2
$\begingroup$

Let $T$ be the complex torus acting on a complex connected algebraic variety $X$ and let $p \colon X\rightarrow Y$ be a good quotient for this action. For any $y\in Y$ we have a sequence $p^{-1}(y) \rightarrow X \rightarrow Y$ which leads to a sequence $\pi (p^{-1}(y))\rightarrow \pi (X)\rightarrow \pi (Y)$ of the corresponding fundamental groups.

Is this sequence an exact sequence?

What i have tried is the following: its easy to see that $im (\pi (p^{-1}(y)) \rightarrow \pi (X))\subset \ker ( \pi (X)\rightarrow \pi (Y)) $ and being fibers of $p$ connected then $\pi(X)\rightarrow \pi(Y)$ is surjective. I've read on a paper that the sequence is in fact exact on $\pi (X)$ but i can't see why $\pi( p^{-1}(y)) \rightarrow \pi(X)$ is injective. sorry if I'm not seeing something trivial.

Thanks in advance.

$\endgroup$
2
  • $\begingroup$ I am not sure what you call a good quotient; I assume this implies that $p$ is smooth and proper, hence topologically a locally trivial fibration; then the result is completely standard (homotopy exact sequence, read any book on algebraic topology). $\endgroup$
    – abx
    Feb 27, 2014 at 15:55
  • $\begingroup$ My definition of good quotient is the following: $p\colon X\rightarrow Y$ is affine and $T$-invariant and the pull-back $p^* \colon \mathcal{O}_Y \rightarrow (p_*\mathcal{O}_X)^T$ is an isomorphism. $\endgroup$ Feb 27, 2014 at 15:59

1 Answer 1

2
$\begingroup$

I assume that in your case, $p^{-1}(y)\rightarrow X\rightarrow Y$ is a fibration. Therefore, there is a long-exact sequence of homotopy groups $$\ldots\rightarrow\pi_n(p^{-1}(y))\rightarrow\pi_n(X)\rightarrow\pi_n(Y)\rightarrow\pi_{n-1}(p^{-1}(y))\rightarrow\ldots.$$ In particular, we have $$\ldots\rightarrow \pi_2(Y)\rightarrow\pi_1(p^{-1}(y))\rightarrow\pi_1(X)\rightarrow\pi_1(Y)\rightarrow\pi_0(p^{-1}(y))\rightarrow\ldots.$$ Since you know $p^{-1}(y)$ to be connected $\pi_0(p^{-1}(y))=0$. Do you happen to know that $\pi_2(Y)=0$?

$\endgroup$
1
  • $\begingroup$ thanks! one case in which i was interested has this condition! $\endgroup$ Feb 27, 2014 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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