4/5 Curved-space supersymmetry
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11:28
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Transcript: English(auto-generated)
00:15
I will start my lecture by going back to something that I was doing yesterday, but it was in
00:24
the last few minutes, and I think I butchered it pretty badly, so it's better to say where I was wrong. First of all, we discussed supersymmetric field theory on the cylinder, and we discovered
00:42
that it has four supercharges, all transforming under the SC2 left isometry of the 3-sphere, but they're not transforming under SC2 right, and I brought these commutation relations, but if you look at your notes, you will discover that I did mess up a sign, I think
01:04
this sign was with a minus, but it is a plus, and that actually has important consequences. So yeah, actually the algebra with a minus, I don't think this even makes any sense, so this is the correct one hopefully, so we did not rewrite the entire thing, but just the
01:24
ones, the more relevant ones. Then there was also a comment about if this is really SC2-1 times E1, and I think indeed as Bruno was saying, if you redefine some operator P as HL plus R, then indeed you can
01:51
replace the commutator between R and Q with the commutator between P and Q, because H commutes with Q, and then this is the operator which appears in the SC2-1 superalgebra.
02:06
So then the SC2-1 will contain the operator P, and the extra R factor or U1 factor is generated by H, so indeed the superalgebra is SC2-1 times U1 times, you also have the
02:24
SC2R which doesn't act on any of the supercharges. So then we discussed what's the representation theory of this superalgebra, and so there
02:41
are two kinds of long multiplets, one happening for when the spin is greater than a half, and then it is comprised, or in the long multiplets happen for H which is greater than 2J plus 2 minus R, or greater or equal, so then you have all this, so you
03:07
have an entire, so in this notation this indicates the spin, the maximum spin of the multiplet under SC2L, and so there is an entire multiplet of spin J with the R charge
03:24
R and the value of H equals to little h, and then there will be two multiplets, one with J plus a half and one with J minus a half, and the R charges are shifted by one but the same value of H, because all the supercharges commute with H, so H always
03:44
stays the same, and then finally there is at level 2 another multiplet with spin J and R charge R minus 2, and so this is fine for J greater or equal than half,
04:00
when J is equal to 0 you lose the multiplet which would have J minus a half, and you're left with this, and then we looked at what happens when the multiplets are short, so the short multiplets happen either for H equals to 2J plus 2 minus R, so again here
04:25
yesterday I had the sign of R wrong, because I had the sign of that R over there wrong, and also there is another thing that I should say, that here I am suppressing L, so I'm setting L to 1, or otherwise you could put an L in front of H I guess.
04:46
So there are two kinds of short multiplets, there is one which has J with R charge R and this value of H, and then another multiplet with the spin J plus a half, so it's basically
05:05
like a long multiplet but it's missing the J minus a half and the level 2 guy, and then there is a very short multiplet which is the singleton which is comprised of something which has H equals minus R and J equals 0 and nothing else.
05:30
Okay so then given the structure you can study how this multiplets can, how the long multiplets split into short multiplets and indeed let's look at what
05:43
happens when this multiplet hits the saturation bound, then you can split it in a short multiplet of this form, so you take the level 0, 1 so that would be JR and then H is going
06:01
to be equal to 2J plus 2 minus R, and so you want this kind of short multiplets, so you will have another J plus a half of 2J plus 2 minus R and R minus 1, but this only takes care of two parts of the long multiplets, you still have two
06:27
multiples to take care of and those have to become another short multiplet which should still be of this kind, so let's check that that is indeed the case, so now the top component of the short multiplet will be J minus a half with R charge R minus 1
06:46
and the same H, 2J plus 2 minus R and the relation between the charges here should be the same as the relation between the charges over there, so in particular indeed you can write
07:01
this as 2J minus a half plus 2 minus R minus 1 because this minus 1 cancels with this plus 1, so indeed the relation of the charges is the same and then you are like left with JR minus 2
07:23
and 2J plus 2 minus R, so indeed this long multiplet here splits into two short multiplets of this kind when J is greater than or equal than a half, then you have to figure out what happens to this long multiplet and there is the same story except that this long multiplet will
07:47
split into a short multiplet of this kind and one singleton, so this you can check by yourself and then you can try to write down the most general index that will count this multiplets up
08:06
to the combination and also here I've messed things up, so this index should be a sum over all the spins and all the H's in your theory of some constant alpha J that in principle can depend
08:32
on H, well let me put the H dependent state, so we have constant alpha J times the number
08:47
of short multiplets of spin J and charge H and then there will be, so by this I mean the number of these short multiplets and then I'll have sum over H, that's sum over J of this, then I
09:24
have some coefficient beta times the number of singletons with charge H and now here I can sum over H and I could put some fugacity for H, so again like as you see in these recombinations
09:50
like because H commutes with everything like the value of H for the multiplets doesn't change, so this is some index and then like as I was discussing last time you can figure out what
10:09
alpha J's and betas have to be in order for this thing to be an index under those sort of combinations and you find that alpha J has to be equal to minus alpha J plus a half
10:21
or J minus a half and then that alpha a half should be equal to minus beta and therefore you can write down what the answer should be, so what you get is the trace
10:43
of minus one to the f times e to the minus beta H. And is it multiplicity or multiplicity? Yeah, see there is still, I cannot write things straight.
