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Let $G$ be a complex reductive group, $X$ a smooth projective variety on which $G$ acts algebraically, and $Y \subseteq X$ a $G$-invariant smooth closed subvariety such that $X\setminus Y$ is also smooth.

If $d$ is twice the complex codimension of $Y$ in $X$, then my understanding of the equivariant Gysin sequence (with coefficients in $\mathbb Q$) is that there is a $\mathbb Q$-module isomorphism $$ H_G^{*}(X, X\setminus Y) \to H_G^{*-d}(Y)$$ which turns the long exact sequence for relative cohomology into the Gysin sequence $$ \rightarrow H_G^{*-d}(Y) \xrightarrow{g} H_G^*(X) \xrightarrow h H_G^*(X\setminus Y) \rightarrow.$$ Of interest to me, is whether the map $g: H_G^{*-d}(Y) \rightarrow H_G^*(X)$ is an $H_G^*(\text{pt})$-module morphism.

As $h:H_G^*(X) \rightarrow H_G^*(X\setminus Y)$ is the dual of an inclusion, it is an $H_G^*(\text{pt})$-algebra morphism, but I cannot seem to work out any details about $g$. In particular, I feel that it is likely not an algebra map, but if anyone has any insight into whether it is a module map it would be greatly appreciated.

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This appears to be an equivariant version of the Thom isomorphism, idem the Thom isomorphism of the normal bundle of $Y_G$ in $X_G$, and the latter is merely the multiplication by the Thom class, i.e., even a morphism of $H^*_G(Y)$-modules.

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  • $\begingroup$ Are you are referring to the Thom class of the normal bundle $N$ of $Y_G$ in $X_G$? In this case, we have a map $\pi:H^{*-d}(Y_G)\rightarrow H^*(N,N-\text{zero-section})$. The Thom isomorphism is then $\varphi:H^{*-d}(Y_G)\rightarrow H^*(N,N-\text{zero-section})$ $$x\mapsto\pi(x)\cup T,$$ where $T$ is the Thom class. Is this generally what you are describing? $\endgroup$ Mar 26, 2014 at 13:37
  • $\begingroup$ Yes, exactly. Plus, use excision. $\endgroup$ Mar 26, 2014 at 13:49

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