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 $q\colon V\to k$, and (b) a linear action $G \to \mathit{GL}(V)$, which are compatible in the sense that $G$ preserves $q$ (viꝫ. $q(g\cdot v)) = q(v)$ for $g\in G$). 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 $(U\oplus U^\vee)^r$ for an irreducible non-self-dual representation $U$.

Anyway, let me concentrate on the important special case where $k=\mathbb{Q}$ and $G=\mathbb{Z}/n\mathbb{Z}$. The irreducible representations of $\mathbb{Z}/n\mathbb{Z}$ 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 $n,r\geq 1$ be integers, let $U_n$ be the faithful irreductible representation of $\mathbb{Z}/n\mathbb{Z}$ over $\mathbb{Q}$ (the one such that the generators act with characteristic polynomial given by the $n$-th cyclotomic polynomial). Can we classify quadratic forms on $(U_n)^r$ which are invariant under the action of $\mathbb{Z}/n\mathbb{Z}$ (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 $\mathbb{Z}/n\mathbb{Z}$-actions (over $V$, linear, preserving $q$) which make $V$ isomorphic to $(U_n)^r$?

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 $n,r$ is there a $\mathbb{Z}/n\mathbb{Z}$-invariant quadratic form on $(U_n)^r$ that is isomorphic to this?

Note: there is a $\mathbb{Z}/n\mathbb{Z}$-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 $\mathbb{Z}/n\mathbb{Z}$-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 $\mathbb{Z}/30\mathbb{Z}$.


The vector space $U_d$ can be viewed as the field $\mathbb Q(\mu_d)$. From this point of view, the group $\mu_n$ acts simply by multiplication by the $d$th roots of unity.

A symmetric bilinear form $Q( a,b)$ which is $\mu_n$-invariant satisifes $Q( \zeta a, \zeta b)=Q(a,b)$ for $\zeta \in \mu_d$, implying $Q(\zeta a,b) = Q( a, \zeta^{-1} b) = Q(a, \overline{\zeta} b)$ which, by linearity, extends to $Q(la , b) Q(a, \overline{l} b)$ for all $l \in \mathbb Q(\mu_d)$, so $Q(a,b)$ depends only on $a \overline{b}$. Because $Q(a,b)$ is bilinear, we can write

$$Q( a,b) =\operatorname{tr} (Q (a \overline{b}c))$$ for some fixed $c \in \mathbb Q(\mu_d)$. (This is because any linear form on a field can be expressed as scalar multiplication times trace).

To be symmetric, we must have $c =\overline{c}$, i.e. $c$ lies in the maximal totally real subfield. To be nondegenerate, we need $c \neq 0$.

The quadratic forms associated to $c_1$ and $c_2$ are $\mu_n$-equivariantly isomorphic if and only if $c_2 = c_1 k \overline{k}$ for some $k \in \mathbb Q(\mu_d)$. So quadratic forms of this type are classified by

$$( \mathbb Q(\mu_d)^+ )^\times/ \mathbb Q(\mu_d)^\times$$ with the embedding given by $k \mapsto k \overline{k}$.

We have local invariants at every place where the extension $\mathbb Q(\mu_d) / \mathbb Q(\mu_d)^+$ is split (by class field theory). Equivalently, these are local obstructions to an element of $\mathbb Q(\mu_d)^+$ being a norm. Expressing the norm form as a quadratic form in two variables, and using weak approximation, we can see that these local obstructions are the only obstructions, or, equivalently, that these local invariants classify the quadratic form with $\mu_n$-action.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.