Linear representation of finite group over reals has invariant symmetric positive-definite bilinear form

For finite-dimensional vector space
Suppose $$G$$ is a finite group and $$\alpha:G \to GL(V)$$ (with $$V$$ finite-dimensional) is a linear representation of $$G$$ over a vector space $$V$$ over $$\R$$, the field of real numbers. Then, there exists a symmetric positive-definite bilinear form $$b: V \times V \to V$$ such that

$$\! b(\alpha(g)v,\alpha(g)w) = b(v,w) \ \forall \ g \in G, \ v,w \in V$$

In other words, we can think of $$b$$ as a dot product invariant under the action of $$G$$.

Further, if we choose a basis for $$V$$ that is an orthonormal basis for $$b$$, then in this basis, all the matrices for $$\alpha(g), g \in G$$, are orthogonal matrices, i.e., as elements of $$O(n,\R)$$ where $$n$$ is the dimension of $$V$$ (and hence the degree of $$\alpha$$). Thus, another formulation is that every finite-dimensional linear representation over the real numbers is an orthogonalizable linear representation.

Related facts for infinite groups

 * Continuous linear representation of compact group over reals has invariant symmetric positive-definite bilinear form

Analogue for complex numbers

 * Linear representation of finite group over complex numbers has invariant Hermitian inner product

Not true for all subfields of reals
It is possible to have a subfield $$K$$ of the real numbers with the property that a linear representation can be realized over that subfield, but it cannot be realized using orthogonal matrices of that subfield, because it does not possess an invariant inner product. Note that all examples of this kind must involve representations of degree more than one.

Here are some examples: