# Hypotheses for Sum of squares of degrees of irreducible representations equals order of group

Consider the group $$G=C_n = \langle r|r^n \rangle$$ (i.e. the cyclic group of order $$n$$). If $$G$$ acts on a complex vector space, then we know that $$r$$ acts on $$v\in V$$ by multiplication by an $$n$$ root of unity (including 1). Furthermore, these correspond to degree 1 characters. We see this from the fact that the sum of squares of degrees of irreducible representations equals order of group (referred to below as "sum of squares formula"; see for instance, Serre, Linear Representation of Finite Groups, p.18 Corollary 2 a. or The Group Properties Wiki).

Then I asked what happens when $$G$$ acts on $$V=\mathbb{R}^2$$ which we can identify with $$\mathbb{C}$$ via the usual vector space isomorphism. In this case, there are two kinds of invariant subspaces. If $$r$$ acts trivially, then $$\text{span}_{\mathbb{R}}((1,0))$$ is a 1-dimensional invariant subspace. However, if $$r$$ acts by rotation, say by rotation by $$120^{\circ}$$ (i.e. in the case $$n=3$$), then our subrepresentation is 2-dimensional because $$r.(1,0) = \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$$.

The sum of squares formula seems to fail in this real case because $$\mathbb{R}$$ is not a splitting field of $$G$$ (in the sense defined in The Group Properties Wiki).

Is there an analogue to the sum of squares formula in the real case?

• Use $\langle X\rangle$ for $\langle X\rangle$. Nov 6 '19 at 22:53