Is the decomposition of finitely dimensional representation of a semisimple Lie algebra into irreducible ones unique? I am reading Humphreys's book GTM9 on Lie algebra representations. The question above seems quite natural but the book doesn't give any clues on this topic, as far as I can see. Let's adopt the settings in the book: algebracally closed field with characteristic 0, finitely dimensional semisimple Lie algebra, and finitely dimensional represetations. Here by unique we mean the decomposition is unique up to order.
 A: The meaning of the question is not unique.
Literally understood I'd say that the answer is no, as soon as one irreducible occurs with multiplicity $\ge 2$. For instance, if $V$ is irreducible, then $V\oplus V$ decomposes both as the given sum, and as the sum of $V\times\{0\}$ plus the submodule $\{(tv,v):v\in V\}$ for each given scalar $t$, yielding, up to order as many decompositions as scalars. 
When there is no multiplicity $\ge 2$, the decomposition is indeed unique up to order (i.e., the (unordered) set of summands is unique: these are precisely the irreducible subrepresentations).
What is true however is a weaker form of uniqueness occurring in general, namely up to isomorphism. Namely for every finite-dimensional representation $V$ with two irreducible decompositions $V=\bigoplus_{i\in I}V_i=\bigoplus_{j\in J}W_j$, there is a bijection $\sigma:I\to J$ such that $V_i$ and $W_{\sigma(i)}$ are isomorphic representations for each $i$.
A: Yes, it is unique up to order. You can see this by observing that the number of times an irreducible representation $S$ appears as a summand in a representation $V$ is exactly $\operatorname{dim} \operatorname{Hom}_{\mathfrak{g}}(S,V)$ (by Schur lemma). And the latter expression is independent of the choice of decomposition of $V$ into simples.
EDIT: As Jason DeVito suggested in the comments, uniqueness does not mean that you can find unique simple submodules of $V$ that make the decomposition of $V$ into simples. But the simples are unique up to isomorphism, of course. (And in fact unique in the strong sense if and only if all the multiplicities are $1$.)
