Let $\mathcal{X}$ be a smooth Deligne-Mumford stack. Then there is an associated stack $I\mathcal{X}$, called the inertia stack of $\mathcal{X}$. 


Why is the inertia stack called "inertia"?


We can also assign a rational number to each object $(x,g)$ of $I\mathcal{X}$, called $\text{age}(x,g)$.Then we can shift the usual grading on $H^\*(I\mathcal{X})$ by age to obtain a so-called age grading on the Chen-Ruan orbifold cohomology $H^\*\_{\text{orb}}(\mathcal{X})$. As discussed [here](https://mathoverflow.net/questions/76044/intuition-behind-the-age-grading-in-quantum-cohomology-of-orbifolds), the age grading is motivated by orbifold GW theory, which is closely related to orbifold string theory.


So another related question is:
why is the age grading called "age"?



Why is the inertia stack of a smooth Deligne-Mumford stacks called inertia?
I think of the word "inertia" in "inertia stack" as representing the same idea as the "inertia" in "inertia group" (which presumably came first). This latter group typically comes up when one has a ramified Galois cover $X\rightarrow Y$ (say, of algebraic varieties over an algebraically closed field $k$ of characteristic 0). If the cover has Galois group $G$, then $Y = X/G$, and the ramification points are precisely those points $x\in X$ with nontrivial stabilizer in $G$. Such a stabilizer $G\_x := \text{Stab}\_G(x)$ is called the *inertia group* at $x$. (Of course such stabilizers have meaning for any group $G$ acting on a thing $X$, so we may as well define the inertia groups of a group action as the stabilizer subgroups.)


As mentioned above, a Galois cover $X\rightarrow Y$ identifies $Y$ with the scheme quotient $X/G$. What happens when we take the stacky quotient instead? Well in this case the map $X\rightarrow X/G$ will factor through the stacky quotient $X\rightarrow [X/G]\rightarrow X/G$, and the second map identifies $X/G$ with the coarse scheme of $[X/G]$. Now we may consider the inertia stack of $[X/G]$. By definition it is a stack $I$ over $[X/G]$ whose fiber over a geometric point $\text{Spec }k\rightarrow [X/G]$ is the automorphism group of that point. In our case, if we let
$$\pi : X\rightarrow [X/G]$$
be the quotient map, and if $x\in X$ is a geometric point, then the fiber $I\_{\pi(x)}$ is precisely the inertia group $G\_x$ of $x$, which to me seems to justify its name.


In general, I believe the following is a correct statement: If a stack $\mathcal{X}$ is locally a quotient of a scheme by a faithful group action, then the fiber of the inertia stack of $\mathcal{X}$ above a geometric point $x$ of $\mathcal{X}$ is precisely the inertia group of a local presentation of $\mathcal{X}$ at (a point above) $x$.


Here's my favorite "real world" example. Consider the covering $\mathcal{H}\rightarrow\mathcal{H}/SL(2,\mathbb{Z})$, where $\mathcal{H}$ is the upper half plane. The stacky quotient $[\mathcal{H}/SL(2,\mathbb{Z})]$ is naturally isomorphic to the moduli stack of elliptic curves, where to a point $\tau\in\mathcal{H}$, we associate the elliptic curve $E\_\tau := \mathbb{C}/\langle 1,\tau\rangle$. It's a simple calculation that for $\gamma\in SL(2,\mathbb{Z})$ and $\tau'\in\mathcal{H}$, the set of isomorphisms $Isom(E\_\tau,E\_{\tau'})$ is precisely the set of $\delta\in SL(2,\mathbb{Z})$ satisfying $\delta\tau = \tau'$. From this we see explicitly how the fibers of the inertia stack give precisely the inertia groups of the (nonfaithful) action of $SL(2,\mathbb{Z})$ on $\mathcal{H}$, and their images in $PSL(2,\mathbb{Z})$ give precisely the inertia groups of the Galois covering $\mathcal{H}\rightarrow\mathcal{H}/SL(2,\mathbb{Z})$.


