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Let $f : C\rightarrow S$ be a proper smooth morphism of relative dimension 1 over a connected scheme $S$.

Let $G$ be a finite group of order invertible on $S$ which acts faithfully and $\mathcal{O}_S$-linearly on $C/S$. Let $\omega_{C/S} = \Omega^1_{C/S}$ denote the dualizing sheaf of $C/S$, then the $G$-action on $C/S$ makes $\omega_{C/S}$ into an $(\mathcal{O}_C,G)$-module, or equivalently it makes $f_*\omega_{C/S}$ into a locally free $\mathcal{O}_S[G]$-module (is it projective in general?).

Proposition 3.2.5 of "Champs de Hurwitz" claims that:

"If $\mathcal{L}$ is an finite locally free $(\mathcal{O}_C,G)$-sheaf, then the Lefschetz trace $L_G(\mathcal{L}_s)$ is constant along geometric fibers ($s\in S$ a geom. point). In particular, the representation $H^0(C_s,(\omega_{C/S}^{\otimes m})_s)$ is independent of $s\in S$"

Here, if $k(s)$ is an algebraically closed field in which the order of $G$ is invertible, and $C_s/k(s)$ a smooth projective curve with an action by $G$, then for a locally free $(\mathcal{O}_{C_s},G)$-module $\mathcal{L}_s$, its "Lefschetz trace" is: $$L_G(\mathcal{L}_s) := \sum_{i=0}^1(-1)^i[H^i(C_s,\mathcal{L}_s)]$$ where the brackets "$[H^i(\cdots)]$" means the element of the representation ring (or Grothendieck group) $R_{k(s)}(G)$ corresponding to the representation of $G$ on $H^i(\cdots)$.

I'm trying to make sense of this proposition. The first question is - what does it mean to say that two representations over different fields are the same?

Ie, in general two geometric points $s_1,s_2\in S$ will have different "residue fields" - so the representations of $G$ on $H^i(C_{s_j},\mathcal{L}_{s_j})$ will be representations over the different residue fields $k(s_1),k(s_2)$.

From the perspective of character theory, we'd want to somehow identify the $n$th roots of unity in $k(s_1)$ and $k(s_2)$. If $H^0(S,\mathcal{O}_S)$ contains a primitive $n$th root of unity $\zeta_n$, then we can "identify" the images of $\zeta_n$ in $k(s_1),k(s_2)$, and since the character of any representation takes values in the additive monoid generated by $\zeta_n$, it would make sense to ask if $[H^i(C_{s_1},\mathcal{L}_{s_1})] = [H^i(C_{s_2},\mathcal{L}_{s_2})]$. However, I don't see why $\zeta_n$ need be a global section on $S$.

Perhaps there is a better way to think about this?

The second question is - Why is the proposition true? The authors do not give a proof or even a reference, but instead provide the cryptic comment: "...the following result (i.e., Prop 3.2.5) is an extension, to the equivariant case, of the classical result which establishes the constancy of the Euler characteristic $\chi(\mathcal{L}_s)$".

Another way to think about it, might be via the "cde" triangle in modular representation theory (c.f. Serre's book "Linear Representations of Finite Groups chapters 14-16, or this thesis). The idea here is to note that for any two points $x,y\in S$, by connectedness, one should be able to find a finite sequence: $$x = x_0 \leftarrow x_1\rightarrow x_2\leftarrow x_3\rightarrow \cdots \leftarrow x_{n-1} \rightarrow x_n = y$$ where the relation $a\rightarrow b$ means $a$ is a generization of $b$, or $b$ is a specialization of $a$. (One can definitely find such a sequence if $S$ is locally noetherian. Does anyone know if it exists for any connected scheme?)

Then, for every relation $a\rightarrow b$ in $S$, one should be able to transport a representation over the residue field $k(a)$ to a representation over $k(b)$ in the following manner: Firstly, we may assume that $a$ is "codimension 1" in $b$. Let $B$ be the closure of $\{b\}\subset S$, which is irreducible and contains $a$. Let $\widehat{O_{a,B}}$ be the complete local ring of $a$ inside $B$, then IF $\widehat{O_{a,B}}$ is a DVR, then $\text{Frac }O_{a,B}$ is the completion of $k(b)$, then by the "cde" triangle, we get maps $$P_{k(a)}(G)\rightarrow R_{k(b)}(G)\rightarrow R_{k(a)}(G)$$ where $P_{k(a)}(G)$ denotes the representation ring generated by projective $k(a)[G]$-modules. Here, the first map is "lift to $\widehat{O_{a,B}}$ and then localize", and the second map is "restrict to a $\widehat{O_{a,B}}$-lattice and then reduce modulo the maximal ideal. The fact that $n := |G|$ is invertible on $S$ implies that both maps are isomorphisms, which are inverse to each other on projective modules, and so we may view the representation $[H^i(C_a,\mathcal{L}_a)]$ as a representation over $k(b)$, and see if it agrees with the representation $[H^i(C_b,\mathcal{L}_b)]$. From this perspective, it seems that Proposition 3.2.5 becomes almost tautological, since both $[H^i(C_a,\mathcal{L}_a)]$ and $[H^i(C_b,\mathcal{L}_b)]$ come from restricting the representation on the free $\widehat{O_{a,B}}$-module $R^if_*\mathcal{L}_{\widehat{O_{a,B}}}$, and the method of transporting the representation from $a$ to $b$ is precisely via this module as an intermediary.

