 I was thinking that these two extra zero most are was responsible for the minus one instead of the plus one But then it converted the one to minus minus one because one minus two Yes, I agree. But but but then but then at the end you said that we have to replace that formula With something and then at that point I have not understood. Okay. Yeah, so the string theory right doesn't know about this replacement Okay, string theory gives you a gauge fix form for all DP brands Which works most of the time except for D instantons that gauge fixing is wrong if you do carry out the gauge fixing as for a DP brand you get this you and he the ghosts Right the further up of a ghost that you get by gauge fixing phi equal to zero Those are the extra to zero modes, right which are reflected in these two states that I wrote down the I wrote down these two states zero and C1 C minus one on zero These are the further up of ghosts That would have gotten by gauge fixing that phi to zero So this in the original expression, right these are the why responsible for is minus two But now what you see from this analysis is that that was because the gauge fixing itself was wrong Right, so you have to remove those integrals over you and me and instead replace it by this ratio over here But the formula with the minus one is correct. Yes. Yes. The formula the minus one was correct. Okay, exactly that minus one was actually one minus two That minus two Was correct as given in the annulus partition function except that the minus two came because you had done it on the wrong gauge fixing So once you have rewritten everything in terms of the integrals, right? So let me write this to the annulus as given by What is its theory? The annulus partition function had integral Let us call it d psi b 1 d psi b 0 This is the one at hb equal to minus one Okay, this is the Bosonic zero mode that we found du dv and exponential of half Psi b1 square This is what would have what the annulus partition function would have given. Okay, that is the result of minus two So minus one came because one zero mode here and two formula x zero modes Now what we are saying is that this integral of course you do by steepest descent That gives the I Psi b zero integral you change variable to y right that gave q1 times dui du dv integral We shouldn't do because this came from the goal the ghost integral right and we shouldn't gauge fix So you remove this Instead we integrate over the gauge invariant variable phi that we had right so integrate over integral d phi Sorry there is also Yeah, so once so this is this was in the gauge fixed form right In the gauge fixed from phi was set equal to zero that's why you didn't have the phi But in the gauge invariant form we have In e to the minus phi square and then you have to divide by the volume of the gauge And now this gives you root pi and This gives you two pi root k That's so that annulus partition function now has been reinterpreted in this language Okay, where everything is finite Okay, except that q1 and k2 is still have to calculate the question from Julian and also if you have question for Julian So you've alluded several times at the symmetry breaking right it's the boundary conditions that break the symmetry Yes, so we should think of this as explicit symmetry breaking yet in the full string field theory kind of You sum over all the boundary conditions so you restore the symmetry Yeah, so this is explicit symmetry breaking from the point of view of the open string that are living on the D instant on but from the Point of view of the closed string theory. It's a spontaneous symmetry breaking. It's like that if you have a DP brand, right? Yeah, good that is transverse scalars. Yeah those transverse scalars you can think of that I mean from the DP brand viewpoint You have put a boundary condition so you have explicitly broken the symmetry But in the case that it's spontaneous then could you think about developing a theory of these Non-perturbative effects that just comes from let's say an effective theory that follows from that symmetry For the for the zero more CS for the for the psi B, right? In this case effective field theory is in fact trivial because it's a flat direction But on Calabi of etc. They are the effective field theory will have a Calabi or target space and indeed you are right that you have to Think of this as a field theory on the Calabi or target space and how much of the non-perturbative effects would that determine? That would only determine in this particular example that will only determine the 2 pi Delta E term right there That will give you the Q1 and the 2 pi Delta E term That these parts are separate right these came from because you had done a wrong gauge fixing the wall sheet The way the wall sheet gives a formula is uses the long wrong gauge fixing Yeah, some kind of UV input for this for this effective theory that you still have to explicitly calculate exactly. Yeah Okay, thanks any other questions for our show called for Julian Ask one more. So are you gonna repeat this discussion also for the fermions in to be then or or well I may not have the time to do it for to be but if you have some specific question You can ask me I'm wondering if it's the discussion is very analog or it gets yeah, I discuss and very analogous exactly So the ghost problem is not there. They go the only ghost problem is in the name also our sector Right, and it's the same as what I described Right. They are this 10 becomes 8 because of this minus again the yeah UV goes that will fall And the zero modes are also similar great the only extra feature is that there is formulaic zero modes And which basically means that when you calculate the Amplitude up to expressly insert those formulaic zero mode operators in the amplitude just like a normal instance in quantum field theory So that part is similar to quantum filter the only feature which you normally don't see is for instance in quantum field theory Is this problem with a post? I do a normal instance don't have this wrong gauge fixing problem. Any other question? Okay, if not, maybe let's thank our speakers again And now there is an announcement Yeah, so I know we everyone's keen to get to lunch, but we wanted to briefly take this opportunity Mention physics without frontiers