Let $X$ be a smooth, projective complex variety and $j \colon D \hookrightarrow X$ a smooth divisor. Then we have a Gysin morphism in singular cohomology

$$ j_\ast \colon H^{\bullet}(D) \to H^{\bullet+2}(X) $$

Now assume that $X$ is acted upon by a finite group $G$ and that $D$ is stable under this action. Then we get actions $g^\ast$ on the cohomology of $D$ and $X$.

Is is true that $j_\ast g^\ast=g^\ast j_\ast$ for any $g$ in $G$?

This looks like some projection formula, but I'm unable to prove it.

  • $\begingroup$ The answer is yes if $G$ acts by diffeomorphisms, since then each $g$ is transverse to $j$. I don't have these books to hand to check, but I'm sure you'll find a proof in either E. Dyer's "Cohomology Theories" or W. Fulton's "Intersection Theory" $\endgroup$
    – Mark Grant
    Jul 4, 2015 at 15:32
  • $\begingroup$ Thanks! I will have a look at Fulton's book. Can you explain what "g is transverse to j" means? $\endgroup$
    – ter
    Jul 4, 2015 at 15:40
  • $\begingroup$ Smooth maps $g: M\to X$ and $j: N\to X$ are transverse if whenever $g(m)=j(n)=x$ then the images of the differential of $g$ at $m$ and the differential of $j$ at $n$ together span the tangent space at $x$ of $X$. $\endgroup$
    – Mark Grant
    Jul 4, 2015 at 15:50
  • $\begingroup$ Alternatively, you can argue that both the restriction $j^*$ and Poincare duality isomorphisms are equivariant (by naturality). $\endgroup$ Jul 4, 2015 at 18:35
  • $\begingroup$ Oh thanks, Donu, that looks simpler. Could you please develop a little bit? $\endgroup$
    – ter
    Jul 4, 2015 at 18:36

1 Answer 1


Let me expand my comment slightly into an answer. To simplify matters, suppose that the coefficients are $\mathbb{Q}$ or $\mathbb{C}$. The Gysin map is Poincaré dual to the restriction map $j^*$. In more detail, we have isomorphisms $H^i(X)\cong H^{2n-i}(X)^\vee$ and $H^i(D)\cong H^{2n-2-i}(D)^\vee$, where $n=\dim X$, given by Poincaré pairings $(\alpha,\beta) =\int_X\alpha\cup\beta$ etc. Under these isomorphism $j_*$ corresponds to the dual to restriction $j^*$. This is $G$-equivariant by functoriality. I'll let you check that $(g^*\alpha,g^*\beta)=(\alpha,\beta)$, and this implies $G$-equivariance of the duality isomorphisms, and that's all you need.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.