However, I'm still not sure what to do if $\widehat{O_{a,B}}$ is not a DVR. For example, what if $S$ is the affine plane, $b$ is the generic point of a singular planar curve on $S$, and $a$ is a singular point in $\overline{\{b\}}$?

Perhaps there is a better way of thinking about this?

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  • $\begingroup$ For finite groups whose order is invertible in the ground field, the isomorphism class of a finite-dimensional (necessarily semisimple) representation is determined by the characteristic polynomials (Brauer-Nesbitt) and hence is determined by knowing it after any extension field. Thus, one is really comparing over algebraically closed fields, and every local noetherian ring $A$ admits a local injection into a discrete valuation ring (with the same fraction field, though bigger residue field by a finitely generated extension). So the formalism in Serre's book is sufficient. $\endgroup$ – nfdc23 Mar 6 '17 at 2:37
  • $\begingroup$ Of course nfdc23 is correct. If you want a reference for constancy, Lemma 4.2.3 of my article with Olsson also implies this. After forming the base change to fields that contain enough roots of unity, the irreducible representations stabilize and are independent of the field. For every irreducible representation $V$, the pushforward of $\mathcal{L}\otimes V$ is flat over the base. So the dimension function $s\mapsto (H^0(C_s,\mathcal{L}_s)\otimes V)^G$ is constant. $\endgroup$ – Jason Starr Mar 6 '17 at 8:14
  • $\begingroup$ @JasonStarr On an intuitive level, I understand what you're saying, so my difficulty is technical. My confusion in comparing representations over different fields is that it's important to make a consistent choice of $n$th roots of unity - Ie, if $G$ is cyclic of order 5 generated by $g$, then the character sending $g\mapsto \zeta_5$ determines a nonisom. representation compared to the character sending $g\mapsto\zeta_5^2$, even though both $\zeta_5,\zeta_5^2$ are primitive 5th roots of 1. In order to make a consistent choice, are you suggesting (in your 3rd sentence) that we pass to an... $\endgroup$ – stupid_question_bot Mar 6 '17 at 19:51
  • $\begingroup$ @JasonStarr etale extension of $S$ by adjoining $\zeta_n$? But then, if e.g. $S = \text{Spec }\mathbb{Z}$ and I want to compare the representation at $\mathbb{F}_p$ with the representation at the generic point, then there are now generally many points in $\mathbb{Z}[\zeta_n]$ which lie above $p$, and picking a point basically corresponds to choosing a (frobenius orbit of a) primitive $n$th root of unity, and a priori it seems possible that depending on the point you choose lying above $p$, you'll get different answers for whether the rep "at $p$" agrees with the one at the generic point. $\endgroup$ – stupid_question_bot Mar 6 '17 at 19:56
  • $\begingroup$ @JasonStarr For your last two sentences, I really like where I think you're going, but what do you mean by "$V$" exactly? What are the tensor products in $\mathcal{L}\otimes V$ and $(H^0(C_s,\mathcal{L}_s)\otimes V)^G$ over? Also, in your paper with Olsson, do you mean this one? If so, by Lemma 4.2.3, do you mean Lemma 4.3, or do you mean Lemma 4.2(3)? $\endgroup$ – stupid_question_bot Mar 6 '17 at 20:13
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Based on the comments, it is clear that this question is as much about modular representation theory (in the tame case) as about curves and pluricanonical sections.

