Full Transcript
https://www.youtube.com/watch?v=tY7jac-iPA4
[00:00] Today I'm going to tell you how to understand all Lee algebbras.
[00:03] Well, maybe that's not totally true, but we are going to go over the core idea that drives the structure for all Lee algebbras.
[00:13] So, let's look at a bit of an outline of what I mean here.
[00:15] And so, let's start with the notion of a league group.
[00:20] So, a league group is just a group that's also a smooth manifold.
[00:22] And so, you can think about like matrices whose determinant is non zero.
[00:27] That's kind of obviously a smooth manifold and it's a group.
[00:30] And if you're scared of the word manifold, we really just mean a smooth surface.
[00:37] So the interesting thing about Lee groups is that they have algebraic properties because they are groups and they've got analytic properties.
[00:44] So you can like do calculus on them because they are smooth surfaces.
[00:49] And then well what's the Lee algebra?
[00:50] Well the Lee algebra is the tangent space of the lead group at the identity.
[00:55] So if we've got a smooth surface, then we can look at a tangent plane or more
[01:00] we can look at a tangent plane or more generally a tangent space.
[01:03] And well, we might as well do that at the identity,
[01:05] which is maybe the most important element of the group.
[01:07] So here's a little picture of that going on right here.
[01:08] You might say, well, why would we want to do that?
[01:11] Well, it turns out that you can understand a lot of what's going on in the Lee group by studying the Lee algebra using the really powerful tools inside of linear algebra.
[01:13] And so in fact we can't only just do that but we can break apart Lee algebbras into their so-called semi-imple pieces and it turns out that those pieces come from a fairly small list.
[01:15] Well the list is actually infinitely long but there are only a few families and every algebra in one of these families behaves very very similarly.
[01:17] And inside of all of these algebbras, there is a common component.
[01:20] And that component is the simplest Lee algebra, SL2.
[01:23] And we'll see exactly what SL2 is
[02:01] SL2. And we'll see exactly what SL2 is in a minute.
[02:04] And if you can understand what's going on with SL2, then that's what's going on with SL2, then that's going to inform the structure of the whole Lee algebra.
[02:09] Okay. So, just as a bit of a summary of what we've said so far, understanding the representations of SL2, we'll talk about what the representations are, but that's going to be the main goal of the video.
[02:14] So, understanding the representations of SL2 is the seed to understanding the whole subject.
[02:17] And so, today we want to look at or explore all finite dimensional representations of the complex SL2.
[02:34] But before we do that, let's do a bit of a history lesson.
[02:37] So it all starts with Gowwa.
[02:39] So in 1832 he was studying finite groups and polomial equations.
[02:41] Very classical stuff.
[02:45] He died in a duel at 20.
[02:47] But his ideas changed mathematics forever.
[02:51] Studying solvability through symmetry.
[02:53] Next up we'll look at Sophus Lee.
[02:56] So in the 1870s
[03:01] Lee.
[03:04] So in the 1870s he was a Norwegian geometer who saw the same strategy could work for differential equations.
[03:09] But instead of a finite group, we needed a continuous group.
[03:13] So that would be rotations, scalings, and flows.
[03:16] And instead of the group itself, his breakthrough was to linearize, to look at the tangent space of the group at the identity.
[03:25] That's the Lee algebra.
[03:27] Everything we're looking at today will start here.
[03:30] And next up, we'll mention killing from the 1880s.
[03:33] He classified all of the simple Lee algebra.
[03:36] So those were those building blocks I was talking about earlier.
[03:38] There are four infinite families and five exceptional algebbras.
[03:43] But his proof had gaps.
[03:46] Then Carton came onto the scene and filled in the gaps and developed the representation theory that we're going to be looking at today.
[03:53] And nowadays Lee algebbras are everywhere from quantum mechanics, particle physics, and number theory.
[03:59] And all of it traced back to one idea, infinite
[04:02] it traced back to one idea, infinite decimal symmetry.
[04:05] And all of it starts computationally with SL2.
[04:08] And now what we're going to do is derive what the Lee algebra SL2 looks like by looking at its Lee group.
[04:14] And so let's recall we want to look for the tangent space at the identity.
[04:19] But we can do that by looking at curves that go through the identity stacking together their tangent vectors into a tangent space.
[04:27] So now let's recall that capital LL2, the group SL2, we're going to be working over the complex numbers all day.
[04:33] That's going to be the group of 2x two matrices with determinant equal to one.
[04:39] We could also write that as this matrix right here abcd where a d minus bc is equal to 1.
[04:43] That's the determinant formula for a 2x2 matrix.
[04:49] But now what I want to do is I'll take a curve in this space.
[04:54] Remember that this is also like a some sort of surface.
[04:57] So we can take a curve in this space and find its tangent vector.
[04:59] Then we'll package all of those together into little SL2, the Lee
[05:04] together into little SL2, the Lee algebra.
[05:09] So let's take a of T inside of SL2 C.
[05:13] I'll probably append the C from now on, but just right now we'll leave it for just a second.
[05:18] And I want it to go through the identity.
[05:20] So that means I need A evaluated at zero to be the identity matrix.
[05:25] Let's recall that that's just the matrix read rowwise 1 0 0 1.
[05:32] But now I'm going to write this A of T, this matrix, well, this curve that is built out of matrices inside of SL2 with its components.
[05:41] So in other words, I'm going to write capital A of T as the matrix little A of T here, little B of T here.
[05:50] We have C of T in this entry and then D of T in this entry.
[05:53] And so each of those are functions from perhaps the real numbers to the real numbers.
[06:01] So that's making our curve here.
[06:03] But now what do we know? We know the determinant of this a of t is always
[06:06] the determinant of this a of t is always equal to one because it's an SL2 C.
[06:09] So equal to one because it's an SL2 C.
[06:11] So that means we have the following.
[06:11] We have a d minus b c must be equal to 1.
[06:16] And I'm going to append the of t just because it'll get kind of messy.
[06:20] But each of these is a function of t.
[06:23] But next up, what I want to do is I'll take the derivative here with respect to t.
[06:28] Why the derivative?
[06:29] Because we're looking for the tangent vector at the identity.
[06:33] In other words, we want the tangent vector at zero.
[06:38] So uh let's do that.
[06:40] So taking the derivative of this with respect to t will give us a prime uh time d and then plus a * d prime.
[06:49] That's just by the product rule on the first term.
[06:50] And then minus b prime * c minus b * c prime.
[06:58] That's got to be equal to zero because the derivative of one is zero because it's a constant.
[07:03] But next up, what I want to do is evaluate this at t= 0.
[07:06] So let's see what I get
[07:10] this at t is equal to zero.
