§ Even and odd functions through representation theory

Consider the action of Z/2Z\mathbb Z/ 2\mathbb Z on the space of functions RR\mathbb R \to \mathbb R. given by ϕ(0)(f)=f\phi(0)(f) = f, and phi(1)(f)=λx.f(x)phi(1)(f) = \lambda x. f(-x). How do we write this in terms of irreps?
  • On the even functions, since e(x)=e(x)e(x) = e(-x) for ee even, we have that,ϕ(0)(e)=e\phi(0)(e) = e and ϕ(1)(e)=e\phi(1)(e) = e [since e(x)=e(x)e(-x) = e(x)], or ϕ(x)(e)=id(e)\phi(x)(e) = id(e), hence the actionof ϕ\phi is that of the trivial representation on the subspace spanned by even functions.
  • On the odd functions, since o(x)=o(x)o(-x) = -o(x), we have that ϕ(1)(o)(x)=o(x)=o(x)=sgn(o)(x)\phi(1)(o)(x) = o(-x) = -o(x) = sgn(o)(x) hence ϕ(1)(o)=o\phi(1)(o) = -o, hence ϕ(x)(o)=sgn(x)(o)\phi(x)(o) = sgn(x)(o) where sgnsgnis the sign representation!
Since the even and odd functions span the space of all functions, as we can write any function ff as the sum of an even part ef(x)[f(x)+f(x)]/2e_f(x) \equiv [f(x) + f(-x)]/2 and an odd part of(x)[f(x)f(x)]/2o_f(x) \equiv [f(x) - f(-x)]/2. So, we have described the action of ϕ\phi in terms of subspaces which span the space, so we've found the irrep decomposition.