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Let $k$ be an algebraically closed field and $G_m$ an algebraic one-dimensional torus. Let $X$ be a smooth irreducible variety over $k$, $Y$ an affine scheme of finite type over $k$ and $f\colon X\rightarrow Y$ a proper morphism.

Assume that there exist actions of $G_m$ on $X$ and $Y$ such that $f$ is $G_m$-equivariant. Moreover, the fixed point locus $Y^{G_m}$ of $Y$ consists of only one point $y.$

Edit (after Allen's answer): Moreover, we assume that for any point $x\in X$ there exists $\lim_{t\rightarrow 0} t\cdot x.$

Question: Are $X$ and $f^{-1}(y)$ homotopy equivalent?

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  • $\begingroup$ What do you mean by "homotopically equivalent"? $\endgroup$
    – Angelo
    Commented Apr 22, 2012 at 11:49
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    $\begingroup$ I don't know what is meant by "homotopically equivalent" either, but my guess is that the identity map of $\mathbb{G}_m \amalg \mathbb{A}^1$, with the action on each factor by multiplication, is a counterexample. $\endgroup$
    – Tyler Lawson
    Commented Apr 22, 2012 at 12:49
  • $\begingroup$ To Tyler: I assume that Francesco wants $X$ to be irreducible. $\endgroup$
    – Angelo
    Commented Apr 22, 2012 at 15:26
  • $\begingroup$ I edited my question. To Tyler: $X$ is irreducible. $\endgroup$
    – user17778
    Commented Apr 23, 2012 at 7:38

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Let $Y = P^2 \setminus ${two of its fixed points}, and $X = Y$ with $f$ the identity. Then the fiber over $y$ is a point, but $Y$ is not contractible, I don't think.

I'm pretty sure your "has one fixed point" isn't the condition you want, but rather, "every $\lim_{z\to 0} z\cdot x$ exists and is $y$".

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  • $\begingroup$ This $Y$ is not affine. In any case you are probably right about the condition. $\endgroup$
    – Angelo
    Commented Apr 24, 2012 at 4:47
  • $\begingroup$ Oops, I didn't notice the condition $Y$ affine. I'll think some more. $\endgroup$
    – Allen Knutson
    Commented Apr 26, 2012 at 1:38

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