Let $\mathfrak{g}$ be the complexified Lie algebra of a simple compact real Lie group. Then the adjoint and coadjoint representation are isomorphic (see Under what conditions can we find a basis of a Lie algebra such that the adjoint representation acts by skew-symmetric matrices?).

In the representation category of $\mathfrak{g}$, we naturally have an intertwiner $\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ (viewing $\mathfrak{g}$ as the adjoint representation) given by the Lie bracket. My question is wether simplicity and compactness are sufficient conditions for the Lie bracket to be the only such intertwiner (up to a constant factor). Phrased differently, when does $F: \mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ with \begin{align} F([x,y] \otimes z) + F(y \otimes [x,z]) = [x,F(y \otimes z)] \end{align} for all $x,y,z \in \mathfrak{g}$ imply $F(y \otimes z) = c \cdot [y,z]$, with $c \in \mathbb{C}$?

I already tried to show that $[[y,z],F(y \otimes z)]=0$, but I can't even get that done.

I also tried working in the basis mentioned in the page I reference. Then the formula becomes \begin{align} \forall r \in \{1 \ldots \dim \mathfrak{g}\}: \sum_{n=1}^{\dim \mathfrak{g}} \langle [e_k,e_i]f^{(nj)}_r + [e_k,e_j]f^{(in)}_r + [e_k,e_r]f^{(ij)}_n , e_n \rangle = 0 \end{align} where $F(e_i \otimes e_j) = \sum_{k=1}^{\dim \mathfrak{g}} f^{(ij)}_k e_k$, and the inner product is defined as $\langle x,y \rangle = \text{Tr}(\text{Ad}_x \text{Ad}_y^*)$, where the star is just taking a matrix adjoint in the basis $\{e_i\}$. However, I don't know how to proceed from here.

Any help is welcome.

  • 1
    $\begingroup$ For semisimple $\mathfrak g$, in principle you should be able to use highest weight theory to see how many copies of $V =\mathfrak g$ (adjoint representation) lie in the tensor product. If it's one, then the result follows from Schur. If it's more, there's more. $\endgroup$ Dec 24, 2020 at 17:30
  • $\begingroup$ Following up on my earlier comment, for a simple complex Lie algebra, the multiplicity of the adjoint rep in its tensor product with itself is $1$, except in the case of $\mathfrak{sl}_{n \ge 2}$ where it is $2$, as discussed in mathoverflow.net/q/129857/27465 and mathoverflow.net/q/130185/27465. So the result is true for all cases except that one. In the $\mathfrak{sl}_{n \ge 3}$ case, the first MO post suggests that the extra dimension of intertwiners is given by something like $x\otimes y \mapsto xy + yx - \frac1n tr(xy+yx)\cdot Id_n$. $\endgroup$ Dec 27, 2020 at 22:34

1 Answer 1


Turning comments into an answer:

For any semisimple complex Lie algebra $\mathfrak g$, if $W$ is any finite-dimensional representation of $\mathfrak g$, then $W$ decomposes into a finite direct sum $W \simeq \bigoplus_i V_i^{m_i}$ where the $V_i$ are mutually isomorphic irreducible (=simple) representations with multiplicities $m_i \ge 1$. Further, if $V$ is any irreducible representation of $\mathfrak g$, then either $$Hom_{\mathfrak g} (V,W) = Hom_{\mathfrak g} (W,V) = 0,$$ in which case $V \not \simeq V_i$ for all $i$; or, if $V \simeq V_i$, then $$dim(Hom_{\mathfrak g} (V,W)) = dim(Hom_{\mathfrak g} (W,V)) = m_i.$$ (This is essentially the definition of semisimple representation plus Schur's Lemma.)

Now in our case we let $\mathfrak g$ be simple, and $V$ the adjoint representation (well-known to be simple). According to https://mathoverflow.net/q/129857/27465 and https://mathoverflow.net/q/130185/27465 (or case-by-case check in LiE), the multiplicity of $V$ in $V \otimes V$ is $1$, except in the case of $\mathfrak g \simeq \mathfrak sl_{n \ge 3}$, in which it is $2$.

(Funnily, the MO posts leave a bit of a gap for showing the multiplicity is $\ge 1$: which is exactly what your question shows, namely, the Lie bracket is an obvious non-trivial element of $Hom_{\mathfrak g}(V \otimes V, V)$.)

So in all cases except $\mathfrak g \simeq sl_{n \ge 3}$, the result is true, again due to Schur's Lemma.

In the case $\mathfrak g \simeq sl_{n \ge 3}$, the missing dimension is given by the intertwiner

$$F_{extra}: x\otimes y \mapsto xy + yx - \frac1n Tr(xy+yx)$$

(the "traceless part of $xy +yx$", as Allen Knutson writes in the MO post; note that this is the composition of the map $x\otimes y \mapsto xy+yx$, whose codomain is $\mathfrak{gl}_n$, with the projection $\mathfrak{gl}_n \twoheadrightarrow \mathfrak{sl}_n, z \mapsto z - \frac1n Tr(z)$).

A little computation indeed shows that $F_{extra}$ is an element of $Hom_{\mathfrak g}(V \otimes V, V)$, and is linearly independent from the intertwiner given by the Lie bracket for all $n \ge 3$ (it does vanish for $n=2$, matching earlier results). What happens there seems to be a decomposition of $V \otimes V$ into symmetric versus alternating tensors and what's special in this case is that there is an extra copy of $V$ within the symmetric ones. If, on the other hand, one tries to imitate the map $F_{extra}$ e.g. in the case of $\mathfrak{so}_n$, it is not even well defined i.e. the image does not lie in $\mathfrak{so}_n$ etc.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .