**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 $mathbb{Z}/30mathbb{Z}$.