Let $G$ be a finite group of order $\ell$. Let $\mathbb{Z}[1/\ell]\to R$ be a ring homomorphism. Let $M$ be an $R$-module. Let $G\to \text{Aut}_{R-\text{mod}}(M)$ be a group homomorphism. A pair $(e,f)$ of $G$-equivariant $R$-module homomorphisms $M\to M$ is a pair of $G$-typical idempotents if $e+f=\text{Id}_M$, $e\circ f = f\circ e=0$, and for every $R$-module $I$ (with trivial $G$-action), $$\text{Hom}_R(\text{Coker}(e),\text{Coker}(f)\otimes_R I)^G = \{0\}, $$ $$\text{Hom}_R(\text{Coker}(f),\text{Coker}(e)\otimes_R I)^G = \{0\}.$$ The basic example is the pair $(\epsilon_M,\phi_M)$, $$\epsilon_M(m):= \frac{1}{\ell}\sum_{g\in G}g\bullet m, \ \ \phi_M = \text{Id}_M-\epsilon_M.$$ For every $G$-equivariant $R$-module homomorphism $u:M\to M''$, $\epsilon_{M''}\circ u$ equals $u\circ \epsilon_M$. From this it quickly follows that the functor $M\mapsto \epsilon_M(M)$ is exact.

There is a set-valued covariant functor $S_M$ that associates to every commutative,unital $R$-algebra $A$ the set of pairs of $G$-typical idempotents of the $A$-module $M\otimes_R A$ with its induced $G$-action. When $M$ is a projective $R$-module of finite rank, this functor is representable by a locally closed subscheme of the affine space parameterizing pairs $(e,f)$ of elements of $\text{Hom}_R(M,M)$. The basic observation is that this $R$-scheme is étale.

To see this, let $I\subset R$ be a square-zero ideal. Denote $\overline{R}=R/I$. Denote $\overline{M}=M\otimes_R \overline{R}$. Let $(\overline{e},\overline{f})$ be a $G$-typical pair for the $\overline{R}$-module $\overline{M}$. Denote by $\overline{E}$, resp. $\overline{F}$, the image of $\overline{e}$, resp. the image of $\overline{f}$. These $\overline{R}$-submodules of $\overline{M}$ give a direct sum decomposition that is $G$-equivariant. Denote by $\pi:M\to \overline{M}$ the quotient homomorphism. Then $\pi^{-1}(\overline{E})$ fits into a short exact sequence, $$0 \to I\otimes_\overline{R} \overline{M} \to \pi^{-1}(\overline{E}) \to \overline{E} \to 0.$$ The quotient by the submodule $I\otimes_\overline{R} \overline{F}$ is a projective $R$-module $E$ with an induced $G$-action. Similarly define $F$. Because the $R$-modules are projective of finite rank, the quotient homomorpisms $$\text{Hom}_R(M,E) \to \text{Hom}_{\overline{R}}(\overline{M},\overline{E}), \ \text{Hom}_R(E,M)\to \text{Hom}_{\overline{R}}(\overline{E},\overline{M}),$$ are both surjective, and similarly for $F$. For an arbitrary lift of a $G$-equivariant $\overline{R}$-module homomorphism, after applying $\epsilon$, the lift is also $G$-equivariant. Thus, the $G$-equivariant direct sum decomposition, $$\overline{M}\twoheadrightarrow \overline{E}\hookrightarrow \overline{M}, \ \ \overline{M}\twoheadrightarrow \overline{F}\hookrightarrow \overline{M},$$ lifts to a $G$-equivariant direct sum decomposition of $M$, $$M\to E\oplus F.$$ Moreover, by hypothesis, $$\text{Hom}_\overline{R}(\overline{E},I\otimes_\overline{R}\overline{F})^G = \{0\}, $$ $$\text{Hom}_\overline{R}(\overline{F},I\otimes_\overline{R}\overline{E})^G = \{0\}. $$ Thus, this $G$-invariant direct sum decomposition is unique. Therefore the functor $S_M$ is étale.

The functor is also proper. Since the functor is representable by a finite type, locally closed subscheme of an affine space, it suffices to verify the valuative criterion of properness. Thus, assume that $R$ is a discrete valuation ring with maximal ideal $\mathfrak{m}$, residue field $k=R/\mathfrak{m}$ and fraction field $K$. For a direct sum decomposition $(E_K,F_K)$ of $M\otimes_R K$, define $E$ to be the kernel of the composition $$M \hookrightarrow M\otimes_R K \twoheadrightarrow (M\otimes_R K)/E_K.$$ In particular, $E$ is saturated. So $E$ is a free $R$-module and $F=M/E$ is also a free $R$-module. By construction, these are $G$-equivariant. By the same argument as above using $\epsilon$, every $G$-equivariant $k$-linear map from $E\otimes_R k$ to $F\otimes_R k$ or vice versa lifts to a $G$-equivariant $R$-module homomorphism from $E$ to $F$. This induces a $G$-equivariant $K$-linear map between $E_K$ and $F_K$. By hypothesis, every such map is the zero map. Thus, the same is true for $E\otimes_R k$ and $F\otimes_R k$. In particular, there is a unique $G$-equivariant, $k$-linear splitting of $M\otimes_R k \twoheadrightarrow F\otimes_R k$. This lifts to a $G$-equivariant $R$-linear splitting of $M\twoheadrightarrow F$. Thus, the generic splitting extends to a $G$-equivariant $R$-module splitting of $M$. So $S_M$ is proper over $R$. Therefore $S_M$ is finite over $R$.