[07:12] So let's see what I get for each of these terms when t is equal for each of these terms when t is equal to zero.
[07:17] So notice if I evaluate D at zero, I'm going to get one.
[07:20] And that's because we need to be able to evaluate this at zero and get the identity matrix.
[07:26] So that means a of 0 is 1, b and c of 0 are zero, and d of 0 is 1.
[07:34] We don't know what the primes are, but we at least know what those are.
[07:38] So anyway, here we're going to get a prime evaluated at zero because again d of 0 is one.
[07:46] And then for this next term, we'll get d prime evaluated at zero.
[07:51] And then finally, notice that these two objects right here just cancel out or they zero out.
[07:58] That's because c of 0 is 0 and b of 0 is also zero.
[08:01] So bringing the zero from the right hand side of the equation down, we'll see that this is equal to zero.
[08:09] So in other words, we have the following setup.
[08:09] We have a
[08:12] have the following setup.
[08:15] We have a prime of 0.
[08:15] So that's going to be equal to little a prime of 0, little b prime of 0, little c prime of 0, little d prime of 0.
[08:22] So we know that this thing is going to be in the Lee algebra of SL2 because it's a tangent vector at the identity.
[08:30] That's the definition of the Lee algebra.
[08:31] And then we also know that a prime of 0 plus D prime of 0 must be equal to 0.
[08:41] And in fact, that's the entire condition of the Lee algebra SL2 is that the sum of the diagonal elements must be zero.
[08:51] So what we could do is do a little bit of a summary here and we'll have little SL2 and I'll put a C here perhaps for the last time.
[08:59] So that's going to be a 2x2 matrix A B C D such that A + D is equal to zero.
[09:05] So in other words, it's a 2x2 matrix where the trace is zero.
[09:09] The trace is simply equal to
[09:14] is zero.
[09:14] The trace is simply equal to the sum of the diagonal elements.
[09:18] And the sum of the diagonal elements.
[09:21] And now I'd probably be making a mistake if I didn't mention this.
[09:24] All of this is attached to a formula involving the determinant, the trace, the exponential of a matrix, and the logarithm of a matrix.
[09:26] attached to a formula involving the
[09:29] determinant, the trace, the exponential
[09:31] of a matrix, and the logarithm of a matrix.
[09:34] In fact, it goes like this.
[09:34] The determinant of the exponential of a matrix.
[09:37] matrix.
[09:39] So I'll just write that e to the a where a is a matrix is going to be equal to e to the trace of that matrix.
[09:41] a where a is a matrix is going to be
[09:46] equal to e to the trace of that matrix.
[09:48] But then we can look at a logarithmic version of this and we'll see that the log of the determinant of another matrix I'll call it b is going to be the same thing as the trace of the logarithm applied to the matrix.
[09:50] version of this and we'll see that the
[09:53] log of the determinant of another matrix
[09:56] I'll call it b is going to be the same
[10:00] thing as the trace of the logarithm
[10:02] applied to the matrix.
[10:02] So in other words the matrix logarithm.
[10:04] And in fact, this formula with the matrix exponential, the matrix logarithm, as well as the determinant and the trace gives motivation for what the actual operation
[10:07] formula with the matrix exponential, the
[10:09] matrix logarithm, as well as the
[10:12] determinant and the trace gives
[10:15] motivation for what the actual operation on the Lee algebra will be, which we on the Lee algebra will be, which we haven't talked about yet.
[10:19] And it actually comes from this formula that actually comes from this formula that you can derive by expanding the logarithm of x * y * x inverse * y inverse.
[10:29] And so that's like the group commutator of matrices x and y.
[10:31] And so if you expand the logarithm of that, you get x * y - y * x plus a bunch of other stuff.
[10:41] But if you package xy - yx into this bracket notation, that gives you what's known as the lee bracket.
[10:48] And that is the standard operation that we use on a lee algebra.
[10:50] Now we can visualize that as well.
[10:52] So here we've got the Lee group, we've got the Lee algebra.
[10:55] We've got two elements of the Lee algebra.
[10:56] So those are tangent vectors of the identity.
[10:58] They're associated with two elements in the Lee group.
[11:01] So we'll call those little X, little Y, big X, and big Y.
[11:03] And then we can calculate this thing which is the
[11:17] can calculate this thing which is the group commutator of the matrices X and group commutator of the matrices X and Y.
[11:20] And then that's going to be pushed back up to this commutator.
[11:23] this Lee algebra bracket of the Lee algebra elements X and Y.
[11:26] And this intertwining of the group operations and the Lee algebra bracket is what allows us to study the curved group via its flat shadow.
[11:29] In other words, the group via its Lee algebra.
[11:32] So looking back at our Lee algebra SL2.
[11:35] So those are all 2x2 matrices with trace zero.
[11:38] Now we can take an obvious basis for this algebra.
[11:40] So here we've got this diagonal matrix 1 0 0 - 1.
[11:42] This upper triangular matrix which we'll call E which is 0 1 0 0 and then this lower triangular matrix which we'll call F which is 0 0 1 0.
[11:45] But then we just saw what the correct operation was on this Lee algebra.
[11:47] So let's look at those so-called bracket relations between these basis elements.
[11:50] We'll
[12:18] between these basis elements.
[12:20] We'll derive one carefully and then I'll leave it to you as a bit of a homework exercise to derive the others.
[12:24] So we've got this H E.
[12:28] We can write this as H E minus E H.
[12:31] That's that bracket or the commutator.
[12:34] And so expanding that out as matrices, we've got 1 0 0 -1.
[12:37] That's going to be multiplied into the upper triangular matrix E.
[12:43] And then minus 0 1 0 0 * 1 0 0 - 1.
[12:48] Now we just have to remember how we do matrix multiplication.
[12:49] So let's see for this first one we'll have 0 1 and then 0 0.
[12:57] So notice that all the bottom row is just zero and we get a one in the upper right.
[13:01] And then let's see from that we need to subtract.
[13:04] Well that turns into 0 - 1 0 0.
[13:07] But now if we put that together we're going to get 2 * 0 1 0 0.
[13:14] I think that's pretty clear.
[13:17] But that's equal to 2 * e.
[13:17] And that gives us our bracket
[13:21] And that gives us our bracket relation of h with e.
[13:24] So let's bring that up here.
[13:26] So we've got this bracket h with e is equal to 2 * e.
[13:31] And then via a really similar calculation.
[13:33] We have the bracket of h with f is min - 2 * f.
[13:37] And then the bracket of e with f will be equal to h.
[13:43] And that'll bring us up to the main things that we want to look in this video.
[13:46] And those are the representations of a le algebra or in other words using a synonym the modules of a le algebra.
