# Tensor product of simple $sl_2$ modules

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!

• Note the last one should be $M(|l-k|)$ it doesn't really matter which way round you put $l$ and $k$ Commented Dec 23, 2023 at 12:35
• @MarianoSuárez-Álvarez Thank you for your clarification! But the following question arised: I know that each summand $M(l)$ of the direct sum is $l+1$ dimensional. Furthermore, it holds that $M(l)=\langle v_0,...,v_m\rangle$, where $v_0$ is the highest weight vector and $v_i=\frac{1}{i!}f^i.v_0$ holds. How does considering the sum of highest weight vectors help me? Commented Dec 23, 2023 at 13:51

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.

• Thank you for this great answer! But I have never heard of the raising (lowering) operator. I read the Wikipedia article but I do not understand how these operators work here. Commented Dec 23, 2023 at 16:24
• @john_psl1298 Sorry about using possibly confusing words (I have been editing the answer heavily). The elements $x,y,h$ are the usual matrices in $\mathfrak{sl}_2$. It turned out that I only really need $h$. Ok, I do need $y$ at the end also :-) Commented Dec 23, 2023 at 16:26
• A nice argument @Mariano. I think it can be made to work even more generally. When there are two isomorphic summands, we need to modify the choice of a generating element, and use a pair of components of unequal weights. By applying a suitable number of a ladder operator we can annihilate one component but not the other etc. Commented Dec 23, 2023 at 19:31
• @Mariano I, too, was surprised about that. For it would lead to a conclusion that all finite dimensional modules are cyclic, which should not be true. Ok, the argument fails for the sum of two trivial modules, because we cannot find distinct weights. What about $M(1)\oplus M(1)$? If $v_0,v_1$ are weight vectors of the former summand (of weights $\pm1$) and $w_0,w_1$ of the latter. Then $v=v_0+w_1$ seems to generate it all. For we have $h\cdot v=v_0-w_1$. Hence $v_0$ and $w_1$ would both be in there. I don't see the error :-) Commented Dec 24, 2023 at 4:19
• @Mariano That makes more sense! There had to be something :-) Commented Dec 24, 2023 at 8:11

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.

• @MarianoSuárez-Álvarez I disagree, the OP finds what seems like a contradiction with the Clebsch-Gordan based off of the dimensions on both sides. I explain why this is not actually a contradiction. Commented Dec 23, 2023 at 10:26