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Take for example, $SU(2)$. There are two well known invariant forms in the fundamental representation, namely (by the Frobenius-Schur indicator) a skew-symmetric bi-linear form: $$ \varepsilon = \left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right) $$ Since for all $S \in SU(2)$: $\quad S^T \varepsilon S = \varepsilon$

and a hermitian sesquilinear form: $$ \mathbb{I}_2 = \left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right) $$ Since for all $S \in SU(2)$: $\quad S^\dagger \mathbb{I}_2 S = \mathbb{I}_2$

Is there any way to show that these are/aren't the only non-degenerate invariant 2-forms for this representation (up to a scalar factor)?

My suspicion is that there can only ever be two, since one can act with the dual space, or the conjugate dual space. (Multilinear forms are also interesting but it seems more productive to constrain the question)

Craig
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    To be clear: By "form" you mean only bilinear forms, i.e. on $V \times V$, or any multilinear forms, i.e. on some $V \times ... \times V$? Also, I guess you want to say "up to scaling" (any scaled version of an invariant form, in either sense, would again be one). – Torsten Schoeneberg Feb 26 '22 at 20:48
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    Yes sorry for the loose language. I guess for the purposes of this question I'm looking for a set of independant equivariant maps from $V\times V \rightarrow \mathbb{F}$. I was wary to use bilinear since the usual inner product is sesquilinear, and I didnt want to potentially rule out anti-linear maps – Craig Feb 26 '22 at 23:44

1 Answers1

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The case of bilinear forms is taken care of by:

Proposition (Bourbaki, Lie Groups and Algebras, ch. VIII § 7 no. 5 prop. 12):

Let $k$ be a field of characteristic $0$, $\mathfrak g$ a split semisimple $k$-Lie algebra (with a fixed choice of CSA and simple roots), $V = V(\lambda)$ a simple $\mathfrak g$-module of highest weight $\lambda$ (each simple $\mathfrak g$-module is of this form).

Then the space of all $\mathfrak g$-invariant bilinear forms on $V$ has dimension $0$ or $1$; meaning that either the only such form is $0$ (case I), or there is up to scaling exactly one non-zero one. If this is the case, this unique one up to scaling is non-degenerate, and either symmetric (case II) or alternating (case III).

Case I happens if and only if $V$ and its dual $V^*$ are not isomorphic as $\mathfrak g$-modules, if and only if $w_0(\lambda) \neq -\lambda$ where $w_0$ is the longest element of the Weyl group.

Case II happens if and only if $V \simeq V^*$ as $\mathfrak g$-modules, and a certain invariant $c(\lambda)$ is even.

Case III happens if and only if $V \simeq V^*$ as $\mathfrak g$-modules, and the invariant $c(\lambda)$ is odd.

Where that invariant $c(\lambda)$ can e.g. be defined as twice the sum of coefficients of $\lambda$ written in the chosen root basis $\Delta$,

$$c(\lambda) := 2 \cdot \sum c_\alpha \text{ where } \lambda = \sum_{\alpha \in \Delta} c_\alpha \alpha$$

(The $2$ is there to make the invariant an integer, matching the math convention in enumerating highest weights e.g. of $\mathfrak{sl}_2$ by integers.)

The basic idea of the proof is to realize that the space of $\mathfrak g$-invariant bilinear forms on $V$, via typical duality concepts (https://math.stackexchange.com/a/3781591/96384, https://math.stackexchange.com/a/3701479/96384) identifies with the space of $\mathfrak g$-module homomomorphisms $Hom_{\mathfrak g}(V, V^*)$. Now use that $V$ is simple, and a strengthened version of Schur's Lemma (proved a little earlier in Bourbaki) which says that in the case at hand, indeed $Hom_{\mathfrak g}(V,V) \simeq k$ (not just some skew field extension of $k$). From this follows the "dimension zero or one" assertion, and non-degeneracy in the latter case, which in turn implies via a proportionality of $B(x,y)$ with $B(y,x)$ that the form must be either alternating or symmetric. The final distinction with the invariant between these cases II and III is a bit more subtle, but follows from consideration of certain subsets of the root spaces.

You'll notice that the whole case distinction is similar to the one of complex versus real versus quaternionic representations. That is of course no coincidence. In the case that $\mathfrak g$ is a compact real semisimple Lie algebra, we have a perfect analogue of the above in Bourbaki loc.cit. ch. IX no. 2 prop. 1 (cited e.g. in https://math.stackexchange.com/a/2774741/96384 and https://math.stackexchange.com/a/4026224/96384), and the proof relies on the above, together with the striking fact that in this (compact!) case, complex conjugation operates (just like dualizing) via that Weyl group element $-w_0$. I.e. where in the above one wrote the dual $V^*$, one now writes the conjugate $\overline V$, and things go through. In Appendix II of this chapter IX of Bourbaki, it is also alluded to how to translate between (non-degenerate) bilinear forms and ("separating" = non-degenerate) hermitian sesquilinear forms. It is not stated explicitly there, but it seems to me that that translation entails analogous uniqueness results for the hermitian forms.

Torsten Schoeneberg
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  • I can not thank you enough for how much I appreciate your contributions to this site Torsten! (A comment: your first set of links are a duplicate pair, I imagine this was not intentional) – Craig Feb 27 '22 at 08:30
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    @Torsten I think in Case III you meant to say that $c(\lambda)$ is odd – Callum Feb 28 '22 at 13:33
  • This proof holds for irreducible representations of the algebra, right? For reducible representations I beleive I could have a greater dimension of invariant bi-linear forms – Craig Mar 23 '22 at 06:06
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    @Craig: Yes. In this context "irreducible" = "simple" which is the word I use several times in this answer. – Torsten Schoeneberg Mar 23 '22 at 06:09
  • @TorstenSchoeneberg while you're here, could you expand why you start with split-semi-simple lie algebras? You address the compact case, but does this carry to quasi-split forms (surprise surprise this question is secretly about the Lorentz group)? – Craig Mar 23 '22 at 06:18
  • Because otherwise there is (in general) no CSA, no weights $\lambda$ etc. to begin with. Surely for any real semisimple $\mathfrak g$, one could look at the complexification $\mathfrak g_\mathbb C$; the result here applies to the $\mathfrak g_\mathbb C$-invariant forms on irreducible (complex) representations $V$ then; maybe that easily gives a corresponding result for $\mathfrak g$-invariant forms on those $V$, maybe not. – Torsten Schoeneberg Mar 24 '22 at 02:54
  • If we are looking at an irreducible representation of a linear group over a complex vector space, and we have two hermitian forms $g, h \in V^* \otimes \overline{V}^*$, then in some basis, the matrices corresponding to $g$ and $h$ must satisfy the Lie algebra identity: $g \lambda g^{-1} = -\lambda^\dagger $ (and the same for $h$). From this it is clear that $(h^{-1}g)$ is a linear intertwiner. By Schur's lemma this must be a multiple of the identity and so $g \sim h$. If one has two independent hermitian products either the space is reducible, or conjugation acts trivially and $V$ is real :) – Craig Jan 25 '24 at 06:37