21 at 01:53
  • Then one can easily to verify this in local. But how about the second and the third? – WakeUp-X.Liu Aug 07 '21 at 02:50
  • Well, in order to see the second problem, we can consider as follows: Let $s=(s_1,...,s_r)$ and $s'=(s_1',...,s_r')$ are two frame field such that $s=s'g$. Then we consider the connection form of $\nabla^E$ is $\omega,\omega'$ respect to $s,s'$. Then $s\omega=\nabla^Es=\nabla^E(s'g)=(\nabla^Es')g+s'dg=s'\omega'g+s'dg=s(g^{-1}\omega'g+g^{-1}dg)$, so $g^{-1}dg=\omega-a^{-1}\omega'g$ is $C^{\infty}-$linear. – WakeUp-X.Liu Aug 08 '21 at 00:53

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