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Given a representable, surjective morphism of Artin stacks $\phi:\mathcal{F}\to\mathcal{G}$, is it true (like it happens for schemes) that $\dim\mathcal{G}\leq\dim\mathcal{F}$?

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By surjective, do you mean that the functors making up the pseudonatural transformation $\phi_X:\mathcal{F}(X)\to \mathcal{G}(X)$ for each affine scheme $X$ are full and essentially surjective? –  Harry Gindi Apr 19 '11 at 19:04
I mean what is explained in the answer –  ginevra86 Apr 20 '11 at 8:41

2 Answers 2

up vote 4 down vote accepted

No. $B\mathbb{G}_m\to\mathrm{pt}$ is surjective with source of dimension -1.

As commented below this is nonsense, since this is not representable.

Let me try to make amends for my flip response to the question. Of course what I am about to write could be equally stupid. However, $\phi$ surjective means that the induced map $\mathcal{F}\times_{\mathcal{G}}G\to G$ is surjective, where $G\to \mathcal{G}$ is smooth surjective with $G$ a scheme. Suppose that the dimension of $G$ is $n$ and that of the scheme $\mathcal{F}\times_{\mathcal{G}}G$ is $n+p$ where $p\geq 0$. Then the dimension of $\mathcal{G}$ is $n-q$ where $q$ is the relative dimension of the smooth surjective morphism $G\times_{\mathcal{G}}G\to G$. Now $\mathcal{F}\times_{\mathcal{G}}G\to\mathcal{F}$ is smooth and surjective. Furthermore: $$(\mathcal{F}\times_{\mathcal{G}}G)\times_\mathcal{F}(\mathcal{F}\times_{\mathcal{G}}G)\simeq (\mathcal{F}\times_{\mathcal{G}}G)\times_G(G\times_\mathcal{G}G)$$ and hence is of relative dimension $q$ over $(\mathcal{F}\times_{\mathcal{G}}G)$. Therefore $$\mathrm{dim}(\mathcal{F})= n+p-q\geq n-q=\mathrm{dim}(\mathcal{G}).$$

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Your map is not representable. –  Mike Skirvin Apr 19 '11 at 18:58
Good point, I wasn't paying attention. –  user10849 Apr 19 '11 at 19:36
I think your argument completes the one I was trying to give, although I think I still need to read your answer more carefully. The key point I seemed to be missing is in your first formula, which indicates that the automorphisms over $\mathcal{F}$ and $\mathcal{G}$ have the same dimension. –  Mike Skirvin Apr 19 '11 at 21:09
The argument does not assert that the \emph{automorphisms} over $\mathcal F$ and $\mathcal G$ have the same dimension. You should think of the fiber of $G\times_{\mathcal{G}}G\to G$ over a point $g$ of $G$ as the morphisms with source (or target) $g$. That of course combines that dimension of the automorphisms with the dimension of the objects -- and is better behaved, since the dimension of the space of automorphisms can (and does) jump. –  user10849 Apr 19 '11 at 22:24
Thank you! Your answer and your comment are very clear! –  ginevra86 Apr 20 '11 at 8:46

This is not a complete answer, but should hopefully be a start. Given a point $x \in \mathcal{F}$, the dimension of $\mathcal{F}$ at $x$ is given by picking an atlas $X$ of $\mathcal{F}$, and then computing \begin{equation} dim_x(\mathcal{F}) = dim_x(X) - dim(Aut_{\mathcal{F}}(x)). \end{equation}

The dimension of $\mathcal{F}$ is then the supremum of its dimension at all its points. Given surjective, representable $\phi:\mathcal{F} \to \mathcal{G}$, we obtain a surjective map of atlases $X \to Y$. Therefore $dim_x(X) \geq dim_{\phi(x)}(Y)$ for all points $x$.

So, we're reduced to comparing $Aut_{\mathcal{F}}(x)$ to $Aut_{\mathcal{G}}(\phi(x))$. There is necessarily a surjection $Aut_{\mathcal{F}}(x) \to Aut_{\mathcal{G}}(\phi(x))$, so there is a question of how much larger the automorphisms of $x$ in $\mathcal{F}$ are compared to those of $\phi(x)$ in $\mathcal{G}$. That is, we would need to verify that \begin{equation} dim_x(X) - dim_{\phi(x)}(Y) \geq dim(Aut_{\mathcal{F}}(x)) - dim(Aut_{\mathcal{G}}(\phi(x))) \end{equation}

holds for all points $x$. I feel like this should be true just because my possibly faulty intuition says that the relative dimension of the automorphisms of a surjective, representable map won't exceed the relative dimension of the atlases. However, I don't see how to prove it right now.

Does anyone know how to prove this, or provide a counterexample?

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