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Jul 9 at 19:54 answer added Nicholas Kuhn timeline score: 2
Jul 9 at 13:22 answer added Andy Putman timeline score: 1
Jul 9 at 10:51 answer added Jason Starr timeline score: 3
Jul 7 at 12:05 comment added Jason Starr Also, this $\mathbb{S}^1$-bundle over $BG$ is pulled back from a $\mathbb{S}^1$-bundle over $B\mathbb{S}^1$, and this space is simply connected. So the derived pushforward of the constant coefficient sheaf $\mathbb{Z}$ is (quasi-isomorphic) to the complex of constant sheaves of the cohomology of the fiber. So the hypotheses for the Thom-Gysin sequence are satisfied also on the pullback.
Jul 7 at 11:40 comment added Jason Starr The $\mathbb{S}^1$-bundle over $X_G$ is pulled back from $BG$. So the cohomology class is the pullback of the “first Chern class” of that circle bundle over $BG$.
Jul 7 at 9:54 comment added 0hliva @JasonStarr Also, if we have this Thom-Gysin sequence, the map $H_G^r(X)\to H_G^{r+2}(X)$ is given by the cup-product $t\smile (-)$ with $t\in H^2_G(X)$. But Tom-Dieck says that this map is the cup-product with some $t\in H^2(BG)$ (page 198 in Transformation Groups).
Jul 7 at 9:50 comment added 0hliva @JasonStarr so the Thom-Gysin sequence associated to the bundle $S^1\to X\times_G S^1\to X_G$? If yes, why is the action of $\pi_1X_G$ on $X\times_G S^1$ trivial so that we can apply the Thom-Gysin sequence?
Jul 6 at 20:59 comment added Jason Starr Actually it is the Thom-Gysin Sequence.
Jul 6 at 18:31 comment added Jason Starr That looks like the Wang Sequence.
Jul 6 at 14:33 history edited YCor CC BY-SA 4.0
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Jul 6 at 13:42
S Jul 6 at 12:53 history asked 0hliva CC BY-SA 4.0