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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 for yourself, but I'll search for reference).

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).

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 reductive complex Lie group, each bi-invariant differential form is closed by the unitary trick, but for this special example, there is of course a direct computation (which you can do for yourself, but I'll search for reference).

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 for yourself, but I'll search for reference).

I think your guess is correct and the integral is zero, at least if you assume that $M$ is a closed oriented manifold. 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 reductive complex Lie group, each bi-invariant differential form is closed by the unitary trick, but for this special example, there is of course a direct computation (which you can do for yourself, but I'll search for reference).

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 reductive complex Lie group, each bi-invariant differential form is closed by the unitary trick, but for this special example, there is of course a direct computation (which you can do for yourself, but I'll search for reference).