 Good house. Okay, so maybe I'll continue so. So again, as I said, our goal is to so we have, we have this, we have until so. We deserve continuing we have until I don't know I guess I'll temporarily call it was a name until the. I'm going to move to pie to end. And so I want to let until the of X lambda, the inverse image of it's lambda so now it's so now so by the way when I talked about before with this generally spare as mission isn't is in all types and that's not this is this is not just in. This is actually in all works in all types. But I mean, you know some of what I'm doing now could be extended to other types to I believe but but as far as you know the comment or it's isn't has worked out yet so I'm going to just be in type A for the rest of this talks, you know, so most of the rest of this talk. And so I want to let until the next land, the inverse image of X lambda and wanted to find describe our orbit paving of until the. But I think I guess from the tour picture you can sort of believe why you might need an orbit paving because you know you sort of are naturally putting in this variety which is CR model finding group so you may end up meeting something like that so. Let me tell you what the situation is here so. So, let me make a definition here. So a tableau Sigma is divisible by D. If the entries of Sigma appear in blocks in horizontal blocks of D consecutive integers. D divides Sigma. So for, for example, I could have a tableau that looks something like this. So this is a tableau, which is diesel why to because every, because you can see it's divided up into blocks of two of pairs of integers which are, you know, consecutive so. This tableau is a Y2 but if I, you know, if I interchange the, the two and the three, it won't be divisible Y2 anymore because I've gotten rid of the consecutive integer condition. So I hope it's clear what I mean by divisible. This is divisible by two. So then, so then we can call what, what D Sigma and be the largest, be the largest D dividing Sigma, because obviously the tableau is a little by four it's a little by two as well so we may as well talk about the largest divisor of the tableau. And that for this to happen of course all the parts of build does it have to be increasing consecutive or could it be decreasing consecutive increasing. At least at least it will see an R in for me I mean one could make them decrease I guess but but I'm only interested in very strict tableau anyway so. So they're going to be only concerned about them being increasing but I suppose you could make them. So, but for this to exist all parts of lambda of course must then be your 5634 isn't very strict. Yeah, I guess you're right this is an honest you're okay well I anyway I do want them to be increasing you're right though you're right I mean this isn't a very strict one. Well, I guess I was only thinking of them being increasing let me put it that way there's no reason they would have to be actually all increasing. So maybe we'll cross that off. So, so then, let me so then here's the theorem. And this is, this is with Martha pre cup, Amber Russell. So just try this is unfortunately off the page here but so so we have just to this location. You know we have I'll just write it down here again so we have until the going down to Ada with until they're here. And so that's the math and so then what's so let. Let's see Sigma be, you know in until the of X lambda be the cell of the paving corresponding to to the restrict, then. The inverse of C Sigma. It's a district union of I'll call them C still the Sigma one. See till the Sigma, the Sigma, where each. See till the Sigma I is isomorphic to CR mod this finite abelian group. And moreover, is he transitively permutes. And so what you know how to how would one prove something like this so the proof. So the proof. I mean, you have to look at the way the paving of the springer fiber is constructed so you have to look at the construction of the paving until the of X lambda, and then basically you have to reduce the reduce the theorem is basically you're after all you're starting of of until the of X lambda so you're starting off with that and then you reduce the theorem to result about the map of torque varieties each of the. Add. Well so you so you actually and then you can actually just, you can work in the land of torque varieties and you actually you do want to bring that right until the here as well and they'll put until the here just for that occurs in the proof so. But maybe I won't say anything more about that but but it ultimately boils down you can imagine that it could be something like this because it ultimately boils down to a result about toward varieties. But maybe I won't say anything else about that right now. But let me. Let me. So using this one. So I, one can describe the Poincare polynomial, Pt of until the land and let me. So to get the nicest such description, it's want to bring in the center so bring in the center. So. So let's, so I'll bring in. So the center is G is just isomorphic to Z in here. And so let's let Z hat be the characters of Z. I'll call me the characters is the energy is chi zero. The chi and minus one. So I mean if I think of Z and is being an influence of unity, then the characters are these were chi I of a is a root of unity I just take a raise it to the power. So then, because the accident things the X, I can decompose the co homology of them to the X lambda, as a direct sum of its chi J isotropic components. And I guess I should take co homology with with complex coefficients here. So let me do that with a comment. And then so what I can do is I can so, so let. And T chi of J of M tilde of X lambda be the dimension of the of the chi J isotropic component of each to I have until the X lambda times T to the I. So then, then I just have the PT of M tilde. And so then the theorem is just the sum as J goes from goes from zero to N minus one of P of T chi J of until the X lambda. And so then the theorem is that, which is again from this is that is in is that let E in. So here is actually quite elementary. It's just an elementary, it's an elementary consequence of this previous stuff it's not it's, it's, you know, but, and I'll let, I'll let D be equal to N over the GCD of any. The concrete polynomial of T, the chi E isotopic component of this concrete polynomial. It actually just it's actually, it's a power of T to some power here times the Poincare polynomial of just the springer fiber, where I take the partition and I divide it by D. And this power here that the exponent is actually D minus one times the dimension of B of X, B of X of lambda. So what's happening here is that is that in this in this co homology you're actually seeing these co homology of the smaller springer fibers sitting is the isotopic components in there. So I'm out of time but the point is that maybe not the point but it in type A, we proved earlier that this that we know that this variety until the map until it's M is closely connected to lose things, generalized springer correspondence. And if you make that connection, this result about this Poincare polynomials can actually is actually sort of in the literature in a way I mean it's it's it's known to experts I guess that, because you can compute some of these Poincare polynomials via an algorithm called the algorithm and it's known that the part of the generalized springer correspondence that this part is picking up does look like, you know, representations of smaller symmetric groups, you know, we're as a symmetric in terms of size and over D. So in a sense this is, this is this, if you make the connection of this variety with the generalized springer correspondence than this result here could be deduced from this generalized springer correspondence fact but from another point of view, this this this result doesn't rely at all on the generalized springer correspondence and it's kind of a very geometric manifestation of this fact that is sort of the this takes work in this much more indirect way in this example so, but I think I'm out of time so maybe I will just stop here so thank you. Okay, thanks very much let's thank the speaker.