# Motivic Semiorthogonal Decompositions for Abelian Varieties

Bronson Lim  BL: Department of Mathematics
California State University San Bernardino
San Bernardino, CA 92407, USA<EMAIL_ADDRESS>and  Franco Rota  FR:
Department of Mathematics
Rutgers
Piscataway, NJ 08854, USA<EMAIL_ADDRESS>
###### Abstract.

Suppose $G$ is a finite group acting on an Abelian variety $A$ such that the
coarse moduli $A/G$ is smooth. Using the recent classification result due to
Auffarth, Lucchini, and Quezada, we construct a motivic semiorthogonal
decomposition for $\mathcal{D}[A/G]$ provided $G=T\rtimes H$ with $T$ a
subgroup of translations and $H$ is a subgroup of group automorphisms.

###### Key words and phrases:

Derived Categories, Fourier-Mukai Functors

###### 2010 Mathematics Subject Classification:

Primary 14F05; Secondary 13J70

##  1\. Introduction

###  1.1. Motivic semiorthogonal decompositions

Suppose $G$ is a finite group acting effectively on a smooth quasi-projective
variety $X$. If each of the quotients
$\bar{X}^{\lambda}=X^{\lambda}/C(\lambda)$ are smooth, where $X^{\lambda}$ is
the fixed loci of $\lambda\in G$ and $C(\lambda)$ is the centralizer of
$\lambda\in G$, then there is an orthogonal decomposition

$HH_{\ast}[X/G]=\bigoplus_{\lambda\in G/\sim}HH_{\ast}(\bar{X}^{\lambda})$

where $G/\sim$ is the set of conjugacy classes. This decomposition holds for
any additive invariant, see [TVdB18]. In [PVdB19], the authors conjecture that
this remains true on the derived (or dg) level. Since the derived category
decomposes orthogonally if and only if the variety is not connected, we should
not expect a fully orthogonal decomposition but only a semiorthogonal
decomposition, see Definition 1.

######  Motivic Semiorthogonal Decomposition Conjecture.

Suppose $G$ is a a finite group acting effectively on a smooth variety $X$ and
that for all $\lambda\in G/\sim$ the coarse moduli
$\bar{X}^{\lambda}=X^{\lambda}/C(\lambda)$ is smooth. Then there is a
semiorthogonal decomposition of $\mathcal{D}[X/G]$ where the components
$C_{[\lambda]}$ are in bijection with conjugacy classes and
$C_{[\lambda]}\cong\mathcal{D}(\bar{X}^{\lambda})$.

A semiorthogonal decomposition satisfying the conditions of the Motivic
Semiorthogonal Decomposition Conjecture is called a motivic semiorthogonal
decomposition, see Definition 2. It is motivic in the sense of [TVdB18] and
[Toe00].

######  Remark 1.

In [BGLL17] it is shown that an effective action is necessary. Roughly, they
take an ineffective group action on an elliptic curve without complex
multiplication and, assuming the conjecture is true, construct an
endomorphisms with cyclic kernel.

There are several known cases of the conjecture which we list now.

  * •

The natural actions of complex reflection groups of types $A,\ B,\
G_{2},F_{4}$, and $G(m,1,n)$ acting on $\mathbb{A}^{n}$, [PVdB19, Theorem C].

  * •

Semidirect product actions of type $G\rtimes S_{n}$ on $C^{n}$ where $C$ is a
curve and $G$ is an effective group acting on $C$, [PVdB19, Theorem B].

  * •

Dihedral group actions on $\mathbb{A}^{2}$, [Pot17, Corollary 6.5.5].

  * •

Curves, [Pol06, Theorem 1.2].

  * •

Cyclic group quotients, [KP17, Theorem 4.1].

###  1.2. Main result

Our main result is an affirmative answer to the Motivic Semiorthogonal
Decomposition Conjecture when $A$ is an Abelian variety and $G=T\rtimes H$
with $T$ a translation subgroup and $H$ a subgroup of group homomorphisms.

######  Theorem 1.

Let $A$ be an Abelian variety and $G=T\rtimes H$ a finite group of
automorphisms of $A$ such that the coarse quotient $A/G$ is smooth. Then there
exists a motivic semiorthogonal decomposition for $\mathcal{D}[A/G]$.

######  Remark 2.

We did not have to add the hypothesis that $A^{\lambda}/C_{G}(\lambda)$ is
smooth for all $\lambda$. This is because it is automatically satisfied
provided $A/G$ is smooth by the classification results in [AL17, ALQ18] which
we review in Section 2.

As an application, we get another proof of the orthogonal decomposition of the
noncommutative motive for $[A/G]$, see [TVdB18, Theorem 1.24, Theorem 1.27].
Let $U$ be the universal additive invariant.

######  Corollary 1.

Let $A$ be an Abelian variety and $G=T\rtimes H$ a finite group of
automorphisms of $A$ such that the coarse quotient $A/G$ is smooth. Then the
motive $U([A/G])$ decomposes orthogonally:

$U([A/G])\cong\bigoplus_{\lambda\in G/\sim}U(\bar{A}^{\lambda}),$

where $\bar{A}^{\lambda}=A^{\lambda}/C(\lambda)$.

###  1.3. Outline of Proof

We rely on the recent work in [AL17, ALQ18] characterizing smooth quotients of
Abelian varieties $A$ by finite groups. In particular, there are only finitely
many indecomposable types with $A/G$ smooth. With this classification, we
prove Theorem 1 by using the known motivic semiorthogonal decomposition for
products of curves, using the results of [LP17], and by constructing a new one
for the stacky surface $[E\times E/G(4,2,2)]$ where
$E=\mathbb{C}/\mathbb{Z}[i]$ is an elliptic curve with automorphism group
$\mu_{4}$.

###  1.4. Outline of Paper

Section 2 is a preliminary section. Section 3 is devoted to the construction
of a motivic semiorthogonal decomposition for
$\mathcal{D}[\mathbb{A}^{2}/G(4,2,2)]$. This is a local model of the
exceptional case in the classification of smooth quotients of Abelian
varieties. The proof of the global exceptional case is Section 4. In Section
5, we complete the proof of the main result. Lastly, in Section 6 we
explicitly compute motivic semiorthogonal decompositions for Abelian surfaces
in the indecomposable cases.

###  1.5. Conventions and notation

We work over $\mathbb{C}$. Unless otherwise stated, all functors are assumed
to be derived. For $X$ a scheme or stack, by $\mathcal{D}(X)$ we mean the
bounded derived category of coherent sheaves on $X$.

###  1.6. Acknowledgements

The first author would like to thank Robert Auffarth, Aaron Bertram, Giancarlo
Lucchini Arteche, and Alexander Polishchuk for helpful conversations. This
research was part of the RTG grant at the University of Utah.

##  2\. Preliminaries

We recall preliminaries on equivariant derived categories, motivic
semiorthogonal decompositions, and smooth quotients of Abelian varieties.
Throughout $X$ is a smooth quasi-projective variety and $G$ is a finite group
acting on $X$. In particular, $\bar{X}=X/G$ is a quasi-projective variety. We
refer to [BKR01, Section 4] for a review of $G$-equivariant derived categories
and [Kuz14] for a survey of semiorthogonal decompositions in algebraic
geometry.

###  2.1. Equivariant derived categories

For computational purposes it is often convenient to work with the
$G$-equivariant derived category of $X$ instead of the derived category of
$[X/G]$. By [vistoli] These are equivalent:
$\mathcal{D}[X/G]\cong\mathcal{D}(X)^{G}$.

The structure map $\pi\colon X\to\mathrm{pt}$ is trivially $G$-equivariant and
therefore descends to the stack quotients: $\pi\colon[X/G]\to BG$. Here
$BG=[\mathrm{pt}/G]$ is the classifying space of $G$-bundles. Any finite
dimensional representation $V$ of $G$ determines a coherent sheaf on $BG$ and
hence an object in $\mathcal{D}(BG)$. For any object
$\mathcal{F}\in\mathcal{D}(X)^{G}$ we therefore have an operation of tensoring
by finite dimensional representations:

$\mathcal{F}\otimes V:=\mathcal{F}\otimes\pi^{\ast}(V).$

For any two $G$-equivariant sheaves on $X$, say
$\mathcal{E},\mathcal{F}\in\mathcal{D}(X)^{G}$, the set of morphisms is a
complex of $G$-representations

$\mathbf{R}\mathrm{Hom}(\mathcal{E},\mathcal{F})\in\mathcal{D}(BG)$

and we have

$\mathrm{Ext}^{\ast}_{[X/G]}(\mathcal{E},\mathcal{F})\cong\left(H^{\ast}\mathbf{R}\mathrm{Hom}(\mathcal{E},\mathcal{F})\right)^{G}.$

###  2.2. Motivic semiorthogonal decompositions

######  Definition 1.

Let $\mathcal{T}$ be a triangulated category. A semiorthogonal decomposition
of $\mathcal{T}$ is a pair $\mathcal{A,B}$ of full triangulated subcategories
of $\mathcal{T}$ such that

  * •

$\mathrm{Hom}_{\mathcal{T}}(b,a)=0$ for all $a\in\mathcal{A},b\in\mathcal{B}$;

  * •

for all $t\in T$, there exists $a_{t}\in\mathcal{A}$ and $b_{t}\in\mathcal{B}$
such that

$b_{t}\to t\to a_{t}\to b_{t}[1]$

is an exact triangle.

In this case, we write $\mathcal{T}=\langle\mathcal{A},\mathcal{B}\rangle$.

We can iterate Definition 1 to get semiorthogonal decompositions with an
arbitrary finite number of subcategories:

$\mathcal{T}=\langle\mathcal{A}_{1},\ldots,\mathcal{A}_{r}\rangle.$

As suggested in the introduction, the following definition is motivated by a
motivic origin of the conjecture.

######  Definition 2.

Suppose a finite group $G$ acts effectively on a smooth variety $X$. A motivic
semiorthogonal decomposition of $\mathcal{D}[X/G]$ is the data of

  * •

a total order $\lambda_{1},\ldots,\lambda_{r}$ on conjugacy classes $G/\sim$;

  * •

embedding functors

$\Phi_{\lambda_{i}}\colon\mathcal{D}(X^{\lambda_{i}}/C(\lambda_{i}))\hookrightarrow\mathcal{D}[X/G]$

for each $\lambda_{i}\in G/\sim$ which are linear over $\mathcal{D}(X/G)$;

  * •

a semiorthogonal decomposition

$\mathcal{D}[X/G]=\langle C_{1},\ldots,C_{r}\rangle;$

where $C_{i}$ is the image of $\Phi_{\lambda_{i}}$.

Given semiorthogonal decompositions
$\mathcal{T}=\langle\mathcal{A,B}\rangle=\mathcal{C},\mathcal{A}\rangle$, we
can produce a new semiorthogonal decomposition by mutating $\mathcal{A}$
either to the left through $\mathcal{C}$ or to the right through
$\mathcal{B}$. That is, there exists mutation functors111We need $\mathcal{A}$
and $\mathcal{B}$ to be admissible but in our setting this will be satisfied.
$\mathbf{L}_{\mathcal{C}},\mathbf{R}_{\mathcal{B}}\colon\mathcal{T}\to\mathcal{T}$
such that

$\mathcal{T}=\langle\mathcal{B},\mathbf{R}_{\mathcal{B}}(\mathcal{A})\rangle=\langle\mathbf{L}_{\mathcal{C}}(\mathcal{A}),\mathcal{C}\rangle.$

In the case $\mathcal{T}$ is a derived category of a smooth projective variety
or Deligne-Mumford stack, the mutation functors are given by the Serre functor
and its inverse, see [BK89].

######  Example 1.

The easiest example is when $\mu_{2}$ acts on a smooth variety $X$ fixing a
smooth divisor $Y$ pointwise. Then $\bar{X}=X/\mu_{2}$ is smooth. Set
$\iota\colon Y\hookrightarrow X$ to be the inclusion, which is
$\mu_{2}$-equivariant, and $\pi\colon X\to\bar{X}$ the quotient mapping. Then
it is straightforward to check that there is a semiorthogonal decomposition

$\mathcal{D}[X/\mu_{2}]=\langle\pi^{\ast}\mathcal{D}(\bar{X}),\iota_{\ast}\mathcal{D}(Y)\rangle.$

This semiorthogonal decomposition is $\mathcal{D}(\bar{X})$-linear and hence
defines a motivic semiorthogonal decomposition.

