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Let us take an N-dimensional (N odd) irreducible representation V of SL(2,R).

It is known that (e.g., Lie groups and Lie algebras III by Vinberg and Onischik, 1994 p. 94) in V there is an invariant symmetric bilinear form $b$ for the action of SL(2,R). Thus, SL(2,R) is embedded into $O(V,b)$ - the orthogonal group of V with respect to the form $b$.

Consider a spin representation of $O(V,b)$, this is a representation in the space of dimension $2^N$. One can restrict this representation to $SL(2,R)\subset O(V,b)$.

The question is: how to decompose this representation of SL(2,R) of dimension $2^N$ into irreducibles?

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I'm not sure whether there is a nice answer to this question, since random restrictions to rank 1 subgroups get unreasonably complicated. But in any case it might be helpful to know something about what motivates the question. –  Jim Humphreys Oct 19 '10 at 16:03
    
Try use characters. –  Melania Oct 19 '10 at 16:37
    
I am a little confused about what is meant by the spin representation in this question. If the dimension of $V$ is $N=2k+1$, then the dimension of the irreducible spinor module $\Sigma V$ is $2^k$. It is the Clifford algebra which has dimension $2^N$ but as a Spin module this is just the exterior algebra $\Lambda V$. So which representation is the question about? $\Lambda V$ or $\Sigma V$? –  José Figueroa-O'Farrill Oct 19 '10 at 18:15
    
José: spinor module over what do you mean? I mean the representation of $SL(2,R)$: take the spin representation of $O(V,b)$ and restrict it to $SL(2,R)$ - you get a $2^N$-dimensional representation of $SL(2,R)$. –  Leonid Petrov Oct 19 '10 at 19:50
    
Jim: I try to identify a certain (other) representation of SL(2,R) which has dimension $2^N$. Well, I know its decomposition into irreducibles, but I do not know which type of "familiar" reps it is. This other representation (in the exterior algebra $\wedge V$) has the following action of the $sl_2$ diagonal generator: $H(e_{i_1}\wedge\dots\wedge e_{i_k})=(2(i_1+…+i_k)−N(N+1)/2)∗(e_{i_1}\wedge\dots\wedge e_{i_k})$. The "spin" representation that is in the question is just a candidate to be equal to this other one. –  Leonid Petrov Oct 19 '10 at 19:58
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2 Answers

up vote 7 down vote accepted

Yes this question is a bit old, but I can never resist some fun character theory. Maybe you have already figured out a satisfactory answer to your original question, but in case not, here is a purely computational method for finding the decomposition you seek.

By general weight theory, the weights of the representation of $SL_2$ of dimension $2n+1$ are $\langle(2n)\beta, (2n-2)\beta, \ldots,(-2n+2)\beta, (-2n)\beta\rangle$ where $\beta$ is the highest weight of the defining 2-dimensional representation (Although $\omega$ is more standard for weights, I use $\beta$ here to avoid confusion with the $\omega$'s I use for the weights of $SO_{2n+1}$).

The weights of the $2n+1$ dimensional representation of $SO_{2n+1}$ are $\langle\omega_1, -\omega_1+\omega_2, -\omega_2+\omega_3,\ldots, -\omega_{n-2}+\omega_{n-1},-\omega_{n-1}+2\omega_n,0,\ldots,\rangle$ where the weights after the $0$ weight are just the negatives of the weights before the $0$ weight.

The embedding $SL_2\hookrightarrow SO_{2n+1}$ that you are interested in takes weights of $SL_2$ to weights of $SO_{2n+1}$; in both lists above I ordered the weights starting from the highest weight and working down.

