Sumários

The representations of sl(2,C) and sl(3,C)

3 fevereiro 2022, 12:30 • John Huerta

The representations of sl2(ℂ )

Remember where we were: we fixed a nice basis of sl 2(ℂ), consisting of matrices E, H and F, which satisfy:

[H, E] = 2E,  [E,F] = H,  [H, F] = -2F.

We decompose any irrep V into the eigenspaces of H:

V = ⊕ a V a

In Lie theory, the eigenvalues occuring here have a special name: they are called the weights of V. The eigenspaces V a are called weight spaces. An eigenvector v in V a is called a weight vector.

Last time we saw that if v is a weight vector in V a , then Ev is a weight vector in V a+2 , while Fv is a weight vector in V a-2. So we get this picture of the representation:

V = ... ⊕ V a-2 ⊕ V a ⊕ V a+2 ⊕ ...

where H multiplies by the weight, E rasies the weight, and F lowers the weight. Because V is finite dimensional, this cannot go on forever, so there must be a largest weight in this sequence:

V = ... ⊕ V a-2 ⊕ V a ⊕ V a+2 ⊕ ... ⊕ V a_max

The weight a_max is called the highest weight, and a nonzero vector v in V a_max is called a highest weight vector. A highest weight vector has the following property: if we try to raise the weight with E, we get 0: Ev = 0.

Similarly, there must be a lowest weight:

V = V a_min ⊕ ... ⊕ V a-2 ⊕ V a ⊕ V a+2 ⊕ ... ⊕ V a_max

and for any vector v in V a_min, we must have Fv = 0.

Now, if we start with a highest weight vector v in V a_max and keep lowering the weight with F, we will eventually get a vector in V a_min. Let us suppose this happens in m steps. Then m is a natural number such that:

F mv is nonzero, but F m+1v = 0.


Proposition. The vectors {v, Fv, F 2v, ..., F mv} form a basis of V.

Proof.
These vectors are linearly independent because they are eigenvectors (weight vectors) with distinct eigenvalues (weights).

To show they span V, let W = span{v, Fv, F 2v, ..., F mv}. Then W is a nozero subspace of V, and it is preserved by the action of E (which raises the weight),  F (which lower the weight) and H (which multiplies by the weight). Thus, W is a nonzero subrepresentation of V. Because V is irreducible, we conclude W = V. QED.

In this basis, we know exactly how H acts:

H F kv = (a_max - 2k) v

and we know exactly how F acts

F F kv = F k+1v.

But it is less clear how E acts:

E F kv = ?

Let us derive a formula for the action E. First, we know Ev = 0. For EFv, we compute:

EFv = FEv + [E,F]v = 0 + Hv = a_max v.

And for EF 2v, we compute:

EF 2v = FEFv + [E,F]Fv = a_max Fv + HFv = (a_max + (a_max - 2)) Fv.

Continuing in this way, we can discover the pattern is:

EF k+1v = (a_max + (a_max - 2) + (a_max - 4) + ... + (a_max - 2k)) F kv.

This formula simplifies to:

EF k+1v = (k+1)(a_max - k) F kv.

We learn something magical when we apply this formula for k = m:

EF m+1v = 0 = (m+1)(a_max - m) F mv.

It vanishes because F mv is in the lowest weight space, so F m+1v = 0. But on the right hand side, the m+1 factor is nonzero, and the vector F mv is nonzero. So the only way for the right hand side to vanish is if:

a_max = m.

Look at this. The highest weight a_max is natural number! Specifically, it is exactly the number of steps we need to go from the highest weight vector v to the lowest.

Theorem. For each natural number m (including zero), there is a unique irreducible representation (V (m), π m) of sl 2(ℂ) with highest weight m.

Exercises. You should prove that:

  • (V(0), π0) is the trivial representation ℂ.
  • (V(1), π1) is the standard representation ℂ2.
  • (V(2), π2) is the adjoint representation.
See also exercises 3 and 7 in the final exam study guide.

The representations of sl3(ℂ).

NONE OF THE FOLLOWING WILL BE ON YOUR EXAM. 

