In these notes, we say a representation is rational, real or complex accordingly as it is defined over the reals or the complex numbers (that is, every entry of every matrix belongs to the corresponding field). Given such a representation V, the character $$\chi$$ is a class function; that is, a function (real or complex valued, accordingly) on the group which is constant on conjugacy classes. It is defined by $$\chi(g) = \mathrm{tr}(g; V)$$, where $$\mathrm{tr}(g;V)$$ is the trace of the operator g on V.

The inner product of two class functions $$\chi_1$$ and $$\chi_2$$ is defined to be $\left<\chi_1,\, \chi_2\right> = \frac1{|G|} \sum_{g\in G} \chi_1(g)\chi_2(g)^*$ where $$\chi(g)^*$$ is the complex conjugate. Consequently, the norm is defined by $\|\chi\|^2 = \left<\chi,\,\chi\right> = \frac1{|G|} \sum_{g\in G} |\chi(g)|^2.$

One also defines the Frobenius-Schur indicator associated to $$\chi$$ by $\nu = \frac1{|G|} \sum_{g\in G} \chi(g^2)$ of which more below.

### Basic properties:

If V1 and V2 are (real or complex) representations, with characters $$\chi_1$$ and $$\chi_2$$, then

1. The character of the direct sum $$V_1\oplus V_2$$ is $$\chi_1+\chi_2$$
2. The character of the tensor product $$V_1\otimes V_2$$ is $$\chi_1\chi_2$$
3. The character of the dual (contragredient) representation $$V^*$$ is $$\chi^*$$ (complex conjugate)
4. The character of the representation Hom$$(V_1,\; V_2)$$ is the product $$\chi_1^* \chi_2$$
5. The characters of any two distinct irreducible representations are orthogonal.

All but the last of these are proved simply by diagonalizing each $$g \in G$$ in turn.

### Consequences:

1. If W is irreducible (real or complex) with character $$\chi_W$$, and V is any representation with character $$\chi$$ then the multiplicity of W in V is equal to $$\left<\chi_W, \chi\right> / \|\chi_W\|^2$$.
2. In particular, let $$\mathrm{Fix}(G, V) = \{v \in V | g\cdot v = v, \forall g \in G\}$$, the subspace of $$V$$ consisting of those points fixed by all elements of G. Then $$\dim \mathrm{Fix}(G,V) = \left<\chi,\, 1\right>$$ where 1 is the character of the trivial representation; that is $$1(g)=1$$ ($$\forall g\in G$$).
3. The character of the symmetric and antisymmetric tensor products of a representation V are $$\chi_s(g) = \frac12 \left( \chi(g)^2 + \chi(g^2) \right), \quad \chi_a(g) = \frac12 \left( \chi(g)^2 - \chi(g^2) \right)$$,
(again, to see this, just choose a basis that diagonalizes g).

### Irreducible Representations

Theorem A complex representation is irreducible if and only if $$\|\chi\|^2=1$$.

However, for real representations this is not true, as can be seen for example from the cyclic group acting on the plane (representations E). A real representation is irreducible if there is only one invariant quadratic form, up to scalar multiples. So this becomes,

Theorem A real representation is irreducible if and only if $$\|\chi\|^2+\nu(\chi)=2$$.

Proof: The first of these theorems is standard but the second less so, so here is an outline proof.
To begin, notice that

$$\|\chi\|^2 + \nu(\chi) = 2 \left<\chi_s,\, 1\right> = 2 \dim \mathrm{Fix}(G, V \otimes_s V)$$

(the last being the symmetric tensor product of V with itself, which is the space of quadratic forms on V - we are assuming the representation is real so that VV*).
The theorem therefore reduces to showing that V is irreducible if and only if $$\dim \mathrm{Fix}(G, V \otimes_s V) = 1$$, which means that up to scalar multiples there is only one invariant quadratic form.
Now, any real representation of a finite group has an invariant positive definite quadratic form: take any positive definite quadratic form Q and average it over the group:

