# SU(N) symmetry and its representations

If a Lagrangian containing an N-multiplet of fields is invariant under global $\mathbf{SU}(N)$ transformations, does that necessarily imply it is invariant under $\mathbf{SU}(N-1)$, $\mathbf{SU}(N-2)$,... all the way down to $\mathbf{U}(1)$ as well, where the representations in each case are $N$x$N$ matrices?

Also, given a symmetry group $\mathbf{SU}(N)$, do representations involving $M$x$M$ matrices exist for all $M \geq N$?

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The latter question about representations is something that would be better asked on math.SE. I'd suggest that you edit it out and repost it over there. – David Zaslavsky♦ Apr 11 '11 at 20:38
I believe you are asking: Is an (N+M)-dimensional representation of an element of SU(N) also an element of SU(N+M)? – Matt Apr 12 '11 at 4:45
To answer the last part of the question(v1) with the understanding that OP is only interested in irreps: No, one counterexamples would be $su(3)$, which does not have a $4$-dimensional irrep, corresponding to $M=4\geq 3=N$. – Qmechanic♦ Feb 19 '12 at 14:09

## 1 Answer

Well, if something is invariant under the action of some group it means that every element leaves it invariant. In particular, every element that belongs to some subgroup. Also, you can get an obvious representation of the subgroup by restriction. If $\rho: G \to {\rm Aut}(V)$ is a representation then so is $\rho_{H}: H \to {\rm Aut}(V)$ for $H < G$. Naturally, this need not be irreducible anymore (obvious example being representation of a trivial subgroup which is reducible to $N$ components).

Well, they surely do because one can take $T\oplus \dots \oplus T$ where $T$ is trivial one dimensional representation. Let's pretend that you meant irreducible representations so that this question is actually non-trivial. Groups of $A_l$ type (with $l = N-1$) are of rank $l$. This means that finite-dimensional representations (as is the case for the irreducible representations of compact $\mathbf {SU}$ forms of these groups) are indexed by $l$ non-negative integers $\{\lambda_i\}_{i=1}^l$. E.g. for $l=1$ (${\mathbf {SU}(2)}$) this integer is connected to spin by $\lambda_1 = 2s$ (and dimension of these representations is $\lambda_1 +1 = 2s+1$). So we are interested in determining dimensions of the irreducible representations indexed by $\lambda$. This is not so straight-forward; in principle one can determine these dimensions by looking at the weight space of the given representation and counting the number of ways one can arrive at any given weight starting from highest weight and applying annihilation operators.

Instead, one can exploit the fact that $\mathbf {SL}$ is basically the same thing as $\mathbf {GL}$ and use Schur-Weyl duality between $\mathbf {GL}(N,\mathbb C)$ and $\mathfrak{S}_k$ (i.e. the symmetry group of $k$-element set) which states that the representation of the above groups on $(\mathbf C^n)^{\otimes k}$ (with obvious actions) decomposes as sum over partitions $\lambda \in Par(k,N)$ (e.g. $(3,2,1) \in Par(6,3)$) of products of irreducible representations $F^{\lambda} \otimes G^{\lambda}$ (with $F$ resp. $G$ being a representation of $\mathbf {GL}(N,\mathbb C)$, resp. $\mathfrak{S}_k$). The dimensions of these representations can then be computed by fancy combinatorial formulas derived from Young tableaux (which enter due to partitions, of course). Basically one writes down some numbers related to $n$ and number of rows under and columns to the right into each box of the tableaux and multiplies all those numbers together. As for whether every integer greater or equal to $N$ can be obtained in this way, I don't know. I suppose by shear amount of these tableaux this should be true but I'll come back to check it later.

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I don't think that, given some $N$, there are $M$-dimensional irreps for $SU(N)$ for all $M > 1$. For $SU(2)$ it's true, but even for $SU(3)$ I don't believe there are irreps of dimension 2, 4 and 5 (for example). I haven't really thought about it, but if these exist, do tell! – Vibert Dec 20 '12 at 19:06
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