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Let $M$ be a smooth manifold of 2n-dim, $v$ be a map from $M$ to the matrix of order $m\times m$. We call $p\in M$ is the singularity, if $v(p)$ is non-invertible. Suppose $v$ is smooth and the singularity is submanifold. Let $C$ be a connected component of singularity, $U$ is the tubular neighborhood of $C$. How to computing the integration $$\int_{\partial U}{\rm Tr}[(v^{-1}dv)^{2n-1}]$$

I guess its result is zero, may be that is wrong.

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You may refer to the book Lectures on Chern-Weil theory and Witten deformation section 1.8. – Guangbo Xu Oct 23 '10 at 23:06
Thanks very much! – Chen Oct 24 '10 at 6:13

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up vote 3 down vote accepted

I think your guess is correct and the integral is zero, at least if you assume that $M$ is a closed oriented manifold and that U contains all singularities. Here is the proof, assuming closedness of M. Let $N = M - U$, a compact manifold with boundary. Let $G=Gl_n (C)$ (in your problem, it does not matter whether you take real or complex matrices) and denote your map by $f: N \to G$. Consider the form $\omega= Tr((g^{-1}dg)^{2k-1})$ on $G$. Then your integrand is $f^{\ast} \omega$ and since $d\omega=0$ (see below), an application of Stokes gives you the answer you suspect. Why is $\omega$ closed? I am tempted to argue that on a compact Lie group, each bi-invariant differential form. Of course $Gl_n (C)$ is not compact, but it is the complexification of the compact $U(n)$ and the unitary trick works. For this special example, there is of course a direct computation (which you can do by yourself). It should be contained in Chern-Simons "Geometric invariant and characteristic forms).

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Thanks very much! – Chen Oct 22 '10 at 2:06

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