By definition a representation $\rho$ of a lie algbera $L(G)$, $\rho: L(G) \rightarrow End(V)$ for some vector space $V$ must be a lie algebra homomorphism.

-So to say that two representations are equivalent is to say that they generate lie algebra's in $End(V)$ which are isomorphic?

-But why does it follow from this that they have the same set of weights?

Attempt at Solution:

  • Let a semisimple lie algebra be $L$, with a Cartan subalgebra $H$

  • A set of weights is a set $S$ of functions, $S=\{\lambda_{i}...\lambda_{k}\} \, \lambda_{i}: H \rightarrow \mathbb{R}$, and associated with each we have a weight space: $$ V_{\lambda_{j}} = \{ v \in V : \forall h \in H, \rho(h)v =\lambda(h)v \} $$

    • If we have two equivalent representations $\rho_{1} \cong \rho_{2}$ , then,by the above there is a LA isomorphism $\phi$ such that: $\phi(\rho_{1}(h)) = \rho_{2}(h)$ so : $$ \rho_{2}(h) = \phi(\lambda(h)v) $$
    • Can we then just conclude - since $\phi$ must be linear, and maps identities to identities: $$ \rho_{2}(h) = \lambda(h)v$$ $$ \implies \forall \lambda_{j} \; (V_{\lambda j})_{\rho_{1}} = (V_{\lambda_{j}})_{\rho_{2}} $$
    • However I'm not very convinced by this argument. More fundamentally need to show that the set of weights for $\rho_{1}$ is the same as the set of weights for $\rho_{2}$.. I guess this somehow follows from the isomorphism, but can't show it explicitly.


  • 1
    $\begingroup$ No, that is not what it means for representations to be isomorphic. $\endgroup$ Commented Apr 11, 2018 at 15:32

1 Answer 1


Two representations $\rho_1:\mathfrak{g}\to \mathfrak{gl}(V)$ and $\rho_2:\mathfrak{g}\to \mathfrak{gl}(W)$ are isomorphic if there is a linear isomorphism $$\phi:V\to W$$ such that $\rho_2(x)\phi(v)=\phi(\rho_1(x)v)$. Given this definition, suppose that $v\in V_\lambda$ is nonzero, so $\rho_1(h)v=\lambda(h)v$. Then, $$ \rho_2(h)\phi(v)=\phi(\rho_1(h)v)=\phi(\lambda(h)v)=\lambda(h)\phi(v).$$ it follows that $\phi(v)\in W_\lambda$ is nonzero, so $\lambda$ is a weight of $W$. Hence the weights of $V$ are a subset of the weights of $W$.

Repeating this argument with $\phi^{-1}$ shows that the weights of $W$ are a subset of the weights of $V$, hence the weights coincide.


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