I am trying to understand why two-dimensional representation of irreducible nonlinear character of $Q_8$ is not "real".

I know that there is exactly one two-dimensional representation of $Q_8$, which associated with this nonlinear character, which, up to equivalence, has the following form: \begin{equation*} \begin{pmatrix} 1 & 0\\ 0 & 1\\ \end{pmatrix} \rightarrow 1; \end{equation*}

\begin{equation*} \begin{pmatrix} -1 & 0\\ 0 & -1\\ \end{pmatrix} \rightarrow -1; \end{equation*}

\begin{equation*} \begin{pmatrix} i & 0\\ 0 & -i\\ \end{pmatrix} \rightarrow i; \end{equation*}

\begin{equation*} \begin{pmatrix} -i & 0\\ 0 & i\\ \end{pmatrix} \rightarrow -i; \end{equation*}

\begin{equation*} \begin{pmatrix} 0 & -1\\ 1 & 0\\ \end{pmatrix} \rightarrow j; \end{equation*}

\begin{equation*} \begin{pmatrix} 0 & 1\\ -1 & 0\\ \end{pmatrix} \rightarrow -j; \end{equation*}

\begin{equation*} \begin{pmatrix} 0 & -i\\ -i & 0\\ \end{pmatrix} \rightarrow k; \end{equation*}

\begin{equation*} \begin{pmatrix} 0 & i\\ i & 0\\ \end{pmatrix} \rightarrow -k \end{equation*}

Why are there no matrices that, when conjugating the matrices presented above, would not give them all with real components?

In general, to simplify this problem, we can assume that one of the matrices associated with the element $i, j$ or $k$ is in such a basis when it has exactly the form that I indicated above. And the rest of the matrices are conjugated by some non-degenerate matrices.But it didn't help me simplify the task much.

Any help?


2 Answers 2


By the sounds of it you want an explanation rather than a proof, because the proof is just the fact that such a representation has to be irreducible, and there is only one by counting the squares of dimensions of irreducible representations.

For an explanation, suppose that we could, in fact, find three real matrices $A$, $B$ and $C$ such that $A^2=B^2=C^2=-1$ and $AB=C$, $BC=A$, etc., so they look like the unit quaternions. What about all $\mathbb R$-linear combinations of $A$, $B$ and $C$, (and $1$)? These would have to form a copy of $\mathbb H$, the quaternions. And then $M_2(\mathbb R)\cong \mathbb H$ because they are both $4$-dimensional $\mathbb R$-vector spaces, but that's certainly not true.

(Now, along the way, you would have to prove that the map from $\mathbb H$ to $M_2(\mathbb R)$ that sends $i$ to $A$, $j$ to $B$ and $k$ to $C$ is injective, which is why this is an explanation rather than a proof.)

Edit: Perhaps you want to simply know why this irreducible complex representation cannot be realized over the real numbers, despite it having a real character. This is because the Frobenius--Schur indicator of the representation is $-1$, which means that the character is real but the representation is not.

  • $\begingroup$ Of course! I didn't think of using the idea of considering vector spaces at all. Thank you very much! I have heard about the Schur indicator, but I did not want to use it to prove this statement, since this task was proposed before studying this object. $\endgroup$
    – Mr. Nobody
    Commented Nov 1, 2022 at 19:04
  • $\begingroup$ @Mr.Nobody If you are trying to prove this, just write down all matrices that square to $-1$ in $M_2(\mathbb R)$. All of $\pm i$, $\pm j$ and $\pm k$ have to lie in that set, but it isn't really big enough for all of them. $\endgroup$ Commented Nov 1, 2022 at 20:46
  • $\begingroup$ Thank you for your help! But I have already dealt with this problem. $\endgroup$
    – Mr. Nobody
    Commented Nov 1, 2022 at 20:54
  • $\begingroup$ You can calculate it basically comes down to having a solution to $x^2 + y^2 = -1$, which is impossible for $x, y \in \mathbb{R}$. $\endgroup$
    – spin
    Commented Nov 2, 2022 at 8:44

There are many $2$-dimensional real representations of $Q_8$; take the trivial one, for example. If you're specifically referring to irreps, then it's not hard to write down (or lookup) them all and verify that there aren't any real ones. If you're referring to faithful reprsentations, then any (complex) representation of a finite group is unitary, and thus any $2$-dimensional real representation $\rho$ of $Q_8$ lies in $SO(2)$. But any finite subgroup of $SO(2)$ is cyclic.

  • 1
    $\begingroup$ Yes, I apologize for not clarifying this detail. It was just about the irreducible nonlinear character of the group $Q_8$, whose dimension is equal to 2. $\endgroup$
    – Mr. Nobody
    Commented Nov 1, 2022 at 18:51
  • $\begingroup$ @Mr.Nobody: No worries. You might also want to take a look at the Frobenius-Schur indicator for the general problem of finding real representations (and it's particularly easy to compute for a group as small as $Q_8$), but bear in mind that idea of a "real" representation in that context may not be exactly what you have in mind. $\endgroup$
    – anomaly
    Commented Nov 1, 2022 at 18:56

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