Associating corresponding weights on the lists, one gets the following correspondence of the fundamental weights:

$\omega_1 = (2n)\beta$

$\omega_2 = (4n-2)\beta$

$\omega_3 = (6n-6)\beta$

$\ldots$

$\omega_{n-1} = (n^2+n-2)\beta$

$\omega_n = (\frac{n^2+n}{2})\beta$

In general, for $k\neq n$, one gets $\omega_k = (2kn-(k^2-k))\beta$

So how do we use this to decompose the spinor representation $\Sigma_n$ (with highest weight $\omega_n$) of $SO_{2n+1}$? Let $X_i = exp(\omega_i)$; then the character of the spinor representation of $SO_{2n+1}$ can be written as:

$\chi(\Sigma_n) = (X_1X_2\ldots X_n)^{-1}(1+X_1)(X_1+X_2)\ldots(X_{n-2}+X_{n-1})(X_{n-1}+X_n^2)$.

From the weight correspondence above, letting $Y = exp(\beta)$ (hence $X_1 = Y^{2n}, X_2 = Y^{4n-2}$ and so on) one gets the restriction of this character back to $SL_2$:

$Res^{SO_{2n+1}}_{SL_2}\chi(\Sigma_n) = Y^{-\frac{1}{6}(4n^3+3n^2-n)}(1+Y^{2n})(Y^{2n}+Y^{4n-2})\ldots(Y^{n^2+n-6}+Y^{n^2+n-2})(Y^{n^2+n-2}+Y^{n^2+n})$

Expanding this product out, one can read off the weights of the restriction as the exponents of $Y$.

For example, when $n=2$ one gets:

$Res^{SO_5}_{SL_2}\chi(\Sigma_2) = Y^{-7}(1+Y^4)(Y^4+Y^6) = Y^3+Y+Y^{-1}+Y^{-3}$

This has weights $3\beta$, $\beta$, $-\beta$, and $-3\beta$, and hence corresponds to the $4$-dimensional irreducible representation of $SL_2$.

When $n=3$, one gets:

$Res^{SO_7}_{SL_2}\chi(\Sigma_3) = Y^{-22}(1+Y^6)(Y^6+Y^{10})(Y^{10}+Y^{12}) = Y^6+Y^4+Y^2+2+Y^{-2}+Y^{-4}+Y^{-6}$

This has weights $6\beta$, $4\beta$, $2\beta$, $0$ (multiplicity 2), $-2\beta$, $-4\beta$, and $-6\beta$ and so this is the sum of the $7$-dimensional irreducible representation with the trivial representation.

Continuing this, for small $n$, the restriction of $\Sigma_n$ splits as $11\oplus 5$ (n=4), $16\oplus 10\oplus 6$ (n=5), and $22\oplus 16\oplus 12\oplus 10\oplus 4$ (n=6).

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Thank you very much for this accurate computation. Actually, my question started from the combinatorial description of the decomposition of restrictions of $\Sigma_n$. This goes as follows. Take the sum $f(\xi)=\xi_1+2\xi_2+\dots+n\xi_n$, where each $\xi_i=\pm1$. Then for $n=5$ you get 16 = the number of possible values this sum can take (from 0 to $n(n+1)/2$); 10 = the number of values it can take for 2 different $\xi$'s, 6 = for three different $\xi$'s, etc. This rep of $SL_2$ appears in study of strict partitions... Then I tried to guess the construction of rep but I was not sure. –  Leonid Petrov Feb 26 '11 at 7:43
    
Incidentally, since the highest weight of the restriction only grows quadratically (it is $\frac{n^2+n+2}{2}\beta$), it follows that for $n\geq 9$ the restriction is not multiplicity free. In fact, for $n=9$, there are 5 distinct irreducible pieces of multiplicity 2 –  ARupinski Feb 26 '11 at 13:59
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I am also unclear about the definition of the spin representation. But if it is the exterior algebra of $V=Sym^{N-1}(W)$ with $W$ the two dimensional defining rep, it is enough to know the decomposition of the graded pieces $$ \wedge^k(Sym^{N-1}(W))\simeq Sym^{k}(Sym^{N-k}(W))\ . $$ This is known since around 1856 and is called the Cayley-Sylvester formula (see this MO post). As for the isomorphism above also called the Wronskian isomorphism it is explained e.g. in this article, section 2.5.

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Well, José in a comment to the question clarifies the matter. –  Leonid Petrov Oct 20 '10 at 5:59
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