11:01
Okay, so this is indeed the, so it does the form of some with an index and actually if we had been even more, well you could have put even more labels because the multiplets can also have spin under the other acitu, the aciturite, so I
11:22
could in principle add also fugacities for the aciturite and if I have conserved flavour charges then I can put even more fugacities here. As long as I have operators which commute with all of the supercharges I can always add fugacities for them and H over L.
11:55
What do you mean it's not really H? So this one is the one that tells what is the BPS condition?
12:00
No, that's H, so this is H which is the operator which commutes with all of the Qs, so it's not the one which sets the BPS condition. The operator which sets the BPS conditions, you could for instance select one particular supercharge, say Q1 and then you
12:21
could sum only over states which have Q bar 1 Q1 equal to zero and that basically gives rise to this kind of condition. But this H is not the operator that sets the BPS condition,
12:43
it's just like an operator which commutes with all of the supercharges. Okay, so we have this object, now we can... Sorry, I think you said it, but the equality, so the conditions alpha j equals minus alpha j minus
13:04
one-half and alpha... Yeah, and alpha half equals minus... These conditions are improved so that when two short MOSFETs can combine into a normal configuration not contribute. Exactly. So we can indeed write down the more general index, supposing that we have some other
13:28
conserved charges, then we can write trace of minus one to the f, we can introduce two complex numbers p and q, then we can write p to the j3 left plus j3 right
13:47
minus r over two plus one times q times j3 left minus j3 right minus r over two plus one and we can also insert some fugacity for some flavor charges
14:09
that I have in my system and I am supposed to take the trace over the Hilbert space on S3.
14:21
And you can see that if p is equal to q, e to the minus beta over L, then this just becomes this guy over here.
14:46
Well, you have to use the fact that h is equal to 2j plus 2 minus r. Okay, so now we can interpret this index also as the partition function of our theory on S3 times
15:13
S1 where the radius of S3 is L and the radius of the S1 is beta. So this should also be
15:23
the partition function of our theory on S3 times S1 and this is radius L and the radius beta. And from this discussion we see that the answer can only depend on the ratio between the two
15:42
radii. So in particular it does not depend on the size of the manifold but only on the I want to make sure I answered. p, q and u. u is some other fugacity if you have some other flavor symmetries.
16:04
Okay, but are they related to r over j and beta? Yeah, beta, yes. So if you take p, so this is more general because I've turned on also the fugacity for the j3 right which is the other rotation. Is that expression more general than?
16:21
Yeah, so this expression is more general than that and it reduces to that expression if you take q equal 1 and then you take p equal q equals e to the minus b over L. But then can you motivate that expression using the same logic? Yes, I said because you have other charges that commute with all the q's,
16:43
you can introduce fugacity for those. So you can choose a carton in the S3 to write and you can add it. So that will be p over q and then you can introduce flavor charges.
17:02
So I will have more to say about the interpretation of p and q later but the interpretation of u, we can also give a different interpretation for u instead of saying we can add fugacity into the trace. We can also say we could add a background gauge field
17:22
which couples to the conserved current corresponding to the flavor symmetry and we can put it along the S1 and then we can compute this partition function where besides having the background fields that we talked about last time turned on, we also have this other background gauge field which wraps around the S1.
17:45
And this is a background gauge field for the would-be conserved flavor symmetry. So this gives a geometrical interpretation for this u and for the geometrical interpretation of p and q you have to wait a little bit. But there is going to be a geometrical
18:04
interpretation also for those. And another comment is that again, these fugacities correspond to adding background gauge fields and these background gauge fields can be complex and indeed this u can be some complex number.
18:27
Then an important fact is that you can see from here that the values of h of the states
18:44
that do contribute at the index is fixed by the angular momentum and the R charge. So h is fixed by L and R. So in particular, if I have some
19:06
RG flow from the UV to the IR of some theory and along the RG flow I can conserve a u1 R symmetry. So there is a u1 R symmetry which is conserved all along. Then the values of h
19:30
of the states that contribute are fixed in terms of charges and angular momentum so they don't change and the index is invariant along the RG flow. That's another way of saying that
19:46
indeed it does not depend on the size but only on the ratio of L and beta.
20:03
So you might worry about all sorts of bad things happening to this index but you can actually check that for instance the spectrum of the theory on the cylinder provided that the R
20:21
charges of the fields are in certain bounds is discrete so it actually is a pretty well behaved object.
20:43
They are fixed in terms of the spin and the R charges. So as an example we can just write down what the... So this in particular means that we can just compute this index
21:06
in the UV where we can start from some asymptotically free theory so we can actually do the computation for the index. It's a computation that you can do in free field theory and then it tells you something about the IR. So that's an important comment.
21:24
The other interesting fact is that you can actually check, so then you can ask what does this index do compute in the IR. So you can actually check that this SU2 slash 1 times U1 etc times S2R
21:49
is a sub-algebra of the super conformal algebra in 4D. So it's a sub-algebra of what is it?
