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Suppose $V,W$ are two finite-dimensional representations of a Lie algebra $\mathfrak{g}$.

Is it true that if their formal characters coincide, $$\mathrm{ch}_V=\mathrm{ch}_W ,$$ then the representations are isomorphic?

Context:

I have read in a couple of sources the derivation of the Clebsch-Gordan formula for $\mathfrak{sl}(2, \mathbb{C})$ using the formal character. These texts simply conclude checking that the two sides of the Clebsch-Gordan formula have the same formal characters. However as long as I know the central character gives rise to a ring isomorphism $$ \mathrm{ch}: K(\mathsf{Mod}_{\mathfrak{g}}) \overset{\cong}{\to} \mathbb{Z}[\Lambda]^{\mathcal{W}}$$ where $K(\mathsf{Mod}_{\mathfrak{g}})$ denotes the representation ring of $\mathsf{Mod}_{\mathfrak{g}}$ (aka the Grothendiek construction of the semiring of representations), $\Lambda$ is the set of integral weights and $\mathbb{Z}[\Lambda]^{\mathcal{W}}$ is the set of invariants under the action of the Weyl group $\mathcal{W}$.

The issue I have is that this isomorphism via the formal character then identifies the classes $[V]$ and $[W]$ in $ K(\mathsf{Mod}_{\mathfrak{g}}) $, and in a Grothendiek construction, these are the same if and only if there is some representation $Z$ such that $V \oplus Z \cong W \oplus Z$. Not sure why this would imply that $V \cong W$, I don't think that the cancellation property holds here.

Minkowski
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1 Answers1

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Yes, it's true. The cancellation property does hold for finite-dimensional representations of $\mathfrak{sl}_2$, and more generally for finite-dimensional representations of any semisimple Lie algebra, because finite-dimensional representations are completely reducible in this case. The formal character is only defined for semisimple Lie algebras anyway, and in this case it is true that two finite-dimensional representations are isomorphic iff they have the same formal character.

One way to think about what's going on is to reduce to the case of compact groups. Namely, every complex semisimple Lie algebra has a compact real form $\mathfrak{g}$ (so the original Lie algebra is the complexification $\mathfrak{g}_{\mathbb{C}}$), which here is $\mathfrak{su}(2)$. We have a three-way equivalence of categories between

  • the category of finite-dimensional complex representations of $\mathfrak{g}_{\mathbb{C}}$ (which here is $\mathfrak{sl}_2(\mathbb{C})$),
  • the category of finite-dimensional complex representations of $\mathfrak{g}$, and
  • the category of finite-dimensional complex representations of the unique simply connected compact Lie group $G$ with Lie algebra $\mathfrak{g}$ (which here is $SU(2)$).

(General Lie theory implies that $G$ exists but it takes extra work to show that $G$ is compact, although in this case we can just verify it explicitly.)

Representations of $G$ are completely reducible because of the usual business with averaging inner products (this is Weyl's unitary trick), and passing through the equivalences implies this is also true of representations of $\mathfrak{g}_{\mathbb{C}}$. It also follows from the orthogonality relations that a finite-dimensional representation of a compact group $G$ is determined up to isomorphism by its character, and the data of the formal character of a representation of $\mathfrak{g}_{\mathbb{C}}$ is equivalent to the data of the character of the corresponding representation of $G$, as we discussed previously for $\mathfrak{sl}_2$.

Qiaochu Yuan
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    Without passing to compact groups you could do the argument more purely Lie-algebraically by arguing that the formal character determines the highest weights and appeal to highest weight theory. This should be fairly explicitly doable for $\mathfrak{sl}_2$ but I'm less familiar with how this works for general semisimple Lie algebras. – Qiaochu Yuan Jul 18 '24 at 17:34
  • Thank you for the elaborated answer! Just to make sure: what you explain about the cancellation property also applies to any semisimple category, right (e.g $k[G]$-Mod)? That is, in every semisimple category $\mathcal{C}$, two objects $X,Y$ are isomorphic if and only if their virtual classes in the Grothendieck ring $K(\mathcal{C})$ are the same, right? – Minkowski Jul 18 '24 at 18:47
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    @Minkowski: with the caveat that we only consider finite-dimensional representations throughout, yes. The Grothendieck ring construction doesn't work well if we consider infinite-dimensional representations, e.g. the Grothendieck ring of $\text{Vect}$ is trivial because of the "Eilenberg-Mazur swindle": https://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle – Qiaochu Yuan Jul 18 '24 at 18:57
  • yes, thanks for pointing that out, I definitely meant something like "semisimple and pivotal" or "semisimple and locally finite". – Minkowski Jul 18 '24 at 20:11