Since $\mu_{2}$ acts trivially on $\iota_{\ast}\mathcal{D}(Y)$, we can tensor
by irreducible representations to get different objects. Let
$\chi\colon\mu_{2}\to\mathbb{C}^{\ast}$ be the unique nontrivial
representation. Then for any object
$\iota_{\ast}\mathcal{F}\in\iota_{\ast}\mathcal{D}(Y)$, we can consider the
object $\iota_{\ast}\mathcal{F}\otimes\chi$. This is still an object in
$\mathcal{D}[X/\mu_{2}]$ but no longer in $\iota_{\ast}\mathcal{D}(Y)$. More
precisely, the $G$-equivariant adjunction formula shows the Serre functor
comes with an additional tensoring by $\chi$. Thus, by mutating, we have
another motivic semiorthogonal decomposition

$\mathcal{D}[X/\mu_{2}]=\langle\iota_{\ast}\mathcal{D}(Y)\otimes\chi,\pi^{\ast}\mathcal{D}(\bar{X})\rangle.$

This example indicates one of the central problems in solving this conjecture.
A canonical ordering on embeddings doesn’t exist: one could have fully-
faithful embedding functors but they (or any reordering) are not necessarily
semiorthogonal.

###  2.3. Smooth quotients of Abelian varieties

We recall the main results of [AL17, ALQ18]. Let $G$ be a finite group of
automorphisms of an Abelian variety $A$ such that $G=T\rtimes H$ with $T$ a
group of translations and $H$ a group of group homomorphisms.

######  Theorem 2.

Suppose $G$ fixes the identity and $\dim(A^{G})=0$. If $G$ doesn’t act
irreducibly on $T_{e}(A)$, then there exists direct product decompositions
$A\cong A_{1}\times\cdots\times A_{k}$ and $G\cong G_{1}\times\cdots\times
G_{k}$ where the action is diagonal and $G_{i}$ acts irreducibly on
$T_{e}(A_{i})$.

Since

$\mathcal{D}[A_{1}\times\cdots\times A_{k}/G_{1}\times\cdots\times
G_{k}]\cong\mathcal{D}[A_{1}/G_{1}]\otimes\cdots\otimes\mathcal{D}[A_{k}/G_{k}]$

we can assume that $G$ acts irreducibly.

######  Theorem 3.

Suppose $G$ fixes the identity, $\dim(A^{G})=0$, $G$ acts irreducibly on
$T_{e}(A)$, and $A/G$ is smooth. If $\dim(A)\geq 3$, then there exists an
elliptic curve $E$ such that $A\cong E^{n}$ and either

  1. (A)

$G\cong C^{n}\rtimes S_{n}$ where $C$ is a finite cyclic group acting on $E$
so that $C^{n}$ acts diagonally and $S_{n}$ permutes the factors; or

  2. (B)

$G\cong S_{n+1}$ and acts on $A=\\{(x_{1},\ldots,x_{n+1})\in E^{n+1}\mid
x_{1}+\cdots+x_{n+1}=e\\}$ by permutations.

If $\dim(A)=2$, then both (a) and (b) can happen and there is a third case:

  1. (C)

$E\cong\mathbb{C/Z}[i]$ and $G=G(4,2,2)\cong H\rtimes S_{2}$ where
$H=\\{(\xi^{a},\xi^{b})\mid a+b\equiv_{2}0\\}$. Here $\xi$ is a primitive
fourth root of unity.

We will refer to these as irreducible quotients of Types (A), (B), and (C).
Motivic semiorthogonal decompositions of Type (A) have been constructed in
[PVdB19]. Types (B) and (C) have not yet been considered. We will construct
one for Type (C) in Section 4. We will show they exist for type (B) in Theorem
6.

##  3\. Motivic semiorthogonal decomposition for $[\mathbb{A}^{2}/G(4,2,2)]$

Before constructing a motivic semiorthogonal decomposition for type (C), we
start by studying the local case of $G(4,2,2)$ acting naturally on
$\mathbb{A}^{2}$.

###  3.1. Representation Theory of $G(4,2,2)$

We will need the representation theory of $G(4,2,2)$. We recall this
information now. The complex reflection group $G(4,2,2)$ is constructed as
follows. Fix $\xi=\sqrt{-1}$ to be a primitive fourth root of unity, let
$\mu_{4}^{2}=\langle(\xi,1),(1,\xi)\rangle$ and $K$ the subgroup given by

$K=\\{(\xi^{a},\xi^{b})\in\mu_{4}^{2}\mid a+b\equiv_{2}0\\}.$

The symmetric group $S_{2}=\langle\sigma\rangle$ acts on $K$ by

$\sigma(\xi^{a},\xi^{b})=(\xi^{b},\xi^{a})$

and we set

$G\coloneqq G(4,2,2)=K\rtimes S_{2}.$

The natural action of $G$ induced by that of $\mu_{4}^{2}\rtimes S_{2}$ on
$X\coloneqq\mathbb{A}^{2}$ is an irreducible two dimensional representation
that we call $V$. Note that $V$ is not self-dual, as
$\chi_{V}(\xi,\xi,1)=2\xi$ is purely complex. We have
$X=\mathbb{A}^{2}\cong\mathrm{Spec}(\mathrm{Sym}(V^{\vee}))$.

######  Proposition 1.

There are ten irreducible representations of $G$. Eight of the irreducible
representations are one dimensional and the remaining two are two dimensional.
Moreover, the one dimensional irreducible representations are 2-torsion.

