Is the highest weight constructed irreducible representiation of a Lie Algebra unique

Question

In Georgis book I stumbled across the sentence:
"It is true that every irreducible representation is finite dimensional and equivalent to one of the constructions we found with the highest weight construction."
Since there can be in-equivalent irreducible representations that means that the highest weight constructed irreducible representation of a Lie Algebra is not unique, right?

Answer 1

As written, the claim is wrong.

There are examples of irreducible representations which are infinite dimensional: Let $V$ be any $\mathbb{C}$ vector space for which a basis of the form $(e_k)_{k \in \mathbb{Z}}$ can be written. Let's look at the lie Algebra $\mathfrak{sl}_2\mathbb{C}$ with the (usual) basis $\{H,X,Y\}$, see this article. Let $t \in \mathbb{R} \setminus \mathbb{Z}$. Define a representation $$\pi: \mathfrak{sl}_2 \mathbb{C} \to \mathfrak{gl}(V)$$ by setting: $$\pi(H)e_k=(t-2k)e_k, \quad \pi(Y)e_k=(k+1)e_{k+1}, \quad \pi(X)e_k=(t-k+1)e_{k-1} \quad \forall k \in \mathbb{Z}.$$ Now, the linear hull of $\{e_k\}_{k < 0}$, call it $V_{-}$, is an invariant subspace of $V$ (under the given action) and is, indeed, irreducible.

Edit: As for your actual question: If we restrict ourselfs to the finite dimensional case and only look at representations of (complex) semisimple Lie Algebras the theorem of highest weight states that the highest weight does determine the representation (up to equivalence), in general there can be irreducible inequvalent representations on the same vector space (if $t-t' \notin 2 \mathbb Z$, we get two inequivalent represntations on $V_{-}$).