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So good morning and welcome back. It's about half time for this lecture course and I decided that I would like to present one central physical example that uses about everything we did so far.

So it will be a very good exercise. It's a very simple example but if you really carry this through properly you'll see we'll really meet almost everything we did.

And so this is formally speaking still part of this section on Lie groups in Lie algebras because we haven't finished this section yet.

But in a sense it stands there as a half time example. So it goes in this consecutive numbering 4.4 and it's the relativistic spin group.

Which goes by the name of SL2,C. That's what the thing is called. And we will now really want to understand this guy and its Lie algebra in every detail very properly.

Now although I say it's a relativistic spin group and it is, in order to really apply it to physics we also need to study representations.

We haven't done that yet but I hope we'll get there. But alone to understand this group as a Lie group and in order to derive the Lie algebra and so on we've got to work a little bit.

But again it's a very good exercise so let's see whether we can carry this through just with what we have.

Okay so let's first start with the definition of this guy and first and foremost before it's a Lie group or whatever it's a set.

And this set has a definition SL2,C is the set of four tuples, quadruples. But I follow tradition and instead of writing ABCD with commas in between I write ABCD as such a square scheme of numbers.

But obviously this is a four tuple because I always know the first, the second, the third, the fourth. Nothing new.

And it comes with a condition and that condition is that, aha, and it's, well if it's a four tuple it's an element of, I take the complex numbers to the fourth.

But that obviously means Cartesian product, Cartesian product, Cartesian product.

But not all of these four tuples. I have a condition, I impose the condition that AD minus BCB1. That's my condition.

So obviously C is a set, this is a set, this is restricted comprehension, there is a condition here. That's a set, right?

And it's a subset obviously of what we denote by C4 which is the short form of this. And this is SL2C as a set. These are the elements of the set.

There is nothing to be said beyond that about the set. Well, but now we make the set into a group.

And the group is then SL2C. A group is first and foremost a set. But then it has a group operation which we call blob.

And this blob is an operation that takes two copies, no, that takes two elements of SL2C and makes them into one new element of SL2C.

And this blob is defined as follows. We take one element which we decided to denote in this funny fashion.

We take another one, ABCDEFGH. And how do we define the blob? We define the blob by the following arcane construction.

It's AEBG, AFBH, CEBG, CFBDH.

Of course, it's easy to remember this because you have been trained to perform an operation called matrix multiplication.

And it is exactly this. But what on earth is a matrix? It's a four-tuple and another four-tuple. We produce from it this four-tuple.

Don't think about matrix multiplication. It's just this rule of how to combine these tuples.

And of course, what we've got to check is whether this rule obeys this restriction, so whether the result is really again in SL2C.

Now, it is because you know this is matrix multiplication. We could now check this by hand.

Well, yeah, you could. You multiply this with this and you subtract this and this and you reduce it.

If you know that AD minus BC is one, you know that EH minus FG is one, you will find that also here this rule is satisfied.

So this needs to be checked, but it of course applies.

Or if you want to think in terms of determinants, of course, the product of two determinants, not the determinant of a matrix product, it's the product of the determinants.

So it will stay one. OK, so obviously we have here a set and we defined a very funny operation, but it provides such a map.

And then you check that the whole thing is associative. This is now really boring. You believe me it is.

You'll check that there exists a neutral element, neutral element, and this neutral element is 1, 0, 0, 1, as you can also quickly check.

And that has AD is one, it lies in SL2C. Very good.

So there is a neutral element that looks like this, and you can also check that for every element small g in SL2C, for every matrix ABCD, you find an inverse matrix.

Let's write it down. OK.

So for every ABCD in SL2C, I claim, and you will quickly check, that 1 over AD minus BC plus minus plus plus minus like this.

D, A, minus BC. I think this is the inverse. Otherwise you please look it up, but I think this is the inverse.

This is the inverse. So we would denote this ABCD inverse.

Of course, the formula 1 over the determinant, and then this funny chessboard pattern.

But you take the transpose first, and then you do these minors and stuff like this. OK, you know how that works.

OK, anyway, so for every element that exists, an inverse element, you can quickly check that the inverse element has again AD minus BC is one, because this is AD minus minus minus BC.

But it's divided by this, so it's again one. So also the inverse, of course, is an element of SL2C.

Ah, so at this point, we learned indeed SL2C, this set together with this operation, is a non-commutative, a non-abelian, but nevertheless it's a group.

OK, fair enough.

Well, fine. So, but now we come to the next step.

I want to think about this group now, about the set that underlies the group, as a topological space.

So let's think about the topological space, SL2C. Well, that's the set, so we need to equip it with some topology before anything else.