Studying the representations of $\mathfrak{su}(3)$ I ran in the following issues: the Cartan subalgebra is
$$\mathfrak{h}=\left\{
\begin{pmatrix}
a_1 & 0 & 0
\\0 & a_2 & 0
\\0 & 0 & a_3
\end{pmatrix}
:a_1+a_2+a_3=0\right\} $$
hence the eigenvectors of the adjoint representations are given by $E_{ij}$ which are matrices with $0$ everywhere and $1$ at the position $(i,j)$, the corresponding "eigenvalues" in $\mathfrak{h}^*$ are $L_i-L_j$ where for $H\in \mathfrak{h}$, $L_i(H)=a_i$. Let $V$ be any finite dimensional irreducible representation, one can show that
$$V=\bigoplus_{\alpha}V_\alpha $$
where $\alpha\in \mathfrak{h}^*$ and for $H$ in $\mathfrak{h}$, $v\in V_\alpha$ we have $Hv=\alpha(H)v$. We can find an $\alpha$ such that for $v\in V_\alpha$
$$E_{12}v=E_{23}v=E_{13}v=0 $$
then $\alpha$ is the highest weight with respect to some weight choice. In the treatments I could find the authors then go on to say that we can find the rest of the representation by applying $E_{21}$ and $E_{32}$ to $v$, and proceed to examine the vector
$$(E_{21})^kv $$
claiming that there must be some $n$ such that $(E_{21})^nv=0$ due to finite dimensionality, and apparently this $n$ can tell us something about the representation.
I don't understand how this is the case, since it seems to me that $(E_{21})^2=0$ hence $(E_{21})^kv=0$ for $k\geq 2$. This seems like such an obvious point that I must be missing something huge, but I can't see it.
The only way out I can find is that possibly $E_{21}^2$ need not be $0$ in any representation, but if $\rho$ is a Lie algebra isomorphism then $\rho$ must be linear hence $\rho(E_{12}^2)=\rho(0)=0$. What is going on?
Best Answer
First a minor point: You're probably looking at representations of the complexification, because the $E_{ij}$ are not even in $\mathfrak{su}_n$; but they are in the complexified version $\mathfrak{su}_n \otimes \mathbb C$, which I will call $\mathfrak{g}$ from now on.
The crucial point is: When you write $E_{21}^2=0$, with the square "$^2$" you probably mean matrix multiplication. But that multiplication is not even part of the Lie algebra structure. Elements of the Lie algebra have no "squares" in that sense.
On the other hand, when your source writes "$(E_{21})^k v$", this is shorthand notation for $(\rho(E_{21}))^k (v)$, where $\rho: \mathfrak{g} \rightarrow End(V)$ is the actual representation on the vector space $V$; so now, $\rho(E_{21})$ is some endomorphism of $V$, and this endomorphism does have higher powers "$^k$" for $k \ge 2$, by which we mean $k$-fold composition of that endomorphism with itself.
So the problem is that we do not have $\rho(E_{21}^k)\stackrel{?}=\rho(E_{21})^k$, for the striking reason that the left hand side makes no sense to begin with. And even if it did, $\rho$ would have no reason to respect that kind of multiplication. The right hand side makes perfect sense though.
To see an example, look at the adjoint representation on $V=\mathfrak{g}$, where $\rho(E_{12}) = ad_{E_{12}}$, meaning that
$$\rho(E_{12}) (v) = [E_{12}, v] \quad \text{ hence}$$
$$\rho(E_{12})^2 (v) = [E_{12},[E_{12}, v]].$$
Can you find a $v$ in the Lie algebra $\mathfrak{g}$ such that this is $\neq 0$, showing that the endomorphism $\rho(E_{12})^2$ is non-zero?