I talked about the Lie algebra {\mathfrak{sl}_2} a while back.  Now I’m going to do it more properly, and using the tools developed.  This is going to feature prominently in some of the proofs in the sequel.

Now, let’s see how all this works for the familiar case of {\mathfrak{sl}_2}, with its usual generators {H,X,Y}. This is a simple Lie algebra in fact. To see this, let’s consider the ideal {I} of {\mathfrak{sl}_2} generated by some nonzero vector {aX + bH + cY}; I claim it is all of {\mathfrak{sl}_2}.

Consider the three cases {a \neq 0, b \neq 0, c \neq 0}:

First, assume {a} or {c} is nonzero. Bracketing with {H}, and again, gives

\displaystyle -2aX + 2 c Y \in I , \ (-2)^2 a X + 2^2 cY \in I, \ (-2)^3 a X + 2^3 cY \in I.

Using a vanderMonde invertibility of this system of linear equations, we find that either {X} or {Y} belongs to {I}. Say {X} does, for definiteness; then {H = [X,Y] \in I} too; from this, {Y = -\frac{1}{2} [H,Y] \in I} as well. Thus {I = \mathfrak{sl}_2}.

If {a=c=0}, then from {b \neq 0}, we find {H \in I}, which implies {X = \frac{1}{2}[H,X] \in I} and similarly for {Y}. Thus {I= \mathfrak{sl}_2}.

I claim now that the algebra {\mathbb{C} H} is in fact a Cartan subalgebra. Indeed, it is easily checked to be maximal abelian. Moreover, since {H} acts by a diagonalizable operator on the faithful representation on {\mathbb{C}^2}, it follows that {H \in \mathfrak{sl}_2} is (abstractly) semisimple.

So given a finite-dimensional representation of {V}, we have

\displaystyle V = \bigoplus_{\lambda} V_{\lambda}

where {V_{\lambda}} is the {\lambda}-eigenspace for {H}, and {\lambda} ranges over the complex numbers! We are now interested in looking at finite-dimensional irreducible representations {V}. In particular, {V} will have a highest weight {\mu}, which in this case is just a largest eigenvalue, and a corresponding eigenvector {v}. Then {X v = 0}, {Hv = \mu v}.

Proposition 1 {\mu \in \mathbb{Z}_{\geq 0}}, and the lowest weight in {V} is {-\mu}.


To see this, we shall prove three commutation relations on {U(\mathfrak{sl}_2)}, the enveloping algebra.

\displaystyle [H, Y^n] = -2n Y^n, \quad [H, X^n] = 2nX^n,

\displaystyle [X, Y^n] = -n(n-1)Y^{n-1} + n Y^{n-1} H .

The first two follow because {[H,Y]=-2Y} (resp. {[H,X]=2X}) and bracketing by {H} is a derivation on the algebra {U(\mathfrak{sl}_2)}.

The third follows by a somewhat tedious inductive argument. It is obvious for {n=0}. Assume it for {n}. Consider

\displaystyle [X,Y^{n+1}] - Y[X, Y^{n}] = [X, Y^{n}]Y + Y^{n} [X,Y] - Y[X, Y^{n}]

because bracketing is a derivation. Now simplifying this using the inductive hypothesis yields:

\displaystyle -n(n-1)Y^{n} + n Y^{n-1} H Y + Y^nH + Y n(n-1) Y^{n-1} - n Y^{n}H.

Simplifying futher gives:

\displaystyle n Y^{n-1} HY - (n-1) Y^n H = Y^{n} H - 2n Y^{n}

So, if we define {Z_n := [X, Y^n]}, then we see that {Z_n} satisfies the recurrence relation

\displaystyle Z_{n+1} - YZ_n =Y^{n} H - 2n Y^{n}

which is also satisfied by the other term at the end of the third identity.

Now, we can prove the proposition. As above, {\mu} was the highest weight and {v} a vector with weight {\mu}. Then by the Poincare-Birkhoff-Witt basis theorem, the family

\displaystyle Y^i X^j H^k \in U(\mathfrak{sl}_2), \quad i,j,k \geq 0

is a basis. Therefore since {v} is annihilated by {X},

\displaystyle \{ Y^i v \}

spans a sub-{\mathfrak{sl}_2}-module of {V}, which is all of {V} by simplicity. Note that {Y^i v} is a weight vector of weight {\mu - 2i}; for large {i} it follows that {Y^i v = 0}.

Let {t} be the largest value such that {Y^t v \neq 0}. I claim now that {t = \mu}.

First of all, we have

\displaystyle [ X, Y^{t+1}] v = X Y^{t+1} v - Y^{t+1} X v = 0.

Now, in view of the third identity:

\displaystyle -(t+1)t Y^{t} v + (t+1) Y^{t} \mu v = 0 ,

and since {Y^t v \neq 0}, it follows that

\displaystyle \mu = t.

In particular, {\mu} is a positive integer and {t = \mu}!