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This is a very basic question about how definitions in homology carry over to the easiest example of stacks. Let $G$ be a finite cyclic group. Consider the classifying stack $\mathcal{B}G$. This has a (nonrepresentable) morphism to a point: $st:\mathcal{B}G\to P$. What do people mean when they say the "fundamental class" $[\mathcal{B}G]$ of $\mathcal{B}G$? (I don't know how homology carries over to stacks.) For example, is it true that $st_*[\mathcal{B}G]=[P]$? Or that $st^*[P]=[\mathcal{B}G]$? Are these equivalent?

My guess: It seems to me that the map $st$ should be considered as "degree" $1/|G|$. Because of this it seems like we should have $st^*[P]=|G|[\mathcal{B}G]$. I'm not sure what this would say about $st_*[P]$.


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up vote 1 down vote accepted

I'm going to write $\text{Spec}(k)$ in stead of $P$. I am going to write $\mathcal{X} = [\text{Spec}(k)/G]$ for the classifying stack of $G$ over $\text{Spec}(k)$. Here $G$ is any finite group. Then we have the morphisms $\text{Spec}(k) \to \mathcal{X} \to \text{Spec}(k)$ whose composition is the identity. The first morphism comes from the trivial torsor over $G$ and is a representable, surjective, finite \'etale morphism of degree $|G|$. Hence it seems natural to guess that the second morphism (which is not representable, but is proper, surjective, and \'etale) has degree $1/|G|$.

By analogy with topology, for any reasonable theory of fundamental classes the first morphism would map the fundamental class of $\text{Spec}(k)$ to $|G|$ times the fundamental class of $\mathcal{X}$ and hence the second morphism would map the fundamental class of $\mathcal{X}$ to $1/|G|$ times the fundamental class of $\text{Spec}(k)$. In other words, the "degree" of $\mathcal{X}$ over $k$ is $1/|G|$.

This is indeed what happens in intersection theory for algebraic stacks, \'a la Vistoli, etc, etc.

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