Question about $sl_2$-representation. Let $V$ be finite-dimensional $sl_2$-representation with the operators $X,Y,H$ where
$$
[H,X]=2X, [H,Y]=-2Y, [X,Y]=H
$$
Let $v$ be a vector from $V$ such that $X^2(v)=0$  but $X(v) \neq 0$ .
Question. Is it true that $Y(X(v))\neq 0?$
My attempt. If $V$ is irreducible representation then  the answer is positive  --  if were $Y(X(v))= 0$  then the one-dimensional vector space generated by $X(v)$  be an invariant subspace of $V$  and we   got a contradiction.
But it is true in the case when $V$  is direct sum of irreducible representations ?  I cant find any counterexample.
 A: It is true and can be understood with standard representation theory of $\mathfrak{sl}_2\mathbb{C}$.
If $V$ is a $\mathfrak{sl}_2\mathbb{C}$-rep, we can decompose $V=\bigoplus V_\lambda$ as eigenspaces wrt the operator $H$, which we call weight spaces. We can use the relations for $X,Y,H$ to prove $XV_\lambda\subseteq V_{\lambda+2}$ and $YV_\lambda\subseteq V_{\lambda-2}$, which is why you might see $X$/$Y$ called raising/lowering or creation/annihilation operators in certain contexts.
Thus, any rep $V$ can be represented in the following fashion: draw a bunch of dots in a bunch of rows. The set of dots represents a vector space basis; every vector is a formal linear combination of the dots. Every dot is an eigenvector for $H$ (the eigenvalues called weights), applying $X$ to a dot yields a scalar multiple of the dot to its right, and applying $Y$ to a dot yields a scalar multiple of a dot to its left. Feel free to draw a picture.
A little more work reveals more: up to iso the unique dim-$d$ irrep is $\mathrm{Sym}^{d-1}\mathbb{C}^2$, spanned by
$$ \{ \, y^{d-1}, \, y^{d-2}x, \, \cdots \, , \, yx^{d-2} \, , \, x^{d-1} \, \} $$
and acted on by $X=x\frac{\partial}{\partial y}$, $\, Y=y\frac{\partial}{\partial x}$, $\, H=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}$. This implies that when we use $YX$ on a dot (basis weight vector), as long as the dot is not on the far right.
For us this means if $X^2v=0$ but $Xv\ne 0$, then $v$ is a linear combination of the second-to-last dots from each row, and applying $YX$ just multiplies the coefficients in this combination by nonzero values, so it can't be $0$.
