What is the intuition behind the inertia orbifold (or stack)? 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:


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*What is the intuition behind inertia orbifolds? How should one think of them? 

*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.  
 A: 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. 