###### Proof.

The center of $G$ is generated by $(-1,-1,1)$ and it is easy to see that the
quotient group is an elementary Abelian 2-group of order 8:

$G/\langle(-1,-1)\rangle=\mu_{2}^{3}.$

Indeed, define a homomorphism $\rho\colon G\to\mu_{2}^{3}$ by

$\rho(\xi^{a},\xi^{b},\sigma)=(\xi^{a+b},\xi^{a-b},\mathrm{sgn}(\sigma)).$

It is a homomorphism as $\xi^{a-b}=\xi^{b-a}$ in this case. The kernel is
$(-1,-1,1)$ and it is evidently surjective. It follows that there are eight
irreducible one dimensional representations. Since the sum of the squares of
the dimensions of the irreducible representations must be $|G|=16$ we have

$8+\dim(V_{1})^{2}+\cdot+\dim(V_{k})^{2}=16$

and thus $k=2$ and $\dim(V_{1})=\dim(V_{2})=2$. These two dimensional
representations are $V$ and $V^{\vee}$. ∎

Let $\rho\colon G\to\mu_{2}^{3}$ to be the surjective homomorphism from the
proof of Proposition 1 and $\pi_{i+1}\colon\mu_{2}^{3}\to\mathbb{C}^{\ast}$
the projection onto the $i$th factor character. Then set

$\chi_{i}=\pi_{i}\circ\rho\colon G\to\mathbb{C}^{\ast}.$

Explicitly

$\displaystyle\chi_{2}(\xi^{a},\xi^{b},\sigma^{c})$ $\displaystyle=\xi^{a+b}$
$\displaystyle\chi_{3}(\xi^{a},\xi^{b},\sigma^{c})$ $\displaystyle=\xi^{a-b}$
$\displaystyle\chi_{4}(\xi^{a},\xi^{b},\sigma^{c})$
$\displaystyle=\mathrm{sgn}(\sigma^{c}).$

The characters $\chi_{2},\chi_{3},\text{ and }\chi_{4}$ generate the character
group of $G$. We also have

$\Lambda^{2}V=\chi_{2}\chi_{4}.$

######  Proposition 2.

There are isomorphisms of representations

$\displaystyle V\otimes V$
$\displaystyle\cong\Lambda^{2}V\oplus\chi_{2}\chi_{4}\oplus\chi_{2}\chi_{3}\chi_{4}\oplus\chi_{2}$
$\displaystyle V^{\vee}$ $\displaystyle\cong V\otimes\Lambda^{2}V\cong
V\otimes\chi_{2}\chi_{3}\cong V\otimes\chi_{2}\chi_{3}\chi_{4}\cong
V\otimes\chi_{2}$ $\displaystyle V\otimes V^{\vee}$
$\displaystyle\cong\chi_{3}\chi_{4}\oplus\mathbf{1}\oplus\chi_{4}\oplus\chi_{3}.$

###### Proof.

For the first isomorphipsm we exhibit an explicit decomposition. Pick the
standard basis $e_{1},e_{2}$ for $V$, then we have the correspondence

$\displaystyle\Lambda^{2}V$ $\displaystyle=\mathrm{Span}\\{e_{1}\otimes
e_{2}-e_{2}\otimes e_{1}\\}$ $\displaystyle\chi_{2}\chi_{3}$
$\displaystyle=\mathrm{Span}\\{e_{1}\otimes e_{1}+e_{2}\otimes e_{2}\\}$
$\displaystyle\chi_{2}\chi_{3}\chi_{4}$
$\displaystyle=\mathrm{Span}\\{e_{1}\otimes e_{1}-e_{2}\otimes e_{2}\\}$
$\displaystyle\chi_{2}$ $\displaystyle=\mathrm{Span}\\{e_{1}\otimes
e_{2}+e_{2}\otimes e_{1}\\}$

The second isomorphism follows since if $f_{1},f_{2}$ are the bases dual to
$e_{1}$ and $e_{2}$ we have that $\sigma(f_{1})=f_{2}$ and converseley.
Moreover,

$(\xi^{a},\xi^{b})(f_{1})=\xi^{-a}f_{1}\text{ and
}(\xi^{a},\xi^{b})(f_{2})=\xi^{-b}f_{2}.$

The last isomorphism follows from the first two. ∎

The following is straightforward. We include it for labeling purposes.

######  Proposition 3.

The conjugacy classes of $G$ are:

$\displaystyle D_{1}$ $\displaystyle=\\{\mathrm{Id}\\}$ $\displaystyle D_{2}$
$\displaystyle=\\{(1,1,\sigma),(-1,-1,\sigma)\\}$ $\displaystyle D_{3}$
$\displaystyle=\\{(1,-1,1),(-1,1,1)\\}$ $\displaystyle D_{4}$
$\displaystyle=\\{(\xi,-\xi,\sigma),(-\xi,\xi,\sigma)\\}$ $\displaystyle
D_{5}$ $\displaystyle=\\{(-1,-1,1)\\}$ $\displaystyle D_{6}$
$\displaystyle=\\{(\xi,\xi,1)\\}$ $\displaystyle D_{7}$
$\displaystyle=\\{(\xi,\xi,\sigma),(-\xi,-\xi,\sigma)\\}$ $\displaystyle
D_{8}$ $\displaystyle=\\{(-\xi,-\xi,1)\\}$ $\displaystyle D_{9}$
$\displaystyle=\\{(1,-1,\sigma),(-1,1,\sigma)\\}$ $\displaystyle D_{10}$
$\displaystyle=\\{(\xi,-\xi,1),(-\xi,\xi,1)\\}.$

The centralizers of each conjugacy class are:

$\displaystyle H_{1}$ $\displaystyle=G$ $\displaystyle H_{2}$
$\displaystyle=\langle(\xi,\xi,1),(1,1,\sigma)\rangle\cong\mu_{4}\times\mu_{2}$
$\displaystyle H_{3}$ $\displaystyle=K$ $\displaystyle H_{4}$
$\displaystyle=\langle(\xi,\xi,1),(\xi^{2},1,\sigma)\rangle\cong\mu_{4}\times\mu_{2}$
$\displaystyle H_{5}$ $\displaystyle=G$ $\displaystyle H_{6}$
$\displaystyle=G$ $\displaystyle H_{7}$
$\displaystyle=\langle(\xi,\xi,1),(1,1,\sigma)\rangle\cong\mu_{4}\times\mu_{2}$
$\displaystyle H_{8}$ $\displaystyle=G$ $\displaystyle H_{9}$
$\displaystyle=\langle(\xi,\xi,1),(1,\xi^{2},\sigma)\rangle\cong\mu_{4}\times\mu_{2}$
$\displaystyle H_{10}$ $\displaystyle=K.$

For all $i=1,...,10$, let $g_{i}$ denote a representative of $D_{i}$. Then,
the fixed loci $X^{i}\coloneqq X^{g_{i}}$ are

(1) $\displaystyle X^{1}$ $\displaystyle=X,$ $\displaystyle
X^{5}=...=X^{10}=\\{0\\},$

and $X^{2},X^{3},X^{4}$ are lines intersecting at the origin.

