# Symmetric part of the $n$th tensor power of $SU(q)$

Let $$V=\mathbb{C}^q$$. Consider the sequence of tensor powers $$V^{\otimes 2} = V\otimes V, \quad V^{\otimes 3} = V\otimes V\otimes V,\dots$$ It's well-known that there is a "braiding" representation of $$S_n$$ on $$V^{\otimes n}$$ given by permuting the tensor factors; i.e., $$\sigma\in S_n$$ acts linearly and on basis vectors via $$\sigma(v_{i_1}\otimes\dots\otimes v_{i_n}) = v_{i_{\sigma(1)}}\otimes\dots\otimes v_{i_{\sigma(n)}}.$$ The symmetric tensors are defined as the subspace $$\text{Sym}^n (V)\subseteq V^{\otimes n}$$ on which $$S_n$$ acts trivially.

There is a natural representation of $$SU(q)$$ on $$\text{Sym}^n(V)$$. For $$q=2$$ this representation is always irreducible. Is this true for larger values of $$q$$? If so, do all irreps of $$SU(q)$$ appear as $$\text{Sym}^n(V)$$ for some $$n$$ (as is the case when $$q=2$$)?

• From the top of my head I would say that the answer is yes and that this is explained very well in the book 'Representation Theory, a first course' by Fulton and Harris. But I don't fully trust my memory and I currently don't have time to look it up. Hence a comment rather than an answer, but maybe you have access to this book somewhere? May 3 at 16:24
• The symmetric powers are all irreducible but not every irrep is a symmetric power. For a start, we have the exterior powers as well. What is true is that every irrep is contained in some tensor power of the standard rep Jun 14 at 23:21

Dunno what the "standard" proof of irreducibility is, but here's one I devised which is in the same spirit as the one I know in the case of $$q=2$$. First, I'll discuss some representation theory background, then I'll show how it applies to the $$q=2$$ case, and lastly I'll generalize it to the case of arbitrary $$q$$.

Suppose $$V$$ is a complex representation of a real Lie group $$G$$. Differentiating, we get a representation of the real lie algebra $$\mathfrak{g}$$. Complexifying, we get a representation of the complex lie algebra $$\mathfrak{g}_{\mathbb{C}}$$. "Enveloping," we get a representation of the universal enveloping algebra $$\mathcal{U}(\mathfrak{g}_\mathbb{C})$$. This sounds fancy, but has a simple interpretation for us: the range of the representation of $$\mathcal{U}(\mathfrak{g}_{\mathbb{C}})$$ is generated by the range of the representation of $$\mathfrak{g}_{\mathbb{C}}$$ under composition of transformations (remember, the range of a lie algebra homomorphism is only closed under the lie bracket); irreducibility of a representation depends only on the representation's range.

Note if $$V$$ is reducible as a representation of $$G$$, then it is also reducible as a representation of $$\mathcal{U}(\mathfrak{g}_{\mathbb{C}})$$, so by contraposition to show $$V$$ is irreducible as a $$G$$-rep we may show it is simple as a $$\mathcal{U}(\mathfrak{g}_{\mathbb{C}})$$-module. A module is simple iff every nonzero element is a cyclic generator (exercise). For this, it suffices to find a nice generating set for the module, where "nice" means: (i) each generator generates all the others, and (ii) any nonzero element generates one of these generators. Here, "generate" means "obtain via linear combinations of compositions of elements of the range of the representation of $$\mathfrak{g}_{\mathbb{C}}$$."

Compare with showing a group action is transitive. Transitive means a group element can get us from any element to any other element, but it suffices to find a particular element and show it is possible to go to and from that particular element to every other element. Also compare with the proof that the permutation rep of a $$2$$-transitive group action is the direct sum of a trivial rep (the invariant subspace) and an irreducible "augmentation" subrep (where coefficients of generators sum to $$0$$).

