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It is clear from the symplectic group isomorphism $$SL(2,\mathbb{C}) \cong Sp(2, \mathbb{C}) $$ that there is an $SL(2,\mathbb{C})$ invariant symplectic form on $\mathbb{C}^2$.

My question is whether or not this invariant form is enough to show the existence of an anti-linear $SL_2$ equivariant map, as it is in the compact case.

Another way to ask this, is whether or not the two irreducible 2d representations of the group are self-conjugate reps? I know for non-compact representations duality and conjugacy properties bifurcate.

EDIT: The answer to this question is that the existance of an invariant form and the existance of an invariant anti-linear map, are only "the same fact" when the relevant group is compact.

Craig
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    Beyond Callum's answer, compare my answer to your question https://math.stackexchange.com/q/4389806/96384, as well as https://math.stackexchange.com/q/4024163/96384 for background. – Torsten Schoeneberg Jan 21 '24 at 17:00
  • It seems my confusion arose in part from conflating, despite my best efforts, the roles of duality and conjugation. Thanks again Torsten. – Craig Jan 21 '24 at 18:19

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Note importantly that reps being conjugate to each other and being quaternionic are features of real Lie algebras and their representations. However, $\mathfrak{sl}(2,\mathbb{C})$ is famously isomorphic to $\mathfrak{so}(3,1)$ as a real Lie algebra so we can look at this in two ways.

As a complex Lie algebra $\mathfrak{sl}(2,\mathbb{C})$ has a single complex 2-dimensional rep which is naturally self-dual as the presence of an invariant bilinear form forces. There is no sense here in which we can talk about this rep being self-conjugate or quaternionic.

Meanwhile as a real Lie algebra $\mathfrak{so}(3,1)$ has two complex 2-dimensional reps (the half-spin reps). They are each self-dual but they are conjugate to each other. As mentioned on Wikipedia, there is a Hermitian structure over the pair of them together (but not individually).

Callum
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  • I think between Torsten & yourself my confusions have been cleared. I'd been holding on to compact intuitions from $SU(2)$ and being sloppy with complexifications. Can you confirm the following is correct here? For a real rep of a (generally non-compact) semi-simple Lie group, it is self-duality which dictates the existence of a G-invariant (symmetric/alternating) bi-linear form, and it is self-conjugacy which dictates the existence of a G-invariant anti-linear map (real/quaternionic structure). In the compact case, these facts imply one another. – Craig Jan 21 '24 at 18:50
  • Additionally, because I find it cool: when these are both satisfied, one finds a rep with an invariant Hermitian form, by composing the anti-linear map and bi-linear form. – Craig Jan 21 '24 at 18:58
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    @Craig Yes self-duality implies the existence of an invariant nondegenerate bilinear form which must be symmetric or alternating (the converse is also true). Self-conjugacy is equivalent to a real or quaternionic structure. When the dual is equal to the conjugate you get a Hermitian form (so self-dual plus self-conjugate falls into this category) – Callum Jan 22 '24 at 15:47