22:02
SU4 slash 1 and actually you can do this for exercise. It's a maximal sub-algebra which does not involve on the right hand side any actual conformal transformation of the cylinder. So if
22:21
you take the n equal 1 super conformal algebra and interpret it as the super conformal algebra on the cylinder then you can ask what is the maximum, well what is a maximum sub-algebra that on the right hand side involves only isometries and not conformal isometries and so
22:46
this is the answer. So then by doing a little bit more work you can check that indeed this index would flow in the IR to what is called the super conformal index of whatever super
23:05
conformal field theory should live in the deep IR in this RG flow. So let me give an example. The examples are going to be a little bit trivial but you can find more complicated ones
23:20
in the literature. No, the index is always the same but in the IR it has the interpretation of this other index. It doesn't flow because it's actually invariant along the flow.
23:41
So let's consider a chiral field of R charge R so then you can compute this index and what you find is the following. It's a product over j and k, maybe I should put m and n because they're
24:08
not related to, of 1 minus pq to the minus R over 2 times p to the m plus 1 q to the
24:25
n plus 1 divided by 1 minus p times q to the R over 2 times p to the m q to the n and here I've suppressed the possible fugacities in case you want to charge this chiral field
24:46
under some other symmetry. So that's the expression for the index. And now we can check that these statements are right, at least in some very simple case. So for instance, let's suppose that the R charge of our chiral field is 1.
25:07
Then that means that we could in principle add a mass term to the super potential because the super potential will have R charge 2. So we can add mass and then we would expect
25:28
the field theory to just go to the empty to some, I mean it's gapped in the IR. So let's check what happens if R is equal to 1 and indeed if R is equal to 1 you can see that
25:51
these guys will become m plus a half and also here you will get m plus a half from this and so the numerator and the denominator are the same and well if you believe that you can
26:04
cancel them one by one, I mean it's an infinite product but you'll have to regularize it in some way, then indeed the index is equal to 1 which means that you only have the vacuum, so that's correct. So then we can also consider another maybe stupid case.
26:27
What's the range for m and n, 0 to infinity? Yes, so that you should say m greater than 0. I think it's equal density. Now we can do another example which is R equal 2.
26:50
But R equal 2 means that we can just add to the super potential a linear piece, so we can take the super potential to be f times phi and then supersymmetry is broken
27:01
and therefore we would expect to find the no supersymmetric vacuum and so let's check what the index is for R equal 2 but for R equal 2 you can see that for m and n equals 0 and R equal 2 in the numerator there is a piece which is 1 minus 1 that's 0 and therefore the index is 0, so it works also in this case.
27:29
Well naturally this is not maybe too satisfying but indeed by using this, so you can compute, you can take some gauge theory of your choice, compute the index in the
27:43
PUV. So suppose your gauge theory is some number of vector multiplets, some number of chiral multiplets, then you will have factors like this for each of the chiral multiplets but there will also be fugacity corresponding to the gauge charges and then in order to impose
28:01
the gauge symmetry you'll have to integrate over these fugacities. There will also be some factors for the fields in the vector multiplets but in total the index will be some integral over the gauge field fugacities of some expression like this, it will be infinite products
28:22
of p's and q's and indeed these integrals can be, there is some theory of how to handle them and you could check for instance that if you have two theories which are IR dual they indeed have the same index. So that's a very non-trigger check, it's much much more non-trivial than
28:42
these very simple examples. So I just want to give you some reference, so the original paper on this index are the ones by Romerzberger, so 0 5 1 0 0 6 0 and 0 7 0 7 0 3 7 0 2.
29:19
So this is actually, the second one is actually the more readable paper.
29:26
Any questions? This expression looks like a particular case of some of these five-dimensional partition functions, the p and q being the omega deformation function. So you would like the p and q to be the omega deformation parameters?
29:47
So p and q are kind of two, are they the sphere? Well yeah, well yes, yes they do. I mean at least one certainly can be interpreted some
30:01
rotationally. They do correspond to j left and j right, yes, with some combinations. I don't know of an actual realization in 5D where these become the... 5D has more parameters, so there is for example the velocity for the incident on particles, so it looks like if you specify... Yeah, no, yeah, it might be that there is some
30:26
background which could be reduced to this. Yes, I don't know, that's an interesting... It might be that this is 3, it's actually 3 inside R4. Inside R4, yeah. Yeah, so there should be some background which reduces to this. Like some kind of boundary.
30:47
That's a very interesting comment, but I don't know of any precise relation. Okay, so I'll just make some extra comment on what, well in the spirit of this last
31:03
comment by Nikita, we can think about using this background that we have on the S3 times S1, also to say something about the theories in 3D. So in particular if you have an n equal 1
31:21
theory in 4D with a U1R symmetry and you reduce down to 3D, then you will get an n equal 2 3D theory with U1R symmetry. And so we can think about, well we have this background with
31:41
all the seismometers, maybe we can reduce it and get some interesting things in 3D. So one thing we can do is to take our S3, and the S3 is a vibration over S2, so we can reduce along the off-fibre. So because nothing depends on S2 right,
32:03
then we can take the off-fibre aligned with S2 right and reduce along that. All the supercharges are going to stay and we are going to get a background which is the 3D, the so-called 3D index, so that would be S2 times S1. So this is S3 times S1,
32:25
we have S2 times S1, so it's usually called the index. There is another way of preserving supersymmetry on S2 times S1, so that corresponds to doing a twist on the S2 with
32:45
U1R symmetry. So that's a different one. And then we can also reduce just along, we can just place our theory on S3 times S1, and then we can reduce along this one and get a theory on S3. And again this theory, both these theories will have four supercharges.
