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Section 2.12 Real, pseudoreal and complex representations

In this course we focus on group representations on complex vector spaces. However, sometimes such a representation may be real, just like complex numbers may be real. So a natural question is: given a representation on a complex vector space, can we determine easily when a representation is real or not?

First, how do we know whether a complex number \(z\) is actually real? Well, it is easy, you look at it, and see whether its imaginary part is zero! More precisely, the statement is that \(z\) is real if and only if \(z = z^*\text{,}\) that is, \(z\) is equal to is complex conjugate, which means that its imaginary part vanishes.

We can do the same thing with matrices. We say that a matrix \(M\) is real if and only if \(M^*\text{.}\) Note that this is really the complex conjugate, not the complex conjugate transpose. So this means that all the matrix entries are real numbers.

Now what about representations? When is a representation real? This is of great interest in physics; for instance, once one understands particles in terms of representations, then anti-particles transform in the complex conjugate representations. So understanding the relation between a representation and is complex conjugate is crucial.

Subsection 2.12.1 Real, pseudoreal and complex

Let \(T : G \to GL(V)\) be a representation of a group \(G\) on a complex vector space \(V\text{.}\) Let us first define the complex conjugate representation.

If \(T\) is a group homomorphism, then \(T(g h) = T(g)T(h)\) for any \(g,h \in G\text{.}\) Taking the complex conjugate of the matrices, \(T^*(g h) = T(g)^* T(h)^*\text{,}\) and hence \(T^*\) is also a group homomorphism.

We would like to say that a representation is real if the entries of the corresponding matrices are real numbers. Or, more precisely, the matrices should be equivalent (by a similarity transformation) to matrices with only real numbers. We would also like to translate this condition into the statement that a real representation is equivalent to its complex conjugate. But this is subtle; real representations are indeed equivalent to their complex conjugate, but the converse is not true. Let us be a little more careful.

Definition 2.12.2. Complex representations.

We say that \(T\) is complex if it is not equivalent to its complex conjugate representation \(T^*\text{.}\) We say that it is non-complex otherwise. That is, \(T\) is non-complex if there exists a matrix \(S\) such that for all \(g \in G\text{,}\)

\begin{equation*} T^*(g) = S T(g) S^{-1}. \end{equation*}
Remark 2.12.3.

In terms of characters, it follows that if the characters \(\chi_i^{(T)}\) are complex, i.e. not equal to their complex conjugate, then \(T\) is complex. The converse is not true however; if the characters are real, then we cannot conclude that \(T\) and \(T^*\) are equivalent.

We now show that we can divide non-complex representations further, when we focus on irreducible representations. We first need a lemma that tells us there are only two types of possible similarity transformations between an irreducible representation and its complex conjugate.

We start with \(T^*(g) = S T(g) S^{-1}\text{.}\) We take the transpose, to get:

\begin{equation*} T^\dagger(g) = (S^{-1})^T T(g)^T S^T. \end{equation*}

Since \(T\) is unitary, the left-hand-side is equal to \(T(g)^{-1} = T(g^{-1})\text{.}\) Thus we get:

\begin{equation*} T(g^{-1}) = (S^{-1})^T T(g)^T S^T, \end{equation*}

which shows that \(T(g)\) and \(T(g^{-1})\) are related. Now this equality is true for any \(g \in G\text{,}\) so we can write it for \(g^{-1}\) instead of \(g\text{.}\) We get:

\begin{equation*} T(g) = (S^{-1})^T T(g^{-1})^T S^T. \end{equation*}

But \(T(g^{-1})^T = S T(g) S^{-1},\) and hence, substituting back,

\begin{align*} T(g) =\amp (S^{-1})^T \left(S T(g) S^{-1} \right) S^T\\ =\amp \left(S^{-1} S^T \right)^{-1} T(g) \left(S^{-1} S^T \right). \end{align*}

What this means is that \(S^{-1} S^T\) commutes with all matrices \(T(g)\) of an irreducible representation. By Schur's lemma Lemma 2.5.2, we know that \(S^{-1} S^T = \lambda I\) for some \(\lambda \in \mathbb{C}\text{.}\) Thus \(S^T = \lambda S\text{.}\) This means that

\begin{equation*} S = (S^T)^T = (\lambda S)^T = \lambda S^T = \lambda^2 S, \end{equation*}

and hence \(\lambda^2 = 1\text{.}\) It follows that \(S = \pm S\text{,}\) that is, it is either symmetric or anti-symmetric.

