 So that's the first thing. So we didn't clear all this out. Now, once again, we have exactly as many zero-modes as in decodes, so we get something non-zero. But each of the brackets choose a different zero-mode in this machine, because you have to have the only way to get something non-zero is having one product between zero-mode. Is that clear? If you choose, but it's even simpler. If you chose the same zero-mode in search twice, then it's just zero. It's the last one you should be able to get. This is the last one you should be able to get. So the only way you get something non-zero is choosing a different zero-mode. But you can choose the same one. You could choose the MF zero-mode from your name. MF one from there, and so there are many possible ways in which you can do this choosing. Each possible way will give you an answer. Imagination I want you to take. You can check this with 2 cross 2. Can somebody guess what the answer to this? But the answer to this is father-neighborness. So suppose we chose the first zero-mode from here, the second one from here, the third one from here, the fourth one from here. What would we get? We get the product of all these factors. Where this one was replaced by B01. This one was replaced. So we would get a factor like B01, B01, there you watch G, thanks, B02, there you do G. And so we chose instead the second zero-mode from here, the first zero-mode from here, everything else, the same. Then what would we get? We get B02 del 1 G, B01 del 2 G, times the same stuff from here. But we'd also get a minus sign, because we've done the integral, because it seems it's different order. And so on, we get many such steps. Now, somebody gets what you're looking at. Actually, there are only two insertions that's the full answer. What is this? There's it now. Exactly. So that's to find the matrix Mkk prime as equal as, 0k doesn't get prime G. This part of the integral is the determinant of that. These are the zero-modes. You take the inner product of this thing, but M of these, M of these, it's an n-prosect matrix, take its determinant, and that's the answer. The net result of doing the zero-mode integral over B, including the B insertions, the net result of that. So that's the determinant of B0k del k prime. Let's also quickly look at what we get with the C's. And when you get the C's, it's also equally simple. Remember what we had? That's what we're going to get the left-movers. You get C sigma i product over all i. We're doing zero-mode expansion. So text somebody to tell me what we're going to get. Only the zero-modes, and what's the final answer going to be, determinant of something, or what is that going to be something. What's looking about these different insertions with different points? There is a different point to pick up, particularly a different zero-mode. You know, so this is now, again, a V cross B matrix with the indices being between zero-mode and which point? The part of the integral over the zero-mode of C, and during the combination from the insertions. That means what we get in the part of the integral that we looked at by doing the integral over all the other stuff. They're not zero-modes. Well, that is just some overall factor. See, because none of the insertions is null, dependent B and C anymore. That's just some overall factor, you know, some determinant, it's product of gammas which you can regulate in something. Throw it out. It's gone. So the whole result of doing the cross path integral is to do the matter. So this cross path integral, we're going to derive these two determinants. Once you have these determinants, no more courses. Now this will not always be the most convenient way to calculate it. Sometimes you can see how you actually evaluate the course of the path integral that you calculated and determined it. But you should remember that this is an alternative. Okay. Now, well, if we start with this discussion by saying that we wanted to show that the measure on that this integral over one digitized space was invariant was invariant on the field redefinition of one digitized space. Now it's obvious that this is the case. But it's simpler than that. It's easy. It's dmT times, or the only thing that depends on T is this thing, right? That's determinant of bk0 ln k prime g. First, we make some transcribing, some in my, some in free period. T is equal to, T is equal to f a of t. What is this major strategy? It's the Jacobian. The determinant of the Jacobian matrix. How is this determinant? By the inverse of that determinant of the Jacobian matrix. You see, because this matrix picks up a product of the inverse Jacobian. And then you use the fact that the determinant of the product of matrices is the product of determinants. So the space of party i is not important. You can use any layer you want. You can get the same answer. I should say that different matrices have to be related by one to one transcribing. Some something here, in changing the curve, measure you get an absolute value. And here you don't get that absolute value. So if the, if you don't make one to one, you run into problems. But of course you do want to make one to one, because you want your coordinates to be a phase full set of coordinates of modulo space. So if you start with a phase full set of coordinates of modulo space, there's only a main phase full, if it's one to one. So that's not this. Okay, so once you have a reasonable parameterization of modulo space, you've got your answer, doesn't that? Doesn't that? It's a step. This is our first, our first tip, nice day of it. We've seen that a possible value you might have had is not one, you know? Okay, great. So, so we've, we've described, we've understood the variation under, on equities, we will have the next five seconds. Okay, okay, so now, questions, comments, or we've got to the next? We'll do that. You know, like, you know, like looking at the, how we got to the, after we got the goes, yeah, yeah, you see, if our original procedure hadn't been completely consistent, we were guaranteeing that all these properties are true, you know, because, if we did everything correctly, we've got a sensible answer. But this is now a check on the answer. You know, there are, in the end, there are a lot of formal manipulations that go out when you do this derivation. And you might, so I think a sensible approach today is to create a, well, that derivation as heuristic, but check on the answer for every property that is physically. So, it's, you know, that is a formal argument, starting from our procedure, but then that's actually true. But this is how we, now we just, by explicit evaluation, show them the next. So, what do we do then? Is there a variation of the path we talked about? Okay, that's a huge, it sounds like a huge infinite dimensional matrix. For example, this determinant, basically path we talked about, because, right, I mean, which we exponentiated into goes, right. Yeah, but you see what we've