# nt.number theory – Classification of vector spaces with a quadratic form and an order n automorphism

Introductory general nonsense (for motivation: feel free to skip): Let $$G$$ be a finite group and $$k$$ be a field of characteristic $$0$$. Consider the set $$mathcal{S}$$ of isomorphism classes of finite dimensional vector spaces $$V$$ over $$k$$ endowed with (a) a nondegenerate quadratic form $$qcolon Vto k$$, and (b) a linear action $$G to mathit{GL}(V)$$, which are compatible in the sense that $$G$$ preserves $$q$$ (viꝫ. $$q(gcdot v)) = q(v)$$ for $$gin G$$). Note that we can take direct sums and tensor products of such data, giving $$mathcal{S}$$ a semiring (“ring without subtraction”) structure; we can also form the Grothendieck group $$mathcal{R}$$ of $$mathcal{S}$$, which is a ring.

Classifying (a) alone and (b) alone is well studied: (a) gives the Grothendieck-Witt ring of $$k$$, and (b) gives the representation ring of $$G$$ over $$k$$. I’m curious about what can be said about both data simultaneously (and compatibly). We have obvious ring homomorphisms from $$mathcal{R}$$ to the Grothendieck-Witt ring of $$k$$ and to the representation ring of $$G$$ over $$k$$, but I think $$mathcal{R}$$ (generally) isn’t a fiber product of them, and I suppose there isn’t much we can say at this level of generality (though I’d be happy to be wrong!).

I might still point out that if $$V = V_1 oplus V_2$$ is a decomposition of $$V$$ as representations of $$G$$ and there is no irreducible factor common in $$V_1$$ and the dual $$V_2^vee$$ of $$V_2$$, then necessarily $$V_1$$ and $$V_2$$ are orthogonal for $$q$$ (proof: apply Schur’s lemma to the $$G$$-invariant linear map $$V_1 to V_2^vee$$ obtained from $$q$$). So we are reduced to classifying elements of $$mathcal{R}$$ (or $$mathcal{S}$$) whose underlying representation is of the form $$U^r$$ for $$U$$ an irreducible self-dual representation of $$G$$, or of the form $$(Uoplus U^vee)^r$$ for an irreducible non-self-dual representation $$U$$.

Anyway, let me concentrate on the important special case where $$k=mathbb{Q}$$ and $$G=mathbb{Z}/nmathbb{Z}$$. The irreducible representations of $$mathbb{Z}/nmathbb{Z}$$ over $$mathbb{Q}$$ are of the form $$U_d$$ (self-dual) for $$d$$ dividing $$n$$ where $$U_d$$ splits over $$mathbb{C}$$ as sum of one-dimensional representations on which a chosen generator acts through each of the primitive $$d$$-th roots of unity. So I ask:

Actual question: Given $$d,n,rgeq 1$$ be integers such that $$d$$ divides $$n$$, let $$U_d$$ be the irreductible representation of $$mathbb{Z}/nmathbb{Z}$$ over $$mathbb{Q}$$ such that the generators act with characteristic polynomial given by the $$d$$-th cyclotomic polynomial. Can we classify quadratic forms on $$(U_d)^r$$ which are invariant under the action of $$mathbb{Z}/nmathbb{Z}$$ (i.e., describe the corresponding elements of the (known) Grothendieck-Witt ring of $$mathbb{Q}$$)? Or equivalently, in the other direction, given a quadratic form $$(V,q)$$ over $$mathbb{Q}$$ (through its image in the G-W ring), can classify $$mathbb{Z}/nmathbb{Z}$$-actions (over $$V$$, linear, preserving $$q$$) which make $$V$$ isomorphic to $$(U_d)^r$$?

I don’t even know the answer when $$q$$ is the standard Euclidean form (viꝫ. $$mathbb{Q}^m$$ with the quadratic form $$x_1^2 + cdots + x_m^2$$): for which $$d,n,r$$ is there a $$mathbb{Z}/nmathbb{Z}$$-invariant quadratic form on $$(U_d)^r$$ that is isomorphic to this?

Note: there is a $$mathbb{Z}/nmathbb{Z}$$-invariant standard Euclidean structure on $$mathbb{Q}^n = bigoplus_{d|n} U_d$$ with cyclic permutation of the coordinates, which induces a quadratic form on each of the $$U_d$$ so that this direct sum is orthogonal (the class of this form in the G-W ring can be computed by the Möbius inversion formula; because there is a scaling involved, it depends on $$n$$, not just $$d$$). It might be tempting to think that all $$mathbb{Z}/nmathbb{Z}$$-invariant quadratic forms on $$U_d$$ are obtained in this way: if my Witt ring calculations are correct, this is not the case: $$U_{30}$$ does not get a standard Euclidean structure that way; but there is a standard Euclidean structure on $$U_{30}$$, namely, take the Coxeter element of the Weyl group of $$E_8$$ as acting on $$mathbb{Q}^8$$ with its standard Euclidean structure, which is then $$U_{30}$$ as a representation of $$mathbb{Z}/30mathbb{Z}$$.