Now apply all of this to the case that $R=\mathbb{Z}[1/\ell]$ and $M=R[G]$, the group algebra of $G$. This defines a finite, étale extension of $\mathbb{Z}[1/\ell]$, i.e., a product of finitely many irreducible domains each of which is a finite étale extension of $\mathbb{Z}[1/\ell]$. Define $K_G$ to be the Galois closure of the compositum of these finitely many finite extensions of $\mathbb{Q}$, and define $\mathfrak{o}_G$ to be the integral closure of $\mathbb{Z}[1/\ell]$ in $K_G$. (In fact, $\mathfrak{o}_G$ already equals the irreducible factor of maximal $R$-rank.) Then we can define the $G$-isotypic decomposition of $\mathfrak{o}_G[G]$, and this will be compatible with arbitrary base change.

Now all of the usual theory of irreducible representations, characters, orthogonality, etc., goes through for algebras over $\mathfrak{o}_G$. Now you can apply Lemma 4.2(3) from my article with Olsson (there are other references) to see that for every irreducible representation $V$ of $G$ over $\mathfrak{o}_G$, for the locally free sheaf $\omega_{C/S}^{\otimes n}\otimes_{\mathfrak{o}_G}V$ on the stack $[C/G]$ has locally free pushforward to the coarse moduli space $C//G$.

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  • $\begingroup$ Dear Professor Starr - Thank you for this response!! Some questions: Are you defining $E$ as the quotient $\pi^{-1}(\bar{E})/I\otimes_{\bar{R}}\bar{F}$? How does one see that this quotient is projective? A few lines later, when you say "after applying $\epsilon$, the lift is also $G$-equivariant" - do you mean $\epsilon = \epsilon_M$? Or do you mean $\epsilon = e$? On the next line, perhaps what you want is $\bar{E}\hookrightarrow\bar{M}\twoheadrightarrow\bar{E}$ to be the identity? $\endgroup$ – stupid_question_bot Mar 7 '17 at 20:43
  • $\begingroup$ @Amy "How does one see that this quotient is projective?" I am defining $E$ as the quotient of $\pi^{-1}(\overline{E})$ by the $R$-submodule $I\otimes_{\overline{R}}\overline{F}$ of the $R$-submodule $\text{Ker}(\pi) = I\otimes_{\overline{R}}\overline{M}$. This quotient is flat over $R$ by the "equational criterion of flatness", cf. Theorems 7.6 and 7.7, pp. 49-50 of "Commutative Ring Theory" by Matsumura. For the $R$-module $H=\text{Hom}_R(M,E)$ with its $G$-action, for every element $u\in H$, $\epsilon_H(u)$ is a $G$-invariant element, i.e., a $G$-equivariant map $M\to E$. $\endgroup$ – Jason Starr Mar 7 '17 at 20:53
  • $\begingroup$ I'm sorry I'm being incredibly dense, but I don't see how to use this "equational criterion of flatness" to show flatness of $E$... $\endgroup$ – stupid_question_bot Mar 7 '17 at 22:05
  • $\begingroup$ Also, when lifting the $G$-equivariant direct sum decomposition of $\bar{M}$ to $M$, you need a lot more than just surjectivity of the reduction map on Hom right? Ie, you need to know that you can choose a lift of $\bar{E}\hookrightarrow\bar{M}$, and a lift of $\bar{M}\twoheadrightarrow\bar{E}$, such that the composition of the lifts is the identity. That this is possible doesn't seem so clear to me. $\endgroup$ – stupid_question_bot Mar 7 '17 at 22:27
  • $\begingroup$ Replacing $R$ by the open affine $R[1/r]$, assume that $\overline{E}\cong \overline{R}^{\oplus r}$. This lifts to an $R$-module homomorphism $R^{\oplus r} \to E$. This induces a commutative diagram from the short exact sequence $0\to I\otimes_{\overline{R}} \overline{R}^{\oplus r} \to R^{\oplus r} \to \overline{R}^{\oplus r} \to 0$ to the short exact sequence $0\to I\otimes_{\overline{R}} \overline{E} \to E \to \overline{E} \to 0$. By the Snake Lemma, $R^{\oplus r}\to E$ is an isomorphism. So $E$ is locally free. $\endgroup$ – Jason Starr Mar 8 '17 at 0:40

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