$Q'(x) := |G|^{-1}\sum_g Q(gx).$

Firstly, if the representation is reducible, write V = U + W. Then there is a non-zero invariant quadratic form on U which can be extended to V by making it zero on W, and there is a non-zero invariant quadratic form on W which can be extended to V by making it zero on U. These two extended forms on V are invariant and linearly independent.
Conversely, consider two linearly independent invariant quadratic forms with at least one of them positive definite. Choose a basis so that the positive definite one is represented by the identity matrix. Then a linear combination of the two forms is aI-S, and this is degenerate iff a is an eigenvalue of S, and such eigenvalues are real as S is symmetric. For such a choice of a, the quadratic form aI-S is invariant and degenerate, and the null-space is an invariant subspace of V, contradicting the fact that V is irreducible.

### Commuting Linear Maps

For any representation, the set of commuting linear maps is a vector space $$\mathrm{Hom}_G(V,V)$$ (more than that, it's an algebra). If V is a complex irreducible representation, then $$\mathrm{Hom}_G(V,V)\simeq \mathbb{C}$$. However, for real representations the result is more complicated.

Theorem Let V be a real irreducible representation. Then $$\mathrm{Hom}_G(V,V)$$ is isomorphic to $$\mathbb{R,\,C}$$ or $$\mathbb{H}$$.

Here $$\mathbb{H}$$ is the set of quaternions.

Proof: Every non-zero element of $$\mathrm{Hom}_G(V,V)$$ is invertible, as otherwise its kernel would be an invariant subspace of V. In other words, $$\mathrm{Hom}_G(V,V)$$ is a division algebra over $$\mathbb{R}$$. By a theorem of Frobenius, any division algebra over $$\mathbb{R}$$ is isomorphic to $$\mathbb{R,\,C}$$ or $$\mathbb{H}$$, as required.

Accordingly, we say that a real representation is of real, complex or quaternionic type.

A natural question arises: can we determine the type from the character? Of course (!), the answer is yes.

Theorem: A real irreducible representation is of:

• real type if $$\|\chi\|^2=1$$
• complex type if $$\|\chi\|^2=2$$
• quaternionic type if $$\|\chi\|^2=4$$

This statement is sometimes made in terms of the Frobenius-Schur indicator defined above. The three types of irreducible representation then correspond to $$\nu=1,\, 0$$ and $$-2$$, respectively

Example: For the 4-dimensional real irreducible representation of the quaternion group Q8, one has $$\|\chi\|^2=4$$ and $$\nu = -2$$.

Proof: This is simply due to the fact that $$\|\chi\|^2 = \left<\chi\otimes\chi,\;1\right> = \dim \mathrm{Hom}_G(V,V)$$.

### Relation between real and complex representations

Given a real representation $$\rho:G\to \mathrm{Aut}(V)$$, one can complexify it, by defining $$\rho^{\mathbb{C}}:G\to \mathrm{Aut}(V^{\mathbb{C}})$$, where $$\rho(g)$$ and $$\rho^{\mathbb{C}}(g)$$ have the same matrix, the second considered as a complex matrix, even though all the entries are real. In particular the two representations have the same characters.

Conversely, given a complex representation, $$\rho:G\to \mathrm{Aut}(V)$$ one can consider the underlying real representation, by considering V as a real vector space rather than a complex one (so of twice the dimension).

Properties:

1. If a complex representation has character $$\chi$$, then the character of the real underlying representation is $$2\mathrm{Re}(\chi)$$. This is because an eigenvalue $$\exp(i\theta)$$ becomes a 2x2 matrix with trace $$2\cos(\theta)$$.
2. Starting with a real representation, complexifying it and then taking the underlying real representation, one ends up with 2 copies of the original representation.
3. If V is a complex irreducible representation, then its underlying real representation is either reducible, and equal to two copies of the same irreducible of real type, or it's irreducible of complex or quaternionic type.