Decompose $V \otimes V \otimes V$ into irreducible representations of $SL_2(\mathbb{R})$

Question

Let $V=\mathbb{C^2}$ be the standard representation of $SL_2(\mathbb{R})$

Decompose $V \otimes V \otimes V$ into irreducible representations of $SL_2(\mathbb{R})$

I will just consider $SL_2(\mathbb{C})$ since there is a 1:1 correspondence between $SL_2(\mathbb{R})$ and $SL_2(\mathbb{C})$

Weight vectors of $V$ are $v_1$ and $v_{-1}$ with weights 1 and -1 respectively

Weight vectors of $V \otimes V \otimes V$:

Weight 3: $v_1 \otimes v_1 \otimes v_1$

Weight 1: $v_1 \otimes v_1 \otimes v_{-1}$, $v_1 \otimes v_{-1} \otimes v_1$ and $v_{-1} \otimes v_1 \otimes v_1$

Weight -1: $v_1 \otimes v_{-1} \otimes v_{-1}$, $v_{-1}\otimes v_{-1} \otimes v_1$ and $v_{-1} \otimes v_1 \otimes v_{-1}$

Weight -3: $v_{-1} \otimes v_{-1} \otimes v_{-1}$

How do I use this information to find the irreducible representations?

Answer 1

The representation $\mathrm{T}^2V = V \otimes V$ plits into $\mathrm{T}^2 V = \Lambda^2 V \oplus \mathrm{Sym}^2V$, where $\Lambda^2V$ is of dimension $1$ where $SL_2$ acts trivially, and $\mathrm{Sym}^2V$ is the $SL_2$-representation on the homogenuous polynomials of degree $2$ which is of dimension 3.

Hence $\mathrm{T}^3V$ splits into $\mathrm{T}^3V \simeq V \oplus (\mathrm{Sym}^2V \otimes V)$. There is the sub-representation $\mathrm{Sym}^3V \subset \mathrm{Sym}^2V \otimes V$, where $\mathrm{Sym}^3V$ is the $SL_2$-representation on the homogenuous polynomials of degree $3$ which is of dimension 4. Since $SL_2$ is semi-simple, there is a complementary sub-representation of dimension $2$ which is not trivial, hence isomorphic to $V$.

So $$V \otimes V \otimes V \simeq V^2 \oplus \mathrm{Sym}^3V,$$ as $SL_2$-representations.

More explicitely : let $(X,Y)$ be the canonical basis of $V$. Then $$V \otimes V \otimes V = W_1 \oplus W_2 \oplus W_3$$ where :

1. $W_1$ ($\simeq V$) has basis $XYX-YXX$ and $XYY-YXY$.
2. $W_2$ ($\simeq V$) has basis $XXY-YXX$ and $XYY-YYX$.
3. $W_2$ ($=\mathrm{Sym}^3V$) has basis $XXX$, $XXY+XYX+YXX$, $XYY+YXY+YYX$ and $YYY$.

Answer 2

One such decomposition is

$ A=\begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\cr -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0\cr 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\end{pmatrix}$

$ B= \begin{pmatrix}0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\cr -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\cr 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1\cr 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\end{pmatrix}$

$C= \begin{pmatrix}0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\cr -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\cr 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0\end{pmatrix}$

Then you have $A.A=B.B=C.C= -I(8)$