We don't have time for much more. The theory for sl 3(ℂ) starts the same way: choose a Cartan subalgebra h in sl 3(ℂ). As with sl 3(ℂ), we choose h to be the subalgebra of diagonal matrices. Now h is 2-dimensional:

h = span{ H 1, H 2}

where H 1 and H 2 are the matrices:

H 1 =

1
   0
  0
0
 -1
  0
0
  0   0

and

H 2 =

0
  0   0
0
  1   0
0
  0  -1

As before, the Cartan subalgebra h is maximal abelian, and ad H is diagonalizable for all H: recall, from last lecture, that we have the formula ad H(E ij) = (a i - a j) E ij, where H is the diagonal matrix

a1
   
  a2
 
     a3

and E ij is the elementary matrix, with 1 in the ijth entry and 0 elsewhere.

The idea is again to apply

Deep theorem. Given any finite-dimensional complex representation (V, ∏) of sl n(ℂ), then ∏ H is diagonalizable for all H in h.

We use this, analogously to sl 2(ℂ), to write a representation V of sl 3(ℂ) as the direct sum of weight spaces:

V = ⊕ a V a

But now we meet a real difference. Because, there was only one H, and a weight was simply an eigenvalue of H. Now there are many Hs. What is a weight? What is a weight space?

Definition. A weight a is a linear map a : h -> ℂ. A weight vector v with weight a in a representation V is a vector v in V such that Hv = a(H) v, for all H in h. The weights with nonzero weight vectors in a representation V are called the weights of V.

Weights are a generalization of eigenvalues, and weight vectors are a generalization of eigenvectors, that allow us to diagonalize all the H in h at once. And we really can diagonalize all the H in h simultaneously, because they commute!

To get a feel for the weights of a representation of sl 3(ℂ), let us consider an example.

Example. The weights of the adjoint representation.

We already know the weight vectors - they're the elementary matrices E ij, at least for i not equal to j. This is because of the equation:

ad H(E ij) = (a i - a j) E ij

(When i = j, E ii has trace 1 and is not in sl 3(ℂ))

So, let us define the weight α ij: h -> ℂ by the formula α ij(H) = a i - a j. Then the above equation reads:


ad H(E ij) =  α ij(H) E ij

To really get a picture of the weights of the adjoint representation, note that:

α ij = -α ji,    α ii = 0,    and  α ij + α jk = α ik .

The above relations imply that all the weights of the adjoint representation can be expressed as a linear combination of two weights. Let us write:


α = α 12, and β = α 23.


Then the other nonzero weights are α 13 = α + β, α 21= -α, α 32 = -β, and α 31 = -α - β. To get a picture, let us plot these weights on the αβ-plane:



In the picture, we draw one dot for each weight space in the adjoint representation, except in the middle: the weight space of weight zero is two-dimensional, and we depict this by adding the extra circle around the zero weight. We have six dots around the outside and two in the middle. The total is eight, as it must be - dim(sl 3(ℂ)) = 8.


This picture is usally drawn with more symmetry, as a regular hexagon with two dots in the middle:



That's just the adjoint representation, but it turns out that all representations of sl 3(ℂ) have pictures like this.


The Eightfold Way

In the 1940s and 1950s, physicists built particle accelerators and began to collide protons together at high energies. Protons are strongly interacting particles, and the collisions produced many new particles. Such strongly interacting particles are called hadrons - the Greek root hadros means strong.

No one expected such a zoo of new particles, and so a search was on for some kind of order, some system to classify the hadrons. Many properties of the particles were measured. Each particle X had an electric charge q(X), which is an integer in suitable units. But many other kinds of "charge" were measured. It turned out each particle had a property, called strangeness s(X), which was also an integer.

When you plot the charge and the strangeness of hadrons on a plane, certain patterns emerge. For instance, here is the spin-0 meson octet (mesons are a type of hadron):



(Image by Laurascudder for Wikipedia.)

And here is the spin-1/2 baryon octet. (Baryons are another type of hadron, which includes the proton and neutron. In fact, the proton and neutron are the particles n and p at the top):



(Image by Laurascudder for Wikipedia.)

As you can clearly see, both of these are pictures of the adjoint representation of sl 3(ℂ)! This led the American physicist Murray Gell-Mann and the Israeli physicist Yuval Ne'emann to propose independently the eightfold way hypothesis. The name comes from appearance of eight particles in the octets, and was Gell-Mann's allusion to the eightfold path to enlightenment in Buddhism. Here is the hypothesis:

Hypothesis. (The eightfold way)
The hadrons are classified by representations of sl3(ℂ).

