 Cool, thank you. So quick thank you to the organisers for having me speak and everyone for coming out on a Saturday. It's quite late where I am, but I imagine it's quite early for some of you as well, so without further ado. I'm going to be talking about the Gromovluss and Rosenberg conjecture on positive scalar curvature. So scalar curvature, if we have a manifold m with a Romanian metric g, the scalar curvature essentially is kind of a local property, which is going to compare the volume of balls in this manifold to the volume of balls in Euclidean n-space. So precisely we're going to take the trace of the Ricci tensor, but if you don't know what this is, just kind of think of this idea of we're looking at how curved the space is. So it's measuring at every single point the curvature basically. So two easy examples, the Euclidean space has curvature equal to zero and the sphere of radius r, the n-sphere of radius r has curvature equal to n-1 over r squared. So we can normalise this so that we can just say the sphere has constant scalar curvature equal to one. And notice here that the scalar curvature is constant. In general this won't be true to be a given value at a specific point, but we can kind of ask is it always greater than zero or always less than zero or always equal to zero. So for instance the Euclidean space there has it always equal to zero. And this kind of gives a very natural question, which is when does a manifold admit a metric of positive scalar curvature? And we're going to denote the curvature by kappa. And we're going to think of this as kind of like an integral over the manifold and it's the sum of all the curvature. Okay, so in dimension two this is actually completely solved. So if we go all the way back to gaspone, the gaspone theorem, if m is a compact surface then the scalar curvature is equal to 4 pi times the Euler characteristic. And this is possibly quite surprising because the Euler characteristic is a topological feature of a space rather than a geometric feature because scalar curvature is definitely something out of differential geometry. And yet here the obstruction to it lies very deeply rooted in topology. I mean the Euler characteristic is a alternating count of cells. So this is quite a surprising fact. And so here you can see immediately if it's a hyperbolic surface, so one of genus at least two, then this value is negative because the Euler characteristic is negative. And if you have a torus then this value is zero so again you can't have positive scalar curvature. As we saw this fear does admit a metric of constant positive scalar curvature and here that invariant becomes positive. So maybe we can try and generalize this idea of finding topological obstruction through a geometric problem and you might ask what about higher dimensions. So before we can do that we kind of have to go take a trip through functional analysis very quickly. Don't worry too much about the details but essentially we're going to define a couple of things. We're going to get some machinery out and then the machinery is going to get increasingly complicated but the goal is still the same to find a topological obstruction. So if M is a closed spin manifold and X is a spinal bundle we want to consider the space of L2 sections from M to X. So this is the space of functions from M to X which are a section to the bundle and they're square integrable. Don't worry too much about what the subject is but it's some kind of space which functional analysis like and it's quite easy to get a handle on. We're going to let D be the Dirac operator on this space, this alpha 2 space. Again if you don't know what this is don't worry too much but a very important fact is that the square of it is equal to the Laplacian plus kappa over 4 where this is our curvature. Now if the Laplacian is always greater than or equal to 0. So essentially we found some object, we've taken some object and we've got some operator which is kind of picking up curvature. At the moment this is still very much in the realm of geometry and we're not really looking at this topology but we're going to get to how they connect. So if kappa is greater than 0 this D squared is now invertible so D is invertible. So we can define an index which is going to be the dimension of the kernel of D minus the dimension of the kernel and of course if D is invertible then this is 0 because the dimension of the kernel and the dimension of the kernel are 0. So the corollary of this originally due to Lick-Narrow-It is that if the index is non-zero then M cannot admit a metric with a positive scalar curvature. But we're still in the realm of functional analysis here, we're not in the realm of topology yet. So quite a famous result due to Atea and Singer is the Atea-Singer index theorem which states that if M is a closed spin 4k manifold then the index is equal to this A hat genus of the manifold. Now this is some quite high-powered topological gadget which you get out of the characteristic classes in the corollary of M. But the point is is it's something you can compute. If you're given like a CW structure on a manifold you can just sit there and you can in principle you can work out what this thing is. You can just sit down and compute it and then you know whether you're going to get a positive scalar curvature or not. Admittedly it's a lot harder to compute than the other characteristic but it is computable. So that's kind of what we want. We want something which is very canonical to the topology of the space and not to the geometry. Okay so how do we get more general? Because we know that this was only in dimensions 4k. So Rosenberg introduced something called the Alfven variant and this one's a bit more a bit more tricky to digest. But essentially we're going to take a closed spin M manifold and G a discrete group. We're going to take any map from the manifold to BG the classifying space and then if M admits a metric or positive scalar curvature the class which somehow ends up in this real k group of the group C star algebra is going to vanish. Now this class begins life in the spin borders and group and it factors through a bunch of different maps. But the key point is there is some class which has to vanish. So to break this down a bit more nicely we have a class which lives in some homologically invariant of a group and it vanishes whenever there's a metric or positive scalar curvature. So it is an obstruction and in fact it coincides with the index of that direct operator we had before. Okay so this is great. When is this enough? Oh and actually one more thing to properly get back into the realm of topology because the k north k of a real C star algebra is still kind of an analytical thing. Bonk runs identifies this with this interesting looking thing which is essentially a kind of the finite subgroup version of EG. And we're going to take the aquavariant k theory and this is something which you can't compute. It's just the difficulty becomes more because the invariance more complicated but it's something which again you can sit down you can work this out for a given group. You can work out what the aquavariant k o of e e under bar