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This is probably trivial but I am trying to find whether or not the degree of a map $f:X\to Y$, with $X$ and $Y$ compact oriented manifolds can be extended to the case where $Y$ is non compact like $GL(N,\mathbb{C})$. In particular I am dealing with $f:S^1\to GL(N,\mathbb{C})$ and it seems to me that since $\pi_1(GL(N,\mathbb{C}))\cong \pi_1(U(N)) = \mathbb{Z}$, the degree should exist and be an integer. Am I wrong? However, I also know that it must be deg$(f) \int_Y\omega=\int_X f^*\omega$, with $\omega$ an invariant differential form and $f^*\omega$ the pull-back of $\omega$. I could choose $\omega = g^{-1}dg$ the Maurer-Cartan form since I am working with a Lie group but the fact that it is non compact means that its volume is infinite. What is the correct way of computing the degree in this case?
I apologize in advance if I am saying something wrong but I am new to some of these concepts. Any kind of help is appreciated!

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I actually found a reference that talks about maps $f:S^{n-1}\to GL(N,\mathbb{C})$ in an old paper by Atiyah https://aareyanmanzoor.github.io/assets/images/atiyah1967.pdf where he says that for $n$ even we can define an integer called deg$f$ (Bott's theorem). This is explained at the bottom of the second page (238) and in the following page he mentions

There is also a differential definition of deg$f$. We put deg$f = \int_{S^{n-1}}f^*\omega$, where $\omega$ is a certain explicitly defined invariant differential form on $GL(N,\mathbb{C})$ and $f^*\omega$ is the induced form on $S^{n-1}$.

Doesn't this mean that the invariant I am looking for is indeed the degree of that map? Can $\omega$ be the Maurer-Cartan form?

TRay
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  • You need to work with proper maps to define degree in this setting. You should explain what $S_1$ and $f$ are. – Moishe Kohan Mar 04 '24 at 03:06
  • See also https://math.stackexchange.com/questions/4635004/why-degree-of-a-proper-map-is-not-a-homotopy-invariant-on-noncompact-manifolds – Eric Towers Mar 04 '24 at 03:12
  • Thank you @MoisheKohan. By $S_1$ I mean the 1-sphere and $f$ is a continuous map that takes values in $\mathbb{R}\cup\lbrace \infty\rbrace = S_1$ and gives back an element of $GL(N,\mathbb{C})$. – TRay Mar 04 '24 at 03:36
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    Then this is not the degree in any sense but a different invariant. Please, replace $S_1$ with $S^1$, which is the standard notation for the circle. – Moishe Kohan Mar 04 '24 at 03:51
  • I see thank you @MoisheKohan. I'll change the notation for the circle . Do you know what kind of invariant it is then? And how to compute it? – TRay Mar 04 '24 at 03:56
  • It is just the induced map of 1st homology groups, dually, 1st cohomology groups. You can compute it by integrating over the circle the pull back of a suitable invariant 1-form. – Moishe Kohan Mar 04 '24 at 04:02
  • Compute $det\circ f: S^1\to {\mathbb C^*}$ and then compute its winding number around $0$. – Moishe Kohan Mar 04 '24 at 04:06
  • Thank you @MoisheKohan. I was actually thinking about this myself but I wasn't sure – TRay Mar 04 '24 at 04:22

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