Notice that the centralizers $H_{2},H_{4},H_{7},H_{9}$ are index two and hence
normal. Consequently the quotients by these normal subgroups define
characters:

$\rho_{2},\rho_{3},\rho_{4}.$

Explicitly,

$\displaystyle\rho_{2}$ $\displaystyle=\chi_{3},$ $\displaystyle\rho_{3}$
$\displaystyle=\chi_{4},$ $\displaystyle\rho_{4}$
$\displaystyle=\chi_{3}\chi_{4}.$

###  3.2. Fourier-Mukai kernels

For $i=1,\ldots,10$, we define Fourier-Mukai kernels by constructing
subschemes $H_{i}\times G$-invariant subschemes $Z_{i}$ of

$X^{i}\times X$

as follows. Define

$Z_{i}=\bigcup_{g\in G}g^{\ast}\Gamma_{i}$

where $\Gamma_{i}$ is the graph of the inclusion $X^{i}\hookrightarrow X$, and
we give $Z_{i}$ its reduced scheme structure. Then $Z_{i}$ is an $H_{i}\times
G$\- invariant subscheme and so $\mathcal{O}_{Z_{i}}$ can be taken as a kernel
for the Fourier-Mukai functor

$\Phi_{i}=\Phi_{\mathcal{O}_{Z_{i}}}\circ\pi_{i}^{\ast}\colon\mathcal{D}(\bar{X}^{i})\to\mathcal{D}[X/G(4,2,2)],$

where $\pi_{i}\colon X^{i}\to\bar{X}^{i}$ is the quotient mapping.
Equivalently, let $\bar{Z}_{i}$ be the image of $Z_{i}$ under the quotient
$(\pi_{i},\mathrm{Id})\colon X^{i}\times X\to\bar{X}^{i}\times X$, then

$\Phi_{i}=\Phi_{\mathcal{O}_{\bar{Z}_{i}}}.$

We note that if $X^{i}$ is a point, then $Z_{i}$ is also a point and the
corresponding $\Phi_{i}$ is simply the pushforward along the inclusion of the
origin which is $G$-invariant. If $X^{i}$ is a hyperplane, then $Z^{i}$ is the
union of two lines in $X^{i}\times X\cong\mathbb{A}^{3}$ as the centralizer is
index two. If $X^{i}=X$, then $\Phi_{1}=\pi^{*}$, the pullback functor from
the coarse space.

######  Proposition 4.

  1. (1)

We have

$\Phi_{1}(\mathcal{O}_{0})=\mathbb{C}[x,y]/(x^{2}y^{2},x^{4}+y^{4})$

and for $p\neq 0$ every other point in $X$ we set $L_{p}$ to be its orbit. If
$p$ has trivial stabilizer, then $L_{p}$ consists of 16 distinct points and

$\Phi_{1}(\mathcal{O}_{p})=\mathcal{O}_{L_{p}}.$

If $p$ has a $\mu_{2}$ stabilizer. In this case the orbit $L_{p}$ of $p$
contains 8 points, and $\Phi_{1}(\mathcal{O}_{p})$ is the sheaf supported on
$L_{p}$ with a nilpotent of order 2 at every point.

  2. (2)

For each $i=2,3,4$, we have

$\Phi_{i}(\mathcal{O}_{0})=\begin{cases}\mathbb{C}[x,y]/(x^{2}y^{2},x^{2}-y^{2})&i=2\\\
\mathbb{C}[x,y]/(xy,(x+y)^{4})&i=3\\\
\mathbb{C}[x,y]/(x^{2}y^{2},x^{2}+y^{2})&i=4\end{cases}$

and for $p\neq 0$ any other point in $\bar{X}^{i}$ we set $L_{p}$ to be the
orbit of the preimage of $p$ under the inclusion $X^{i}\hookrightarrow X$.
That is, let $p_{1},p_{2},p_{3},p_{4}$ be the four distinct points in the
preimage of $p$, then

$L_{p}=\\{p_{1},p_{2},p_{3},p_{4},\gamma p_{1},\gamma p_{2},\gamma
p_{3},\gamma p_{4}\\}$

where $\gamma$ is a nontrivial coset representative. Then

$\Phi_{i,1}(\mathcal{O}_{p})\cong\mathcal{O}_{L_{p}}.$

###### Proof.

  1. (1)

The functor $\Phi_{1}$ is pullback $\pi^{*}$ from the coarse moduli. The
maximal ideal of the origin pulls back to the ideal of $G$-invariant
polynomials $(x^{2}y^{2},x^{4}+y^{4})$, and a point $p\neq 0$ to its orbit.

If $p$ has trivial stabilizer, then the pull back of its image under
$X\to\bar{X}$ consists of 16 points, all belonging to the locus where the
quotient map is a local isomorphism.

If $p$ has a $\mu_{2}$ stabilizer, then there are 8 other points in its orbit,
and then the pull-back $\pi^{*}(\mathcal{O}_{\pi(p)})$ is non-reduced of
length 2: locally around $p$ the quotient map $X\to\bar{X}$ is given by
$\mathbb{C}[u,v^{2}]\to\mathbb{C}[u,v]$ and $\pi(p)$ pulls back to
$\mathbb{C}[u,v]/(u,v^{2})$.

There are no other possibilities for the stabilizer of a point
$p\in\mathbb{A}^{2}$.

