Let $G_1$ and $G_2$ be compact Lie groups. We know that each finite-dimensional complex irreducible representation of $G_1\times G_2$ is the tensor product of an irreducible representation of $G_1$ and an irreducible representation of $G_2$.

But for real representation of $G_1\times G_2$, what is the general form of an irreducible real representation of the product group?

I don't think tensor product of a real irreducible representation of $G_1$ and one of $G_2$ works.

For example, if $G_1=\mathbb{Z}/p\mathbb{Z}$, $G_2=\mathbb{R}$, let $\rho_1: \mathbb{Z}/p\mathbb{Z}\longrightarrow O(2)$ send $m$ to the matrix

$\cos \frac{2\pi m}{p}, -\sin \frac{2\pi m}{p}$

$\sin \frac{2\pi m}{p}, \cos \frac{2\pi m}{p}$

and let $\rho_2: \mathbb{R}\longrightarrow O(2)$ send $t$ to the matrix

$ \cos at, -\sin at$

$\sin at, \cos at$

then the representation $G_1\times G_2\longrightarrow O(2)$ sending $(m, t)$ to

$\cos (at+\frac{2\pi m}{p}), -\sin (at+\frac{2\pi m}{p})$

$\sin (at+\frac{2\pi m}{p}), \cos (at+\frac{2\pi m}{p})$

is a real irreducible representation of $G_1\times G_2$ with representation space $\mathbb{R}^2$. It is definitely NOT $\rho_1\otimes \rho_2$, whose representation space is $\mathbb{R}^2\otimes\mathbb{R}^2=\mathbb{R}^4$. It's NOT the tensor product of any real irreducible representation $G_1$ and one of $G_2$ either.

I'm curious for any compact Lie group $G_1$ and $G_2$, whether there is a general form of an (irreducible) real representation of $G_1\times G_2$. So far I haven't found any reference on this.

  • $\begingroup$ Do you know how to deduce the classification of irreducible real representations from the classification of irreducible complex representations? $\endgroup$ May 18, 2015 at 23:42
  • $\begingroup$ Your example representation is non-faithful. It factors as $G_1 \times G_2\rightarrow (SO(2) \times G_2)/(\mathbb{Z}_p)_\text{diag}\rightarrow SO(2)$. You could say the missing tensor factor is a trivial representation of the kernel. $\endgroup$ May 19, 2015 at 10:27
  • $\begingroup$ @Megan: I'm confused by your example, since there is no obvious structure of "compact Lie group" on $\mathbb{Z}_p$. How are you defining compact Lie groups? $\endgroup$ May 20, 2015 at 14:56
  • $\begingroup$ @Jim: presumably $\mathbb{Z}_p$ refers to the cyclic group of order $p$, not the $p$-adic integers. $\endgroup$ May 20, 2015 at 18:56
  • $\begingroup$ @Qiaochu: Yes, but finite groups are usually misleading when regarded as compact Lie groups. Maybe the definition allows for this, but books treating representations of compact Lie groups don't usually cover finite groups and their rich theory beyond perhaps noting that some ideas for the compact (usually connected) groups arise by analogy from the Frobenius-Schur theory. Fine points of the representations of compact Lie groups usually involve maximal tori, Weyl groups, roots, etc. $\endgroup$ May 20, 2015 at 20:40

1 Answer 1


(Later comment- see analysis at the end): Not a complete answer, though I think this method should generalize to finite dimensional representations of compact Lie groups. (The Frobenius-Schur indicator is not generally available, but its role is to identify the nature of an invariant bilinear form, which is the key issue): The answer for finite groups is relatively easy to analyse, with the aid of the Frobenius-Schur indicator $\nu(\chi) = \frac{1}{|G|}\left( \sum_{g \in G} \chi(g^{2})\right).$ Let $\chi,\mu$ be complex irreducible characters of $G_{1},G_{2}$ respectively. If one of $\chi$ or $\mu$ is not real-valued, then $G_{1} \times G_{2}$ has an irreducible (over $\mathbb{R}$, but not over $\mathbb{C}$) representation affording character $(\chi \otimes \mu) + (\overline{\chi} \otimes \overline{\mu}).$ (This corresponds to the construction implicitly used in one of the examples in the question: given an $n$-dimensional complex representation of any group $G$, construct a $2n$-dimensional real representation of $G$ by replacing the complex matrix entry $a+bi$ by the $2 \times 2$ matrix $\left(\begin{array}{clcr} a & -b\\b& a \end{array}\right)$. If the original representation is irreducible with non-real character, then the resulting representation is irreducible as a real representation,but not as a complex representation).

If $\chi$ and $\mu$ are both real-valued, then there is an absolutely irreducible real representation affording character $\chi \otimes \mu$ if $\nu(\chi) = \nu(\mu)$ (for note that $\nu(\chi \otimes \mu) = \nu(\chi)\nu(\mu))$, and if $\nu(\chi) = -\nu(\mu)$, there is an irreducible (over $\mathbb{R}$, but not over $\mathbb{C}$) real representation of $G_{1} \times G_{2}$ affording character $2(\chi \otimes \mu).$

Note that it is only in the case that $\nu(\chi) = \nu(\mu)= 1 $ that the given irreducible real representation is the tensor product of irreducible real representations of the factors.

Later edit: Perhaps it is best to view everything from the viewpoint of analyzing the endomorphism algebra of an irreducible finite-dimensional real representation, in which case the same arguments apply in the finite case and the compact Lie group case. This is a finite dimensional (associative) division algebra over $\mathbb{R}$, hence is one of $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H},$ (the latter being the algebra of quaternions). In the first case, the representation is absolutely irreducible, so irreducible when the scalars are extended to $\mathbb{C}.$ In the second case, the representation is (after extension of scalars) the sum of two inequivalent irreducible complex representations which must afford mutually conjugate irreducible characters, so be dual to each other. In the last case, the representation is (after extension of scalars) a direct sum of two isomorphic complex irreducible representations, affording the same (real-valued) character. It is then quite easy to see that if the group is a direct product, it is only in the first case that it is possible for the representation to be a tensor product (in any fashion) of two smaller dimensional real representations of the whole group (noting of course that an irreducible representation of one factor may be viewed as an irreducible representation of the whole group with the other factor in its kernel).

  • $\begingroup$ BTW, is there a practical way to compute the Schur indicator of representations of compact groups? Especially, the tensor product of representations of compact groups? Thanks. $\endgroup$
    – Megan
    May 29, 2015 at 23:05
  • $\begingroup$ The end of the answer tells you how to obtain for compact groups what the indicator does in the finite case (but the representation is irreducible as real representation, and we are not in the trivial $1$-dimensional case). Calculate the dimension of the space of matrices commuting with everything in the group. If that dimension is greater than 1, the given representation is not a tensor product of real representations of smaller dimension. If the dimension is 1, and the group is a direct product, then the representation is a tensor product of irreducible real representations of the factors. $\endgroup$ May 30, 2015 at 8:39

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