Special linear Lie algebra

By definition, the Lie algebra ${\displaystyle {\mathfrak {sl}}_{2}\mathbb {C} }$ consists of two-by-two complex matrices with zero trace. There are three standard basis elements, ${\displaystyle e}$,${\displaystyle f}$, and ${\displaystyle h}$, with

${\displaystyle e={\begin{bmatrix}0&1\\0&0\end{bmatrix}}}$, ${\displaystyle f={\begin{bmatrix}0&0\\1&0\end{bmatrix}}}$, ${\displaystyle h={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}}$.

The commutators are

${\displaystyle [e,f]=h}$, ${\displaystyle [h,f]=-2f}$, and ${\displaystyle [h,e]=2e.}$

The Lie algebra ${\displaystyle {\mathfrak {sl}}_{2}\mathbb {C} }$ can be viewed as a subspace of its universal enveloping algebra ${\displaystyle U=U({\mathfrak {sl}}_{2}\mathbb {C} )}$ and, in ${\displaystyle U}$, there are the following commutator relations shown by induction:[1]

${\displaystyle [h,f^{k}]=-2kf^{k},\,[h,e^{k}]=2ke^{k}}$,

${\displaystyle [e,f^{k}]=-k(k-1)f^{k-1}+kf^{k-1}h}$.

Note that, here, the powers ${\displaystyle f^{k}}$, etc. refer to powers as elements of the algebra U and not matrix powers. The first basic fact (that follows from the above commutator relations) is:[1]

From this lemma, one deduces the following fundamental result:[2]

The first statement is true since either ${\displaystyle v_{j}}$ is zero or has ${\displaystyle h}$-eigenvalue distinct from the eigenvalues of the others that are nonzero. Saying ${\displaystyle v}$ is a ${\displaystyle {\mathfrak {b}}}$-weight vector is equivalent to saying that it is simultaneously an eigenvector of ${\displaystyle h,e}$; a short calculation then shows that, in that case, the ${\displaystyle e}$-eigenvalue of ${\displaystyle v}$ is zero: ${\displaystyle e\cdot v=0}$. Thus, for some integer ${\displaystyle N\geq 0}$, ${\displaystyle v_{N}\neq 0,v_{N+1}=v_{N+2}=\cdots =0}$ and in particular, by the early lemma,

${\displaystyle 0=e\cdot v_{N+1}=(\lambda -(N+1)+1)v_{N},}$

which implies that ${\displaystyle \lambda =N}$. It remains to show ${\displaystyle W=\operatorname {span} \{v_{j}|j\geq 0\}}$ is irreducible. If ${\displaystyle 0\neq W'\subset W}$ is a subrepresentation, then it admits an eigenvector, which must have eigenvalue of the form ${\displaystyle N-2j}$; thus is proportional to ${\displaystyle v_{j}}$. By the preceding lemma, we have ${\displaystyle v=v_{0}}$ is in ${\displaystyle W}$ and thus ${\displaystyle W'=W}$. ${\displaystyle \square }$

As a corollary, one deduces:

• If ${\displaystyle V}$ has finite dimension and is irreducible, then ${\displaystyle h}$-eigenvalue of v is a nonnegative integer ${\displaystyle N}$ and ${\displaystyle V}$ has a basis ${\displaystyle v,fv,f^{2}v,\cdots ,f^{N}v}$.
• Conversely, if the ${\displaystyle h}$-eigenvalue of ${\displaystyle v}$ is a nonnegative integer and ${\displaystyle V}$ is irreducible, then ${\displaystyle V}$ has a basis ${\displaystyle v,fv,f^{2}v,\cdots ,f^{N}v}$; in particular has finite dimension.

The beautiful special case of ${\displaystyle {\mathfrak {sl}}_{2}}$ shows a general way to find irreducible representations of Lie algebras. Namely, we divide the algebra to three subalgebras "h" (the Cartan Subalgebra), "e", and "f", which behave approximately like their namesakes in ${\displaystyle {\mathfrak {sl}}_{2}}$. Namely, in an irreducible representation, we have a "highest" eigenvector of "h", on which "e" acts by zero. The basis of the irreducible representation is generated by the action of "f" on the highest eigenvectors of "h". See the theorem of the highest weight.

When ${\displaystyle {\mathfrak {g}}={\mathfrak {sl}}_{n}\mathbb {C} ={\mathfrak {sl}}(V)}$ for a complex vector space ${\displaystyle V}$, each finite-dimensional irreducible representation of ${\displaystyle {\mathfrak {g}}}$ can be found as a subrepresentation of a tensor power of ${\displaystyle V}$.[4]