[13:55] So a vector space v is called a g module or a representation of g if there is some sort of bilinear action from g cross v to v.
[14:05] So what I really mean by that is we can take any element of the Lee algebra and it will attack any vector and give us a new vector and it'll do that in a linear way.
[14:16] So it'll do that in a very nice way and then uh furthermore we have this rule that it
[14:24] furthermore we have this rule that it satisfies or it preserves the bracket.
[14:28] satisfies or it preserves the bracket.
[14:29] So in other words, we have the bracket of XY acting on V is the same thing as X
[14:35] acting on V is the same thing as X acting on Y which is acting on V minus Y
[14:38] acting on Y which is acting on V minus Y which is acting on X which is acting on
[14:40] which is acting on X which is acting on V.
[14:43] So notice it essentially just turns this bracket into a commutator kind of.
[14:47] this bracket into a commutator kind of.
[14:51] So let's look at what this uh means inside of SL2 since that's mostly what
[14:54] inside of SL2 since that's mostly what we're interested in here.
[14:56] So that means if we take 2 * e and we act on v, that's going to be the same thing as taking h
[15:01] going to be the same thing as taking h and e and acting on v.
[15:04] Because notice that we know the bracket of h with e is
[15:07] that we know the bracket of h with e is 2 * e.
[15:09] But then we can expand this out.
[15:13] That's going to be h do e dov and then
[15:18] and then minus e do.h.v.
[15:22] minus e do.h.v.
[15:25] So like that.
[15:27] And now we'll often just abuse notation here and we'll rewrite this as h e dov minus e h do.v just to keep it a little bit simpler.
[15:36] And now we can state our main goal which is to classify all finite dimensional irreducible representations of SL2.
[15:44] And let's start with some really simple examples.
[15:47] Now let's start with the simplest example of a representation of SL2.
[15:52] It's the so-called trivial representation.
[15:54] So it's one-dimensional which means it's spanned by a single vector and then everything acts on that vector as zero.
[16:02] So h dov is the same thing as e do.v which is the same thing as f.v which is zero.
[16:07] So this is so simple that it's kind of boring.
[16:10] Now let's move on to the first interesting example which is the so-called standard representation.
[16:17] And for the standard representation, what I mean is that we'll take two-dimensional complex vectors.
[16:21] And this is the standard or sometimes the
[16:25] this is the standard or sometimes the natural representation because our SL2 are 2x2 matrices.
[16:28] natural representation because our SL2 are 2x2 matrices.
[16:29] And what do they naturally do?
[16:32] They multiply into two-dimensional vectors.
[16:35] And so we can take the maybe standard basis of C2 to be 1 0 and 01.
[16:39] So if we know what everything is doing to these two vectors, then we're pretty much good to go.
[16:45] We know everything about this representation.
[16:48] So let's look.
[16:49] So let's notice that if we were to do E times the vector 1 0, so that's going to be the same thing as the matrix that we have for E multiplying into the vector 1 0.
[17:05] But that's pretty clearly equal to the vector 0.
[17:07] But then let's look at E acting on the vector 01.
[17:13] So again, that's going to be the matrix for E multiplied into the vector 01.
[17:19] But check it out.
[17:20] What we get there is back to the vector 1 0.
[17:23] So now we know
[17:26] Back to the vector 1 0.
[17:26] So now we know what E does to these two vectors.
[17:28] Now what E does to these two vectors.
[17:28] Now let's see what F does these two vectors.
[17:32] Let's see what F does these two vectors.
[17:32] So F multiplied into the vector 1 0.
[17:37] And so F multiplied into the vector 1 0.
[17:37] And so we can do matrix vector multiplication here.
[17:40] We'll see that we get 01.
[17:42] Get 01.
[17:42] And then if we look at f multiplied into the vector 01, I'll let you check that you get the vector 0 0.
[17:50] In other words, it zeros the whole thing out.
[17:50] Now let's see what h does.
[17:53] So if we do h on the vector 1 0.
[17:56] So recall that h was the matrix 1 0 - 1.
[17:59] So we need to multiply that into the vector 1 0.
[18:03] So let's observe that we just get the vector 1 0 out again.
[18:07] And then if we do h onto the vector 01, we will get negative what we started with.
[18:14] I'll let you work out the details if you want to.
[18:16] And so here's a picture that we'll draw based off of these actions.
[18:19] And what we'll see is that this picture will be like a general
[18:30] that this picture will be like a general framework for understanding all of the representations of SL2.
[18:35] So let's consider this dot to be corresponding to the vector 01.
[18:41] And let's say this dot right here is corresponding to the vector 1 0.
[18:47] And let's do a little bit of color coding here.
[18:55] Let's say that a line segment or an arrow that is yellow corresponds to acting by E.
[19:01] And then maybe a line segment or an arrow that is pink corresponds to acting by f and one that is blue is corresponding to acting by h.
[19:12] So let's do the actions by h first because notice they take the vector back to themselves.
[19:17] So I can just put a little loop right here to show that that is the action of acting by h.
[19:24] And then I can put one and a negative one in here because notice that those two vectors were Higgen vectors with igen value one
[19:32] were Higgen vectors with igen value one and negative 1.
[19:34] So now let's see what E does.
[19:37] Well by this right here, notice that E takes 01 to 1 0.
[19:40] So that brings us up this chain right here.
[19:44] But then E takes this vector to zero.
[19:46] So we could just write that as an arrow to zero.
[19:49] And then well what about F?
[19:51] Well, it kind of does the opposite of E.
[19:53] So it'll take our vector 1 0 back to our vector 01 and then it'll take this to 0.
[19:56] So before we talk about a general structure, I want to work through one more example that will highlight the fact that this isn't just a cute picture.
[20:09] This is actually something that gets generalized.
[20:11] So for our next representation, we'll call we'll take the so-called adjoint representation that is SL2 acting on itself via the bracket.
[20:17] So in other words, if we have x doy, that's just simply the bracket of x and y.
[20:27] And so now we could just calculate some commutators and then we'll have the whole picture.
