Consider the Lie algebra $\mathfrak{sl}_2(\mathbb{C})$, with standard basis. Let $$ X = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ be one of these basis vectors. Let $w$ be a lowest weight vector with lowest weight $\beta$ in the weight space $V_\beta$. Then we can generate an irreducible representation with basis $$ \{ w, X(w), X^2(w), \dots \}. $$ My question is as follows: by the simple properties of the matrix $X$, its square is the zero matrix. Then how come the vectors which form the irreducible representation's basis above which have been formed by application of $X$ twice or more times to them are non-trivial? The answer to this is very obvious I feel, but I can't tell what I'm missing.
I feel my error is that really I should be writing something like $\rho(X)w$ instead of $X(w)$ where $\rho$ is the representation. In this case, I see why my above reasoning is wrong ($\rho(X) \neq X$ as matrices), but then here's my problem: $\rho$ is a representation so by definition a homomorphism. So then if we take something like $\rho(X)^2$ which shouldn't be trivial in general, we can write $\rho(X)^2 = \rho(X^2) =\rho (0)$ by the properties of a homomorphism. Or is this expected, and the truly interesting thing to consider is $\rho^2(X)$ which is a different story?
Thanks.
Edit: When talking about Lie algebras, for $\rho$ to be a Lie algebra homomorphism we need it to preserve the bracket and be $\mathbb{K}$-linear where $\mathbb{K}$ is a field. So me writing $\rho(X^2)=\rho(X)^2$ doesn't make sense for Lie algebras (though it would if I were talking about Lie group homomorphisms!)
Representation of a lie algebra is not a representation of associative algebra! There is no reason that $\pi(XY)$ should be equal to $\pi(X)\pi(Y)$. What holds instead is $$\pi([X,Y]) = \pi(X)\pi(Y)-\pi(Y)\pi(X).$$
In $\mathfrak{sl}_2(\mathbb C)$ we have the base $H$, $X$, $Y$ such that $[H,X] = 2X, [H,Y]=-2Y, [X,Y] = H$. It is enough to demand that representation respects these relations, and all the other brackets follow from bilinearity.
The key here is that for any representation of $\mathfrak{sl}_2(\mathbb C)$, $X$ sends eigenspace of weight $\alpha$ to eigenspace of $\alpha + 2$, and similarly $Y$ sends it to eigenspace of $\alpha - 2$. $X^2$ is not necessarily $0$, it maps eigenspace of $\alpha$ to $\alpha + 4$ and so on.
Concretely, the lowest weight is some non-positive integer $-n$ and the weights are $$\{-n,-n+2,-n+4,\ldots , n-2, n\}.$$ It turns out that all the eigenspaces are necessarily one-dimensional, so one can indeed generate the whole module by applying $X$ repeatedly to the lowest weight vector $w$.