Representations of Lie groups

31 janeiro 2022, 16:30 • John Huerta

Let G be a Lie group, and g = Lie(G) its Lie algebra. In the first half of this course, we studied representations for finite groups. We now wish to study them for Lie groups.

Definition. A representation of G is a homomorphism π: G -> GL(V), where V is a finite-dimensional vector space over ℝ or ℂ.

Example. Every Lie group G has a god-given representation on its Lie algebra g, called the adjoint representation of G, and denoted Ad: G -> GL(g). If G is a matrix Lie group, then Ad is given by conjugation of matrices:

Ad A(X) = AXA -1 , for A in G and X in g.

More generally, if G is any Lie group, Ad comes from differentiating the conjugation action of G on itself.

We can also study representations of Lie algebras.

Definition. Let g be a Lie algebra over 𝕂 = ℝ or ℂ. A representation of g is a homomorphism of Lie algebras ∏: g -> gl(V), where V is a finite-dimensional vector space over 𝕂.

Example. Every Lie algebra g has a god-given representation on itself, called the adjoint representation of g, and denoted ad: g -> gl(g). An element X in g acts on Y in g by bracketing: ad X(Y) = [X,Y].

Exercise. Show that ad is a representation of g for any Lie algebra g. You will need the Jacobi identity!

Now we want to ask, how are representations of a Lie group G related to those of its Lie algebra g = Lie(G)? To understand this, let us reach into our toolbox in differential geometry: given any smooth map f: M -> N between manifolds M and N, and a point p of M, we get a linear map between tangent spaces, f': T pM -> T f(p)N, the pushforward.

Now apply this tool to a homomorphism of Lie groups, φ: G -> H. Let us take the pushforward at the identity:

φ': T eG -> T eH.

The second tangent space is at e because φ(e) = e - it's a group homomorphism! But we know these tangent spaces by another name; they're the Lie algebras of G and H, respectively. So we get a linear map:

φ': g -> h

Fact. The linear map φ' is a Lie algebra homomorphism.

Thus, from Lie group homomorphisms, we get Lie algebra homomorphisms by pushforward at e. Finally, let us apply this idea to representations. A representation of G is a homomorphism π: G -> GL(V). Taking the pushforward at e, we get a Lie algebra representation π': g -> gl(V).

Key question: can we go backwards, and get Lie group representations from Lie algebra representations?

It turns out the answer is yes, if we assume that G satisfies a topological condition: G must be "simply connected".

Definition. A topological space X is a simply connected if

  • for any two points x and y, there is a continuous curve connecting them: that is, there is a map c: [0,1] -> X such that c(0) = x and c(1) = y.
  • if c1 and c2 are two curves connecting x and y, c1 can be continously deformed into c2 through a family of curves that connect x and y: we say that c1 and c2 are homotopic.

Proposition. If G is a simply-connected Lie group with Lie algebra g = Lie(G), then any finite-dimensional representation ∏: g -> gl(V) of the Lie algebra comes from a unique representation π: G -> GL(V) of the Lie group via pushforward: ∏ = π'.

This propostion is a big help! Representations of Lie algebras are essentially just linear algebra, so they are much easier to study that representations of Lie groups. Easiest of all is linear algebra over the complex numbers ℂ, because then we can always find eigenvalues and try to diagonalize matrices.

In light of this, we will focus on one family of examples. We will study the representations of the complex special linear group, SL n(ℂ).

Fact. The complex special linear groups SL n(ℂ) is simply connected for all n ≥ 1. Hence the complex, finite-dimensional representations of the Lie group SL n(ℂ) are the same as those of the Lie algebra sl n(ℂ).

SL 1(ℂ) is the trivial group. So the first interesting example is SL 2(ℂ), and its Lie algebra sl 2(ℂ). It is this case that we will focus on for this lecture and most of the next.

In any case, the idea is always to try to diagonalize as many matrices as possible:

Definition. A Cartan subalgebra h of sl n(ℂ) is a maximal abelian subalgebra such that the adjoint action ad H of any H in h can be diagonalized.


