Direct sum decomposition of weight spaces and relation to Tensor products.

Question

There are 3 parts to the question that I am trying to understand, and while it is not homework it seems instrumental in decomposition modules into weight spaces and their relation to tensor products.

Let $L= sl(2,\mathbb{C})$, and let $V = \mathbb{C}^2$ be the "natural module" with the standard basis $v_1,v_2$. Now let $W = \mathbb{C}^2$ be another copy of the "natural module":

$\textbf{Question 1}$: What is this "natural module"? What is the basis?

Now with respect to the basis $v_1 \otimes v_1$, $v_1 \otimes v_2$, $v_2 \otimes v_1$, $v_2 \otimes v_2$ one finds that the matrices of e, f, and g are:

$\rho (e) = \left(\begin{array}{cc} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right)$ $\rho (f) = \left(\begin{array}{cc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right)$ $\rho (h) = \left(\begin{array}{cc} 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 \end{array}\right)$

$\textbf{Question 2}$: Where do these come from? The only thing I can think of is that they are using e, f, and g which are elements of $sl(2,\mathbb{C})$ where $e = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)$ $f = \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right)$ $h = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$ as these are used as a basis for an exercise in the first chapter.

Now I can see that the eigenvalues of the 4 x 4 matrices are 0 and 2 hence finding the weights we have $V_0$ and $V_2$ and so we are building up the weight space and have $V_0 \oplus V_2$, I however am unsure of why I am told that $V \otimes W \simeq V_0 \oplus V_2$

$\textbf{Question 3}$: How do I find, in general an explicit direct sum decomposition of $V \otimes W$ into a direct sum of irreducible decompositions?

Any help on any 3 parts is very helpful, this is a roadblock for me. Also note that the 3rd part of the question is Exercise 15.2 from Erdmann and Wildon's book "Introduction to Lie Algebras".

Answer 1

Since the matrix lie algebras ${\frak gl}_n(k)$, ${\frak sl}_n(k)$, etc. are by definition the lie algebras comprised of certain linear transformations of the vector space $k^n$, we call $k^n$ the "natural" space (or module) on which the matrix lie algebras act.

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You're correct; the matrix representations of the standard basis $e,f,g\in{\frak sl}_2(\Bbb C)$ are given by

$$e=\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad f=\begin{pmatrix}0&0\\1&0\end{pmatrix}, \qquad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

The basis $\{v_1,v_2\}$ on $V$ induces a basis on the tensor product $V\otimes V$, say $\{w_1,w_2,w_3,w_4\}$:

$$\begin{array}{ll} w_1=v_1\otimes v_1 & w_2=v_1\otimes v_2 \\ w_3=v_2\otimes v_1 & w_4=v_2\otimes v_2 \end{array}$$

If a lie algebra element acts by $A$ on $V$, then it acts by $x\otimes y\mapsto Ax\otimes y+x\otimes Ay$ on $V\otimes V$. If we use the relations $ev_1=0$, $ev_2=v_1$ we can compute how $e$ acts on $w_{1,2,3,4}$:

$$ew_1=e(v_1\otimes v_1)=ev_1\otimes v_1+v_1\otimes ev_1=0\otimes v_1+v_1\otimes 0=0.$$

$$ew_2=e(v_1\otimes v_2)=ev_1\otimes v_2+v_1\otimes ev_2=0\otimes v_2+v_1\otimes v_1=w_1 $$

and then you should get $ew_3=w_1$ and $ew_4=w_2+w_3$. Since we know how $e$ acts on the basis vectors $w_1,w_2,w_3,w_4$ of $V\otimes V$, we can write down its matrix. You should get the matrix

$$\rho(e)=\begin{pmatrix}0&1&1&0\\0&0&0&1\\0&0&0&1\\0&0&0&0\end{pmatrix}.$$

You can use the same procedure to compute the matrices for $f$ and $h$.

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Finding the irreducible representation decomposition of $V\otimes W$ is generally a hard problem. One often exploits specific features of $V$ and $W$ to get the job done, rather than use some uniform procedure. One can use bilinearity of the tensor product to reduce the problem of decomposing $V\otimes W$ for general $V,W$ to decomposing it for irreducible $V$ and $W$. And for that, I believe we can numerical invariants to compute he multiplicities. I know that for group representations (namely semisimple ones), we can compute the multiplicity of the irrep $U$ inside $V\otimes W$ as

$$\dim_{\Bbb C}\hom_{\Bbb C[G]}(U,V\otimes W)=\dim_{\Bbb C}\hom_{\Bbb C[G]}(\Bbb C,U^*\otimes V\otimes W)$$

$$=\dim_{\Bbb C}(U^*\otimes V\otimes W)^G=\frac{1}{|G|}\sum_{g\in G}\chi_{U^*\otimes V\otimes W}(g)=\frac{1}{|G|}\sum_{g\in G}\chi_U(g^{-1})\chi_V(g)\chi_W(g).$$

Answer 2

$\def\sl{\mathfrak{sl}} \def\C{\mathbb{C}}$

Answer 1: The natural module of $\sl_{2}$ is the two dimensional vector space $\C^{2}$ on which $\sl_{2}$ acts as $2 \times 2$ matrices of trace $0.$ The basis is the standard basis of $\C^{2}$.

Answer 2: So you notice that you have matrices that give the action of $e, f, h$ on the standard basis vectors of $V$ and $W$. Then, the action of (say) $e$ on (say) $v_{1} \otimes v_{2}$ is defined as

$$\rho(e)(v_{1} \otimes v_{2}) = e(v_{1}) \otimes v_{2} + v_{1} \otimes e(v_{2}).$$

You should be able to use the basis $v_{1} \otimes v_{1}, v_{1} \otimes v_{2}, v_{2} \otimes v_{1}, v_{2} \otimes v_{2}$ to compute the new matrices of $e, f, h$.

Answer 3: In general, taking two irreducible representations $V$ and $W$ of a semisimple lie algebra and finding the irreducible decomposition of $V \otimes W$ can be somewhat tricky to compute but there is an algorithm of sorts and it's nice and easy for $sl_{2}.$ I will elucidate using a more general example.

So, hopefully, you know that each irreducible representation of $sl_{2}$ is represented by a positive integer $n$ and that if $V$ corresponds to $n$, then the weight spaces of $V$ are one dimensional and go from $n$ to $-n$ in steps of $2$.

So, suppose $V$ corresponds to $n$ and $W$ corresponds to $m$. Then $\{V_{-n}, \ldots, V_{n}\}$ and $W_{-m}, \ldots, W_{m}\}$ are the weight spaces of $V$ and $W$, and we can write the weight space decomposition of $V \otimes W$ as

$$V \otimes W = \oplus_{i \in \{-n, - n+ 2, \ldots n\}, j \in \{-m, -m+2, \ldots, m\}} V_{i} \otimes W_{j}.$$

Now, we begin the decomposition by picking off highest weight vectors. The highest weight in $V \otimes W$ is $m+n$. So this gives us one irrep corresponding to $m+n$. We now delete one multiplicity from each of the the weights that belong to $V_{m+n}$ (one weight from $-(m+n)$ to $m+n$ increasing in multiples of $2$.) We then look at the highest weight remaining in $V \otimes W$. This gives us another irreducible which we can peel off of $V \otimes W$. We proceed this way until we exhaust all the weights of $V \otimes W$.

Note that at this point, you can actually construct combinatorial formulas which give you the irreducible decomposition but I just wanted to outline the general algorithm without obscuring it in details.

Let's go through the specific example with your $V$ and $W$. In your case, $V$ and $W$ are irreducible representations of dimension $2$ and hence highest weight $1$ (dimension is 1 + highest weight always for an irrep of $sl_{2}$.) Thus, the $V$ and $W$ have a total list of weights (with multiplicities) of $\{-1, 1\}$. Hence, $V \otimes W$ has weight spaces of weight -2, 0 (with multiplicity 2) and $2.$. The highest weight is $2$ so it contains $V_{2}$ as an irrep. This leaves just the weight $0$ with multiplicity 1 which corresponds to $V_{0}$.