[20:30] So notice if we have e do
[20:34] whole picture. So notice if we have e do e, that's e acting on itself. That's the
[20:36] e, that's e acting on itself. That's the bracket of e with e, but that's equal to
[20:38] bracket of e with e, but that's equal to zero just based off the fact that this
[20:40] zero just based off the fact that this bracket was the commutator. This is e *
[20:42] bracket was the commutator. This is e * e minus e * e. But now we could look at
[20:46] e minus e * e. But now we could look at perhaps f e. So that's going to be the
[20:50] perhaps f e. So that's going to be the bracket of f with e, which that's going
[20:52] bracket of f with e, which that's going to give us minus h. So we didn't talk
[20:54] to give us minus h. So we didn't talk about this, but if you look at xy versus
[20:58] about this, but if you look at xy versus yx, they differ by a sign. And so that's
[21:01] yx, they differ by a sign. And so that's why we were able to only write down the
[21:03] why we were able to only write down the bracket of ef and we got h. And then we
[21:06] bracket of ef and we got h. And then we get this one immediately. Okay. So now
[21:09] get this one immediately. Okay. So now let's look at h. E. And notice that that
[21:13] let's look at h. E. And notice that that will give us the bracket of h with e,
[21:15] will give us the bracket of h with e, which was 2 e. We saw that before. So
[21:19] which was 2 e. We saw that before. So now let's move on to perhaps acting on
[21:23] now let's move on to perhaps acting on H. So we'll need E.H. So no notice
[21:28] H. So we'll need E.H. So no notice that's going to be the bracket of E with
[21:30] that's going to be the bracket of E with H, but that's going to be equal to min
[21:33] H, but that's going to be equal to min -2 * E using essentially the one right
[21:37] -2 * E using essentially the one right above. And then that symmetry.
[21:40] above. And then that symmetry. Now we'll do what's next? F.H.
[21:43] Now we'll do what's next? F.H. So that's going to be the bracket of F
[21:45] So that's going to be the bracket of F with H. That's going to give us 2F
[21:49] with H. That's going to give us 2F just by using the fact this is going to
[21:52] just by using the fact this is going to be minus HF, but that's going to cancel
[21:55] be minus HF, but that's going to cancel the minus sign out. And then next we
[21:57] the minus sign out. And then next we have H.H, which is going to be the
[21:59] have H.H, which is going to be the bracket of H with H, which is pretty
[22:02] bracket of H with H, which is pretty clearly equal to zero based off of what
[22:04] clearly equal to zero based off of what we said fairly recently. Okay, so those
[22:08] we said fairly recently. Okay, so those are the actions onto H and the actions
[22:10] are the actions onto H and the actions on to E. Now, what about the actions on
[22:14] on to E. Now, what about the actions on to F? So we'll need EF. So that's going
[22:17] to F? So we'll need EF. So that's going to be the bracket of E with F, which
[22:19] to be the bracket of E with F, which will give us H. We need F.F.
[22:24] will give us H. We need F.F. So that's going to be the bracket of F
[22:25] So that's going to be the bracket of F with F. That will give us zero. And then
[22:28] with F. That will give us zero. And then H do F. That's going to be the bracket
[22:30] H do F. That's going to be the bracket of H with F. That'll give us minus 2F.
[22:34] of H with F. That'll give us minus 2F. So now we know how every element of the
[22:36] So now we know how every element of the Lee algebra acts on every basis vector
[22:41] Lee algebra acts on every basis vector of the representation. It just turns out
[22:43] of the representation. It just turns out that the basis vectors of the
[22:45] that the basis vectors of the representation are simply the elements
[22:47] representation are simply the elements of the Lee algebra. That's why this all
[22:50] of the Lee algebra. That's why this all looks like we've just done the same
[22:51] looks like we've just done the same thing over and over. Now let's draw a
[22:54] thing over and over. Now let's draw a picture just like we had before. So I'm
[22:57] picture just like we had before. So I'm going to lay the basis vectors out in a
[23:01] going to lay the basis vectors out in a certain order. So I'll put H at the end
[23:05] certain order. So I'll put H at the end and then we'll have or sorry H in the
[23:06] and then we'll have or sorry H in the middle, F on the left hand side and E on
[23:08] middle, F on the left hand side and E on the right hand side. And then we'll use
[23:11] the right hand side. And then we'll use our same notation. So we've got arrows
[23:15] our same notation. So we've got arrows that are blue are H, arrows that are
[23:19] that are blue are H, arrows that are yellow are E, and then arrows that are
[23:23] yellow are E, and then arrows that are pink are F. And then we're going to have
[23:26] pink are F. And then we're going to have EN values for all of the H's as we saw
[23:28] EN values for all of the H's as we saw from our calculation.
[23:30] from our calculation. So notice that we had H acting on E gave
[23:34] So notice that we had H acting on E gave us an igen value of two. That's from
[23:38] us an igen value of two. That's from let's see H acting on E right here. H
[23:42] let's see H acting on E right here. H acting on H gave us an igen vector of
[23:45] acting on H gave us an igen vector of zero. Here we got an igen vector of
[23:47] zero. Here we got an igen vector of minus2.
[23:49] minus2. And then furthermore we had E brought us
[23:53] And then furthermore we had E brought us up the chain and then it deposited us to
[23:58] up the chain and then it deposited us to zero after the top if you will. And then
[24:01] zero after the top if you will. And then F brings us down the chain and it
[24:05] F brings us down the chain and it deposits us at zero after we're at the
[24:07] deposits us at zero after we're at the bottom, if you will. And I guess the big
[24:10] bottom, if you will. And I guess the big question here, which I've hinted at
[24:12] question here, which I've hinted at quite a bit, is now that we've seen two
[24:14] quite a bit, is now that we've seen two pictures that look similar like this to
[24:16] pictures that look similar like this to this, does this hint towards some sort
[24:18] this, does this hint towards some sort of general structure inside of
[24:22] of general structure inside of representations of SL2 as a whole?
[24:24] representations of SL2 as a whole? Thanks for sticking around this long in
[24:26] Thanks for sticking around this long in the video. If you're enjoying it, make
[24:27] the video. If you're enjoying it, make sure and give it a thumbs up. And if
[24:28] sure and give it a thumbs up. And if you're not yet subscribed, consider
[24:30] you're not yet subscribed, consider subscribing. It really helps out. Okay,
[24:32] subscribing. It really helps out. Okay, so here's a bit of a fact that we're not
[24:34] so here's a bit of a fact that we're not really going to prove. It's beyond the
[24:36] really going to prove. It's beyond the scope of what we're doing today. And
[24:38] scope of what we're doing today. And that is if V is a finite dimensional
[24:42] that is if V is a finite dimensional representation of SL2, then the action
[24:45] representation of SL2, then the action of H is diagonalizable.
[24:48] of H is diagonalizable. But that means that V has a basis of
[24:51] But that means that V has a basis of igen vectors of H. So if we've got a
[24:56] igen vectors of H. So if we've got a basis of igen vectors of H, then that
[24:58] basis of igen vectors of H, then that means that we can decompose V into
[25:02] means that we can decompose V into pieces that all have the same igen
[25:04] pieces that all have the same igen value. So in other words, we can write V
[25:07] value. So in other words, we can write V as the sum over lambda of V lambda. And
[25:11] as the sum over lambda of V lambda. And here if the V lambda is not equal to
[25:14] here if the V lambda is not equal to just the zero vector space, then lambda
[25:16] just the zero vector space, then lambda is called a weight. And then just what's
[25:19] is called a weight. And then just what's V lambda? Well, v lambda is equal to all
[25:21] V lambda? Well, v lambda is equal to all v and v where hv is equal to lambda v.