Example. Let h be the diagonal matrices in sl n(ℂ). This subalgebra is:
  • abelian, because diagonal matrices commute;
  • maximal, because any additional matrices would be off diagonal, and would fail to commute;
  • adH is diagonalizable: for H the diagonal matrix with entries a1, a2, ..., an, and Eij the elementary matrix with 1 in the ijth entry and 0 elsewhere, we have adH(Eij) = (ai - aj) Eij. This shows that Eij is an eigenvector of adH with eigenvalue ai - aj, and we can use this formula to find a basis of eigenvectors in sln(ℂ).

For sl 2(ℂ), there's a Cartan subalgebra h = span{ H }, spanned by just one element:

H =
 1    0
 0   -1

In fact, we have a standard basis for sl 2(ℂ) including the element H:

E = 
 0    1
 0    0

H =
 1    0
 0   -1

F =
 0    0
 1    0

This basis satisfies the important relations:

[H, E] = 2E,    [E,F] = H,     [H, F] = -2F.

To help us analyze the representations of sl 2(ℂ), we note the following theorem without proof:

Theorem. All complex, finite-dimensional representations of sl n(ℂ) are completely reducible.

So, any representation can be decomposed into a direct sum of irreducible representations (irreps). From now on, we will assume that the representation V is irreducible. The key result for understanding V is a bit deep:

Deep theorem. Given any finite-dimensional complex representation (V, ∏) of sl 2(ℂ), then ∏ H is diagonalizable.

This should be plausible, because H itself is a diagonal matrix, and we already know that ad H is diagonalizable. The miracle is that ∏ H is diagonalizable for all ∏.

We use this diagonalizabity as follows: decompose the irrep V into eigenspaces:

V = ⊕ a V a

where the direct sum is over all complex numbers a which are eigenvalues of ∏ H, and each summand V a is the eigenspace for a. In other words, for all v in V a, we have Hv = av (really ∏ Hv = av, but I am going to suppress ∏).

We know how H acts - it acts by these eigenvalues. Next, we need to know how the other basis elements E and F act:

Proposition. If v is in V a, then Ev is in V a+2 and Fv is in V a-2.  (Really ∏ Ev and ∏ Fv, but I am going to suppress ∏.)


The proof of this proposition is so important, we call it the fundamental calculation:

HEv = EHv + [H, E]v = aEv + 2Ev = (a + 2)Ev.

Similarly, HFv = (a - 2)Fv. This is what we wanted to check.

So, we have arrived at this picture of the representation V:
V = ... ⊕ V a-2 ⊕ V a ⊕ V a+2 ⊕ ...
where:
  • E acts by raising the eigenvalue by 2;
  • F acts by lowering the eigenvalue by 2;
  • H acts by multiplying by the eigenvalue.
We're almost in a position to understand V completely. We will do so in the next lecture.

The exponential map

27 janeiro 2022, 12:30 • John Huerta

Let G be a Lie group with Lie algebra g = T eG. We want to understand the Lie algebra g better, and its relationship to G.

To help us, let us define:
Definition. A one-parameter subgroup of G is a homomorphism φ: ℝ -> G.

Note that this is not actually a subgroup. It's a smooth map φ: ℝ -> G such that

  • φ(0) = e;
  • φ(s + t) = φ(s) φ(t) for all s and t in ℝ.
(In general, for G and H Lie groups, a homomorphism is a smooth map φ: G -> H which is also a group homomorphism: φ(xy) = φ(x) φ(y).)

By differentiating at t = 0, every one-parameter subgroup gives us an element of the Lie algebra: X = φ'(0) is in g. Conversely, though it is not obvious, every element of g determines a unique one-parameter subgroup.

To see how this works, let us assume G is a matrix Lie group. That is, G is a closed subgroup of GL n(𝕂) for 𝕂 = ℝ or ℂ. Consider φ: ℝ -> G a one-parameter subgroup of G, with X = φ'(0) in g.

Note that for any integer N and real number t, we have:

φ(t) = φ(t/N) N.

For large N, the smooth function φ is well-approximated by its Taylor series:

φ(t/N) ≈ φ(0) + t/N φ'(0) + ...

(Note that we can add these terms because they are matrices.)

