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Let $\mathscr{X}$ be a smooth DM-stack with projective coarse moduli space. I am interested in the orbifold cohomology ring $H^\mathrm{orb}(\mathscr{X})$, as defined by Chen-Ruan (for orbifolds) and Abramovich-Vistoli (algebraically). Additively, this is nothing but the ordinary cohomology of the inertia stack, but it has a funny multiplication and a funny grading.

The multiplication I think I can motivate from the point of view of GW theory: if you think (somewhat perversely) of the ordinary multiplication in the cohomology ring of a variety $X$ as arising from 3-pointed genus zero stable maps into $X$, then the multiplication in $H^\mathrm{orb}(\mathscr{X})$ should come from 3-pointed stable orbicurve maps, which it does.

The grading is the part that is a mystery to me. Or rather, I understand that it has the nice properties that the grading of the untwisted sector is the same as the usual one, and that the multiplication preserves the grading.

But I have zero intuition for why the multiplication preserves the grading -- I can read through the proofs, but it looks like magic. Is there any a priori reason to think that this is the "right" grading? I am vaguely aware that physicists have been considering the age grading for some time now, and surely they must have had some reason for this. I also know that the age grading shows up in the Riemann-Roch formula on an orbicurve -- can the age grading and/or its compatibility with the quantum product be motivated from this fact?

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It is my understanding that the original motivation did come from GW theory: Chen and Ruan came across the orbifold cohomology ring, and its product, as a byproduct and first step of defining the orbifold quantum cohomology. The strongest motivation I have for the grading also comes from this perspective, and maybe we have to start with the question of why Chen and Ruan were trying to define orbifold GW theory to begin with.

The fairy tale story I have in my head is that string theorists found that they could do string theory on mildly singular spaces. Often, rather than dealing with singularities, we instead get rid if it, by either a) resolving the singularity, or b) deforming the equations to something smooth. If our singular space is actually a smooth as an orbifold, orbifold cohomology gives us a third option, c).

Now, the physicists also had intuition that, in nice cases and if we look at it the right way, all three approaches should essentially give us the same thing. So, for instance, passing from a) to b) directly can go by the name "conifold transition".

To get back to the age grading: a) and c) are related by what has come to be known as the Crepant Resolution conjecture: if $f:Y\to X$ is a crepant resolution of an orbifold $X$ (i.e. $f$ is a resolution and $f^*(K_X)=K_Y$), then $X$ and $Y$ should have "the same" quantum cohomology. In particular, the base graded vector spaces of $X$ and $Y$ should be the same.

A strong motivation for the age grading of quantumn cohomology is that it makes this true: if $X$ is an orbifold that has a crepant resolution, then the graded vectorspace $H^{orb}(X)$ is isomorphic to the graded vector space $H^*(Y)$, for any crepant resolution $Y$. This statement about the graded vector space was proved using motivic integration by Yasuda, and independently by Lupercio and Poddar, and provides a lot of evidence that the age shifting is the "right" shifting. I should note that the stronger statement about the orbifold quantumn cohomlogy rings being "the same" is still open, and in the general case requires some work to even state correctly, as in this answer by Jim Bryan.

I feel like a really good answer to your very last question should be possible and not that hard, but I don't have it at my fingertips now. So I'll just give a short "yes": the obstruction bundles used to define the orbifold cup product are exactly cohomology spaces of bundles on orbifolds, and so the RR formula is going to pop up there. If you're okay with the age grading showing up in the RR formula, then from the definition of the quantum product it follows that the age grading of $H^{orb}$ has to be what it is. I feel like I'm glib and just sort of restating your question -- I'm trying to say I think you're right.

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  • $\begingroup$ Thanks a lot for the answer. I'm going to reveal my ignorance here but I am slightly confused about the statement that $H^{\mathrm{orb}}(X)$ is supposed to coincide with $H^\ast(Y)$ as graded vector spaces -- isn't the former $\mathbf Q$-graded and the latter $\mathbf Z$-graded? $\endgroup$
    – Dan Petersen
    Commented Sep 22, 2011 at 7:45
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    $\begingroup$ You are correct. The point is that $X$ will not have a crepant resolution unless $H^{orb}(X)$ is actually integrally graded. In fact, for $X$ to have a crepant resolution, all the degree shifting numbers should be even. To see that this is reasonable, note that over a point $x$ in $X$, the isotropy group $G_x$ will act on $K_X$. To be crepant, we want $f^*(K_X)=K_Y$ -- since $K_Y$ has no orbifold structure, it seems that we should have that $K_X$ is the trivial representation of $G_x$, which means that $G_x$ acts on $T_xX$ with determinant 1. $\endgroup$
    – Paul Johnson
    Commented Sep 22, 2011 at 9:50
  • $\begingroup$ It is very nice that the age grading turns out to be compatible with the crepant resolution conjecture, but I don't think this answers the question. There must be some geometric sense in which the twisted sectors of an orbifold have fractional dimension. $\endgroup$
    – euklid345
    Commented Sep 25, 2011 at 18:13
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To my understanding Chern-Ruan cohomologies was an attempt to build orbifolded LG theory. The goal that was eventually achieved in Fan-Jarvis-Ruan theory.

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  • $\begingroup$ Hey Alex! Nice to see you here. $\endgroup$
    – Dan Petersen
    Commented Aug 29, 2012 at 13:44

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