done here, is throw out the part, that this is also infinite dimensional determinant. What we've managed to do very nicely is separate the part that doesn't matter, and the part that plays with the boundary line. So that's what we've done. Taken the path we talked about in the determinant show, then it does depend on the module that would be very simple. In a way that cancels the change of measure, only the few more. Exactly. That's, so that's the first property we were interested in. Now, there are many other things we should check, that we should check about, some of which have been bought in European, I think. Okay, so, the next thing I'm just going to do, you know, there are functions here of the things he checks. Okay, so, the next thing you might want to check is, how do we know that the final answer does not depend on what fiduciary method you used. You used in your constructions. Suppose we take two different methods that are related by wild transformations, we'll be at the same answer. Well, you should, right. Okay, now you might be thinking, well, well, well, well, and this sounds like deja vu, we've been through this before. We had this whole discussion of how the place of the stress tensor, and therefore the partition function and so on, did depend on the wild method, provided we have the critical event. It's a question, but, if we were just doing the integral of a partition function, we know that it doesn't depend on the wild method, but we're not just doing the integral of the partition function, we also have these various insertions. Okay, so we should make sure, we should make sure that our insertions are also quiet. It's a very easy job because it's a very easy job because remember that B and C were themselves wild. You remember that when they wrote down the action, we saw that which of the right index structure for B and C, they were wild animals. Now, these physical vertex operators, these physical vertex operators have definite insertions of V, have definite wild transformations. They have definite wild transformation properties. In fact, the same wild transformation property as the inverse of the metric, because they're one-on-one operator. You know, how can we see that? You see, conformal transformation. These operators are Lorentz scalars, they have no spin. So conformal transformation, okay? A conformal transformation, some compound of a wild transformation, and some difuermapism, okay? So if we understand how these things transform under, we know how about the transformation is under conformal transformations. And we know about the transformation is under wild transformation. Under difuomorphism. Then we know that that transformation is under wild transformation. These things are essentially difuomorphism scalars. All the conformal transformation properties come from wild transformations. And we see that the operators, these v operators have to be one in one operators. How do we see that? We see that because we know that the corresponding state has to have L0 equals 1 and L0 bar equals 1 and under the state operator map that maps the scaling direction of the operators. We have also seen that these operators are primary, that is crucial. We have seen that they are primary, therefore their transformation under conformal transformations is very simple. It is just del z by del z prime to the power 1, del z tilde by del z prime tilde to the power 1 and now if you convert that to a wild transformation, it is exactly the wild transformation of an object which is like the inverse of the nation. I have said this fast but is it clear? But maybe next class. Maybe actually, I will do that. But is it clear? Is the idea clear? There is no need at all. Let us say this is a formula. You remember when we first, when we were discussing conformal field theory long ago, we discussed the transformation of primary operators under conformal transformations. And we said that if an operator has weight h h bar, then the transformation has to be the power h del z bar del z bar prime bar h bar. This is the factor of transformation. But now let us compare this to the matrix. How does the matrix change? Well, yeah. Well, 1 minus 1 minus 1, depending on the dimension. So, the matrix or the inverse of the matrix, we have to get out the matrix. We transform exactly the same way but with factor 1 and 1. The matrix is exactly the same as that of the matrix. Let me say this way. You see, under the same coordinate change, the matrix is transformed by a wire transformation. The matrix is transformed by a wire transformation. And the wire transformation factor is this, when either 1 1 or minus 1 minus 1, which I get straight away. It is exactly the factor of the wire transformation if the primary field and the question is 1 1. So, what we are saying is that, now if you do it correctly, I would have to get the inverse is correct. But if g alpha beta goes to e to the power minus 2 phi g alpha beta, then v goes to e to the power 2 phi, if the operator p goes 1 1. You see, I know this. This is one of the two signs. You see, this is the right answer. Because of that, every insertion, every insertion of square root g times v is y. So, the insertion is fine. b and c is fine. Actually, there is only one thing that is dangerous and that is the b insertion. Let me write it down. Is it b insertion? b insertion was square root g, g alpha mu, g beta mu, del alpha beta, b alpha beta, del k g. Now, I am suppose we have power wire transformation. There are some weights. Then as many square root, you see, this is where suppose this transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. This transform is like e to the power minus 2 phi. It's a choice of fiducian metric for every choice of module. It's a choice of fiducian metric in a module's dependent fashion. The answer should depend on it. You see, B is stressless. Business directly acts on the stu-5. It contracts B with G and you just get stu-5. Again, so this potential problem, this insertion is awesome. I, in gladiants, am caring in, is manifest. It's manifest in the action because the stress-sensitization is in the critical dimension. It's manifest for C insertions because C is vitamin B. It's manifest for square root G times V because V is a one-one-one. It's manifest for this thing because we only potential problems. It's actually completely vital in the sense that if you choose different fiducian metrics that are different from each other and apply some vital transformation, it doesn't matter. Okay. Now, I said I, well, you guys, science is possible. That's the difference. Okay, I want to tell you about DRST invariance, also about, if you want to remember. You know, we've got these things inserted at specific points. Okay, that's very funny. If I choose different points to insert these things on, will I get the same answer? We're gonna do that. We want to prove that the action, that if you change the vertex operator, my vertex operator...