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, 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"?


1 Answer 1


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 $\mathcal{H}\rightarrow\mathcal{H}/SL(2,\mathbb{Z})$.

  • $\begingroup$ Thanks for the answer and sharing with us this interesting example. I thought "inertia" has something to do with the "inertia" in physics: the resistance to change; apparently that's not the case. I was not familiar with the notion of "inertia group", which I think indeed justifies the name "inertia stack" from your explanation. But why do people call the stabilizer subgroup of a group action "inertia group"? Now I recall a little bit the "inertia group" from algebraic number theory. I wouldn't be surprised if the person who coined up the term "inertia group" got inspiration from physics. $\endgroup$ Oct 9, 2019 at 19:02
  • $\begingroup$ The inertia group from number theory is essentially the same as the one described above, except one needs to be more careful about exactly what a fixed point is when the residue field is not algebraically closed. $\endgroup$
    – Will Chen
    Oct 9, 2019 at 19:48
  • $\begingroup$ As for the use of inertia in physics, I think of points with large inertia groups as carrying additional "weight" compared to those with smaller inertia. Perhaps a picture I have in mind is that near a point with nontrivial inertia, the group action moves things around less and less, as if there's additional "physical" mass near that point. Maybe the person who first used the word inertia in the context of fixed points had a similar picture in mind. $\endgroup$
    – Will Chen
    Oct 9, 2019 at 19:49
  • $\begingroup$ This is very intuitive. So attaching an inertia group to a point is like assigning a mass to it. I have another question regarding the example of moduli stack of elliptic curves. Why do you take images in $PSL(2,\mathbb{Z})$ when you consider inertia groups of the Galois covering $\mathcal{H} \to \mathcal{H}/SL(2,\mathbb{Z})$? Every generic point has an inertia group $\{I,-I\} \cong \mu_2$, right? By a generic point, I mean a point not in the orbits of $i$ or $e^{2\pi i/3}$. $\endgroup$ Oct 10, 2019 at 1:17
  • $\begingroup$ @YuhangChen That's just a case where the terminology breaks down. When I said "the inertia groups of the Galois covering $\mathcal{H}\rightarrow\mathcal{H}/SL(2,\mathbb{Z})$, I meant the inertia groups in the traditional sense, which are subgroups of $PSL(2,\mathbb{Z})$ because the Galois group of that covering is $PSL(2,\mathbb{Z})$. In general the traditional inertia groups are a measure of how much the generalized/stacky inertia groups jump. $\endgroup$
    – Will Chen
    Oct 10, 2019 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.