  2. (2)

The second statement follows since the pullup of $\mathcal{O}_{p}$ under the
quotient map $X^{i}\to\bar{X}^{i}$ is

$\pi_{i}^{\ast}(\mathcal{O}_{p})\cong\bigoplus_{i=1}^{4}\mathcal{O}_{p_{i}}.$

Then since $p\neq 0$, the support of $\pi_{i}^{\ast}(\mathcal{O}_{p})$ is
contained in the open locus where the projection is an isomorphism onto its
image. Finally, if $\gamma$ is a coset representative of the only nontrivial
coset, then when we equivariantize it just translates each of the four points
under $\gamma$.

The first statement is more involved. We will only do the computation for
$i=3$ as the other computations are analagous under a linear change of
coordinates. In this case $\lambda=(1,-1,1)$. We can take $\\{1,\sigma\\}$ as
a transversal for $C(\lambda)$ in $G$. Thus
$Z_{\lambda}=\Gamma_{\lambda}\cup\sigma^{\ast}\Gamma_{\lambda}$ with its
reduced scheme structure. As modules we have

$H^{0}(\mathcal{O}_{Z_{\lambda}})\cong\mathbb{C}[z,x,y]/(z-x-y,xy).$

Let $X^{3}\to\bar{X}^{3}$ denote the quotient map. Note that
$C(\lambda)/(\lambda)\cong\mu_{4}$ and acts in the natural way on $X^{2}$. In
particular if $x$ is the coordinate on $X^{4}$, then the quotient map is
$x\mapsto x^{4}$. It follows that for $0\in\bar{X}^{2}\cong\mathbb{A}^{1}$, we
have

$\pi_{3}^{\ast}(\mathcal{O}_{0})=\mathcal{O}_{40}$

where $H^{0}(\mathcal{O}_{40})\cong\mathbb{C}[z]/(z^{4})$ is a non-reduced
zero-dimensional scheme of length 4 concentrated at the origin. Then we pull-
up the resolution $\mathcal{O}_{X^{3}}\xrightarrow{z^{4}}\mathcal{O}_{X^{3}}$
to $X^{3}\times X\cong\mathbb{A}^{1}_{z}\times\mathbb{A}^{2}_{x,y}$. Tensoring
with $\mathcal{O}_{Z_{\lambda}}$ yields

$\mathcal{O}_{Z_{\lambda}}\xrightarrow{z^{4}}\mathcal{O}_{Z_{\lambda}}.$

As this is injective we have that this complex is quasi-isomorphic to

$\mathbb{C}[z,x,y]/(z-x-y,xy,z^{4})$

This pushes forward to $\mathbb{C}[x,y]/((x+y)^{4},xy)$.

∎

######  Corollary 2.

Let $R$ denote the regular representation of $G$. Then:

$\displaystyle\Phi_{1}(\mathcal{O}_{0})\cong$ $\displaystyle R.$

Moreover we have the following isomorphisms of representations:

$\displaystyle\Phi_{2}(\mathcal{O}_{0})\cong$ $\displaystyle\mathbf{1}\oplus
V^{\vee}\oplus V\oplus\chi_{2}\oplus\chi_{2}\chi_{3}\oplus\chi_{3}$
$\displaystyle\Phi_{3}(\mathcal{O}_{0})\cong$ $\displaystyle\mathbf{1}\oplus
V^{\vee}\oplus
V\oplus\chi_{2}\chi_{3}\oplus\chi_{2}\chi_{3}\chi_{4}\oplus\chi_{4}$
$\displaystyle\Phi_{4}(\mathcal{O}_{0})\cong$ $\displaystyle\mathbf{1}\oplus
V^{\vee}\oplus
V\oplus\chi_{2}\oplus\chi_{2}\chi_{3}\chi_{4}\oplus\chi_{3}\chi_{4}$

###### Proof.

This is straightforward. For example, a basis for $\Phi_{1}(\mathcal{O}_{0})$
as a $\mathbb{C}$-vector space is

$\begin{split}1,x,y,xy,x^{2},xy,y^{2},\\\ x^{3},x^{2}y,xy^{2},y^{3},x^{4},\\\
x^{3}y,xy^{3},x^{5},x^{4}y,x^{5}y.\end{split}$

We have isomorphisms

$\displaystyle\mathbb{C}\\{1\\}$ $\displaystyle\cong\mathbf{1},$
$\displaystyle\mathbb{C}\\{x,y,x^{5},x^{4}y\\}$
$\displaystyle\cong(V^{\vee})^{\oplus 2},$
$\displaystyle\mathbb{C}\\{x^{3},x^{2}y,xy^{2},y^{3}\\}$ $\displaystyle\cong
V^{\oplus 2},$ $\displaystyle\mathbb{C}\\{x^{2},y^{2}\\}$
$\displaystyle\cong\chi_{2}\chi_{3}\oplus\chi_{2}\chi_{3}\chi_{4},$
$\displaystyle\mathbb{C}\\{x^{3}y,xy^{3}\\}$
$\displaystyle\cong\chi_{3}\oplus\chi_{3}\chi_{4},$
$\displaystyle\mathbb{C}\\{xy\\}$ $\displaystyle\cong\chi_{2},$
$\displaystyle\mathbb{C}\\{x^{4}\\}$ $\displaystyle\cong\chi_{4},$
$\displaystyle\mathbb{C}\\{x^{5}y\\}$ $\displaystyle\cong\chi_{2}\chi_{4}.$

∎

######  Proposition 5.

For each $i=1,\ldots,10$, the Fourier-Mukai functors

$\Phi_{i}\colon\mathcal{D}(\bar{X}^{i})\to\mathcal{D}[X/G(4,2,2)]$

are fully-faithful.

