 This algebraic geometry video will be about Duval singularities, which are sort of surface singularities. They've been rediscovered several times. They've got numerous other names. They're also called simple surface singularities. Well, they're also called client singularities for rational double points or canonical singularities in dimension two, which as the large number of names suggested, these have been rediscovered many times by lots of different people. So we'll introduce them by Klein's method of introducing them. What you do is you take a group G acting on C squared. So this is going to be a finite group G. And you take the quotient of C squared by G and for reasonable groups G, this gives you a variety which has a singular point at the origin. As you remember from our discussion of quotients, or maybe you don't remember, I don't know, the coordinate ring of C squared over G is given by taking the coordinate ring of C squared and just taking its invariance under the group G. For example, suppose we have a cyclic group of order n, generated by an element sigma such that sigma of X is equal to zeta of X and sigma of Y is equal to zeta to minus one of Y, where zeta to the n equals one is an nth root of unity. Then invariance of this action are given by X, which is equal to X to the n, Y, which is equal to Y to the n, and z, which is equal to X, Y. So the coordinate ring of G is generated by these three elements, X, Y and z. However, there are some relations between them because it's fairly obvious that z to the n equals X, Y. And we can also write this as X plus Y squared over four minus X minus Y. So that should be capital X squared over four. So this gives us one sort of Duval singularity. I'll now list them all. So the Duval singularities are as follows. There's a cyclic ones, which are denoted by a n, which look like X squared plus Y squared plus z to the n plus one equals zero. Then there are the dihedral ones, which correspond to X squared plus z, Y squared plus z to the n minus one equals naught. There's a tetrahedral one, which is given by X squared plus Y cubed plus c to the four equals naught. Octahedral given by X squared plus Y cubed plus Yz cubed equals naught. And icosahedral given by X squared plus Y cubed plus c to the five equals naught. So the alt icosahedral group, the icosahedral one, which we're going to talk about, is not given by acting on the icosahedral group of order 60 on c squared because the icosahedral group of order 60 doesn't act on c squared. However, there is a binary icosahedral group, which is order 120 and is a double cover of the icosahedral group that does act on c squared. And Klein showed that the invariance generated by three elements X, Y and Z satisfying this relation. So for all of these, we're going to work in characteristic zero. And for all of these varieties, the only singular point is the point naught, naught, naught. So what I want to do is to show how to resolve one of these singularities using blow-ups. And I'm going to do this singularity here, if you can do this one, then you can do all of them. So the idea is we're going to blow up the singular point here. And we find that when we blow it up, we produce another singular point. And then we blow up that singular point and that produces more singular points and so on. And we're going to keep going until we run out of singular points. So let's see what happens. So let's start with X squared plus Y cubed equals plus Z to the five equals zero. And we're going to blow up at the origin. This means we introduce coordinates. Here we've got A3 and we have to introduce coordinates for the projective space P squared. And we're going to call these coordinates X1, Y1, Z1. And in order to get the blow-up, we need to add the usual relation. So X, Y1 equals Y, X1, X, Z1 equals Z, X1 and so on. And what we do is P squared is covered by three copies of A squared. So the three copies are given by X1 equals one, Y1 equals one and Z1 equals one. And on each of these three copies, we have to check and see if we get any singular points on the blow-up. So for X1 equals one, what we do is we're really changing, making the change of variables, Y equals Y1, X, Z equals Z1, X. So we get the equation X squared plus Y1 cubed X cubed plus Z1 to the five, X to the five equals zero. So that's just coming from this equation making this substitution. So this is one of the pieces of the blow-up of this. And we can take out the factor of X squared and we get one plus Y1 cubed X plus Z1 to the five X cubed equals naught. And you can easily check this is non-singular. And for Y1 equals one, we get something similar. We get X1 squared Y squared plus Y cubed plus Z1 to the five, Y to the five equals zero. And we can take out the factor of Y squared. And again, this has no singularities. So there are no singularities there. No singularities there. Finally, we have to look at Z1 equals one. So here we put X equals X1 Z, Y equals Y1 Z. And we find the equation becomes X1 squared Z squared plus Y1 cubed Z cubed plus Z to the five equals naught. We take out the factor of Z squared, X1 squared plus Y1 cubed Z plus Z cubed equals zero. And again, we can search for singularities. The singularities are given by taking the derivatives of this. So we get two X1 equals zero, three Y1 Z equals zero. Sorry, three Y1 Z plus Y cubed equals zero and three Z squared equals zero. And this time we have a singular point given by X1 equals Y1 equals Z equals zero. So we've got just one singular point and the singularity is the singular point of this variety here. So what we've done is you've gone from X squared plus Y cubed plus Z to the five to a slightly simpler curve here. So now we do the same thing with this new curve. So we take the curve