33:16
And so the four supercharges for the 3D theory on S3 are going to have the same properties
33:23
of these. They're both going to transform only under S2 left and be inert under S2 right, which means that you can actually squash the S3 in various ways. And people have studied various possible squashings which preserve four or less of the supercharges.
33:48
Can you take a linear combination of S1 and the geometry of S3? Yeah, so what you can do is that you can glue the cylinder after a rotation,
34:02
and then you can reduce on this, if you want it to be reducing along some twisted thing, as you were saying. And then what you get is that the S3 is the squashed version, not rational. You can just decide to take this S3 here and identify it with the S3 here after
34:26
rotation of an arbitrary angle, and then you get an arbitrary squashing. You don't get the most general squashing in this way, but you get a good... So all this squashing will in particular preserve four supercharges because they only involve the S2 right.
34:46
Okay, so there is still one extra comment on this, which is the interpretation in 3D of the background fields for the
35:02
flavor symmetries. So as we said, here we can add background gauge fields which couple, let's call them, I don't know, little a mu flavor, which couple to the conserved flavor symmetries of your theory. And these can be in principle
35:23
complex, so they have a real and an imaginary part. So you can try to interpret what this real and imaginary part will be on S3. And well, the real part of a mu is just something which... So you can imagine like it just shifts the arc current
35:44
by something proportional to the flavor current. So when you write the theory on S3, there will be various couplings to the arc current, in the same way as we add couplings to the arc current in 4D. And the coefficients of these couplings can be shifted by turning on this
36:03
flavor fugacities, and imaginary ones become real masses on the S3. So on the S3 you have the possibility of turning on real masses, and those are encoded in this dimensional reduction, in the imaginary part of the flavor fugacities. Okay, so I think this exhausts what
36:29
I can say about this index, at least for now. So I will jump to a somewhat different subject, unless there are questions. I was told that I'm pretty bad with these blackboards, so I should
36:57
try to be better. Can you say something about generalization to 4D n equals 2?
37:09
Yes, I can say something, but it's not gonna... Okay, this is a complicated topic, but let me just make some remarks. So first of all, it's obvious that if you take an n equal 2 theory,
37:22
it's also n equal 1, and it has a U1 R-symmetry, you can just take the carton in S2, so you can put it on this background. Now you would think that this would break the S2 R-symmetry to U1, but actually if you do things carefully, you discover that that is not the case.
37:44
So the corresponding background in n equal 2 will preserve eight supercharges, and unfortunately it's not particularly useful because... So one thing which makes this background useful is that if you look at the Lagrangian that I brought down, for instance, for the chiral field, it depends on the R-charges of the fields, and in particular,
38:06
if the R-charges are between 0 and 2, then there is an actual potential for the chiral fields, which is some kind of master which goes down like 1 over R-squared. So when the manifold is
38:22
small, there is a large trap for the fields, so they are trapped near the origin in field space. So that makes the computation resilient to all sorts of... There are no flat directions, but if you do the similar computation in n equal 2, then when you go from n equal 2 to n equal 1,
38:47
the R-symmetry of the resulting field, it turns out that there is some field of R-charge 0, and then this field has a flat direction, and it makes using this background much less obvious.
39:02
So that's one comment. There are many more comments that I can make about n equal 2, but we will leave it for later because it will take a long time. Okay, so if there are no other questions on this, I will skip to a somewhat different topic.
39:25
So I said at various points that this idea of coupling to background supergravity was useful because it would allow us to classify what are the possible manifolds on which we can
39:41
place a certain theory preserving some amount of supersymmetry. So in particular, let's see how this works for the case of n equal 1 theories with a U-1 R-symmetry, which as we said,
40:03
coupled to new minimal supergravity. So the generalized Keeling-Spinner equation, in this case, we brought it last time, but I'll rewrite it again as the following form.
40:37
And then there is another equivalent equation for the zeta tildes,
40:43
but because this equation only involves zetas and the other equation only involves zetatildes, we can just forget about the other one for now and look for what manifolds which have one solution, I mean backgrounds, which allow for one solution to this equation.
41:00
Okay, so some comments. Again, so we are working in Euclidean space, so that means that if I take a spinner zeta, alpha and complex conjugate, its components, I don't get the components of zeta tilde, but I get the components of another spinner,
41:24
let's call it zeta dagger, which also transforms as a left-handed one, but with an upper index. Okay, so this is, if you want, by definition, the definition of zeta dagger. And with this definition, then you can check that zeta dagger zeta is just equal to the absolute value
41:45
of zeta one squared plus the absolute value of zeta two squared. So one thing you have a little bit careful about is that if you take the complex conjugate of zeta with upper indices, this is going to give minus zeta dagger with lower indices.