###### Proof.

In the case $i=1$, this is pullback from $X/G(4,2,2)$ which is fully-faithful.
If $i=5,\ldots,10$, this is pushforward of the origin, which is fully-faithful
since the structure sheaf of the origin is exceptional:

(2)
$\mathrm{Ext}^{\ast}(\mathcal{O}_{0},\mathcal{O}_{0})^{G}=\left(\mathbf{1}[0]\oplus
V[-1]\oplus\Lambda^{2}V[-2]\right)^{G}=\mathbf{1}[0].$

The computations for $i=2,3,4$ are all similar. For notational convenience, we
will compute the case of $i=3$. In this case, we have two types of points. If
$p\neq 0$, then we look at the orbit of a preimage under the quotient map. The
orbit is a reduced subscheme of length 8 $L_{p}$. Moreover, we have

$\mathrm{Ext}^{\ast}(\mathcal{O}_{L_{p}},\mathcal{O}_{L_{p}})^{G}=\left(\oplus_{i=1}^{8}\mathrm{Ext}^{\ast}(\mathcal{O}_{p_{i}},\mathcal{O}_{p_{i}})\right)^{G}.$

Clearly when $\ast=0$ the $G$-action permutes the hom-sets and so the
8-dimensional hom-space has only one copy of the trivial representation. If
$\ast=2$, then we have

$\left(\oplus_{i=1}^{8}T_{p_{i}}X^{i}\otimes N_{p_{i}}X^{i}\right)^{G}$

and since $X^{3}$ is codimension 1 it must be that $\lambda_{3}$ acts
nontrivially on the normal bundle. Thus there are no invariants.

For $p=0$, we recall that
$\Gamma(\Phi_{3}(\mathcal{O}_{0}))\cong\mathbb{C}[x,y]/(xy,(x+y)^{4})$. Hence,

$\mathrm{Hom}(\Phi_{3}(\mathcal{O}_{0}),\Phi_{3}(\mathcal{O}_{0}))^{G}\cong\mathbb{C}$

as this is determined by the image of $1$ and
$\Phi_{3}(\mathcal{O}_{0})^{G}\cong\mathbf{1}$. For vanishing of the second
ext group we use duality:

$\mathrm{Ext}^{2}(\Phi_{3}(\mathcal{O}_{0}),\Phi_{3}(\mathcal{O}_{0}))\cong\mathrm{Hom}(\Phi_{3}(\mathcal{O}_{0}),\Phi_{3}(\mathcal{O}_{0})\otimes\Lambda^{2}V)$

and by Corollary 2

$\Phi_{3}(\mathcal{O}_{0})\cong\mathbf{1}\oplus V\oplus
V^{\vee}\oplus\chi_{2}\chi_{3}\oplus\chi_{2}\chi_{3}\chi_{4}\oplus\chi_{4}$

and so $(\Phi_{3}(\mathcal{O}_{0})\otimes\Lambda^{2}V)^{G}=0$. This implies
vanishing of the second ext group and completes the proof. ∎

###  3.3. Assembling the Motivic Semiorthogonal Decomposition

In this section we construct a motivic semiorthogonal decomposition of $[X/G]$
using the functors $\Phi_{i}$. We start by observing that the positive
dimensional loci have semiorthogonal images:

######  Lemma 1.

  1. (1)

Let $i=2,3,4$ and $p\in X^{i}$. We have

$\mathrm{Ext}^{\ast}(\Phi_{i}(\mathcal{O}_{p}),\Phi_{1}(\mathcal{O}_{p}))^{G}=0$

  2. (2)

For $j=2,3,4$, $j\neq i$,

$\mathrm{Ext}^{\ast}(\Phi_{i}(\mathcal{O}_{0}),\Phi_{j}(\mathcal{O}_{0}))^{G}=0.$

###### Proof.

Recall that $\Phi_{1}=\pi^{*}$, and write

$\displaystyle\mathrm{Ext}^{\ast}(\Phi_{i}(\mathcal{O}_{p}),\pi^{*}(\mathcal{O}_{p}))$
$\displaystyle\cong\mathrm{Ext}^{\ast}(\pi^{*}(\mathcal{O}_{p}),\Phi_{i}(\mathcal{O}_{p})\otimes\Lambda^{2}V)^{\vee}$
$\displaystyle\cong\mathrm{Ext}^{\ast}(\mathcal{O}_{p},\pi_{*}(\Phi_{i}(\mathcal{O}_{p})\otimes\Lambda^{2}V))^{\vee}$
$\displaystyle=0,$

where the last equality follows because
$\Phi_{i}(\mathcal{O}_{p})\otimes\Lambda^{2}V$ has no invariants (see
Corollary 2 for $p=0$, and Proposition 4 for $p\neq 0$).

The argument is slightly more involved for the second statement. We only
illustrate the case $i=2$ and $j=3$, the others are analogous. A resolution of
$\Phi_{2}(\mathcal{O}_{0})$ is the Koszul complex

$\mathbb{C}[x,y]\otimes\chi_{2}\chi_{3}\chi_{4}\xrightarrow{\begin{pmatrix}y^{2}-x^{2}\\\
x^{2}y^{2}\end{pmatrix}}\begin{array}[]{c}\mathbb{C}[x,y]\\\ \oplus\\\
\mathbb{C}[x,y]\otimes\chi_{2}\chi_{3}\chi_{4}\end{array}\xrightarrow{\cdot\begin{pmatrix}x^{2}y^{2}&x^{2}-y^{2}\end{pmatrix}}\mathbb{C}[x,y]$

where $\chi_{2}\chi_{3}\chi_{4}$ is the weight of $x^{2}-y^{2}$. Applying
$\mathrm{Hom}(-,\Phi_{3}(\mathcal{O}_{0}))$ yields

(3) $\Phi_{3}(\mathcal{O}_{0})\xrightarrow{\begin{pmatrix}0\\\
y^{2}-x^{2}\end{pmatrix}}\begin{array}[]{c}\Phi_{3}(\mathcal{O}_{0})\\\
\oplus\\\
\Phi_{3}(\mathcal{O}_{0})\otimes\chi_{2}\chi_{3}\chi_{4}\end{array}\xrightarrow{\cdot\begin{pmatrix}x^{2}-y^{2}&0\end{pmatrix}}\Phi_{3}(\mathcal{O}_{0})\otimes\chi_{2}\chi_{3}\chi_{4}.$

Recall from Proposition 4 that
$\Phi_{3}(\mathcal{O}_{0})\cong\mathbb{C}[x,y]/(xy,(x+y)^{4})$. The kernel of
multiplication by $y^{2}-x^{2}$ is the submodule

