 Now the z square n is also dimension 2 because z square is dimension minus 2. This also works because t is a dimension 2, there is also a dimension 1 and 1 over z is a dimension 2. So everything that we have on the right hand side must be a dimension 4. Now the first question I ask is could you have something like for instance 1 over z is 6 times something. If you have something like 1 over z is 6 times something, it will have to be an operator of a negative dimension so that the net dimension had to perform. We will prove at the next class of the class after that that any unitary quantum field theory, any unitary conformable field theory, there is no operator of dimension less than 0, no local operator of dimension less than 0. Moreover, there is a unique operator whose dimension is 0 and that is the identity of it. We will prove this at the next class of the class after that. So that is the view of the moment that we are dealing with a unitary quantum field theory. It may be possible to prove what I have said more generally but anyway, let's put in the example of unitary theory at least. The maximum singularity in this operator product expansion is 1 over z to the 4, right? Moreover, the coefficient of 1 over z to the 4 has to be the identity of it and therefore must be a pure number. So whatever the pure number is, let's call it c by 2, c by 2 z. In principle, we should say q and then we got 2 is d of z by z squared and then plus d of z. We knew is to realize that the statement that we are making should be translational in there. I have talked about d of z, d of z, but I could change coordinates such that this point becomes 0 and then this point could become minus z. For ordinary operators in a path in general, there is no difference between a b and b a because the operator doesn't even have a particular order really. Time is at the side of a path in general. These two statements of 90 years ago, why don't we conclude it? We have concluded that the d of z, d of 0, must be equal to d of 0, d of minus z. That's translation meanings, shifted origin, which must come, which inside the path in general, same thing as d minus z, d of 0. This function, the function that appears in the right hand side, must be an even function. Thank you, my God. Let me, because it's, right, right, right, right. So let's see this as carefully as we can. Actually, we should have had, that is delta at z equal to 0. This is z equal to 0 and this is z equal to 0. Right. So see, suppose we've done it the other way round. Suppose we've done it the other way round, what would we have got? We would have got g of 0, g of, okay, so, let's, let's say this, let's, let's, let's take this as a little clip and a little bit of one thing I said, because you think we didn't distinguish in this expression, we had evaluated the operator at the position and which we had to 0. Okay, so let me say more clearly what I can do, you know, what, what we can do. Okay, suppose we flip these two things, then the, the answer must change, but we must do all our evaluations. So, so, if I, if I make the translation shift, what do I get? I get t of 0, t of minus z, t, t of minus z is equal to 0. Is equal to t of 0, t of minus z is equal to 2t of minus z divided by z squared. Okay, let's just work with the terms that we know first, plus del t of minus z divided by z, k plus, okay. Now, let us take this, this business and. Why? Divide it by minus. Divide it by minus. Divide it by minus. Yes, No, no, no. Yes, if it's equal to minus minus z. Yes, yes, sorry, sorry, sorry. That's what it's all about. What is the general rule? The rule that is. OK, let me do it with z and w. So let's put this as w. This is z minus w over x squared. This is z minus w. This is w. This is w. So that is generally true. OK. Now, what I want to say is that this must be the same thing with z and w. The left hand side is still there in the other side. So this must also be equal to t of z over w minus z over x, where del t of z over w minus z. This is between del t and z. And del t at w is a term that is of order w minus z. Model value of 1 over w minus z will give you a regular, which may not get track of. No, not at all. However, this difference is, because we basically expand this, we, at first order, we get a singular term which will modify this. So this is equal to 2t at w, z minus w into del t of w minus z over x squared. And we should add to this plus del t at z that will make us delta w minus z over delta x. Now, this is not the same, this is not the same as this, because this side is equal to that. But we've also got this, which appears to be the factor of 2. And when I add this to this side, do you see this? Again, consistency check works. For the terms, we can retain it. But you see the way it worked, that when there was an order term in the operator product expansion, it flipped a sign under w goes to minus z, w goes to z, because z minus z goes to w minus z. But a term higher up in the expansion kicks in to compensate for that. Now, we know that there could be a term with z, the form. But they can unclaim that there's no possible z cubed. Why is that? Under z goes to minus w, the z goes to w, the z cubed w flips a sign. But because the coefficient of this is not an operator, that's, I mean, in fact, it cannot compensate for that. Of the requirement that this left hand side, the left or the right hand side is invariant under z goes to w, it rules out, again, the expansion of the stress tensor itself is t of z, t of w, c by 2 over z minus w to the fourth plus 2 t of z, we should be careful, w minus w squared plus del t of w. So from general principles, we have determined all the singular terms in the operator product expansion of the stress tensor with itself, up to one unknown constant, negative c. This constant c is very important. It has a name that's called the central charge of certain form of theory. And we will see in the next class that it's, well, we will see in the next class that as we continue in our study of string theory, that the central charge is crucially quantity defined in the property of the string. But I think we should solve the other different cases. So let's put that camera off, and how do you do it? We'll have a red button. Red button. Oh, this one.