So approximately:

φ(t/N) ≈ I + t/N X

We substitute this into our formula:

φ(t) = φ(t/N) N ≈ (I + t/N X) N

I haven't proved it, but in the limit as N goes to infinity, this approximation becomes exact:

φ(t) = lim N->∞ (I + t/N X) N

Do you recognize the function on the right-hand side? It's the exponential function of matrices:

φ(t) = exp(tX)

So it seems that we have a map from the Lie algebra to the Lie group:

exp: g -> G

In fact, though we have only argued for the matrix Lie group case here, we always have this map. This is the content of the following theorem.

Theorem. For G a Lie group with Lie algebra g, there is a smooth map

exp: g -> G

called the exponential map, satisfying the following properties:
  1. exp: g -> G is a local diffeomorphism at 0 in g.
  2. For any one-parameter subgroup φ: ℝ -> G, with X =  φ'(0), we have φ(t) = exp(tX) for all t in ℝ.
  3. If G is a matrix Lie group, then exp is given by the usual power series for the exponential function: exp(X) = I + X + X2/2! + ... + Xn/n! + ...
  4. If G is a matrix Lie group, then its Lie algebra is g = { X in gln(𝕂) : exp(tX) is in G, for all t in ℝ}.


Property 4 gives us a way to compute the Lie algebras of matrix Lie groups, a tool we have wanted. Here is an example; more examples are in the lectures notes under Exercise 4.1.1!


Example.  Show that sl n(ℝ), the n x n real traceless matrices, is the Lie algebra of SL n(ℝ). In addition, compute the dimension of the Lie group, dim(SL n(ℝ)).


Solution. Let us temporarily denote the Lie algebra by g = Lie(SL n(ℝ)). We want to show that g = sl n(ℝ). We will need a matrix identity:


det(exp(X)) = exp(tr(X)).

This holds for all n x n matrices with entries in ℝ or ℂ. By the fourth property in the theorem, we know that X is in g if and only if exp(tX) is in G for all t in ℝ. Since G = SL n(ℝ), this means that det(exp(tX)) = 1 for all t. By the identity, we have exp(t tr(X)) = 1 for all t. Differentiating in t and setting t = 0, we conclude tr(X) = 0. So, X is in sl n(ℝ). Conversely, if X is in sl n(ℝ), X is traceless, and the identity shows det(exp(tX)) = 1 for all t.

In sum, we conclude g = sl n(ℝ) as desired.


As a bonus, we can easily work out the dimension now. This is because dim(SL n(ℝ)) = dim(sl n(ℝ)), and sl n(ℝ) is a vector space, so we just need to find a basis. As you can check, there is a basis for sl n(ℝ) consisting of n 2 - 1 matrices, so we have dim(SL n(ℝ)) = n 2 - 1.

Lie groups and Lie algebras

24 janeiro 2022, 16:30 • John Huerta

We recalled the definition of Lie group, and gave abundant examples:

  1. (ℝn, +) is an abelian Lie group.
  2. The real general linear group GLn(ℝ) is, first of all, an open set in the set of n x n real matrices, Matn(ℝ). This makes it a manifold with one chart, and it is easy to check the group operations, given by matrix multiplication and matrix inversion, are both smooth. Hence, GLn(ℝ) is a Lie group.
  3. Similarly, the complex general linear group GLn(ℂ) is a Lie group.
  4. (a) The real special linear group SLn(ℝ) is the subgroup of GLn(ℝ) consisting of matrices with determinant 1. Because the determinant is a continous function, SLn(ℝ) is a closed subgroup. Hence, SLn(ℝ) is a Lie group by Cartan's theorem. Similarly, the complex special linear group SLn(ℂ) is a Lie group.
  • (b) The real orthogonal group On(ℝ) is the subgroup of GLn(ℝ) consisting of matrices A that satisfy ATA = I. Because the map sending A to ATA is continuous, this is a closed subgroup. Hence, On(ℝ) is a Lie group by Cartan's theorem. Similarly, the complex orthogonal group On(ℂ) is a Lie group.
  • (c) The Lorentz group O(1,n) is the subgroup of GLn+1(ℝ) consisting of matrices that satisfy AThA = h, where h is the (n+1) x (n+1) diagonal matrix with diagonal [-1, 1, 1, ..., 1].
  • (d) The unitary group Un is the subgroup of GLn(ℂ) of matrices satisfying A*A = I, where A* denotes the conjugate transpose of A. The special unitary group SUn is the subgroup of GLn(ℂ) satisfying A* A = I and det(A) = 1.
  • (e) The symplectic group Spn(𝕂) is the subgroup of GL2n(𝕂) of matrices satisfying ATΩA = Ω. Here, 𝕂 denotes ℝ or ℂ and Ω denotes the 2n x 2n that looks like this (each block is n x n):
     0  -I
     I
    		 0
We also defined a Lie algebra, and gave abundant examples:

