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Recently, I took a course in representation theory at Imperial College, and on the first homework the questions were about certain sneaky relationships when it came to representations of SU(2).

Denoting the $n/2$-representation of SU(2) by [$n$], we were asked to find the relationship between $\text{Sym}^{k} ([n])$ and $\text{Sym}^{n} ([k])$. In the previous questions we had proven among other things that $\text{Sym}^{2} ([n])=\text{Sym}^{n} ([2])$ and $\text{Sym}^{3} ([4])=\text{Sym}^{4} ([3])$. And so I figured, "Ah, they want us to show $\text{Sym}^{k} ([n]) = \text{Sym}^{n} ([k])$! Well, this should be easy!"

I thought it would take half an hour at most, that there was probably some combinatorial formula out there that allowed me to demonstrate it, but after five hours of work, I had to admit defeat. The next day, I admitted I had been defeated to the TA, and asked her to show me the actual proof of the matter, because I was now very interested in seeing it. She was a bit baffled that I had managed to spend 5 hours on the problem, and told me that all I was expected to do was to show that the dimension of $\text{Sym}^{k} ([n])$ equaled to the dimension of $\text{Sym}^{n} ([k])$ (which is rather trivial to prove, as the lectures had given us the derivation of the dimensionality formula of any symmetric power of SU(2)), and from that infer that $\text{Sym}^{k} ([n]) = \text{Sym}^{n} ([k])$.

This answer struck me as rather unsatisfying, and so I set out to try to find the formal proof. My searches on the internet yielded no result, and I cannot find it proven in any mathematics textbook. I am currently trying to go through ring theory and representation theory on my own, but I can see no end to my quest yet.

Consequently, I am now appealing to you ladies and gentlemen online for help on the matter, as the more I learn, the more difficult does it seem to me to prove that

$\text{Sym}^{k} ([n]) = \text{Sym}^{n} ([k])$.

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  • $\begingroup$ I don't understand your TA's advice. These representations aren't irreducible, so you can't hope to infer that they're isomorphic from the fact that their dimensions match. $\endgroup$ – Qiaochu Yuan May 18 '15 at 21:09
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This identification is called Hermite reciprocity, and I find it pretty mysterious. It's a special case of a more general problem called plethysm, which I find extremely mysterious.

Here is a proof. If $V$ denotes the defining $2$-dimensional representation of $SU(2)$, let me write the desired isomorphism as

$$S^n S^m V \cong S^m S^n V.$$

To show that two representations of $SU(2)$ are isomorphic it suffices to show that they have the same character, or equivalently that they have the same weights. If you aren't familiar with this term, "weights" refers to the decomposition of the representation under the action of the diagonal $U(1)$. This decomposition is specified by a multiset of integers, where the integer $n$ corresponds to the one-dimensional representation $z \mapsto z^n$ of $U(1)$. For example, the weights of the irreducible representation $S^n V$ of dimension $n + 1$ are $-n, -n+2, \dots n-2, n$.

Given the weights of a representation $W$, the weights of the symmetric power $S^n W$ correspond to multisets of size $n$ in the weights of $W$. So the question is to exhibit a weight-preserving bijection between multisets of size $n$ in the integers $-m, -m-2, \dots m-2, m$ and multisets of size $m$ in the integers $-n, -n+2, \dots n-2, n$.

It will be much easier to do this after normalizing a bit. First, add $m$ to all of the integers in the first multiset of weights, and add $n$ to all of the integers in the second multiset of weights. This has the effect of shifting the corresponding total weights by $nm$ and so doesn't affect the structure of the problem. Similarly, we can divide all of the weights by $2$.

Now we want to exhibit a weight-preserving bijection between multisets of size $n$ in the integers $0, 1, 2, \dots m$ and multisets of size $m$ in the integers $0, 1, 2 \dots n$. But this is straightforward: they both correspond to Young diagrams fitting in an $n \times m$ box, where we get the first kind of multiset by looking at the lengths of the rows and the second kind of multiset by looking at the lengths of the columns (or maybe the other way around; whichever). In both interpretations the total weight is the total number of boxes, so this is a weight-preserving bijection as desired.

(Pretty mysterious, as I said. The generating function counting such Young diagrams based on the total number of boxes is the q-binomial coefficient ${n+m \choose n}_q$. I have no conceptual explanation of what it's doing here.)

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  • $\begingroup$ Well, this does look impressive! Fear I must disappoint you that much of it is above my head, and so I hope you don't mind a few follow-up questions? I've read some chunks of Georgi, so I'm not entirely unfamiliar with decomposition, weights, etc. but this " "weights" refers to the decomposition of the representation under the action of the diagonal $U(1)$" is something that I've never heard before. Similarly, the rest of that paragraph refers to stuff I sort of feel is reminiscent of what I did in that homework, but I still am not entirely up to speed with the lingo. $\endgroup$ – StormyTeacup May 18 '15 at 21:22
  • $\begingroup$ You wouldn't happen to have any resources you could recommend me to get up to speed with it? $\endgroup$ – StormyTeacup May 18 '15 at 21:28
  • $\begingroup$ @StormyTeacup: well, how did you prove things about representations of $SU(2)$ then? $\endgroup$ – Qiaochu Yuan May 18 '15 at 21:34
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    $\begingroup$ @StormyTeacup: "weights" is another way of talking about the exponents in those polynomials. As you say, symmetric powers don't correspond to a nice operation on polynomials but they do correspond to an operation on weights, which is why I didn't mention polynomials in my answer. $\endgroup$ – Qiaochu Yuan May 18 '15 at 21:45
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    $\begingroup$ @StormyTeacup: that's very unfortunate. I don't know of a great resource that explains symmetric and exterior powers from first principles, but you might try the Wikipedia articles, and if those don't work, maybe Fulton and Harris. $\endgroup$ – Qiaochu Yuan May 18 '15 at 23:07

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