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I'd like to find a reference for the following fact.

First, some background: we can define de Rham cohomology of a smooth manifold $X$ of dimension $d$ using the de Rham complex $$ \Omega^0_X\to \Omega^1_X \to\ldots \to \Omega^d_X $$ of smooth forms. We can equally define it using the distributional de Rham complex $$ D^0_X\to D^1_X\to \ldots \to D^d_X $$ where $D^i$ are the $i$-currents, i.e. linear functionals on $(d-i)$-forms with compact support. Now, suppose we have an inclusion $i:Z\to X$ of a closed submanifold of codimension $r$. Then there is a map of complexes $$ \Omega^i_Z \to \ker(D^{i+r}_X \to D^{i+r}_{X-Z}) $$ given by sending an $i$-form $\eta$ to the functional on $(d-r-i)$-forms $$ \omega \mapsto \int_Z \eta \wedge \omega. $$ This should induce the Gysin/excision isomorphism $H^i(Z) \cong H^{i+r}_Z(X)$ (alternately notated $H^{i+r}(X,X-Z)$). I think I know how to prove this, by using a comparison with singular cohomology and developing the theory of cycle classes, then using Poincare duality, but it's quite unwieldy and requires introducing a lot of auxiliary theory and constructions.

My question is: is there a more direct way to prove that the above map of complexes is a quasi-isomorphism? Or does someone have a reference where any proof (not necessarily direct) is given?

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  • $\begingroup$ Just a suggestion: have you tried considering maps $D^i_Z \to D^{i+r}_X$, rather than $\Omega^i_Z \to \Omega^{i+r}_X$? $\endgroup$ Commented Jul 14, 2023 at 5:50
  • $\begingroup$ I thought about this, but I couldn't figure out how to define such a map. The problem is that you can't really wedge a general current with $\delta_Z$ (I think), but you can wedge a differential form with it, resulting in the formula I wrote in the question. $\endgroup$
    – xir
    Commented Jul 14, 2023 at 16:00
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    $\begingroup$ There's a pullback (restriction) map on differential forms $\Omega^j_X \to \Omega^j_Z$, and it sends compactly-supported forms to compactly-supported forms. Now just define the map $D^i_Z \to D^{i+r}_X$ to be the dual of this. No need to even mention $\delta_Z$. $\endgroup$ Commented Jul 15, 2023 at 16:58
  • $\begingroup$ Oh sheesh, I was being very silly. Thanks! $\endgroup$
    – xir
    Commented Jul 16, 2023 at 21:24

1 Answer 1

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Here is one way to construct your push-forward map. Consider a tubular neighbourhood $U$ of $Z$ and identify it with the normal bundle $\pi\colon\nu\to Z$. We will have to assume that $\nu$ is orientable. Then $\nu$ carries a Mathai-Quillen-Thom form $\theta\in\Omega^r(\nu)$, as explained in the book of Berline, Getzler and Vergne. It depends on a metric and a connection on $\nu$, which you can choose as you like. It decays rapidly, so it induces a smooth form on $X$ supported in $\overline U$. Then the classical push-forward can be realised in de Rham cohomology by $$\Omega^\bullet(Z)\owns\omega\mapsto\pi^*\omega\wedge\theta \quad\in\quad\Omega^{\bullet+r}(X)\;.$$ However, this form is only in $\ker(\Omega^\bullet(X)\to\Omega^\bullet(X\setminus\overline U))$. To remedy this, note that $\theta$ actually depends on an additional parameter $T\in(0,\infty)$, so write $\theta_T$. As $T\to\infty$, the form tends to $0$ everywhere outside the zero section. The limit is still well-defined as a current on $\nu$ (and hence on $X$). Hence $$\lim_{T\to\infty}\pi^*\omega\wedge\theta_T\quad\in\quad \ker\bigl(D^{\bullet+r}(X)\to D^{\bullet+r}(X\setminus Z)\bigr)\;.$$ In fact, this current behaves as a $\delta$-distribution on $Z$. In other words, for any $\alpha\in\Omega^\bullet(X)$, one has $$\int_Z\alpha\wedge\omega=\lim_{T\to\infty}\int_X\alpha\wedge\pi^*\omega\wedge\theta_T\;.$$

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