Definition 2.12.5. Real and pseudoreal representations.

Let \(T : G \to GL(V)\) be a non-complex unitary irreducible representation of a group \(G\) on a complex vector space \(V\text{.}\) Thus \(T\) is equivalent to \(T^*\text{:}\)

\begin{equation*} T^*(g) = S T(g) S^{-1}, \end{equation*}

for some matrix \(S\text{.}\) We say that \(T\) is real if \(S\) is symmetric, and pseudoreal if \(S\) is anti-symmetric.

Now if a representation is given by matrices \(T(g)\) that only have real entries, then it is certainly real according to our definition, since \(S\) can be taken to be the identity matrix. But is it true that all real representations are equivalent to representations with matrices that only have real entries? This is what we show next.

First, we notice that the similarity transformation \(S\) can be taken to be unitary. Indeed, since \(T^*(g) = S T(g) S^{-1}\)

\begin{equation*} S T(g) = T^*(g) S = \left(T(g^{-1}) \right)^T S, \end{equation*}

and so \(S = T(g)^T S T(g)\text{.}\) We can then calculate:

\begin{align*} S^\dagger S =\amp T(g)^\dagger S^\dagger T(g)^* T(g)^T S T(g)\\ =\amp T(g)^\dagger S^\dagger S T(g), \end{align*}

since \(T\) is unitary. Thus \(T(g) S^\dagger S = S^\dagger S T(g)\text{.}\) It follows that \(S^\dagger S\) commutes with all \(T(g)\text{,}\) and hence by Schur's lemma Lemma 2.5.2, we know that \(S^\dagger S = \lambda I\) for some \(\lambda \in \mathbb{C}\text{.}\)

But the similarity transformation \(S\) is defined only up to overall rescaling (and it is non-zero), so we can choose an appropriate rescaling so that \(S^\dagger S = I\text{,}\) that is, it is unitary.

Now that we know that \(S\) is unitary, we can prove the lemma. We assume that the representation is real, so that \(S\) is a symmetric unitary matrix. One can show that any symmetric unitary matrix \(S\) can be written as \(S = W^2\) for some unitary symmetric matrix \(W\) (proof left as exercise). Thus \(T^*(g) = S T(g) S^{-1}\) becomes \(T^*(g) = W^2 T(g) (W^{-1})^2\text{.}\) We multiply by \(W^{-1}\) on the left and \(W\) on the right to get:

\begin{equation*} W^{-1} T^*(g) W = W T(g) W^{-1}. \end{equation*}

Using the fact that \(W\) is unitary symmetric, we have \(W^{-1} = W^\dagger = W^*\text{,}\) and \(W = (W^{-1})^*\text{.}\) So we can write:

\begin{equation*} W T(g) W^{-1} = W^{-1} T^*(g) W = W^* T^*(g) (W^{-1})^* = \left( W T(g) W^{-1} \right)^*. \end{equation*}

It thus follows that \(W T(g) W^{-1}\) has only real entries, thus \(T\) is equivalent to a representation with only real entries.

Subsection 2.12.2 The Frobenius-Schur indicator

We have now defined when an irreducible representation on a complex vector space is real, pseudoreal, or complex. We understand that a real representation is equivalent to a representation whose matrices only have real entries. But in general, the representation may not be given in this form. How do we determine whether an irreducible representation is real, pseudoreal or complex?

Fortunately there is a neat criterion, known as the Frobenius-Schur indicator, which uses our beloved characters. Here goes!

We will leave the proof as an exercise.

So to determine whether an irreducible representation is real, pseudoreal or complex, we simply need to evaluate the Frobenius-Schur indicator. We also note here that if \(g\) and \(h\) are in the same conjugacy class, than \(g^2\) and \(h^2\) also are; thus the Frobenius-Schur indicator is also a class function, and could be written as a sum over conjugacy classes.

Remark 2.12.9.

