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Show that the subspace of $sl(3,\mathbb{C})$ consisting of matrices of the form $$ A = \left( \begin{matrix} * & * & 0 \\ * & * & 0 \\ 0 & 0 & 0 \end{matrix} \right) $$

is isomorphic to $sl(2,\mathbb{C})$.

We can therefore regard $sl(3,\mathbb{C})$ as a module for $sl(2,\mathbb{C})$ with the action given by $x \cdot y = [x,y]$ for $x \in sl(2,\mathbb{C})$ and $y \in sl(3,\mathbb{C})$. Show that as an $sl(2,\mathbb{C})$ module

$$sl(3,\mathbb{C})=V_2 \oplus V_1 \oplus V_1 \oplus V_0$$

Where $V_i$ is the space of all homogeneuos polynomials of degree $i$ in $\mathbb{C}[X,Y]$, i.e. the unique (up to isomorphism) irreducible $(i+1)$-dimensional representation of $sl(2, \mathbb C)$.

Okay, so for this problem the isomorphism between $A$ and $sl(2,\mathbb{C})$ is the obvious one. I'm also understanding quite well how this shows that $sl(3,\mathbb{C})$ is a $sl(2,\mathbb{C})$ module.

How do I show the decomposition into irreducible modules ?? Help appreciated, thanks!

Torsten Schoeneberg
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Eugene
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1 Answers1

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Show if you let $A$ act on (i.e. take brackets $[A,X]$ with $X$ from) the subsets

$$ \left( \begin{matrix} * & * & 0 \\ * & * & 0 \\ 0 & 0 & 0 \end{matrix} \right), \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ * & * & 0 \end{matrix} \right) , \left( \begin{matrix} 0 & 0 & * \\ 0 & 0 & * \\ 0 & 0 & 0 \end{matrix} \right), \left( \begin{matrix} x & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & -2x \end{matrix} \right) $$

of $sl(3, \mathbb C)$ respectively, each is a) stable and b) irreducible under the action. (That is almost trivial (pun intended) for the last one, and you must have basically shown it for the first one when you proved $A \simeq sl_2$.)

Now if you are comfortable with using the fact that irreducible $sl_2$-representations are unique up to iso in each dimension, you are done, because the dimensions of the above spaces are $3,2,2,1$ respectively. As a sanity check and reminder of highest weight theory, you can also check that the element

$$h= \left( \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{matrix} \right) $$

acts with eigenvalues $2,0-2$ on the first, with eigenvalues $1,-1$ on the second and third, and with eigenvalue $0$ on the fourth.

Torsten Schoeneberg
  • 29,325
  • Hmm, when you say stable, you mean the image lies back in $A$, correct? Do you have any tips for showing that the action is irreducible? I wouldn't even know how to begin!! – Eugene Oct 14 '20 at 21:09
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    By "stable" I mean for each of those subsets, call them $X_i$, we have $[A, X_i] \subseteq X_i$. For irreducibility, you can take any non-zero $x \in X_i$ and show that the vector space generated by all $[A,x_i]$ is the entire $X_i$. E.g. for the second one, for non-zero $x$ in there, come up with elements $a, a' \in A$ such that $[a,x] = \pmatrix{0&0&1\0&0&0\0&0&0}$ and $[a',x] = \pmatrix{0&0&0\0&0&1\0&0&0}$. – Torsten Schoeneberg Oct 14 '20 at 22:14
  • As for the "sanity check" part, why does checking that the action of $h$ has those certain eigen values help? – FireFenix777 Oct 18 '20 at 14:16
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    Because the irreducible $\mathfrak{sl}_2$-representation called $V_i$ in your question is characterized by the fact that the element $H =\pmatrix{1&0\0&-1}$ acts on it with eigenvalues $i, i-2, i-4, ... 2-i, -i$. Cf. any introduction to representation theory of Lie algebras, the $sl_2$-case; in a quick search on this site I found this question https://math.stackexchange.com/q/3827979/96384 and this answer https://math.stackexchange.com/a/3288482/96384, surely there's more. – Torsten Schoeneberg Oct 18 '20 at 16:49