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Context
Let $\mathfrak g$ be a Lie algebra over a field $k$. Let $V$ be a $\mathfrak g$-representation. One can prove that $V$ and $V^*$ (the contragredient representation) are isomorphic as $\mathfrak g$-representations if and only if there exists a non-degenerate $\mathfrak g$-invariant bilinear form $\langle \cdot , \cdot \rangle$ on $V$. This shows (by looking at the Killing form) that the adjoint representation of any semisimple Lie algebra is self-dual. On the other hand, by comparing characters one can show that any $\mathfrak sl(2, \mathbb C)$-representation is self-dual.

Questions
What are other examples of self-dual $\mathfrak g$-representations? What are necessary and sufficient conditions (on $\mathfrak g$ or the action on $V$) for $\mathfrak g$-representations to be self-dual? Is it true that for $\mathfrak g$ semisimple any $\mathfrak g$-representation is self-dual? (Maybe one could compare characters again.)

Max Demirdilek
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    Related: https://math.stackexchange.com/q/3064375/96384, https://mathoverflow.net/q/111127/27465, https://math.stackexchange.com/q/212097/96384, https://math.stackexchange.com/a/3280399/96384, https://math.stackexchange.com/q/1731499/96384 – Torsten Schoeneberg Feb 04 '22 at 23:32

1 Answers1

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Firstly, I'll note that not every representation for a semisimple Lie algebra is self-dual but we can classify those that are relatively easily.

Let's assume for simplicity that $\mathfrak{g}$ is complex and we're only talking about finite-dimensional stuff. An irreducible $\mathfrak{g}$-representation is defined by its highest weight and this is a dominant integral weight. In other words the highest weight is of the form $\sum_i m_i \omega_i$, for $\omega_i$ the fundamental weights and $m_i$ a set of non-negative integers. Thus we can classify all irreducible representations by decorating the Dynkin diagram with $m_i$ at the node corresponding to $\omega_i$.

For each Dynkin diagram there is a special diagram automorphism which "implements" the duality. There are a few ways to describe this but one way is take the unique element $w$ of the Weyl group which sends the fundamental chamber to its negative and then $-w$ permutes the simple roots giving a diagram automorphism. However we find this automorphism, we can list how it acts:

  • For $A_n$ it flips the diagram (left to right)
  • For $D_n$, n odd it swaps the end pair of nodes
  • For $E_6$ it flips the diagram (left to right)
  • For everything else ($D_n$, n even, $B_n$, $C_n$, $E_7$, etc.) it is the identity

The dual representation is given by the diagram after this automorphism has been applied. Thus when the Dynkin diagram is in the "everything else" group all representations are self dual. However, for $A_n$, $E_6$ and $D_n$ (n odd) the representation is only self-dual if the diagram is symmetric with respect to that automorphism.

So for example, with $\mathfrak{sl}_{n+1}$ the standard representation is only self dual if $n+1 = 2$. The adjoint representation here has highest weight $\omega_1 + \omega_n$ so we can see that this is self-dual. If $n+1 =2k$ is even there is another easy example the wedge product $\bigwedge^k V^{2k}$.

For $\mathfrak{so}_{2n}$ we see that every representation is self-dual for $n$ even, but for $n$ odd some aren't, including the half-spin representations which are instead dual to each other.

Callum
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