Explicit basis for Clebsch–Gordan decomposition of $V_3\otimes V_2=V_5\oplus V_3\oplus V_1$ in $\mathfrak{sl}(2,\mathbb C)$ Motivations & Informations:
https://www.wikiwand.com/en/Tensor_product_of_representations#/Lie_algebra_representations
Brian C – Lie groups, Lie algebras, and representations, An elementary introduction (2015, Springer) Appendix C.1 Theorem C.1.

Theorem C.1. Let $m$ and $n$ be non-negative integers with $m ≥ n$. If we consider $V_m \otimes V_n$ as a representation of $\mathsf{sl}(2; \mathbb{C})$, then
$$
  V_m \otimes V_n
  \cong V_{m+n} \oplus V_{m+n-2} \oplus \dotsb \oplus V_{m-n+2} \oplus V_{m-n} \,,
$$
where $\cong$ denotes an isomorphism of $\mathsf{sl}(2; \mathbb{C})$ representations.
(Screenshot from the Hall's book)

We were given $$V_3 = \langle \{u_3,u_1,u_{-1},u_{-3}\} \rangle$$ and $$V_2 = \langle \{u_2,u_0,u_{-2}\} \rangle \,.$$
Since there is some ambiguity, I assume these vectors are eigenvectors of the action $H=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}\in \mathfrak{sl}(2;\mathbb{C})$
I know that if I apply the action $Y=\begin{bmatrix} 0&0\\1&0\end{bmatrix}\in \mathfrak{sl}(2;\mathbb{C})$ we decrease the eigenvalues.
But how to write $V_5,V_3,V_1$ explicitly by using tensor product action $$\pi_3\otimes\pi_2=\pi_3\otimes I+I\otimes \pi_2 \,.$$
I have read the proof in the Hall’s book, he derives the result, gives the example but, he did not show how the action creates the explicit bases for each direct summand, so my question how to compute these basis for each direct summand?
 A: Here's a very practical approach. Start with a highest weight vector. This must be an element of the irreducible $V_5$ component so we can use it to generate $V_5$. It is easy to find such a vector: $u_3 \otimes u_2$. The repeated action of $Y$ then gives you (I'll assume $Yu_i = u_{i-2}$):
$$\begin{aligned} u_3 &\otimes u_2\\ u_1 \otimes u_2 &+ u_3 \otimes u_0 \\ u_{-1} \otimes u_2 + 2u_1 &\otimes u_0 + u_3 \otimes u_{-2} \\ u_{-3} \otimes u_2 + 3u_{-1} &\otimes u_0 + 3u_1 \otimes u_{-2} \\ 4u_{-3} \otimes u_0 &+ 6u_{-1} \otimes u_{-2} \\ 10u_{-3} &\otimes u_{-2}
\end{aligned}$$
Note that we can rescale the $u_i$ to make this look a little more symmetrical. Either way though this is a basis for $V_5$. Now we'd like to do the same for the $V_3$ component. We know that the highest weight of this component has $\omega_3(H) = 3$ and the weight space of this is the span of $u_3 \otimes u_0, u_1 \otimes u_2$ but clearly part of this weight space lies in $V_5$ and we need to find the $V_3$ part. Fortunately we can recognise this as the part of the weight space that is killed by our raising operator $X = \begin{bmatrix}0&1\\0&0\end{bmatrix}$. Now, using $[X,Y] =H$, we can compute that $Xu_1 = 3u_3$ and $Xu_0 = 2u_2$ so that $X(au_1 \otimes u_2 +bu_3 \otimes u_0 ) = (3a+2b) u_3 \otimes u_2 $. So to get $0$ we can take $2u_1 \otimes u_2 -3u_3 \otimes u_0 $ as our highest weight vector for $V_3$ then we can rinse and repeat. Repeatedly applying $Y$ now gives:
$$\begin{aligned} 2u_1 \otimes u_2 &-3u_3 \otimes u_0\\
2u_{-1} \otimes u_2 - u_1 &\otimes u_0 -3 u_3 \otimes u_{-2} \\
2u_{-3} \otimes u_2 + u_{-1} &\otimes u_0 -4 u_1 \otimes u_{-2} \\
3u_{-3} \otimes u_0  &- 3u_1 \otimes u_{-2} \\
\end{aligned}$$
And you can hopefully see that $Y$ sends the last one to $0$.
Again my choice of the $u_i$ could be improved a little to make some of this look simpler, but I didn't want to hide the process.
Finally we find $V_1$ using the kernel of $X$ in the $1$-weight space i.e. the span of $u_3\otimes u_{-2}, u_1 \otimes u_0, u_{-1} \otimes u_2$. We find $Xu_{-1} = 4u_1, Xu_{-2} = 2u_2$ (and $Xu_{-3} = 3u_{-1}$ but we won't need this). So  we obtain $$X(au_{-1} \otimes u_2 + bu_1 \otimes u_0 + cu_3\otimes u_{-2}) = 4au_{1} \otimes u_2 + 3bu_3\otimes u_0 + 2bu_1 \otimes u_2 + 2cu_3 \otimes u_0 = (4a+2b)u_{1} \otimes u_2 + (3b+2c)u_3 \otimes u_0.$$
So a natural choice is $u_{-1} \otimes u_2 -2 u_1 \otimes u_0 + 3u_3\otimes u_{-2}$ and applying $Y$ to this gives $u_{-3} \otimes u_2 -u_{-1} \otimes u_0 + u_1\otimes u_{-2}$ as our final basis element.
There are definitely more technical ways to generate bases of tensor products representations (e.g. crystal bases) and probably a neater specialisation of that to this case but hopefully this is helpful.