Consider the standard representation $$\mathbb{C}^2=\mathrm{span}\{x,y\}$$ of $$\mathrm{SU}(2)$$. We can identify $$S^n\mathbb{C}^2$$ with the space of homogeneous degree $$n$$ polynomials in $$x$$ and $$y$$, on which its complexification $$\mathfrak{su}(2)_{\mathbb{C}}$$ $$=$$ $$\mathfrak{su}(2)\oplus i\mathfrak{su}(2)$$ $$=$$ $$\mathfrak{sl}(2,\mathbb{C})$$ acts by derivations, i.e. the "product rule" applies. (Note, if $$\mathfrak{g}$$ is a real lie algebra comprised of complex matrices, we don't always have $$\mathfrak{g}\cap i\mathfrak{g}=0$$, but when we do we can set $$\mathfrak{g}_{\mathbb{C}}=\mathfrak{g}\oplus i\mathfrak{g}$$.) In particular, if $$x$$ and $$y$$ correspond to the standard basis column vectors $$e_1=[\begin{smallmatrix}1\\0\end{smallmatrix}]$$ and $$e_2=[\begin{smallmatrix}0\\1\end{smallmatrix}]$$, then the standard basis matrices $$e_{12}=[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}]$$ and $$e_{21}=[\begin{smallmatrix}0&0\\1&0\end{smallmatrix}]$$ correspond to $$x\frac{\partial}{\partial y}$$ and $$y\frac{\partial}{\partial x}$$ respectively. Note composition of differential operators does not correspond to matrix multiplication, for example $$e_{11}e_{22}=0$$ as matrices but not inside the algebra $$\mathcal{U}(\mathfrak{gl}_2)$$.

The monomials $$\{x^n,x^{n-1}y,\cdots,xy^{n-1},y^n\}$$ form a basis for $$S^n\mathbb{C}^2$$. They are all also eigenvectors of the diagonal operator $$e_{11}-e_{22}=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}$$, and we can order the monomials by eigenvalue (aka "weight"). Applying $$e_{12}=[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}]$$ or $$e_{21}=[\begin{smallmatrix}0&0\\1&0\end{smallmatrix}]$$ will shift a basis vector to the one just higher or lower in weight (in some incarnations, these may be called "ladder" operators, or "creation / annihilation" operators). Thus, this basis is a "nice" generating set at least in sense (i) above, and also in sense (ii) as we observe below.

Suppose $$v(x,y)$$ is any nonzero homogeneous degree $$n$$ polynomial. It has a "highest" and "lowest" monomial in it. (Higher means higher powers of $$x$$.) We can apply $$e_{12}$$ enough times that it raises the lowest monomial of $$v$$ to be $$x^n$$, and annihilates all other monomials in $$v$$. Thus, $$e_{12}^mv$$ will be $$x^n$$ up to scaling.

Since the monomials are a "nice" generating set (as described in the previous section), $$S^n\mathbb{C}^2$$ is an irreducible representation of $$\mathfrak{sl}(2,\mathbb{C})=\mathfrak{su}(2)_{\mathbb{C}}$$, hence of $$\mathrm{SU}(2)$$.

Now consider $$S^n\mathbb{C}^q$$ acted on by $$\mathfrak{sl}(q,\mathbb{C})=\mathfrak{su}(q)_{\mathbb{C}}$$. Setting $$\mathbb{C}^q=\mathrm{span}\{x_1,\cdots,x_q\}$$, we can say identify $$S^n\mathbb{C}^q$$ with degree $$n$$ homogeneous polynomials in $$x_1,\cdots,x_q$$. The "nice" generating set will be monomials, of course, which we can order reverse-lexicographically. That is, for a monomial $$x_1^{m_1}\cdots x_q^{m_q}$$ (with $$m_1+\cdots+m_q=n$$), we can interpret $$m_q\cdots m_1$$ as number with $$q$$ digits in base $$n+1$$, and say larger numbers are "lower" and smaller numbers are "higher."

Suppose $$cx_1^{m_1}\cdots x_q^{m_q}$$ is the "lowest weight" monomial in a nonzero $$v\in S^n\mathbb{C}^q$$. Then we have

$$(e_{12}^{m_2}\cdots e_{(q-1)q}^{m_q})v=cx_1^nm_2!\cdots m_q!$$

That is, the differential operator above will annihilate all but one monomial in $$v$$. This establishes part (ii) of monomials being a "nice" generating set, I'll leave (i) as an exercise.

Since monomials form a "nice" generating set for $$S^n\mathbb{C}^q$$ (as outlined in the rep thry background section), it is irreducible as an $$\mathfrak{sl}(q,\mathbb{C})$$-module, hence also as an $$SU(q)$$-rep.