[25:24] v and v where hv is equal to lambda v. In other words, it's all of the v with
[25:26] In other words, it's all of the v with an igen vector, an higen, sorry, an igen
[25:29] an igen vector, an higen, sorry, an igen value, an higen value of lambda. And
[25:32] value, an higen value of lambda. And this is called a weight space. So if
[25:35] this is called a weight space. So if everything in our basis is an higen
[25:40] everything in our basis is an higen vector, well, what about e and f? So are
[25:44] vector, well, what about e and f? So are those going to be Higgen vectors as well
[25:47] those going to be Higgen vectors as well or will they may be made up of linear
[25:49] or will they may be made up of linear combinations of things from different
[25:51] combinations of things from different weight spaces? Well, let's see. So we
[25:55] weight spaces? Well, let's see. So we can do this via a pretty simple
[25:57] can do this via a pretty simple calculation.
[25:58] calculation. So let's run through it. Let's do H
[26:02] So let's run through it. Let's do H acting on E acting on V. But then I'm
[26:07] acting on E acting on V. But then I'm going to write that as H E acting on V.
[26:10] going to write that as H E acting on V. Kind of abusing notation a little bit.
[26:13] Kind of abusing notation a little bit. And then I'm going to take that H E and
[26:15] And then I'm going to take that H E and I'm going to write it as E H plus the
[26:19] I'm going to write it as E H plus the bracket of H with E. And then that's
[26:22] bracket of H with E. And then that's going to all be acting on V. Now, why
[26:24] going to all be acting on V. Now, why does that work? Well, let's recall that
[26:28] does that work? Well, let's recall that this H E is essentially going to be the
[26:31] this H E is essentially going to be the same thing as H E minus E. So, if we put
[26:37] same thing as H E minus E. So, if we put these two things together, it cancels
[26:38] these two things together, it cancels back to what we wanted it to. But we
[26:41] back to what we wanted it to. But we know what the bracket of H with E is.
[26:44] know what the bracket of H with E is. What was it? It was 2 E. So that's going
[26:47] What was it? It was 2 E. So that's going to give us E H acting on V and then plus
[26:52] to give us E H acting on V and then plus 2E acting on V. I just wrote this H
[26:56] 2E acting on V. I just wrote this H acting on V. I distributed the V through
[26:58] acting on V. I distributed the V through if you will. But now H acting on V is
[27:03] if you will. But now H acting on V is equal to lambda* V just based off the
[27:05] equal to lambda* V just based off the fact where V comes from. And so that
[27:08] fact where V comes from. And so that means we can take this and we can
[27:10] means we can take this and we can replace it with lambda * v. But now we
[27:13] replace it with lambda * v. But now we can factor out the numbers. Notice that
[27:16] can factor out the numbers. Notice that lambda and 2 are just numbers. And we
[27:19] lambda and 2 are just numbers. And we have lambda + 2 times e dov. So let's
[27:24] have lambda + 2 times e dov. So let's see what we have. We have h acting on
[27:28] see what we have. We have h acting on this vector is the same thing as lambda
[27:31] this vector is the same thing as lambda + 2 times this vector. So what that
[27:35] + 2 times this vector. So what that tells us is that the vector e dov is in
[27:39] tells us is that the vector e dov is in the lambda + 2 igen space. And now we
[27:44] the lambda + 2 igen space. And now we can play the same game to see where f is
[27:47] can play the same game to see where f is or sorry where fv is. So we can do h do
[27:51] or sorry where fv is. So we can do h do fv. So that's going to be the same thing
[27:54] fv. So that's going to be the same thing as let's see that's going to be FH plus
[27:58] as let's see that's going to be FH plus the bracket of H with F and then that's
[28:02] the bracket of H with F and then that's acting on V. But then that's going to be
[28:05] acting on V. But then that's going to be equal to 2 lambda * F minus 2 FV. That's
[28:13] equal to 2 lambda * F minus 2 FV. That's because H acting on V is the same thing
[28:15] because H acting on V is the same thing as multiplying by lambda. And then the
[28:17] as multiplying by lambda. And then the bracket of H with F is minus 2F. But
[28:19] bracket of H with F is minus 2F. But then we can factor stuff out here and we
[28:21] then we can factor stuff out here and we have lambda minus 2 * f dov. But that
[28:26] have lambda minus 2 * f dov. But that tells us that fv is in v lambda minus 2.
[28:33] tells us that fv is in v lambda minus 2. So let's look at the seed of what this
[28:35] So let's look at the seed of what this is telling us. So the seed that we've
[28:37] is telling us. So the seed that we've just seen is the following. So if we've
[28:41] just seen is the following. So if we've got this vector right here that has
[28:43] got this vector right here that has higgen value lambda, then if we push it
[28:47] higgen value lambda, then if we push it up with e, we get an igen value of
[28:49] up with e, we get an igen value of lambda plus 2 and so on and so forth. If
[28:52] lambda plus 2 and so on and so forth. If we push it back this way with f, we get
[28:54] we push it back this way with f, we get an igen value of lambda minus2.
