Weight and weight space

Let $\mathfrak{g}$ be a complex semisimple Lie algebra with a fixed Cartan subalgebra $\mathfrak{h}$ (Cartan subalgebra), and let $V$ be a representation of $\mathfrak{g}$ (Representations of Lie algebras), written $H\cdot v$ for $\rho (H)(v)$. No finite-dimensionality of $V$ is assumed here.

For $\mu \in {\mathfrak{h}}^{\ast }$ the weight space of $\mu$ is $V_{\mu }=\{v\in V:H\cdot v=\mu (H)v\text{ for every }H\in \mathfrak{h}\};$ this is a linear subspace of $V$ (Linear subspace of a vector space), being the intersection of the kernels of the endomorphisms $\rho (H)-\mu (H){\mathrm{id}}_V$. A weight of $V$ is a functional $\mu \in {\mathfrak{h}}^{\ast }$ with $V_{\mu }\mathrm{\ne }0$, and a nonzero $v\in V_{\mu }$ is a weight vector of weight $\mu$. The zero functional is allowed as a weight; $V_0$ is the space of vectors fixed by $\mathfrak{h}$.

Distinct weight spaces are independent. Indeed, if $v_1+\cdots +v_m=0$ with $0\mathrm{\ne }{v}_j\in V_{{\mu }_j}$ and pairwise distinct functionals ${\mu }_1,\dots ,{\mu }_m$, choose $H\in \mathfrak{h}$ with the scalars ${\mu }_j(H)$ pairwise distinct; this is possible because the finitely many sets $\{H\in \mathfrak{h}:({\mu }_i-{\mu }_j)(H)=0\}$ are proper subspaces of $\mathfrak{h}$ (the functionals ${\mu }_i-{\mu }_j$ are nonzero for $i\mathrm{\ne }j$) and a finite union of proper subspaces of a vector space over the infinite field $\mathbb{C}$ is proper. Then the $v_j$ are eigenvectors of $\rho (H)$ for the pairwise distinct eigenvalues ${\mu }_j(H)$ and hence are linearly independent (Eigenvectors belonging to pairwise distinct eigenvalues are linearly independent), forcing $v_1=\cdots =v_m=0$, a contradiction. Thus the sum ${\sum }_{\mu }{V}_{\mu }$ is direct.

If $V$ is finite-dimensional, then $V$ has only finitely many weights: the nonzero weight spaces form a direct sum inside $V$, so their number is at most $\dim V$.