e fin G is. This has a simpler statement though of G is torsion free. Here it just says that the real k theory of BG is isomorphic with a natural map to the real k theory of the reduced group C star algebra. Now again this is kind of there's a lot going on here but essentially we have something which lives in a topological invariant a class and we just want to know whether that class is zero or not. So it's it's something we can work out. Just the difficulty increases. The bonk runs conjecture is known to hold for a lot of groups. So this is kind of any group which you might consider the set in for positive scalar curvature. It's probably known that the bonk runs conjecture is satisfied. So for example hyperbolic groups relative to hyperbolic groups cat zero cubicle groups finite groups. These are all known to satisfy this. So most kind of natural groups you would consider. Okay so how can we turn this into something which is more general? The idea is that we take a closed spin and manifold again and we're going to only consider maps which induce an identity on the fundamental group of the manifold. And then the conjecture is that you only need to consider these maps. So if the class vanishes now then G has a metric or positive scalar curvature and if the class doesn't vanish then it's not. So it's kind of like the necessary and sufficient condition. So hopefully the idea is that this is the only topological obstruction we need to consider. So when has this conjecture been verified? Well it's known for all simply connected. It's known when the fundamental group of M is finite with periodic cohomology. So this is kind of things like cyclic groups, generalized quaternion groups, SL2P, so quite usually finite groups but not a huge amount. And also it's known when G is torsion free and the dimension of G is at least is less than nine. And then finally the only infinite groups which have torsion where it seems to be known for are fuchting groups. And there is also a counter example due to Thomas Schick with Pi-1 equal to z to the four direct sums z3. So it's not it's not satisfied by everything. You can you can find manifolds where this isn't it doesn't characterize positive scalar curvature. It is just an obstruction. So if I had more time I would talk about Thomas Schick's example because it's really nice but unfortunately I don't. So my little add-on to this is if G is either PSL2Z1 over P for P congruent to 11, 112 or a lattice in PSL2C with all finite subgroup cyclic then G also satisfies this conjecture. So if you take a manifold with one of these fundamental groups dimension at least five then you can just look at this K real K group and you can look at the image of this class through a bunch of maps which you can work out quite explicitly. And then if this class vanishes then you know that these groups admit a metric of positive scalar curvature. So for a really simple example if you take any manifold with dimension congruent to three or five mod eight then the manifold will admit metric positive scalar curvature if it's fundamental group is PSL2Z1 over P. So that's kind of a quick application of this theorem. So I mean my immediate question is why only P mod 12? Why is P only congruent to 11 mod 12? So I'll say a few words about that and then we'll wrap up. So a group G has property M if every finite subgroup is contained in a unique maximal finite subgroup and NM if every maximal finite subgroup is self-normalizing. So these are kind of group theoretic properties and they turn out to be quite important in the proof. So PSL2Z1 over P only satisfies M and NM when P is congruent to 11 mod 12. Essentially this comes from the way these groups can be realized as fundamental groups of graphs and groups. If you're not familiar with that then just kind of black box this along with everything else I've said. But the idea then is that when G satisfies these two properties M and NM there's a spectral sequence due to Davis and Luke which is quite useful. You can compute stuff very explicitly with it and it kind of collapses quite nicely. So the proof then we get a diagram which gets quite complicated and this diagram is directly follows from having those properties N and NM. But we're not going to worry too much about where this comes from because you could spend a long time discussing this diagram. But I mentioned a couple of things. So the map P is the connective cover. If you know about connective cover as a spectra and the sum is over finite congenital classes. But again this is kind of technical so I don't want to dwell on this too much. Oh and X is the this classifying space for proper actions which we saw earlier but we're going to quotient out by the G action. So in the case of PSR2Z1 over P this actually turns out to be a wedge of spheres which is I mean it's a very nice space it's quite easy to see that P is going to be a niceomorphism once you know about connective covers. Okay so this kind of hints at a more general theorem which is if G satisfies M and the bomb cones conjecture and all finite subgroups of the cyclic. If X is this space E thin G over G then P and this connective map is a isomorphism this connective cover sorry for all n greater than equals to five then G satisfies the GLR conjecture. So this gives a lot of examples of groups with torsion which now satisfy this and it's perhaps worth interested to note that if we go all the way back to Thomas Schick's example with z to the four plus that direct sum z mod three the whole group normalizes z mod three so it doesn't satisfy M yeah and I'm sorry it doesn't satisfy NM. Now whether this is actually the obstruction to this I don't want to say but clearly the example where it's not known this is a it doesn't satisfy NM so maybe that's interesting I'm not sure. Okay so my open question is does PSR2 z 1 over P satisfy the conjecture for other primes and I have no idea whether it does or not but I would love to know the answer so thank you for listening. Oh sorry I forgot to say that part I just started clapping. Any questions for Sam? I have a quick question. So you need like a like a passing comment about something about fundamental groups of graphs of groups could you just say a sentence or two more about that? Yeah sure so PSR2 z 1 over P is an amalgamated free product of PSR2 z over a principal congruent subgroup so apart from for P equals two and three these are just three groups so then it's a push out of like graphs so that's how you get the wedge of spheres basically for E and RG. So could you say something more about like if I if I gave you a group that I knew was a fundamental group of a graph of groups could it like fit into this narrative in some way? So you'd want to be amalgamated if it has torsion you definitely don't want to be amalgamating over a subgroup with torsion so all of your edge groups would have to be torsion free and then it's really a case of whether the groups the space is involved already satisfy this covering thing so if they're all kind of low-dimensional groups then they probably do so so um you kind of get like most three manifold groups and three or before groups from this nice. Great thanks.