[28:58] an igen value of lambda minus2. But then recall that we want to look at
[29:00] But then recall that we want to look at irreducible and finite dimensional
[29:02] irreducible and finite dimensional modules or representations. Irreducible
[29:05] modules or representations. Irreducible just means that we can't break them into
[29:08] just means that we can't break them into smaller pieces. And well, finite
[29:11] smaller pieces. And well, finite dimensional just to keep everything a
[29:13] dimensional just to keep everything a little simpler. And so that means that
[29:15] little simpler. And so that means that the dimension of any of these v mu has
[29:19] the dimension of any of these v mu has to be zero or one. So if it's
[29:21] to be zero or one. So if it's two-dimensional, then we'll have these
[29:23] two-dimensional, then we'll have these two disjoint lines here and it'll no
[29:26] two disjoint lines here and it'll no longer be irreducible. And then the fact
[29:28] longer be irreducible. And then the fact that it's finite dimensional means that
[29:31] that it's finite dimensional means that there's got to be a vector v 0. So that
[29:34] there's got to be a vector v 0. So that if we hit it with e we get zero because
[29:38] if we hit it with e we get zero because if there's no such vector then this
[29:40] if there's no such vector then this chain will just go infinitely to the
[29:42] chain will just go infinitely to the right. And so what we'll do is we'll
[29:44] right. And so what we'll do is we'll call that vector v sub0 and we'll call
[29:48] call that vector v sub0 and we'll call the weight attached to that vector
[29:50] the weight attached to that vector lambda the highest weight. And the
[29:52] lambda the highest weight. And the vector will be called the highest weight
[29:54] vector will be called the highest weight vector. So that means we've got a
[29:56] vector. So that means we've got a highest weight for this thing of lambda
[29:58] highest weight for this thing of lambda and the vector the highest weight vector
[30:00] and the vector the highest weight vector is v 0. So just as a bit of a summary
[30:03] is v 0. So just as a bit of a summary we've got v 0 in v lambda and then if we
[30:06] we've got v 0 in v lambda and then if we go up that way with e we just get zero
[30:09] go up that way with e we just get zero and then v1 will be f * v 0 and that'll
[30:14] and then v1 will be f * v 0 and that'll be in v lambda minus 2. V2 will be F^2 V
[30:18] be in v lambda minus 2. V2 will be F^2 V 0 that'll be in V lambda minus 4 all the
[30:20] 0 that'll be in V lambda minus 4 all the way down VN will be F applied N times 2
[30:25] way down VN will be F applied N times 2 V 0 and that'll be in V lambda - 2N and
[30:30] V 0 and that'll be in V lambda - 2N and then again since this is finite
[30:31] then again since this is finite dimensional we need to hit a place where
[30:35] dimensional we need to hit a place where if we apply F enough we get zero but
[30:38] if we apply F enough we get zero but that's just working with F descending us
[30:41] that's just working with F descending us down the chain and what happens if we
[30:44] down the chain and what happens if we ascend the chain with E. Well, notice
[30:47] ascend the chain with E. Well, notice that E applied to VK will be a multiple
[30:50] that E applied to VK will be a multiple of VK + one because that moves us up one
[30:53] of VK + one because that moves us up one weight. So now the question is is well,
[30:57] weight. So now the question is is well, what's that multiple? So if we call that
[30:59] what's that multiple? So if we call that multiple CK, then maybe we could find
[31:02] multiple CK, then maybe we could find some sort of formula that will define
[31:05] some sort of formula that will define what CK is. Well, notice that we know
[31:08] what CK is. Well, notice that we know that E * V 0 is 0. But that tells us
[31:12] that E * V 0 is 0. But that tells us that C 0 must be equal to zero because
[31:15] that C 0 must be equal to zero because notice that we've got nothing over there
[31:18] notice that we've got nothing over there on the right hand side. But then we can
[31:20] on the right hand side. But then we can also look at this equation which will be
[31:22] also look at this equation which will be very telling and that is if we take CK +
[31:26] very telling and that is if we take CK + 1 time VK and if we subtract CK * VK and
[31:33] 1 time VK and if we subtract CK * VK and expand those out as follows. So this is
[31:36] expand those out as follows. So this is going to be EF * VK. So let's talk our
[31:40] going to be EF * VK. So let's talk our way through that. So F * VK is going to
[31:44] way through that. So F * VK is going to be VK + 1 by what we have over here. And
[31:48] be VK + 1 by what we have over here. And then if we apply E to VK + 1, we'll get
[31:51] then if we apply E to VK + 1, we'll get back to CK + 1 VK by this formula right
[31:54] back to CK + 1 VK by this formula right here.
[31:55] here. And then from that we will subtract F E
[32:00] And then from that we will subtract F E VK. And that's because E VK will be CK
[32:05] VK. And that's because E VK will be CK VK minus one. and then f brings it out
[32:07] VK minus one. and then f brings it out back up to VK. But let's observe that
[32:11] back up to VK. But let's observe that that's simply the bracket of E with F
[32:14] that's simply the bracket of E with F acting on VK. In other words, that's H
[32:17] acting on VK. In other words, that's H acting on VK.
[32:20] acting on VK. But notice that H acting on VK is lambda
[32:23] But notice that H acting on VK is lambda minus 2K * VK, which we can see from our
[32:27] minus 2K * VK, which we can see from our chart right here. And now what we can do
[32:30] chart right here. And now what we can do is extract the coefficients of VK from
[32:33] is extract the coefficients of VK from both sides of this equation and move
[32:35] both sides of this equation and move some things around and we'll see that CK
[32:38] some things around and we'll see that CK + 1 is equal to CK
[32:41] + 1 is equal to CK plus lambda minus 2 * K. And then we've
[32:44] plus lambda minus 2 * K. And then we've got this seed of C 0= 0. It gives us
[32:48] got this seed of C 0= 0. It gives us this nice recursion for our CK's. And
[32:51] this nice recursion for our CK's. And now that we have this recursion, let's
[32:52] now that we have this recursion, let's see if we can guess at the closed form
[32:56] see if we can guess at the closed form for one of these CKs. And I think we can
[32:59] for one of these CKs. And I think we can do that just by looking at a chart. So
[33:02] do that just by looking at a chart. So let's make our chart. We've got K and
[33:04] let's make our chart. We've got K and then we have C subK. And we'll do this
[33:07] then we have C subK. And we'll do this for a couple of entries. And I think
[33:09] for a couple of entries. And I think after a couple of entries, we'll see
[33:11] after a couple of entries, we'll see what we have. So C of 0 is equal to 0.
[33:14] what we have. So C of 0 is equal to 0. That was our seed. And then C sub 1.
[33:18] That was our seed. And then C sub 1. Well, that's going to be C 0 plus lambda
[33:21] Well, that's going to be C 0 plus lambda minus 0. That's simply going to be
[33:22] minus 0. That's simply going to be lambda. And then for C sub 2, we'll have
[33:25] lambda. And then for C sub 2, we'll have lambda plus lambda minus 2. We can write
[33:28] lambda plus lambda minus 2. We can write that as 2 * lambda minus1. And then
[33:33] that as 2 * lambda minus1. And then let's see for c sub3
[33:36] let's see for c sub3 we'll have 2 lambda minus1
[33:39] we'll have 2 lambda minus1 plus lambda minus well that's going to
[33:41] plus lambda minus well that's going to be min -4. So putting all that together
[33:44] be min -4. So putting all that together we're going to have 3 * lambda minus 2.