X1 squared plus Y1 cubed Z plus Z cubed equals naught. And again, this is embedded in A3. So we take it multiplied by a copy of P squared. And this time we're going to call the coordinates X2, Y2, Z2. And as before, we cover P2 with three copies of A2. And the three copies are going to be X2 equals one, Y2 equals one, and Z2 equals one. And for X2 equals one, there are no singularities. I won't write this out in detail because it's much like the previous case. Here again, there are no singularities. For Y2 equals one, we make the substitution Z equals Z2, Y1, X1 equals X2, Y1. And our equation becomes X2 squared Y1 squared plus Y1 to the four Z2 plus Y1 cubed Z2 cubed equals zero. So now we can take out the factor of Y1 squared and we get the equation X2 squared plus Y1 squared Z2 plus Y1 Z2 cubed equals naught. And again, there is now a singularity at naught, naught, naught. And you can check this is the only singularity. And you may sort of look at this and think, you know, what have we actually gained? Because we've gone from this equation here to this equation here. And frankly, this equation doesn't look any simpler than that one here. Here we've got a sum of monomers of degree two, four, three. Here we've got a sum of monomers of degree two, three, four. So we don't seem to make any progress. In fact, we have made some progress, although it's really quite subtle. In fact, it shows that measuring the complexity of a singularity is rather tricky because this singularity really is slightly simpler than this, but it's not entirely obvious what invariant you can use to detect this simplification. Anyway, let's carry on. So again, we're going to blow up this point at a cubed times p squared. So we're going to introduce three more variables X3, Y3, Z3 for this copy of p2. And just as before, we cover this p2 with three copies of affine space. So we get X3 equals one, Y3 equals one and Z3 equals one. I'll start getting a bit fast because writing out every detail of these calculations gets a bit tedious. So here we have no singularities. Here we have a singularity at... Here the equation we get is X3 squared Y1 squared plus Y1 cubed Z3 plus Y1 to the four Z3 equals naught. And we can take out the factor of Y1 squared and we find the only singularity is at X3 equals Y1 equals Z3 equals naught. This one, the equation we get is X3 squared plus Y3 squared Z2 plus Y3 Z2 squared equals zero. And again, we find there's now a singularity at X3 equals Y3 equals Z2 equals zero. So we've actually got two different singularities appearing. We've got one here and one here. So one at this point corresponding to this equation with the... Okay, this one I should have taken out the factor of Y1 squared. So now we've got two singularities to look at. And let's first look at the singularity X3 squared plus Y1 Z3 plus Y1 squared Z3 cubed equals naught. And again, we block introducing variables X4, Y4, Z4. And we again cover it by these three copies of project of space X4 equals 1, Y4 equals 1, Z4 equals 1, X4 equals 1. We have no singularities and Y4 equals 1. We have no singularities and Z4 equals 1. We have no singularities. So blowing up this just once, we actually finally get rid of all singularities. Well, that got rid of one of these singularities here, but now we've got to look at the other one. So the other one, we have a singularity of shape X3 squared plus Y3 cubed Z2 plus Y3 Z2 squared equals naught. And here for the blow up, we're going to introduce variables X5, Y5, Z5. And we find for X5 equals 1, it's non-singular for Y5 equals 1. We find two singularities at X5 equals 0, Y3 equals 0 and Z5 equals 0 or minus 1. At Z5 equals 1, we again get two singularities at X5 equals 0, Z2 equals 0 and Y5 equals 0 or minus 1. So we seem to get four singularities. In fact, one of the singularities here is really the same as the singularity here. So we've got three singularities to resolve. And as the audience, if there's anybody left at this point is probably getting a bit bored of me blowing up singularities. I'll just tell you what happens. If you take each of these three singularities here and blow them up once, you get rid of, you get something that's non-singular. So there are three more blow-ups and I'm not going to write them out. So let's summarize what happened. We had eight blow-ups and we finally ended up with something non-singular. And what you can do is each blow-up introduces an exceptional curve which looks like P1. And you can sort of draw a picture of what these P1s look like and how they intersect. And if you do it, they sort of intersect like this. And one, two, three, four, five, six, seven, two. They intersect in a configuration looking like that. And if you draw a point for each of these lines and connect two points by a line, if the corresponding line is joined, if you see what I mean. So I'm sort of taking the dual of this diagram. You get the E8 Dinkin diagram, which is partly why this is called an E8 singularity. You notice the problem with all this blow-up is, first of all, blow-up involves a tremendous amount of bookkeeping. These blow-ups, we had to bring three new variables. So together with the three variables we started with, we have 27 variables to keep track of, which is a bit of a pain. The other thing you notice is that each blow-up doesn't do very much. A blow-up can be a very mild operation, which is just sort of making a tiny little improvement to your singularity. This example is also a bit misleading because each time we did a blow-up, the singularities only formed a single point. In general, if you do a blow-up, if you blow up a point, you can suddenly find that the singularities form a subset of dimension greater than one. So we'll see an example of that in the next lecture.