42:04
You can check this. Okay, so this will be... So this is one comment. So the other comment is that, as we said last time, this spinner zeta is charged under the u1r symmetry, so that it actually lives into
42:26
the u1, the unitary line bundle corresponding to the u1r symmetry times sc2 plus this left-handed spin bundle. So for a generic manifold, this sc2 plus
42:46
the spin bundle does not exist, but any orientable manifold in 4D is spin C, and in spin C, this line bundle, unitary line bundle times sc2 can be well defined.
43:02
So in principle, we can work on any orientable manifold, but there will be constraints on the kind of theory we can put in. We will comment on that later. Okay, so then there is another
43:21
somewhat trivial point that one can make is that, well, this equation has at most two non-trivial solutions. So how do we see that? Well, the first thing that is clear is that this is homogeneous in zeta, and it has first derivatives only. So if zeta is equal to zero
43:45
at some point, then the derivative of zeta is going to be zero, and then that means that zeta is equal to zero everywhere. Well, at least in some open set. So that means that if you have
44:10
more than two solutions, you can always take linear combinations of them, which are going to be zero, and then that means that they are trivial. Okay, so in particular, this same
44:23
point tells you that if there is a non-trivial solution, then zeta must be different than zero everywhere, and as we'll see, just this fact has some very important consequences.
44:48
So it's up to naught, it's unique. What you just said means that if it exists, it's unique up to naught. The solution? Yes. There can be two solutions. Can be two solutions?
45:00
Yeah, because the... What was the argument to take linear combinations of Spanish at some point? Then French is everywhere. No, but you have two components of zeta. They don't live in a line bundle. Okay, so... I suspect it's useful if you say what spin C structure is.
45:26
Okay, this is gonna take me a while. Let's say if people are interested, I can comment on that separately. I guess I won't be finishing this. So let's see.
45:46
All right, so now... Okay, I'm gonna mess with the blackboards maybe. Let's see, you can put this up and I can write this one. Okay, so now we can, in order to continue to study
46:24
this equation, it is useful to just make some statements which do not even use them, they just use the fact that the solution is nowhere vanishing. So for instance, we can construct the following object for which I've chosen an appropriate name by using...
46:53
Suppose we have a solution, then we can build up this object. So let's make some general comments.
47:04
So first of all, you can notice that there is a zeta at the numerator, but there is also a zeta at the denominator. There is a zeta dagger at the numerator and a zeta dagger at the denominator. So that means that this is actually uncharged under the complexified
47:22
U1R. So it is a good tensor on the manifold and it is a good tensor because zeta is nowhere vanishing, so I can divide by the norm squared. This object has no zeros. So this is a good, it's a good tensor and we can try to figure out some properties of this tensor by just
47:43
using Fierze identities. So you're welcome to check these in your spare time. So you can multiply two of these and what you find is minus the identity and you can also check that
48:01
if I take g mu nu, that's basically a nu lambda. What did I write here? Yeah, there I mean.
48:32
Oh yeah, I have some... So what does this mean? So it means that this thing J is an almost complex structure. This is more or less the definition of an almost complex
48:57
structure and also that the metric is compatible with this almost complex structure. So
49:07
fine. So from this we can already draw an interesting conclusion which is if you have a solution, then there is an almost complex structure. Therefore, for instance, we can exclude that we are ever able to find the force sphere as a solution to our equation because the force
49:26
sphere does not allow for an almost complex structure. So that's out. So as we saw, we can realize the force sphere but for theories which couple to old minimal super gravity and not
49:40
to new minimal super gravity. So now we have an almost complex structure. It's natural to ask does this object actually, is this object actually a complex structure or not? So for that, you have to check that this tensor with a name that I cannot pronounce.
50:05
So can I ask something? It's confusing. When we did, we both suggested the day before the score localization for n equal to 2, we had the killing spinors that they can be 0, the north pole or the south pole. Yes, let me answer that question. So the n equal to
50:23
killing spinner equations or the n equal 1 killing spinner equations in old minimal, they are different in such that they also have zeta tilde on the other side. So that means that it is possible to find solutions where either zeta or zeta tilde vanishes
50:41
at some point, not however both zeta and zeta tilde, because the same comments go through that if both zeta and zeta tilde are equal to 0, then you will find that the solution is everywhere 0. So indeed, for instance, on Peston's theory on S4, the left-handed part of the spinner which
51:03
gives the supercharge vanishes at the north pole and the right-handed vanishes at the other pole or vice versa. So as I said, it's natural when you have an almost complex structure to check if it is actually complex. So for this, you can do two things. Either you can check
51:25
that this tensor, the Neumann noise tensor vanishes. So I can write down what this is and hopefully I get the indices right. So I'm writing this just so that you see that
51:49
it just involves j and the derivatives of j. So if this is a complex structure, then this
52:12
object better be 0. So then you know what j is, you know what equations zeta satisfies, you can find what equations zeta dagger satisfies by taking the complex conjugate,
52:24
then you just can plug in and you can compute. Or maybe if you're good with some computer, you can have the computer compute for you and you find that it is indeed 0. So this is a pretty horrible computation, but there is actually a simple computation that maybe you can try to do,
52:43
which is the following. So you can check that if you have an holomorphic vector, then x mu sigma tilde mu zeta is equal to 0. And then an equivalent characterization
53:05
of a complex structure is something for which the commutator of two holomorphic vectors is also called holomorphic. So if you take x mu and y mu holomorphic,
53:21
which means with only such that 1 minus ij on them is 0, then the commutator should also be holomorphic. And by using that an holomorphic vector satisfies this condition, it's actually
53:43
much easier to check this. You just take a derivative of this and then you dot it with another holomorphic vector, and then you just anti-symmetricize some indices and you get the result pretty easily.
