I'm in the process of learning about representations of Lie Algebras, specifically $\mathfrak{gl}(n)$ (over $\mathbb{C}$), and I am having some trouble wrapping my head around the details of weight vectors.
Here's the passage I'm struggling with: "Let $E_{ij},\,i,j=1,\dots,n$ be the standard basis for $\mathfrak{gl}(n)$. The finite-dimensional irreducible representations of $\mathfrak{gl}(n)$ are in one-to-one correspondence with $n$-tuples of complex numbers $\lambda = (\lambda_{1},\dots,\lambda_{n})$ such that $\lambda_{i}-\lambda{i+1}\in\mathbb{Z}_{+}$ for $i=1,\dots,n$. Such an $n$-tuple is called the highest weight of the corresponding representation, which we shall denote $L(\lambda)$.
$L(\lambda)$ contains a unique, up to a multiple, nonzero vector $\xi$ (the highest vector) such that $E_{ii}\xi=\lambda_{i}\xi$ for $i=1,\dots,n$ and $E_{ij}\xi=0$ for $1\leq i < j \leq n$."
As I understand it (and maybe this is where I'm tripping up), in the standard basis $E_{ij}$ is the $n$ by $n$ matrix with $1$ in the $i,j$ index and $0$ elsewhere. If that is the case, and the statement $E_{ij}\xi=\lambda_{i}\xi$ uses the usual definition of matrix-vector multiplication, then in order for both $E_{11}\xi=\lambda_{1}\xi$ and $E_{12}\xi=0$ to be true, $\xi_{1}$ must be equal to $1$, and all other entries of $\xi$ must be $0$.
This seems too restrictive to be true, and it makes me think I am missing something. Does $E_{ij}\xi=\lambda_{i}\xi$ not use the standard definition of matrix-vector multiplication? Is my logic wrong somewhere? Am I missing something trivial? Or is what I said above correct and I'm overthinking it?
Best Answer
The error in your reasoning is the assumption that the action of $E_{ij}$, w.r.t. the given irreducible representation, is given by the matrix $E_{ij}$ itself: This is not the case. Instead, a representation of ${\mathfrak g}{\mathfrak l}(n)$ on a vector space $V$ corresponds to an arbitrary Lie algebra homomorphism $\rho: {\mathfrak g}{\mathfrak l}(n)\to {\mathfrak g}{\mathfrak l}(V)$, with the action of $E_{ij}$ being given by the matrix $\rho(E_{ij})$.
Note that the notion of a representation makes sense even if ${\mathfrak g}$ is an arbitrary Lie algebra, which may not be defined as a matrix Lie algebra yet -- in a way, that's the point, because you 'represent' abstract elements by concrete matrices. If ${\mathfrak g}$ does already arise as a matrix (sub)algebra -- as is the case for ${\mathfrak g}{\mathfrak l}(n)$ -- then the identity is a special representation, but there are typically still many others.
Finally, you may want to convince yourself that the identity representation of ${\mathfrak g}{\mathfrak l}(n)$ is the irreducible representation corresponding to the highest weight $(1,0,...,0)$.