Trying to confirm few properties for a highest weight vector of a $sl(2,\mathbb{C})$ module I've been trying to study some Lie algebras in my spare time... Not really a mathematician but it's fun to try! Anyway, I have a question for y'all.
Say that $V$ be an $sl(2,\mathbb{C})$ module, not necessarily finite dimensional. Suppose $w \in V$ is a highest-weight vector of weight $\lambda$; that is, $e \cdot w = 0$ and $h \cdot w = \lambda w$ for some $\lambda \in \mathbb{C}$, and $w \neq 0$. I'm trying to understand why the following two things are true:
i) for $k = 1,2,...$ we have $e \cdot (f^k \cdot w) = k(\lambda - k + 1)f^{(k-1)} \cdot w$
ii) $e^kf^k \cdot w = (k!)^2 \binom\lambda kw$
Once I figure this out, I'm supposed to be able to deduce that if $\binom \lambda k \neq 0$ then the set of all $f^j \cdot w$ for $0 \leq j \leq k$ is linearly independent. Hence,if $V$ is finite dimensional, then $\lambda$ must be a nonnegative integer. I haven't been able to figure out (i) and (ii), and so I just assumed them and tried to solve this part, but alas, I can't do this part either :-(.
Here we have that where $e$ is typical basis vector of $sl(2,\mathbb{C})$ that is the  $2 \times 2$ matrix with a $1$ in the top right entry and $0$ elsewhere and $h$ is the basis vector with $1$ in the top left, and $-1$ in the bottom right.
If anyone could walk me through this that would be greatly appreciated. Thank you for your time!
 A: I will write $E, F, H \in \mathfrak{sl}_2$ for the Lie algebra elements, and $e, f, h \in \operatorname{End}_\mathbb{C}(V)$ for the operators on $V$, so that $E \cdot v = e(v)$ for example. It is usually very fruitful to write equations in $e, f, h$ rather than $E, F, H$, since commutators can be rearranged, unlike Lie brackets.
After some fiddling around repeatedly applying the relation $ef = fe  +h$, you can get the general fact that
$$ ef^k = f^k e + \sum_{a + b = k - 1} f^a h f^b \quad \text{ for } k \geq 0. $$
Next, use $fh = (h + 2)f$ repeatedly to rewrite the sum, to get
$$ ef^k = f^k e + k(h + k - 1) f^{k - 1}.$$
Hence if $v \in V$ is weight $\lambda$, the equation above gives that $ef^k v = f^k e v + k(\lambda + k - 1) f^k v$, if $v$ is furthermore highest weight then the $f^k e v$ term is zero. This gets you (i).
Now, repeatedly apply $eh = (h - 2)e$ to get $e^{k-1} h = (h - 2(k - 1))e^{k-1}$, then apply $e^{k-1}$ on the left of the above equation to get
$$ e^k f^k = e^{k-1} f^k e + k(h - k + 1)e^{k-1} f^{k-1}.$$
If $v$ has weight $\lambda$ and is highest-weight, then this gives you
$$ e^k f^k v = k (\lambda - k + 1)e^{k-1} f^{k-1} v, $$
and you can proceed by induction to show (ii).
As for the last part, if you do the induction on the operators, rather than the operators applied to $v$, you find that $e^k f^k = (k!)^2 \binom{h}{k}$ as an equality of operators restricted to $\ker e$: furthermore $f$ is nilpotent, so $\binom{h}{k} = 0$ on $\ker e$ for $k$ large enough. Since $\binom{h}{k}$ is a polynomial in $h$ with distinct integer roots, $h|_{\ker e}$ is diagonalisable with integer eigenvalues.