54:01
So as a conclusion, we have that if there is a supercharge, so zeta satisfying those equations implies that the manifold is complex and that the metric
54:26
is compatible with the complex structure, which means that if we choose appropriate coordinates, we can write it as ds squared is going to be g i i bar d z i d z bar i bar.
54:49
So in particular, the metric does not have any g ij or g i bar j bar components.
55:08
So are there any questions so far? So is a statement that there is a solution to the generator? So if there is a solution to those equations, then from that solution you can build a complex
55:23
structure, therefore your manifold is complex, and the metric is Hermitian. So just a characterization of a complex manifold is that I can cover it in patches, and on each patch I can choose complex coordinates z i's, in which the metric,
55:52
for instance, will get that form. And in going from one patch to another, so this is the important part, the transition functions between one patch and another are
56:08
holomorphic functions of the coordinates. So the new coordinate z i prime are holomorphic functions of the z's. So you can cover your manifold in patches which look like C2,
56:25
and then going from patch to patch, you just change the coordinates with holomorphic functions. Okay, so this is showing that if there is a solution, then the manifold is complex.
56:41
Now we can try to go in reverse and show that if the manifold is complex, then we have a solution. So that is what we'll try to do in the following. So now if the manifold is complex, and we're using species structure, then your
57:02
z i's will become now what, 1-0 forms, or 0-0-0-2? Well, the z i's will become a scalar, the supercharge will correspond to a scalar. With the scalar? Yes, and so with the supercharge I can also build an holomorphic two-form. Give me a second.
57:21
Then can you write that multiplexing in a modulated node? Yes, then one can twist various, I mean, are you talking about trying to write down some twisting? Yes, yes, that can be done. Okay, so that's one thing I could
57:48
try to show tomorrow, but I don't know, depending on what people are interested in hearing about. One thing which is important is that you notice that the
58:13
complex structure that I have written down is self-dual, just because sigma mu is self-dual.
58:26
So that will be somewhat important. That's just a manifestation that it can be built out of zeta, which is a left-handed spinor. If instead you add the solution to the other Killing-Spinner equation, the one from the right-handed spinors, then again you would be
58:43
able to write down a complex structure, but that would be anti-self-dual. But the same comments will go through. Let me make a comment first about, let's look at this equation that I brought there,
59:01
and let's just set v to zero. Then it just looks like d mu minus i a mu of z equals zero. So you might want to ask, when can I solve such an equation? And well, the answer is that you basically want the
59:23
holonomy of the Levi-Civita connection. Well, the holonomy of the Levi-Civita connection in general would be in s2 plus times s2 minus, but you would like the s2 left part, the s2 plus part, the one under which the zeta transforms, to just be u1.
59:45
So if that's true, then you can twist away. So if the holonomy of the Levi-Civita connection is u1 times s2 minus instead of s2 plus
01:00:01
times s2 minus, then you would be able to twist away the u1 part by using the u1 R-symmetry, and indeed this can be done on scalar manifolds, so this happens on scalar manifolds. Again, you want your complex structure to be self-dual so
01:00:26
that the u1 will be in the right place. If it's anti-self-dual, then you will do it with the zeta tilde. Okay, so that's a comment, but it gives in spirit what you would want to accomplish, so this can be accomplished on scalar
01:00:42
manifolds, but we would like to show that if we have the extra freedom of adding this other background field v, then we can do this on any complex manifold. Does it really imply scalar or just commission complex? Well, I think it's true for
01:01:03
scalar. I'm not completely sure about the reverse. Okay, so in order to proceed, the one thing that we can look at
01:01:32
are covariant derivatives of the complex structure. Actually, I think yes
01:01:42
because of what I'm going to say. It's a fact that if the manifold is complex, then the covariant derivatives of the complex structure, they are not all independent, but they are actually all encoded into the
01:02:04
divergence. In order to check this, you have to use the horrible tensor being zero. Okay, so then if this object is zero, then the manifold is
01:02:31
scalar and then you can just set v equals zero and use this trick to solve your Keeling-Spino equation. If this object is nonzero, what we want to show
01:02:45
is that we can make use of v to accomplish the same. This is the divergence. Maybe I should write it more carefully.
01:03:01
The statement is that if the manifold is complex, then all the derivatives of the complex structure are encoded into the divergence. That's a statement that in order to check it, you have to use the fact that j is an actual complex structure and not an almost complex structure.
