# Ch. 7: Representations of $\liesl_2(\CC)$ (Wednesday, September 21)

:::{.remark}
If $L\in \Lie\Alg$ and $s$ is a semisimple element, then $\phi(s)$ is semisimple in any finite-dimensional representation $\phi$ of $L$.
In particular, taking the natural representation, this yields a semisimple operator.
For the same reason, $\ad_s$ is semisimple.
Similar statements hold for nilpotent elements.
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## 7.1: Weights and Maximal Vectors

:::{.remark}
Let $L \da \liesl_2(\CC)$, then recall

- $x = \matt 0100$
- $h = \matt 100{-1}$
- $y= \matt 0010$
- $[xy] = h$
- $[hx] = 2x$
- $[hy] = -2y$

Goal: classify $\lmod^\fd$.
By Weyl's theorem, they're all completely reducible, so it's enough to describe the simple objects.
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:::{.definition title="Weight decomposition and weight spaces"}
Note that $L \leq \liegl_2(\CC) = \liegl(\CC^2)$, and since $h$ is semisimple, $\phi(h)$ acts semisimply on any finite-dimensional representation $V$ with $\phi: L\to \liegl(V)$.
I.e. $\phi(h)$ acts diagonally on $V$.
Thus $V = \bigoplus _{ \lambda} V_{ \lambda}$ which are eigenspaces for the $\phi(h)$ action, where
\[
V_\lambda \da \ts{v\in V \st h.v = \lambda v, \lambda\in \CC}
.\]
If $V_\lambda \neq 0$ we say $\lambda$ is a **weight** of $h$ in $V$ and $V_\lambda$ is the corresponding **weight space**.
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:::{.lemma title="?"}
If $v\in V_ \lambda$, then $x.v \in V_ {\lambda+2}$ and $y.v \in V_{ \lambda- 2}$.
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:::{.exercise title="?"}
Prove this using the commutation relations.
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:::{.definition title="Highest weights"}
Note that if $V$ is finite-dimensional then there can not be infinitely many nonzero $V_\lambda$, so there exists a $\lambda\in \CC$ such that $V_{ \lambda} \neq 0$ but $V_{ \lambda+ 2} = 0$.
We call $\lambda$ a **highest weight** (h.w.) of $V$ (which will turn out to be unique) and any nonzero $v\in V$ a **highest weight vector**.
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## 7.2: Classification of Irreducible $\liesl_2(\CC)\dash$Modules

:::{.lemma title="?"}
Let $V \in \lmod^{\fd, \irr}$ and let $v_0\in V_{ \lambda}$ be a h.w. vector.
Set $v_{-1} = 0$ for $i\geq 0$ and $v_i \da {1\over i!}y^i v_0$, then for $i\geq 0$,

1. $h.v_i = (\lambda- 2i)v_i$
2. $y.v_i = (i+1)v_{i+1}$
3. $x.v_i = ( \lambda- i + 1) v_{i-1}$.

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:::{.proof title="?"}
In parts:

1. By the lemma and induction on $i$.

2. Clear!

3. Follows from $i x.v_i = x.(y.v_{i-1}) = y.(x.v_{i-1}) + [xy].v_{i-1}$ and induction on $i$.

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:::{.remark}
Some useful facts:

- The nonzero $v_i$ are linearly independent since they are eigenvectors of $h$ with different eigenvalues -- this is a linear algebra fact.

- The subspace of $V$ spanned by the nonzero $v_i$ is an $L\dash$submodule, but since $V$ is irreducible the $v_i$ must form a basis for $V$.

- Since $V$ is finite-dimensional, there must be a smallest $m\geq 0$ such that $v_m \neq 0$ but $v_{m+1} = 0$, and thus $v_{m+k} = 0$ for all $k$.
  Thus $\dim_\CC V = m+1$ with basis $\ts{v_0, v_1, \cdots, v_m}$.

- Since $v_{m+1} = 0$, we have $0 = x.v_{m+1} = ( \lambda- m) v_m$ where $v_m\neq 0$, so $\lambda = m \in \ZZ_{\geq 0}$.
  Thus its highest weight is a non-negative integer, equal to 1 less than the dimension.
  We'll reserve $\lambda$ to denote a highest weight and $\mu$ an arbitrary weight.

- Thus the weights of $V$ are $\ts{m, m-2, \cdots, \star, \cdots, -m+2, -m}$ where $\star = 0$ or $1$ depending on if $m$ is even or odd respectively, each occurring with multiplicity one (using that $\dim V_{\mu} = 1$ if $\mu$ is a weight of $V$).
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:::{.theorem title="?"}
Let $V \in \lmod^{\fd, \irr}$ for $L\da \liesl_2(\CC)$, then

1. Relative to $h$, $V$ is a direct sum of weight spaces $V_\mu$ for $\mu \in \ts{m, m-2,\cdots, -m+2, -m}$ where $m+1=\dim V$ and each weight space is 1-dimensional.

2. $V$ has a unique (up to nonzero scalar multiples) highest weight vector whose weight (the highest weight of $V$) is $m\in \ZZ_{\geq 0}$

3. The action $L\actson V$ is given explicitly as in the lemma if the basis is chosen in a prescribed fashion. In particular, there exists a unique finite-dimensional irreducible $\liesl_2\dash$module of dimension $m+1$ up to isomorphism.

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:::{.corollary title="?"}
Let $V \in \lmod^{\fd, \irr}$, then the eigenvalues of $h\actson V$ are all integers, and each occurs along with its negative with the same multiplicity.
Moreover, in a decomposition  of $V$ in a direct sum of irreducibles, the number of simple summands is $\dim V_0 + \dim V_1$.
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:::{.remark}
Existence of irreducible highest weight modules of highest weight $m \geq 0$: 

- $m=0$: take the trivial representation $V=\CC$.
- $m=1$: take $V = \CC^2$ with the natural representation.
- $m=2$: take $V = L$ with the adjoint representation.

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:::{.remark}
The formula in the lemma can be used to construct an irreducible representation of $L$  having highest weight \( \lambda= m \) for any $m\in \ZZ_{\geq 0}$, which is unique up to isomorphism and denoted $L(m)$ (or $V(m)$ in Humphreys) which has dimension $m+1$.
In fact, the formulas can be used to define an infinite-dimensional representation of $L$ with highest weight $\lambda$ for any $\lambda\in \CC$, which is denoted $M( \lambda)$ -- we just don't decree that $v_{m+1} = 0$, yielding a basis $\ts{v_0, v_1,\cdots}$.
This yields a decomposition into 1-dimensional weight spaces $M( \lambda) = \oplus _{i=1}^\infty M_{ \lambda-2i}$ (**Verma modules**) where $M_{ \lambda-2i} = \gens{v_i}_\CC$.
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