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As an example consider the complex $16$-dimensional representation of $\mathfrak{so}(10)$. When $\mathfrak{so}(10)$ is reduced to the subalgebra $\mathfrak{so}(9)$, the complex $16$-dimensional representation of $\mathfrak{so}(10)$ becomes the real $16$-dimensional representation of $\mathfrak{so}(9)$

$$ 16 \rightarrow 16 $$

Another example is the complex $27$ of $\mathfrak{e}_6$, which becomes the real $26\oplus 1$, when $\mathfrak{e}_6$ is reduced to $\mathfrak{f}_4$.

A complex $16$ has $16\cdot 2=32$ degrees of freedom, a real $16$ only $16$. Where do these degrees of freedom go? Surely they can't get lost so do we really get somehow

$$ 16 \rightarrow 16 \oplus i 16 $$

or something like that?


The other way round things are more transparent. When a algebra with only real representations becomes a subalgebra with complex representation, we always have

$$ R= r_1\oplus \bar r_1 \oplus \ldots$$

jak
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    What do you mean reduced? I would have guessed that you meant restricting the representation to the subalgebra, but that does not fit your description of what happens. – Tobias Kildetoft Aug 24 '15 at 12:41
  • @TobiasKildetoft I think I mean restricting the representation to the subalgebra or why does it not fit my description? Another word for the same I think is branching. The $16$ of $SO(10)$ branches to the $16$ of $SO(9)$ etc – jak Aug 24 '15 at 12:45
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    After restricting the action the space is still a 16 (or 27)-dimensional complex vector space. So do you mean that the representation becomes the complexification of a real representation? – Vincent Aug 24 '15 at 12:50
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    @Vincent I think this is what I'm trying to understand. For example page 227 in this pdf http://cds.cern.ch/record/134739/files/198109187.pdf table 49 the branching rules for $E_6 \rightarrow F_4$ are listed. One of these is $27 \rightarrow 26 \oplus 1$. Or on page 221 table 43 the branching rule for $SO(10) \rightarrow SO(9)$ is $16 \rightarrow 16$. I'm trying to understand what this means, because the $16$ of $SO(10)$ is complex, but the $16$ of $SO(9)$ is real. Do the authors mean here implicitly that we get a complexification of the $16$ of $SO(9)$ when we branch from $SO(10)$? – jak Aug 24 '15 at 13:02

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As alluded to in the comments, this is just an error of terminology, and the terminology is indeed unfortunate. Here's what's happening:

  • $\mathfrak{so}(10)$ has two inequivalent irreps on $16$-dimensional complex vector spaces; they are the ones with highest weight vector $\varpi_4$ and $\varpi_5$ respectively, where $\varpi_4$ and $\varpi_5$ are the two fundamental weights which correspond to (i.e. are dual to the simple roots which form) the two "horns" in the Dynkin diagram $D_5$. By the way, they seem to be prominent. They feature in https://mathoverflow.net/q/40564/27465 and recently their lifts to the group level came up in Embed a Spin group to a special unitary group.

  • These two irreps are, as said, not equivalent to each other, but they happen to be conjugates of each other.

  • On the other hand, $\mathfrak{so}(9)$ has up to equivalence just one irrep on a $16$-dimensional complex vector space; it is the one whose highest weight vector is $\varpi_4$, which corresponds to the short root in the Dynkin diagram $B_4$.

  • This representation is self-conjugate (like all representations of something of type $B$).

Now, the branching you cite implies that if you restrict either of those two $16$-dimensional irreps of $\mathfrak{so}(10)$ to $\mathfrak{so}(9)$, you get that $16$-dimensional irrep $V_{\mathfrak{so}(9)}(\varpi_4)$.

What confuses you and others is that some people (especially physicists) would call those two irreps of $\mathfrak{so}(10)$ "complex", and the irrep of $\mathfrak{so}(9)$ "real". However, that $\mathfrak{so}(9)$-representation is still on a $16$-dimensional complex vector space. The terminology "real" here means that this one has a "real structure": when you view that $\mathbb C^{16}$ as a $32$-dimensional real vector space, you can find a $16$-dimensional real subspace which is invariant under the $\mathfrak{so}(9)$-action. (Also and equivalently, this complex representation comes from a real representation by complexifying this $16$-dimensional real subspace, as Vincent suggests in a comment.) Whereas no such real subspace will be invariant under either of the $\mathfrak{so}(10)$-actions; which happens to follow from the fact stated above that they are conjugate but inequivalent; which in this terminology means those irreps are "truly complex". But, stressed again, all these irreps are on $\mathbb C^{16}$, and no dimension is "lost" in the restriction / branching.

Maybe after asking this question you noticed that bad terminology, because it was the content of a question you asked one day later: Are the physics and math definitions of a complex representation equivalent? A related confusion came up in What property of the root system means a Lie algebra has complex structure?


It might be worthwhile to remark that the above situation repeats itself almost verbatim for every inclusion $\mathfrak{so}(4n+1) \subset \mathfrak{so}(4n+2)$, where the bigger Lie algebra has two conjugate "truly complex" irreps of dimension $2^{2n}$ (which are minuscule and apparently called "half-spin"), and both of them "branch" to just one complex irrep, of same dimension, of the smaller group (also minuscule and apparently called "spin"), and the latter one has a real structure.

One can see that explicitly in the case $n=1$, where the inclusion $\mathfrak{so}(5) \subset \mathfrak{so}(6)$ translates via standard isomorphisms to $\mathfrak{sp}(2) \subset \mathfrak{su}(4)$ (where $\mathfrak{sp}(2)$ denotes the compact real form of the symplectic Lie algebra of type $C_2$), which one can both write as certain complex $4\times 4$-matrices, and then one can check that the natural action of the bigger one on $\mathbb C^4$ is non-equivalent to its conjugate, whereas the action of $\mathfrak{sp}(2)$ is equivalent to its conjugate and has a "real structure".