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I am studying orbifolds with view towards Chen-Ruan cohomology. I have been struggling with inertia orbifolds but have no intuition about them at this point. I would appreciate your motivating me by answering the following questions:

  1. What is the intuition behind inertia orbifolds? How should one think of them?
  2. Why are inertia orbifolds important? What are the useful for?

Of course, given an orbifold, inertia orbifold comes for free. So it is indeed a natural mathematical object to think about. I hope to know some down-to-earth answers.

Thank you for sharing your ideas with me.

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A related question:… – David Ben-Zvi Feb 21 '13 at 19:11
up vote 16 down vote accepted

I am not sure what kind of answer you are looking for. But if you have a stack $X$, then the inertia stack $IX$ is basically the gadget parametrizing pairs $(x,\sigma)$ where $x$ is a point of $X$ and $\sigma$ is in the isotropy group at $x$ (an automorphism of $x$). Informally, the locus where you have automorphism group $G$ becomes "doubled" $|G|$ times. Example: for the stack $[\mathbb A^1/\mu_2]$ the inertia stack is $[\mathbb A^1/\mu_2] \sqcup B\mu_2$, the extra $B\mu_2$ corresponding to the origin being doubled because it has an extra automorphism.

One way to think about it is as a kind of "infinitesimal loop space", where instead of taking maps to $X$ from a circle we take maps from the homotopically equivalent object $B\mathbb Z$. This is pleasant because the inertia stack is the fibered product $X \times_{X\times X} X$, and for a topological space $X$ the homotopy fibered product $X \times_{X\times X}^h X$ is the space of free loops on $X$.

You can motivate the inertia stack through Gromov--Witten theory. If $X$ is a variety, then there is an evaluation map from the stack of stable $n$-pointed maps to $X$ to $X^n$. If $X$ is a stack, then the correct notion is that of a twisted stable map, and in this case the evaluation maps do not land on $X$ but in its inertia stack $IX$! (In fact it lands on the rigidified inertia stack, where some automorphisms have been removed from the picture, but nevermind this). So quantum cohomology of a stack is not extra structure on the cohomology ring of $X$ itself, but on the cohomology ring of $IX$.

Since it was discussed in some now deleted comments, let me say a few words about the example $X = BG$. Then $X$ has a single point (point = $\mathbb C$-point), corresponding to the trivial torsor over a point. What are the automorphisms of the trivial torsor? For every $g \in G$ we have an automorphism given by the action of $G$. But these will not correspond to distinct points of the inertia stack $IX$ in general, because these automorphisms may be isomorphic. So we should figure out what is a morphism between automorphisms of the trivial torsor. By thinking a bit one sees that morphisms are given by conjugation in the group, i.e. a morphism between the automorphisms "acting by $g$" and "acting by $g'$" is an element $h$ such that $g = hg'h^{-1}$. In other words, $IX$ is given by the stack quotient $[G/G]$, where the first $G$ is the underlying set of $G$, and the group $G$ acts on itself by conjugation. Equivalently, $IX = \coprod_{[g]} BC_G(g)$ where the disjoint union is taken over conjugacy classes in $G$ and $C_G(g)$ is the centralizer of $g$. Using unnecessarily fancy words, $IX$ is the classifying stack of the "loop groupoid" of the finite group $G$. This illustrates the informal statement earlier, that the locus with automorphism group $G$ becomes "doubled" $|G|$ times: the stack $[G/G]$ is of course $|G|$ times larger than $[pt/G]$, in a natural sense.

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great, thanks! $\phantom{.}$ – Jacob Bell Feb 21 '13 at 22:08
Thank you for the great answer, Dan! – Kim Feb 23 '13 at 10:13

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