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I have read GTM218 Pg466, which argues that the degree of a proper map is not a homotopy invariant by showing the maps $F,G: \mathbb{C}\rightarrow\mathbb{C}$ given by F(z)=z and G(z)=z^2 are homotopy but have different degrees. But I'm wondering how does the compactness act in the proof of the degree of a map on the compact manifold is a homotopy invariant. I have a proof below: Since F is homotopy to G, then the induced map from $H^2_{dR}(\mathbb{C})$ to $H^2_{dR}(\mathbb{C}) F^{*}=G^{*}$. So for any 2-form w, $F^{*}w-G^{*}w=dv$, where $v$ is a 1-form. Let $k=degF$ and $t=degG$, we have $\int_{\mathbb{C}}F^{*}w=k\int_{\mathbb{C}}w$ and $\int_{\mathbb{C}}G^{*}w=t\int_{\mathbb{C}}w$. With Stokes theorem, we have $\int_{\mathbb{C}}(F^{*}-G^{*})w=\int_{\mathbb{C}}dv=\int_{\partial\mathbb{C}}v=0$ , I know this proof must be wrong, but what's the mistake?

Peter
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  • I think it's empty, so the integral must be zero. – Peter Feb 08 '23 at 14:29
  • The Stokes' theorem I know assumes a compact, oriented manifold with boundary. This is not $\mathbb{C}$. Maybe there's some other version you're applying? Either way, you should include this in your question. – Randall Feb 08 '23 at 14:30
  • I check GTM218 again. On page 411, I see the author says that if the boundary is empty, we can interpret the integral on the boundary to be zero. – Peter Feb 08 '23 at 14:33
  • I agree, but this assumes the manifold is compact. Unless you're using a different version of the theorem..... – Randall Feb 08 '23 at 14:34
  • There are versions without compactness, but they make assumptions on the forms you can integrate (and maybe these are violated in your case?). I don't really know, that's why I'm asking. – Randall Feb 08 '23 at 14:37
  • I think the issue comes from what one means by “homotopy invariance”: if you mimic the proof that classical de Rham groups are homotopy-invariant, the analog for de Rham cohomology groups with compact support could be invariance under proper homotopies (ie proper smooth maps $[0,1] \times M \rightarrow N$). What givea some credit to this theory is that the “obvious” homotopy $(t,z) \in [0,1] \times \mathbb{C} \longmapsto (1-t)z+tz^2 \in \mathbb{C}$ is not proper in this sense: the pre-image of $0$ contains $(t,-(1-t)/t)$ for every $t>0$. – Aphelli Feb 08 '23 at 15:14
  • I agree with your opinion. But if we just focus on my proof, what's wrong with it? – Peter Feb 08 '23 at 15:57
  • Based on your opinion, I have an idea that maybe dv is not compact supported. Do you think it is right? – Peter Feb 08 '23 at 15:59
  • If you look at it in the $1$-dim case of $\mathbb{R}$, you're asking for an improper integral to converge to $0$. That's rare. – Randall Feb 08 '23 at 16:00
  • This problem has been talked about in https://math.stackexchange.com/questions/4198921/why-does-stokess-theorem-for-manifolds-without-boundary-not-work-for-int-ax – Peter Feb 09 '23 at 01:57
  • To satisfy the compactness, we can only get f to be zero on the boundary. – Peter Feb 09 '23 at 01:58

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