[33:49] we're going to have 3 * lambda minus 2. But I think you can maybe see what
[33:51] But I think you can maybe see what pattern is being constructed here. And
[33:54] pattern is being constructed here. And that is for the kth term we will have k
[33:59] that is for the kth term we will have k * lambda minus k + 1. Now that may not
[34:04] * lambda minus k + 1. Now that may not seem super helpful but that's actually
[34:06] seem super helpful but that's actually going to allow us to determine that
[34:09] going to allow us to determine that lambda can only take certain values. So
[34:12] lambda can only take certain values. So let's see how that calculation goes. So
[34:14] let's see how that calculation goes. So so far what we saw is if we have an
[34:16] so far what we saw is if we have an irreducible finite dimensional
[34:17] irreducible finite dimensional representation then it has to have this
[34:20] representation then it has to have this basis v 0 v1 up to vn where v 0 is our
[34:25] basis v 0 v1 up to vn where v 0 is our highest weight vector and then vk + 1
[34:28] highest weight vector and then vk + 1 was f applied to vk. So it's this
[34:31] was f applied to vk. So it's this iterative process and then f applied to
[34:34] iterative process and then f applied to vn was zero whereas e applied to v 0 was
[34:38] vn was zero whereas e applied to v 0 was zero and then finally we had this e
[34:42] zero and then finally we had this e applied to vk is k * lambda minus k + 1
[34:45] applied to vk is k * lambda minus k + 1 * v k minus1
[34:49] * v k minus1 okay good but now I said that we could
[34:53] okay good but now I said that we could get some restrictions on what n or sorry
[34:56] get some restrictions on what n or sorry what lambda has to be so let's see how
[35:00] what lambda has to be so let's see how we can do that. So, it's all going to be
[35:02] we can do that. So, it's all going to be built off of this thing that we already
[35:03] built off of this thing that we already have on the board. We have f applied to
[35:06] have on the board. We have f applied to vn
[35:08] vn is equal to zero. But if we were to call
[35:12] is equal to zero. But if we were to call f applied to vn vn +1, that means that
[35:16] f applied to vn vn +1, that means that we need vn +1 to be zero. But then we
[35:20] we need vn +1 to be zero. But then we know that E applied to VN + one has got
[35:24] know that E applied to VN + one has got to be equal to C N +1 * VN just based
[35:30] to be equal to C N +1 * VN just based off of this rule right here where
[35:32] off of this rule right here where remember that this was the C K number
[35:36] remember that this was the C K number that we had before. But then this Vn
[35:39] that we had before. But then this Vn plus one vector is the zero vector. So
[35:41] plus one vector is the zero vector. So we need this also to be the zero vector.
[35:44] we need this also to be the zero vector. But VN is not the zero vector. which
[35:48] But VN is not the zero vector. which means we need cn +1 to be equal to zero.
[35:53] means we need cn +1 to be equal to zero. But then plugging n +1 into here, that's
[35:56] But then plugging n +1 into here, that's going to give this this equation n +1 *
[36:01] going to give this this equation n +1 * lambda minus n has to be equal to zero,
[36:04] lambda minus n has to be equal to zero, which means lambda must be equal to n.
[36:09] which means lambda must be equal to n. And that's actually a really big
[36:10] And that's actually a really big takeaway is if you've got an irreducible
[36:13] takeaway is if you've got an irreducible representation of SL2 that's finite
[36:16] representation of SL2 that's finite dimensional and let's say it has n +1
[36:20] dimensional and let's say it has n +1 dimensions
[36:21] dimensions then this highest weight lambda has to
[36:25] then this highest weight lambda has to be equal to n one less than the
[36:27] be equal to n one less than the dimension that we're working with. So
[36:29] dimension that we're working with. So now let's look at a nice visualization
[36:31] now let's look at a nice visualization of all of this in action. So we can
[36:34] of all of this in action. So we can start with the trivial representation
[36:36] start with the trivial representation and observe that this satisfies this
[36:39] and observe that this satisfies this picture. So h goes back to the original
[36:42] picture. So h goes back to the original vector with the igen value of zero and
[36:44] vector with the igen value of zero and then e and f turn the vector into zero
[36:47] then e and f turn the vector into zero as well. And then we also already saw
[36:50] as well. And then we also already saw the standard representation the
[36:52] the standard representation the two-dimensional representation had this
[36:54] two-dimensional representation had this picture and the adjint representation
[36:57] picture and the adjint representation also had this picture. So these two
[37:00] also had this picture. So these two representations were what motivate us
[37:02] representations were what motivate us motivated us to look down this road in
[37:05] motivated us to look down this road in the first place. But it turns out that
[37:08] the first place. But it turns out that it continues on. So we've got a
[37:09] it continues on. So we've got a four-dimensional representation that is
[37:11] four-dimensional representation that is built via this picture and then 5 6 7 8
[37:14] built via this picture and then 5 6 7 8 9 10 dimensional representations that
[37:17] 9 10 dimensional representations that build this picture. In fact, this turns
[37:19] build this picture. In fact, this turns into a complete classification of all
[37:23] into a complete classification of all finite dimensional irreducible
[37:24] finite dimensional irreducible representations of SL2. So we won't
[37:27] representations of SL2. So we won't check that this is a complete
[37:28] check that this is a complete classification
[37:30] classification carefully at least. But this classifies
[37:33] carefully at least. But this classifies all of these representations. So before
[37:35] all of these representations. So before we do our final task at looking at a
[37:37] we do our final task at looking at a realization of these representations,
[37:39] realization of these representations, I'd like to tell you about my second
[37:41] I'd like to tell you about my second channel, Math Major, which has full
[37:43] channel, Math Major, which has full lecture courses and mostly upper
[37:45] lecture courses and mostly upper division math classes. And in fact, I
[37:47] division math classes. And in fact, I keep that channel ad free thanks to my
[37:49] keep that channel ad free thanks to my support on Patreon and via channel
[37:51] support on Patreon and via channel memberships. If you'd like to contribute
[37:54] memberships. If you'd like to contribute to helping me keeping that ad free,
[37:56] to helping me keeping that ad free, consider becoming a patron or joining
[37:58] consider becoming a patron or joining the channel, but there's no pressure.
[38:00] the channel, but there's no pressure. Okay, so now that we've classified all
[38:04] Okay, so now that we've classified all of these irreducible finite dimensional
[38:06] of these irreducible finite dimensional representations, let's look at a nice
[38:08] representations, let's look at a nice realization of them. So let's take WN to
[38:12] realization of them. So let's take WN to be the span of all degree in well
[38:16] be the span of all degree in well homogeneous degree in polomials in the
[38:19] homogeneous degree in polomials in the variables X and Y. And so we can write
[38:23] variables X and Y. And so we can write down a pretty simple basis like this.