01:03:22
Let's compute this object by using the Keeling-Spino equation. You have to take a derivative of j and then you have to plug in the equation and what you get is that this is v mu plus v bar mu plus i... no, these are mu minus v
01:03:59
bar mu times j mu nu. Okay, so it's not zero and the fact that it's
01:04:15
nonzero is encoded in v. In particular, you can invert this relation and you can
01:04:23
write v in terms of the divergence of the complex structure, but this does not completely determine v. There is still something that you can add, u mu,
01:04:40
and remember that v has to be conserved. It's an auxiliary field in the conserved vector. That means that this piece works, it's conserved, then this piece also needs to be separately conserved. So grad mu, so the divergence
01:05:02
of u mu is zero and then in order for this to solve that, you also need to impose that the anti-eulomorphic components of u are zero.
01:05:24
This encodes some freedom of choosing v, which is not completely determined by the complex structure of your manifold. So you choose a manifold, it will have some complex structure, then this complex structure is gonna...
01:05:42
the manifold is not scalar, it's gonna tell you what v mu has to be, up to changing the anti-eulomorphic components by a conserved part.
01:06:02
To conclude, you have to use another fact about complex manifold, which is... as you see here, the covariant derivative of the complex structure is not zero, it's encoded in v, but if you change... if instead of working with the
01:06:26
Levi-Civita connection, you choose a different connection, you can choose a connection such that it is compatible with the metric. So I'm gonna claim that there exists a connection that is compatible with the metric and it's
01:06:52
also compatible with the complex structure. So this connection goes...
01:07:06
well actually there is more than one, there is an entire one parameter family of such connections, but we are going to choose one in particular which is called the Chern connection. So the definition of the Chern
01:07:20
connection is that, like say we write down the spin connection, omega mu nu rho for the cern connection, that is going to be the Levi-Civita one, the one that you all know about, but then you have to subtract a piece which gives the contortion tensor and the contortion tension is determined by the
01:07:47
complex structure in the following way, where this is just the Levi-Civita spin
01:08:09
connection. So the claim is that if you use this spin connection, then it is compatible with the metric and with the complex structure. So it
01:08:22
makes sense to use this one instead of the Chern connection and why is that? Well it's because this connection is compatible with gem mu nu, its solonomy is going to be U1 times SU2. So what did I call it? Plus. So that's because if you take... when the
01:08:59
manifold is complex you can take frames which are oromorphic and then
01:09:03
like in going from patch to patch you can just... two oromorphic frames are going to be related by a U1 times SU2 transformation and because the connection is compatible with the complex structure, an oromorphic
01:09:23
frame is going to stay oromorphic. So now I can try to just take my equation over there and rewrite it in terms of the Chern connection instead of the Levi-Civita one. So let's do it here and let's see what
01:10:05
this becomes. So we have to make some redefinition. So we have to use first of all that V is now known in terms of the complex structure plus a piece which is undetermined and then we have to use the fact that we want to change the
01:10:23
connection to be the Chern one. So after you do all this you get the following, where A mu C is just the old UNR gauge field A mu but plus a well
01:10:52
defined one form which is some function of the complex structure. Okay
01:11:12
so this means that you're just shifting the UNR connection by a well-defined one form so that's legit and then you can rewrite that equation
01:11:24
in this way and by now it's clear how you want to solve it because the claim is that the holonomy of the Chern connection is in U1 times SU plus but we are just looking at left-handed spinors so I can just twist away the
01:11:40
holonomy by using the U1-R connection yeah the ambiguous part drops from the equation. Okay so so we can we can be a little bit more explicit and how much
01:12:02
time do I have left like 15 minutes okay so in order to be a little bit
01:12:42
more explicit let's introduce another object call it P mu nu and that's just going to be Z sigma mu nu times Z. So this object you can check in these two indices it is holomorphic but then it also has R charge 2 because it
01:13:06
it has two spinors so this guy lives into the square of the UNR line bundle times times the bundle of holomorphic two forms which is also
01:13:30
called the canonical line bundle so you can check that indeed this thing is
01:13:40
is a line bundle in the same way as you check that the determinant line bundle is a line bundle. Okay so but this guy is different than zero everywhere therefore this tells you that this object is trivial and therefore we can
01:14:05
identify L with minus a half of the canonical line bundle so this does not
01:14:22
make it well so then L is not going to be well defined but L times K to the other half will be and that's basically what you use to construct to construct the spinner. So very explicitly we can take some coordinate
01:14:42
patch so in some coordinate patch we can take the component 1 2 of P and we can call this little p then we can define another object s which is little p times g to the minus a quarter where g is the determinant of the metric so we can we can construct this object and so s is clearly
01:15:13
different than zero and it has R charge 2 and now you can ask how does s change under changes of coordinates which are holomorphic so now
01:15:24
you do a holomorphic coordinate change and what you discover is that the new s in the new coordinates is equal to the old s in the old coordinates times
01:15:47
a phase and this phase is the determinant of dz prime over dz divided by the determinant of dz bar prime over dz bar to the power a half so it's
01:16:07
important that this thing is a phase now s has R charge 2 so that means that in going from patch to patch in doing a holomorphic coordinate change you can undo it by doing a U and R transformation so this can be
