Arbitrary Representation of $\frak{sl}(2,\Bbb{C})$

Question

Let $V$ be a representation of $\frak{sl}(2,\Bbb{C})$, and let $C \in End(V)$ be defined by $C = \rho (e) \rho (f) + \rho (f) \rho (e) + \frac{1}{2} \rho(h)^2$, where $e,f,h$ are the standard generators of $\frak{sl}(2,\Bbb{C})$. Show that if $V = V_k$ is an irreducible representation with highest weight $k$, then $C$ is a scalar operator: $C = c_k id$. Compute the constant $c_k$.

What exactly is the set $V_k$? Is it $\{v \mid Cv = \lambda Cv \}$? Or is it $V_k = \{v \mid \rho (x)v = \lambda \rho (x)v \}$? Where exactly do the $v$'s live? Are we supposed to think of $V_k$ as living in some ambient vector space? And what exactly does "highest weight" mean?

Answer 1

$V_k$ being a highest-weight representation with highest weight $k$ means (by definition) that there exists a vector $v_k \in V_k$ such that $\rho(e)v_k = 0$, $\rho(h) v_k = k v_k$, and $V_k$ is generated as a vector space by $v_k, \rho(f)v_k, \rho(f)^2 v_k, \ldots$.

For the highest-weight vector $v_k$, the commutation relation $[e, f] = h$ gives that $$(\rho(e) \rho(f) - \rho(f) \rho(e)) v_k = \rho(e) \rho(f) v_k = \rho(h) v_k = k v_k,$$ and so we can compute $C$ acting on $v_k$: $$ Cv_k = \rho(e) \rho(f) v_k + \rho(f) \rho(e) v_k + \frac{1}{2}\rho(h)^2 v_k = k v_k + 0 + \frac{1}{2} k^2 v_k = \frac{2k + k^2}{2} v_k.$$ So we have that $C$ acts on the highest-weight vector by the scalar $\frac{2k + k^2}{2}$. In order to show that $C$ acts on $V_k = \operatorname{span}(v_k, \rho(f) v_k, \rho(f^2) v_k, \ldots)$ by the same scalar, we show that $\rho(f)$ commutes with $C$: (I write $f$ rather than $\rho(f)$ in the following equation, for clarity). $$ \begin{aligned} Cf - fC &= ef^2 + fef + \frac{1}{2} h^2 f - fef - f^2 e - \frac{1}{2} f h^2 \\ &= ([e, f] - fe)f - f([f, e] - ef) + \frac{1}{2} h([h, f] - fh) - \frac{1}{2} ([f, h] - hf)h \\ &= [e, f] f - f[f, e] + \frac{1}{2} h[h, f] - \frac{1}{2} [f, h]h \\ &= hf + fh + \frac{1}{2} h( -2f) - \frac{1}{2} (2f) h \\ &= 0. \end{aligned}$$ Hence for any $\rho(f)^n v_k$, we have $$C \rho(f)^n v_k = \rho(f)^n Cv_k = \rho(f)^n \frac{2k + k^2}{2} v_k = \frac{2k + k^2}{2} \rho(f)^n v_k.$$

Answer 2

The representation $V_k$ of $\mathfrak{sl}(2,\mathbb C)$ (where $k\in\{0,1,2,\ldots\}$) is the only (up to isomorphism) irreducible representation of $\mathfrak{sl}(2,\mathbb C)$ for which there is a vector $v\neq0$ such that $h.v=kv$ and that $e.v=0$.

Answer 3

My favorite way of viewing $V_k$ is to see it as the space of homogeneous bivariate polynomials in two unknowns $x$ and $y$. In other words, the linear span of $x^{k-i}y^i$ with $0\le i\le k$.

In this case the basis elements of the Lie algebra act by the following partial differential operators $$ \begin{aligned} \rho(e)&=x\frac{\partial}{\partial y}\\ \rho(f)&=y\frac{\partial}{\partial x}\\ \rho(h)&=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}. \end{aligned} $$ Leaving it to you to verify the claims and calculate the constant $c_k$.