[38:25] down a pretty simple basis like this. It'll be x the n x n -1 * y. The next
[38:28] It'll be x the n x n -1 * y. The next one will be x n - 2 * y^2 going all the
[38:31] one will be x n - 2 * y^2 going all the way up to y to the n. And then h will
[38:35] way up to y to the n. And then h will act as x * the derivative with respect
[38:38] act as x * the derivative with respect to x - y * the derivative with respect
[38:41] to x - y * the derivative with respect to y. And then e will act as x and then
[38:45] to y. And then e will act as x and then the partial with respect to y. And f
[38:47] the partial with respect to y. And f will be y and then the partial with
[38:49] will be y and then the partial with respect to x. And so we won't check all
[38:52] respect to x. And so we won't check all of the details to show this is a
[38:53] of the details to show this is a representation, but we will do a sample
[38:55] representation, but we will do a sample calculation. So let's see what happens
[38:57] calculation. So let's see what happens when you do the bracket of E with F
[39:00] when you do the bracket of E with F acting on a test function A A. So that
[39:04] acting on a test function A A. So that means we're going to have E F acting on
[39:06] means we're going to have E F acting on A minus F acting on A. But notice that F
[39:11] A minus F acting on A. But notice that F acting on A is simply going to be equal
[39:14] acting on A is simply going to be equal to Y and then the partial of A with
[39:17] to Y and then the partial of A with respect to X. So I'll just write that as
[39:20] respect to X. So I'll just write that as a subx and then here we'll have f and
[39:23] a subx and then here we'll have f and then e acting on a will be x and then
[39:27] then e acting on a will be x and then the partial of a with respect to y. But
[39:30] the partial of a with respect to y. But now let's act with the remaining things.
[39:32] now let's act with the remaining things. So e means we need to take the partial
[39:35] So e means we need to take the partial with respect to y. We've got to use the
[39:38] with respect to y. We've got to use the product rule here. So that's going to
[39:40] product rule here. So that's going to give us x * and then we'll have a subx
[39:44] give us x * and then we'll have a subx plus y and then a sub xy. That's the
[39:48] plus y and then a sub xy. That's the second partial. Then I just left that x
[39:50] second partial. Then I just left that x out front. And then next we'll have
[39:53] out front. And then next we'll have minus and then we'll have a sub y and
[39:57] minus and then we'll have a sub y and then plus x times this partial xy.
[40:02] then plus x times this partial xy. That's just by taking the derivative
[40:03] That's just by taking the derivative again with the product rule. But notice
[40:06] again with the product rule. But notice that we've got some terms that repeat
[40:07] that we've got some terms that repeat and thus cancel. This will cancel with
[40:10] and thus cancel. This will cancel with this. And in the end, we're going to be
[40:13] this. And in the end, we're going to be left with x and then the partial of a
[40:15] left with x and then the partial of a with respect to x minus y and the
[40:17] with respect to x minus y and the partial of a with respect to y. In other
[40:19] partial of a with respect to y. In other words, that's just h acting on a because
[40:24] words, that's just h acting on a because h is exactly that kind of action here.
[40:27] h is exactly that kind of action here. Okay? So this e fa being the same thing
[40:30] Okay? So this e fa being the same thing as h a is exactly the kind of thing that
[40:34] as h a is exactly the kind of thing that you need to check that this is a
[40:36] you need to check that this is a representation in the first place. And
[40:38] representation in the first place. And then you can do some other sample
[40:41] then you can do some other sample calculations to convince yourself that
[40:44] calculations to convince yourself that this is a realization of the same
[40:46] this is a realization of the same representation that we just hinted at.
[40:48] representation that we just hinted at. Let's say we do h acting on x to the a y
[40:52] Let's say we do h acting on x to the a y to the b. Well, just by taking
[40:54] to the b. Well, just by taking derivatives, it's pretty easy to see
[40:57] derivatives, it's pretty easy to see that this is going to be a minus b * x a
[41:02] that this is going to be a minus b * x a y to the b. So that means that this x to
[41:04] y to the b. So that means that this x to the a y the b is an igen vector of h.
[41:08] the a y the b is an igen vector of h. And then you can look at e acting on
[41:11] And then you can look at e acting on this same monomial. And you'll see that
[41:14] this same monomial. And you'll see that this is going to be b x to the a + 1 y
[41:19] this is going to be b x to the a + 1 y to the b minus one. And then f acting on
[41:22] to the b minus one. And then f acting on this same monomial will give us a x to
[41:26] this same monomial will give us a x to the a minus1 y the b + 1. So that gives
[41:31] the a minus1 y the b + 1. So that gives us a hint at this chain forming. So
[41:36] us a hint at this chain forming. So let's see, we've got x to the a y to the
[41:38] let's see, we've got x to the a y to the b in the middle. We'll put that thing in
[41:40] b in the middle. We'll put that thing in the middle. And then over here we'll
[41:42] the middle. And then over here we'll have x to the a + 1 y to the b minus
[41:45] have x to the a + 1 y to the b minus one. And then over here we'll have x to
[41:46] one. And then over here we'll have x to the a minus one y to the b. And we see
[41:51] the a minus one y to the b. And we see that each of these is an igen vector of
[41:56] that each of these is an igen vector of our action by h. And then we've laid
[42:00] our action by h. And then we've laid this out so that E is pushing us up the
[42:03] this out so that E is pushing us up the chain and then F is pushing us down the
[42:08] chain and then F is pushing us down the chain. Now if you check, you'll see that
[42:11] chain. Now if you check, you'll see that E and F don't act with the same scalers
[42:15] E and F don't act with the same scalers as they did before, but you can reindex
[42:19] as they did before, but you can reindex or maybe renormalize the basis here to
[42:23] or maybe renormalize the basis here to have it act with exactly the same
[42:25] have it act with exactly the same scalers.
[42:26] scalers. Okay. So, well, that's about all we
[42:28] Okay. So, well, that's about all we needed to do today. But before we end,
[42:31] needed to do today. But before we end, I'd like to point out that all of the
[42:34] I'd like to point out that all of the semi-imple algebbras, so that would be
[42:36] semi-imple algebbras, so that would be like SLN, we looked at SL2 today, S O N
[42:40] like SLN, we looked at SL2 today, S O N and SP2N, as well as the exceptional
[42:42] and SP2N, as well as the exceptional ones, have copies of SL2 inside of them,
[42:46] ones, have copies of SL2 inside of them, which gives rise to these chains of
[42:49] which gives rise to these chains of representations in their
[42:51] representations in their representations.
[42:53] representations. And those chains allow you to decompose
[42:56] And those chains allow you to decompose the representations of those bigger lee
[42:58] the representations of those bigger lee algebbras so that you can understand
[43:01] algebbras so that you can understand them in terms of this representation of
[43:04] them in terms of this representation of this simplest one this SL2 which is why
[43:08] this simplest one this SL2 which is why understanding this maybe seed case is
[43:11] understanding this maybe seed case is the most important thing to
[43:13] the most important thing to understanding the representations of Lee
[43:15] understanding the representations of Lee algebbrge as a whole and that's a good
[43:17] algebbrge as a whole and that's a good place to