01:16:24
undone with U and R and therefore under holomorphic coordinate changes followed by appropriate U and R transformations this s transforms like a scalar yes
01:16:50
indeed so under holomorphic coordinate changes plus U and R s is scalar and
01:17:12
now we can just use s to construct our spinner as Nikita was saying okay so
01:17:20
let's see so again this is gonna be very I mean for people who are not interested in the explicit part like you can just forget about it like I mean the the global existence of the spinner is already proved but you can
01:17:44
actually write down expressions in each patch so to write expression in this patch you have to make some choices so you have to introduce some frame so we choose some holomorphic frame which would be the following one Q square
01:18:04
root of g 1 1 bar is 1 plus square root of 2 over g 1 1 bar 2 1 bar is it you and the e 2 is gonna be equal to square root of square root of 2 over g
01:18:28
1 1 bar g to the 1 quarter times dz 2 so then you can check that the metric is just e 1 e 1 bar plus e 2 e 2 bar and that if you do if you change okay
01:18:51
that's not enough and then you can just solve in this frame for the for the killing spinner equation when written in terms of the of the of the
01:19:04
chain connection which I think I just erased and what you get is that the solution is that equals square root of s over 2 times 1 0 and you can also get what the components of the chain connect of the you and our connection
01:19:25
are going to be minus i over 8 di of log g minus i over 2 di of log s and a i bar c is i over 8 di bar log g minus i over 2 di bar log s okay so in summary
01:19:56
like if you have a complex manifold you can always find one supercharged at
01:20:03
least on it and this superchargers correspond to well it's a complex manifold with the self to a complex structure then like you can find the left-handed spinner which solves the new minimal equations and therefore you get at least one supercharged and the superchargers are charged one and it
01:20:21
will square to zero i over 2 yeah sorry that's my horrible okay so that are there any questions on this so in some sense this is just once you
01:20:46
realize that you can change connections to the chain connection and with that change the equation looks exactly like the one Keller manifolds it's the same story it's not very different okay so there are
01:21:04
some more comments I can make about what if we want manifolds with more than which allow more than one just supercharged we could we could try to find manifolds that have two or more and now there are different
01:21:20
choices so you can ask for two independent Zetas so two independent supercharges with the same R charge so I'm not gonna say much about this case here because it's complicated and also I don't know the answer completely so
01:21:42
what we found is that in the case the manifold is compact then this is a very restrictive this is a very restrictive requirement and in the case the manifold is non compact there are some differential equations for the metric
01:22:00
you have to satisfy and it's not clear what the most general solution is so this is not but the other another case which is interesting is to ask for so this is for 2z for 1z and 1z
01:22:20
tilde so one solution to the one supercharged with R charge 1 and one supercharged with R well then it's clear that your manifold will have a complex structure J mu nu but it also have another complex structure J tilde mu nu which is going to be anti-self-dual instead of self-dual but as it turns out this is not the best way
01:22:47
to describe this manifold it's much easier to work with the bilinear of z and z tilde so you can just build out z sigma mu z tilde and as we said in the last lecture this is a killing vector K mu so you can check various things about this killing vector so
01:23:06
first of all you can check that K is homomorphic with respect to J and J tilde and you can also
01:23:23
check that if you take K and you square it you get 0 it's a complex killing vector so that's not unreasonable and finally you also find out that its norm well okay that's not too
01:23:42
important its norm is non-zero because both of the z and z tilde are everywhere non-zero and then there are actually expressions for J mu nu and J tilde mu nu in terms of K
01:24:03
so J mu nu can be written as some object Q plus a half of epsilon mu nu
01:24:24
K mu K bar nu minus K nu K bar nu okay so now there are like two different cases that mu nu
01:24:49
nu mu okay sorry so this is mu this maybe I should just put m n and m is it better yeah
01:25:14
J tilde would be with the opposite sign one is self-dual and the other one is anti-self-dual
01:25:23
okay so now there are two cases case one K commuted with K bar is different than zero okay so then you have to do some work and
01:25:41
you discover that this implies that the manifold is S3 times S1 or S3 times R and so okay we already know about that case number two is the case when they commute
01:26:03
so this is more or less intuitive right because if it's non-zero then you start getting more than two killing vectors and they have to form an algebra so what that could there are not many options okay so in this case they are if they are zero they commute and then you can actually
01:26:21
write down an expression for the metric so what you discover is that your manifold so these two vector fields are going to generate a torus action on your manifold and the torus never shrinks to zero side because the norm of the killing vector is always different than zero
01:26:41
and then you can write down what the metric of the manifold will be there will be some conformal factor depending on some complex coordinate z which is never zero and then we'll have some coordinate dw plus h zz bar dz times dw bar
01:27:07
plus h bar zz bar dz bar plus some function czz bar dz dz bar and this omega square is equal to
01:27:27
q the norm of k squared so in this case i think k is just dw okay so i think i should stop but
01:27:51
i'm yeah i'm gonna say a little bit more about this next time yeah h and h bar and c
01:28:01
are arbitrary functions so this tells you how the torus is fibered over the base and this is the metric over the base so in particular next time we'll see that well s3 times s1 is indeed of this one any other question i think everybody is sweltering