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I know that given an irreducible $\mathfrak sl_2$ module $V$, it is $V\cong V(m), m\in \mathbb Z_+$, whose basis $(v_0,\cdots,v_m$) is such that:

$$H\cdot v_j = (m-2j)v_j, Y\cdot v_j = (j+1)v_{j+1}, X\cdot v_j = (m-j+1)v_{j-1}, \mbox{for } j\geq 0 \mbox{ and }v_{-1} = 0 $$

and $V(m) = \bigoplus^m_{k= -m} V_{m-2k}$.

Now given $V=V(m), W=W(n)$ irreducible $\mathfrak sl_2$ modules, why is that $V\otimes W \cong V({m+n})\oplus \cdots \oplus V({|n-m|})$?

Well, we can choose such a basis for $V$ and $W$, respectively: $(v_0,\cdots, v_m)$ and $(w_0,\cdots, w_n$). Then the set $v_i\otimes w_j, 0\leq i \leq n, 0 \leq j \leq m ,$ is a basis for the tensor product $V\otimes W$. Ok, now applying an element of the cartan subalgebra give us:

$H\cdot v_i\otimes w_j = (m+n-2(i+j))v_i\otimes w_j.$

Since $i+j$ lies in the set $\{1,2,\cdots, n+m\}$, it follows that the eigenvalue for $v_i\otimes w_j$ also lies in the set $\{n+m, n+m-2, \cdots, -n-m+2,-n-m\}$. Now how to proceed from here? Looking for what we got to prove, we see that there are no negative eigenvalue anymore; why is that? any hint on how to proceed?

user2345678
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    You are mixing the notation for the modules of highest weight $m$ and its individual weight spaces. The module with highest weight $m$ has weights $m-2k$, $k=-m,\cdots,m$. But you need to tell us which is denoted which in your textbook, as we don't have a copy. I suspect this mix-up plays a role in your difficulties as well. – Jyrki Lahtonen Oct 07 '19 at 18:43
  • More specifically, if $V(m)$ is that $(m+1)$-dimensional module, then we might denote the weight space spanned by $v_j$ as $V(m)_{m-2j}$. But then the correct claim would read $$V(m)\otimes V(n)\cong V(m+n)\oplus V(m+n-2)\oplus\cdots\oplus V(|m-n|).$$ In other words the summands on the right are modules not weight spaces. For example $V(m+n)$ is an $(m+n+1)$-dimensional module with weights $m+n-2j$ with $j$ ranging from $0$ to $m+n$. – Jyrki Lahtonen Oct 07 '19 at 18:46
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    Like here. – Jyrki Lahtonen Oct 07 '19 at 18:50
  • @JyrkiLahtonen you're right. I was considering it as weight spaces. I will edit the question and try to rethink of it in the correct way – user2345678 Oct 07 '19 at 18:52
  • What's happening is a generalization of the following. Consider the case $m=3$, $n=1$. Then $V(3)$ is spanned by $v_0,v_1,v_2,v_3$ and $V(1)$ by $w_0,w_1$. Then $v_0\otimes w_0$ is the only vector belonging to eigenvalue $3+1=4$, both $v_1\otimes w_0$ and $v_0\otimes w_1$ belong to eigenvalue $2$, both $v_2\otimes w_0$ and $v_1\otimes w_1$ belong to eigenvalue $0$, both $v_3\otimes w_0$ and $v_2\otimes w_1$ to eigenvalue $-2$, and lastly $v_2\otimes w_1$ is the only eigenvector belonging to $-4$. – Jyrki Lahtonen Oct 07 '19 at 18:59
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    You see that the dimensions of weight spaces match those of $V(4)\oplus V(2)$, and this is what you are supposed to verify in general. – Jyrki Lahtonen Oct 07 '19 at 18:59
  • Ok. Now you've got a correct formulation of the result (also known as the Clebsch-Gordon formula) – Jyrki Lahtonen Oct 07 '19 at 19:01
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    You can proceed as follows: The vector $v_0\otimes w_0$ in $V(m)\otimes V(n)$ has the highest weight $m+n$, implying that $V(m+n)$ must be a summand. Accounting for the weights appearing in $V(m+n)$ you the see that the remaining part of the module has a 1-dimensional highest weight space $m+n-2$, and you can proceed to split off a copy of $V(m+n-2)$. You need to keep track of the dimensions of the remaining weight spaces. When you reach $V(|m-n|)$ you will notice that it exhausts all the weight spaces. – Jyrki Lahtonen Oct 07 '19 at 19:09
  • Lord Shark rushed to post an answer that essentially does this, but using the so called (formal) characters. – Jyrki Lahtonen Oct 07 '19 at 19:10
  • @JyrkiLahtonen it seems a good way; arguing with highest vectors like you did and using the formula $(m+1)(n+1) = \sum_{k=0}^m n+m-2k+1$, we can indeed exhaust the dimension of $V(m)\otimes V(n)$ – user2345678 Oct 07 '19 at 19:24

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Define the character $\text{ch}_V$ of a (finite-dimensional) representation of $\frak sl_2$ as $\sum_k \dim E_k(V) t^k$ where $E_k(V)$ is the eigenspace of $H$ acting on $V$ for the eigenvalue $k$. Then $\text{ch}(V\oplus W) =\text{ch}(V)+\text{ch}(W)$ and $\text{ch}(V\otimes W) =\text{ch}(V)\text{ch}(W)$. Also $$\text{ch}(V(n))=t^n+t^{n-2}+t^{n-4}+\cdots +t^{-n}=\frac{t^{n+1}-t^{-n-1}} {t-t^{-1}}.$$ The characters for the $V(n)$ are linearly independent, and the character determines a representation up to isomorphism.

Let $n\ge m$. Then \begin{align} (t-t^{-1})\text{ch}(V(n))\text{ch}(V(m))&=(t^{n+1}-t^{-n-1}) (t^m+t^{m-2}+\cdots+t^{-m})\\ &=(t^{m+n+1}-t^{-m-n-1})+(t^{m+n-1}-t^{-m-n+1})+\cdots +(t^{m-n+1}-t^{-m+n-1})\\ &=(t-t^{-1})(\text{ch}(V(m+n))+\text{ch}(V(m+n-2))+\cdots+\text{ch}(V(m-n))) \end{align} and so $$V(n)\otimes V(n)\cong V(m+n)\oplus V(m+n-2)\oplus\cdots\oplus V(m-n).$$

Angina Seng
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