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?
Best Answer
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).