  1. ℝn with zero bracket, [v,w] = 0, is an abelian Lie algebra.
  2. The real general linear algebra gln(ℝ) is defined on the vector space Matn(ℝ), with Lie bracket given by the commutator: [X,Y] = XY - YX.
  3. The complex general linear algebra gln(ℂ) is defined on the vector space Matn(ℂ), with Lie bracket given by the commutator: [X,Y] = XY - YX.
  4. (a) The real special linear algebra sln(𝕂) is the Lie subalgebra of gln(𝕂) consisting of matrices of zero trace, for 𝕂 = ℝ or ℂ.
  • (b) The real orthogonal algebra on(𝕂) is the Lie subalgebra of gln(𝕂) consisting of matrices X that satisfy XT+ X = 0.
  • (c) The Lorentz algebra o(1,n) is the Lie subalgebra of gln+1(ℝ) consisting of matrices that satisfy XTh + hX = 0, where h is the (n+1) x (n+1) diagonal matrix with diagonal [-1, 1, 1, ..., 1].
  • (d) The unitary algebra un is the Lie subalgebra of gln(ℂ) of matrices satisfying X*+ X = 0, where X* denotes the conjugate transpose of X. The special unitary algebra sun is the Lie subalgebra of gln(ℂ) satisfying X* + X = 0 and tr(X) = 0.
  • (e) The symplectic algebra spn(𝕂) is the Lie subalgebra of gl2n(𝕂) of matrices satisfying XTΩ + ΩX = 0, where Ω is the same matrix as above.
In the list above, I have used lower case letters in place of the conventional Fraktur letters.

Every Lie group G has a Lie algebra g, defined on the tangent space to the identity: g = T eG. Naturally, for each of the Lie groups in the first list, its Lie algebra is the corresponding entry in the second list. So far, we only have enough differential geometry to check that T eG is a subspace of g, as we showed in the example where G = O(1,n), the Lorentz group, and g = o(1,n), the Lorentz algebra. Next time, we will see how to get equality.

Lie groups

20 janeiro 2022, 12:30 • John Huerta

We completed our differential geometry toolkit with the Lie bracket of vector fields: given any two vector fields X and Y are a manifold M, the Lie bracket [X,Y] on M is the operator on smooth functions defined by taking the commutator of X and Y:

[X,Y](f) = X(Y(f)) - Y(X(f)), for any f: M -> ℝ smooth.

The remarkable thing is that

Prop. [X,Y] is a vector field on M.

Using coordinates for the proof makes essential use of the fact that partial derivatives commute.

We then defined a Lie group: a Lie group G is a group which is also a manifold such that the group operations are smooth. We stated Cartan's theorem on closed subgroups, and finished up by giving several examples of Lie groups:

  1. (ℝn, +) is an abelian Lie group.
  2. The real general linear group GLn(ℝ) is, first of all, an open set in the set of n x n real matrices, Matn(ℝ). This makes it a manifold with one chart, and it is easy to check the group operations, given by matrix multiplication and matrix inversion, are both smooth. Hence, GLn(ℝ) is a Lie group.
  3. Similarly, the complex general linear group GLn(ℂ) is a Lie group.
  4. (a) The real special linear group SLn(ℝ) is the subgroup of GLn(ℝ) consisting of matrices with determinant 1. Because the determinant is a continous function, SLn(ℝ) is a closed subgroup. Hence, SLn(ℝ) is a Lie group by Cartan's theorem. Similarly, the complex special linear group SLn(ℂ) is a Lie group.
  • (b) The real orthogonal group On(ℝ) is the subgroup of GLn(ℝ) consisting of matrices A that satisfy ATA = I. Because the map sending A to ATA is continuous, this is a closed subgroup. Hence, On(ℝ) is a Lie group by Cartan's theorem. Similarly, the complex orthogonal group On(ℂ) is a Lie group.