Tensor product of simple $sl_2$ modules

Question

I am working on the following problem:
Let $M(n)$ be the finite-dimensional, simple $\mathfrak{sl}_2(\mathbb{C})$-module with highest weight $n\in\mathbb{N}_0$. Show that the module $M(l)\otimes_{\mathbb{C}} M(k)$ is cyclic for $l,k\in\mathbb{N}_0$.
I am familiar with the Clebsch-Gordan statement. Hence, I am puzzled by how the dimension of the tensor product can be 1: $$M(l)\otimes M(k)\cong M(l+k)\oplus M(l+k-2) \oplus … M(|l-k|)$$ for $k\geq l$. What am I overlooking? Any help is greatly appreciated!

Answer 1

Assume $k\ge\ell$ for simplicity. For all $j$, $0\le j\le \ell$, let $v_j$ be the highest weight vector of the summand $M(\ell+k-2j)$. Following Mariano's suggestion, consider the vector $$ v=v_0+v_1+\cdots+v_\ell. $$ Let $W$ be the submodule generated by $v$. I will show that $W=M(\ell)\otimes M(k)$. Cyclicity follows immediately.

Let $x$ be the raising operator, $y$ the lowering operator, and $h=[x,y]$. Because the vectors $v_j$ are all weight vectors, we have $h\cdot v_j=(k+\ell-2j)v_j$ for all $j$. It follows that for all exponents $m\in\Bbb{N}$ we have $$ h^m\cdot v_j=(k+\ell-2j)^mv_j. $$ Here I abbreviated the $m$-fold operation by $h^m\cdot v_j=h\cdot(h\cdot(\cdots (h\cdot v)\cdots)$ (if you are familiar with the universal enveloping algebra, this is exactly how it acts).

So for all $m$, $0\le m\le \ell$, we have $$ h^m\cdot v=\sum_{j=0}^\ell(k+\ell-2j)^m v_j. $$ Observe that for all $m$ we have $h^m\cdot v\in W$. We can further extend the idea, and let any polynomials $P(h)$ of $h$ act on the module $W$ in the obvious way. Here $P[T]\in\Bbb{C}[T]$ is the polynomial ring.

Claim. For all $i, 0\le i\le\ell$, the vector $v_i\in W$.

Proof. Consider the span $V$ of the vectors $v_0,v_1,\ldots,v_\ell$. The element $h$ acts on $V$ via the diagonal matrix with elements $\lambda_j=k+\ell-2j$, $0\le j\le \ell$ along the diagonal. Consider the interpolation polynomial $$ P_i(h)=\prod_{0\le j\le\ell, j\neq i}(h-\lambda_j). $$ We clearly have $P_i(h)\cdot v \in W$, and $P_i(h)$ acts as the diagonal matrix with entries $P_i(\lambda_j)$, $0\le j\le\ell$. By design $P_i(\lambda_j)=0$ for all $j\neq i$, but $P_i(\lambda_i)\neq0$. Therefore $P_i(h)\cdot v=z_iv_i$ for some non-zero constant $z_i\in\Bbb{C}$. QED

But $v_j$ generates the summand $M(k+\ell-2j)$ (you only need to apply the lowering operator to it a sufficient number of times). Therefore all the summands $M(k+\ell-2j)\subseteq W$, and we are done.

Answer 2

The dimension of $M(\ell)$ is $\ell+1$. Thus the dimension of the left-hand side is $(\ell+1)(k+1)$, and the dimension of the right hand side is $$(\ell+k+1)+\cdots+(\ell-k+1)=\sum_{i=1}^{k+1}(\ell-k+2i-1)=(\ell-k)(k+1)+(k+1)^2=(\ell+1)(k+1),$$ and they are compatible.