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?

  • 1
    $\begingroup$ Use $\langle X\rangle$ for $\langle X\rangle$. $\endgroup$
    – Shaun
    Nov 6 '19 at 22:53

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