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According to this question, it is easy to know whether a (complex, finite-dimensional) representation is self-dual or not: just check if the weight distribution in space is symmetric about the origin.

Now, for self-dual representations, we have two possibilities: the representation being real or quaternionic (see this paper). The question is:

Is it possible to conclude whether the representation is real or quaternionic by seeing the geometric composition of the weights in space, just like the question of self duality?

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    $\begingroup$ Self conjugate irreducible representations are quaternionic or real type. Self duality and self conjugacy coincide when $G$ is compact (as is assumed in the paper you linked), but these conditions are not generally equivalent. $\endgroup$ – Henrik Winther Jun 7 '15 at 18:17
  • $\begingroup$ @HenrikWinther Yes... I mean in the case that $G$ is compact, where self-conjugacy and self-duality are equivalent. $\endgroup$ – Jjm Jun 7 '15 at 18:20
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There is some misleading information in the literature about this issue, due (as mentioned by Henrik Winther) to special properties of compact groups.

Suppose $\pi$ is an irreducible representation of a real group.

If $\pi$ is self-dual then it supports an invariant bilinear form, which is symmetric or skew-symmetric. Which one is given by the Frobenius Schur indicator $\epsilon(\pi)=\pm1$.

If $\pi$ is self-conjugate, it is real or quaternionic. Write the real/quaternionic indicator $\delta(\pi)=\pm1$ respectively.

If $\pi$ is Hermitian (admits an invariant Hermitian form, not necessarily positive definite), then self-dual and self-conjugate are equivalent (a stronger statement than the first comment above). If $\pi$ is furthermore unitary then $\delta(\pi)=\epsilon(\pi)$.

There is an elementary formula for $\epsilon$ when $\pi$ is finite dimensional. See Bourbaki, Lie Groups and Lie Algebras, Chapter 7-9, or see The Real Chevalley Involution, arXiv:1203.1901 for a simpler proof.

Proposition: $\epsilon(\pi)=\chi_\pi(\exp(2\pi i\rho^\vee))$

where $\chi_\pi$ is the central character and $\rho^\vee$ is one-half the sum of the positive coroots. Obviously this is independent of the real form. In general, however, $\delta(\pi)$ is sensitive to the real form.

Corollary: If $\pi$ is unitary, in particular if $G$ is compact, the real-quaternionic indicator is $\delta(\pi)=\chi_\pi(\exp(2\pi i\rho^\vee))$.

This gives the simplest answer to the original question, assuming it was only about compact groups. For non-compact groups the relationship is more complicated, although there is still a closed formula; this is the subject of a forthcoming dissertation by Ran Cui at the University of Maryland.

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It is unlikely that there is a procedure of the kind you would like, because the weight structure of representations depends only on the complexified Lie algebra, while distinguishing between real and quaternionic type depends on which real form is being considered. This is something that is obscured, I think, by restriction to only compact forms, because then you have a bijection between real and complex algebras. For example, the first fundamental (complex) representation of $\mathfrak{sl}_2(\mathbb{C})$ corresponds to a quaternionic type (real) irreducible of the real form $\mathfrak{su}(2)$, but real type irreducible for the other form $\mathfrak{sl}_2(\mathbb{R})$. Both of these have the same weights.

Another piece of evidence is that there is no mention of such a procedure to determine the type in Onishchik - Lectures on Real Semisimple Lie Algebras and Their Representations - EMS (2003). This book is the most comprehensive guide to computing types of fundamental representations from scratch that I know of, but the way described there is rather involved.

If you are willing to accept the types of the fundamental representations as given (e.g. from a table), then there is however a rather simple formula dating to Cartan for determining the type of other arbitrary irreducibles in terms of highest weight. The details for this is also given in Onishchik's book.

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  • $\begingroup$ I'm confused. I thought the issue was whether the invariant vector in $V\otimes V$ was in $Sym^2 V$ or $Alt^2 V$, and I don't see how this would be affected by choice of real form. $\endgroup$ – Allen Knutson Jun 8 '15 at 2:25
  • $\begingroup$ @AllenKnutson Consider the example I gave. Both have invariant skew symmetric forms, but one representation is real and the other quaternionic. So that cannot be the whole story. $\endgroup$ – Henrik Winther Jun 8 '15 at 8:09
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    $\begingroup$ All right, so your take on the question is about trying to extend the action of $G$ on $V$ to an action of $U(n,\mathbb R,J)$ vs. $U(n,\mathbb H,J)$ where the form $J$ need not be definite. Since the questioner made clear in comment that they're interested in the compact case, I believe this is an unnecessary wrinkle for them. $\endgroup$ – Allen Knutson Jun 8 '15 at 10:27
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    $\begingroup$ @AllenKnutson I answered the question as written, maybe not as intended. I worry not only about the person that asked the question, who perhaps wants the compact case only, but also about others who might happen upon this because they want to know the answer to the question that was actually stated. (it's a very interesting question!) $\endgroup$ – Henrik Winther Jun 8 '15 at 11:05
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$\mathrm{Hom}_G(V^*,V) \cong \mathrm{Hom}(V^*,V)^G \cong (V\otimes V)^G \cong (\mathrm{Sym}^2 V\oplus \mathrm{Alt}^2 V)^G$, so the first is nonzero ($V$ is self-dual) iff $V$ possesses a symmetric or alternating invariant form. By Schur's lemma it can have only one of the two.

It's not as easy to do in your head, but you can indeed compute the weight multiplicities of $Alt^2 V$ from those of $V$: $m_\nu(\mathrm{Alt}^2 V) = {m_{\nu/2}\choose 2} + \sum_{\{\lambda,\mu\}, \lambda\neq \mu} m_\lambda(V) m_\mu(V)$.

Now you need to determine whether those multiplicities give a representation with an invariant vector. This is the most annoying part. For each positive root $\beta$, we have a differencing operator "at $\mu$, subtract the value located at $\mu+\beta$", and these commute. Apply all of them and see if the value at the origin is $1$ or $0$. (This is undoing the denominator in the Weyl character formula, just Fourier transformed to deal with weight multiplicities.) If $1$, then $V$ is quaternionic (for $G$ compact); if $0$, then $V$ is real (for $G$ compact, and $V$ assumed self-dual).

The latter half of this answer may not be so different from answering your boldface question with "Yes, in principle you must be able to, since the representation is characterized by its multiplicity diagram."

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To tell whether the representation is real or quaternionic, you need to look at its highest weight. The situation is as follows.

First of all, you need to ensure that the representation is self-dual. This happens if and only if the highest weight lies in some vector subspace of $\mathfrak{h}^*$. (For a lot of groups, this is just the whole space).

After this, you have two cases:

  • For some groups, every self-dual representation is of real type.
  • For the other groups, the highest weights of real representations form a sublattice of index $2$ in the lattice of all self-dual weights. Basically, to determine the type of a representation, you must add together some of the (integer) coordinates of its highest weight, and examine the parity of the sum. If the sum is even, the representation is real; it it is odd, the representation is quaternionic.

I have put together a concise table listing all the existing real forms of all simple Lie groups and giving a recipe for each. It can be found here: http://www.math.u-psud.fr/~smilga/Lie_group_tables/tables.html ; the relevant part is the next-to-last item. The last item can also be of interest: it contains some explicit pictures for low-rank groups.

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