Todd Trimble Notes on SU(2) reps

Contents

Definition of spin representations

The unitary irreducible representations of $SU(2)$ (which is isomorphic to the group of unit quaternions $S^3$) couldn’t be simpler to describe. Start with the tautological representation

If $\exp \colon Vect \to CAlg$ is the free commutative algebra functor, aka the polynomial algebra functor, then by functoriality of $\exp$ we get

so that $SU(2)$ acts by graded algebra automorphisms on $\mathbb{C}[x, y]$. Thus $SU(2)$ acts linearly on each degree $n$ homogeneous component; this component has complex dimension $n+1$. Let $V_n$ denote this component.

We note in passing that $\exp(V) = \sum_{n \geq 0} V^{\otimes n}/n!$, where the homogeneous component $V^{\otimes n}/n!$ is the $n^{th}$ symmetric power of $V$. Hence the spin representation $V_n$ is the $n^{th}$ symmetric power of the tautological representation $V_1$.

Lie algebra representations

It is worth going into more concrete detail. The tautological $\mathbb{C}$-linear action of $SU(2) = S^3$ on $\mathbb{C}^2$ can be described quaternionically as assigning to a unit quaternion $q$ the linear map

(this is a left $\mathbb{C}$-module homomorphism). As a real Lie group, $SU(2) = S^3$ is 3-dimensional. Moreover, it is connected and simply connected, so the continuous linear representations $SU(2) \to GL(W)$ on real vector spaces $W$ will be in natural bijection with Lie algebra representations $su(2) \to gl(W)$.

Elements of $SU(2)$, i.e., $2 \times 2$ unitary matrices of determinant $1$, can be represented as $e^{i A}$ where $A$ is a $2 \times 2$ Hermitian matrix of trace 0. (Similarly, any unit quaternion $q$ is of the form $\cos(\theta) + \sin(\theta)v = e^{\theta v}$ for some purely quaternionic unit vector $v$, seen as generating a subalgebra of the quaternion algebra isomorphic to $\mathbb{C}$.) Thus we identify $su(2)$ with the Lie algebra of Hermitian matrices of trace $0$. A conventional basis is given by

known as Pauli matrices. The tautological Lie algebra representation on $\mathbb{C}^2$ given by these matrices can be written as

By differentiating at the identity the $SU(2)$-action on $\mathbb{C}[x, y]$ by algebra automorphisms, we see the $su(2)$-action on $\mathbb{C}[x, y]$ is by derivations. From the previous display, where $H$ takes $a x + by \mapsto a x - b y$ etc., it is easy to read off the derivation representation:

This gives us the Lie algebra action of $su(2)$ on the spin representation $V_{n}$ (which in our representation is generated by monomials $x^j y^k$ of total degree $n = j + k$). We have

Ladder operators

Since we have been dealing with complex representations of $su(2)$, we may as well complexify $su(2)$, and this will allow us a neater presentation in terms of raising and lowering (aka creation and annihilation) operators. The complexification of $su(2)$ is the complex Lie algebra $sl_2(\mathbb{C})$, and $su(2)$ is known as the real compact form of $sl_2(\mathbb{C})$.

Representations of $sl_2(\mathbb{C})$ are of course very classical, and we relate them to our development above as follows. Define lowering and raising operators by

In matrix form,

and we have the classical Lie algebra relations for $sl_2(\mathbb{C})$:

In terms of derivations, we have

and the action of $sl_2(\mathbb{C})$ on the space $V_n$ of homogeneous degree $n$ polynomials is therefore given by

Thus $A$ lowers the exponent on $x$, and $A^\ast$ raises it.

This representation makes it palpably obvious that the spin $n$ representation is irreducible: starting with any non-zero vector $v$ (i.e., a homogeneous polynomial of degree $n$), if we hit it enough times with a lowering operator $A$, we eventually obtain a lowest weight vector (a nonzero scalar multiple of $y^n$), and from there we can get a nonzero scalar multiple of any monomial $x^j y^k$ by applying the raising operator $A^\ast$ enough times. Hence the submodule generated by any nonzero $v$ is the whole module $V_n$.

Pretty much the same idea shows that any irreducible representation of $sl_2(\mathbb{C})$ is one of the spin $n$ representations. The point is that any irreducible representation of $su(2)$ is finite-dimensional (by a compactness argument), and then the yoga of raising and lowering operators allows us to deduce the existence, in a finite-dimensional representation, of a highest weight vector $v$, or alternatively, of a lowest weight vector. The structure of the submodule generated by $v, A v, A^2 v, \ldots, A^n v$, where $A^n v$ is a lowest weight vector, must be that of $V_n$, just by using the Lie algebra relations between $A$, $A^\ast$, and $H$. (See Fulton and Harris for the more detailed proof.)

Tensor products

The category of $SU(2)$-representations is a symmetric monoidal category, with the action on a tensor product being a diagonal action

At the “infinitesimal” or Lie algebra level, the corresponding action of the enveloping algebra $U sl_2(\mathbb{C})$ on a tensor product is given in terms of generators $x \in sl_2(\mathbb{C})$ by

(corresponding to the fact that the standard coalgebra comultiplication $\delta: U\mathfrak{g} \to U\mathfrak{g} \otimes U\mathfrak{g}$ is determined by its effect on generators $x$, which is $\delta(x) = x \otimes 1 + 1 \otimes x$).

In particular, if $H(v) = \alpha v$ and $H(w) = \beta w$, then $H(v \otimes w) = (\alpha + \beta)(v \otimes w)$. This makes it easy to work out the structure of a tensor product

Namely, if a representation $V$ is specified by a function $\chi_V: \mathbb{Z} \to \mathbb{N}$ where $\chi_V(\alpha)$ is defined to be the dimension of the $H$-eigenspace of $V$ with eigenvalue $\alpha$, then the tensor product of modules is specified by

or in other words by a convolution product:

In particular, if $V, W$ are irreducible, then $\chi_V, \chi_W$ are “uniform distributions” of finite support centered at $0$, and we just work out the convolution product of two such uniform distributions. This has the characteristic shape of an Egyptian or Mayan pyramid, which is a “stacked” sum of uniform distributions. This can be left as an exercise, but anyway here is the answer: if $m \leq n$, then

going from the bottom of the pyramid to the top. And that’s all for now, folks.