The classification between real, pseudoreal and complex irreducible representations also hold for compact groups. Moreover, those can be distinguished using the Frobenius-Schur indicator as above, but with the sum replaced by an integral over the compact group.

As an example, let us look at the irreducible representations of \(S_3\text{,}\) as studied in Section 2.10. To evaluate the Frobenius-Schur indicator we will need the square of the group elements. We note that the square of the identity is still the identity. As for the transpositions (conjugacy class \(C_2\)), their square is the identity. For the cyclic permutations (conjugacy class \(C_3\)), we calculate. Take for instance \((123)\text{.}\) The square is \((123)(123) = (132)\text{,}\) and hence it is still a cyclic permutation of length three. Thus the square of elements in the conjugacy class \(C_3\) are still in \(C_3\text{.}\)

Let us now determine whether the irreducible representations are real, pseudoreal or complex. First, the two one-dimensional representations are manifestly real, since they map all group elements to real numbers. Let us, for fun, evaluate the Frobenius-Schur indicator for these two representations.

For the trivial representation, we get:

\begin{equation*} \frac{1}{|S_3|}\sum_{g \in S_3} \chi^{(1)}(g^2) = \frac{1}{6} \left((1)(1) + (3)(1) + (2)(1) \right)= 1, \end{equation*}

and hence the representation is real, as expected (this is obviously always the case for the trivial representation).

For the non-trivial one-dimensional representation \(T^{(2)}\) (using the notation in Section 2.10). The sum is:

\begin{equation*} \frac{1}{|S_3|}\sum_{g \in S_3} \chi^{(2)}(g^2) = \frac{1}{6} \left( (1)(1) + (3)(1) +(2)(1) \right) = 1, \end{equation*}

and hence this representation is also real, as expected.

Finally, we look at the two-dimensional representation. We get:

\begin{equation*} \frac{1}{|S_3|}\sum_{g \in S_3} \chi^{(2)}(g^2) = \frac{1}{6} \left((1)(2) + (3)(2) + (2)(-1)\right) = 1, \end{equation*}

and thus this representation is also real! This is interesting, because depending on how you think about these \(2 \times 2\) matrices, they may be written as matrices with complex entries. But in fact in Example 2.1.6 we had already seen one way to write these matrices with real entries, and thus the representation must be real indeed.

Let us now look at the two-dimensional representation of the quaternion group given in Example 2.1.7. First, we should check that it is irreducible. Using the criterion for irreducibility Theorem 2.9.1, and calculating the traces of the matrices, we get:

\begin{equation*} (2)^2+(-2)^2+(0)+(0)+(0)+(0)+(0)+(0)= 8, \end{equation*}

which is the order of the quaternion group. Thus it is irreducible. It is the only irreducible representation of dimension greater than one for the quaternion group.

Let us determine whether it is real, pseudoreal or complex. First, we note that for the quaternion group, \(i^2 = (-i)^2 = j^2 = (-j)^2 = k^2 = (-k)^2 = -1,\) and of course \(1^2 = (-1)^2 = 1\text{.}\) Thus the Frobenius-Schur indicator evaluates to:

\begin{equation*} \frac{1}{8} \left( (2) + (2) +(-2)+(-2)+(-2)+(-2)+(-2)+(-2) \right)= - 1. \end{equation*}

We thus conclude that this representation is pseudoreal. That is, it is equivalent to its complex conjugate, but the similarity transformation involves an anti-symmetric matrix. This is an example where all the characters of the representation are real, but the representation is not, it is pseudoreal instead. In particular, the \(2 \times 2\) matrices cannot be brought into matrices with only real entries by a similarity transformation.

As a last example, let us look at the one-dimensional representation for the cyclic group \(\mathbb{Z}_4\) given by the complex numbers \(\{1,-i,-1,i\}.\) We calculate the square of the elements. We get that \(1^2 = (-1)^2 = 1,\) and \((-i)^2 = i^2 = -1\text{.}\) Thus the Frobenius-Schur indicator is:

\begin{equation*} \frac{1}{4} \left(1 + (-1) + 1 + (-1) \right) = 0, \end{equation*}

and hence the representation is complex, as expected.

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