$\left(x^{2}+y^{2}\right)\cong\mathbb{C}\left\\{x^{2}+y^{2},x^{3},y^{3},x^{4}\right\\}\cong\chi_{2}\chi_{3}\oplus
V\oplus\chi_{4}.$

The cokernel of multiplication by $x^{2}-y^{2}$ is

$\mathbb{C}[x,y]/(xy,x^{2}-y^{2})\otimes\chi_{2}\chi_{3}\chi_{4}\cong\chi_{2}\chi_{3}\chi_{4}\oplus
V\oplus\chi_{4}.$

Observe that the middle cohomology is the sum of the above kernel and
cokernel. Then, none of the cohomologies of (3) have invariant summands. ∎

Next, we focus on the zero dimensional loci and compare them to the positive
dimensional ones. We start from this computation:

######  Lemma 2.

For each $i=1,2,3,4$, we have isomorphisms in $\mathcal{D}(BG)$:

$\displaystyle\mathrm{RHom}(\mathcal{O}_{0},\Phi_{1}(\mathcal{O}_{0}))$
$\displaystyle\cong\chi_{2}\chi_{4}[0]\oplus\left(\chi_{2}\chi_{4}\right)^{\oplus
2}[-1]\oplus\chi_{2}\chi_{4}[-2]$
$\displaystyle\mathrm{RHom}(\mathcal{O}_{0},\Phi_{2}(\mathcal{O}_{0}))$
$\displaystyle\cong\chi_{3}[0]\oplus\left(\chi_{3}\oplus\chi_{2}\chi_{4}\right)[-1]\oplus\chi_{2}\chi_{4}[-2]$
$\displaystyle\mathrm{RHom}(\mathcal{O}_{0},\Phi_{3}(\mathcal{O}_{0}))$
$\displaystyle\cong\chi_{4}[0]\oplus(\chi_{4}\oplus\chi_{2}\chi_{4})[-1]\oplus\chi_{2}\chi_{4}[-2]$
$\displaystyle\mathrm{RHom}(\mathcal{O}_{0},\Phi_{4}(\mathcal{O}_{0}))$
$\displaystyle\cong\chi_{3}\chi_{4}[0]\oplus(\chi_{3}\chi_{4}\oplus\chi_{2}\chi_{4})[-1]\oplus\chi_{2}\chi_{4}[-2].$

###### Proof.

We again only compute the case $i=3$. It is technically easier to use duality
first:

$\displaystyle\mathrm{Ext}^{\ast}_{[X/G]}(\mathcal{O}_{0},\Phi_{i}(\mathcal{O}_{0}))$
$\displaystyle\cong\mathrm{Ext}^{2-\ast}_{[X/G]}(\Phi_{i}(\mathcal{O}_{0}),\mathcal{O}_{0}\otimes\Lambda^{2}V).$

We resolve $\Phi_{3}(\mathcal{O}_{0})$ (computed in Proposition 4) by the
Koszul complex noting that $xy$ has weight $\chi_{2}$:

$\displaystyle\mathrm{Ext}^{\ast}_{[X/G]}(\Phi_{i}(\mathcal{O}_{0}),\mathcal{O}_{0}\otimes\Lambda^{2}V)$
$\displaystyle\cong
H^{\ast}(\mathcal{O}_{0}\chi_{2}\chi_{4}\to\mathcal{O}_{0}\chi_{2}\chi_{4}\oplus\mathcal{O}_{0}\chi_{4}\to\mathcal{O}_{0}\chi_{4})$

and hence the statement. ∎

Finally, we study Hom spaces between zero dimensional loci:

######  Lemma 3.

Let $\rho,\sigma$ be irreducible representations of $G$. Then, the position
$(\rho,\sigma)$ is marked by a ✓in the following table if and only if

$\mathrm{Ext}^{*}_{[X/G]}(\mathcal{O}_{0}\otimes\rho,\mathcal{O}_{0}\otimes\sigma)=0.$ $\rho$ $\sigma$ | $\mathbf{1}$ | $\chi_{2}$ | $\chi_{3}$ | $\chi_{4}$ | $\chi_{2}\chi_{3}$ | $\chi_{2}\chi_{4}$ | $\chi_{3}\chi_{4}$ | $\chi_{2}\chi_{3}\chi_{4}$ | $V$ | $V^{\vee}$
---|---|---|---|---|---|---|---|---|---|---
$\mathbf{1}$ |  | ✓ | ✓ | ✓ | ✓ |  | ✓ | ✓ | ✓ |
$\chi_{2}$ | ✓ |  | ✓ |  | ✓ | ✓ | ✓ | ✓ |  | ✓
$\chi_{3}$ | ✓ | ✓ |  | ✓ | ✓ | ✓ | ✓ |  | ✓ |
$\chi_{4}$ | ✓ |  | ✓ |  | ✓ | ✓ | ✓ | ✓ | ✓ |
$\chi_{2}\chi_{3}$ | ✓ | ✓ | ✓ | ✓ |  | ✓ |  | ✓ |  | ✓
$\chi_{2}\chi_{4}$ |  | ✓ | ✓ | ✓ | ✓ |  | ✓ | ✓ |  | ✓
$\chi_{3}\chi_{4}$ | ✓ | ✓ | ✓ | ✓ |  | ✓ |  | ✓ | ✓ |
$\chi_{2}\chi_{3}\chi_{4}$ | ✓ | ✓ |  | ✓ | ✓ | ✓ | ✓ |  |  | ✓
$V$ |  | ✓ |  |  | ✓ | ✓ |  | ✓ |  |
$V^{\vee}$ | ✓ |  | ✓ | ✓ |  |  | ✓ |  |  |

###### Proof.

Twisting the (2) we obtain

$\mathrm{Ext}^{\ast}_{[X/G]}(\mathcal{O}_{0}\otimes\rho,\mathcal{O}_{0}\otimes\sigma)=\left[(\mathbf{1}[0]\oplus
V[-1]\oplus\Lambda^{2}V[-2])\otimes(\rho^{\vee}\otimes\sigma)[0]\right]^{G}.$

and one checks which choices of $(\rho,\sigma)$ produce invariant summands,
using Proposition 2. ∎

It follows from Lemma 2 that $\mathcal{O}_{0}$,
$\mathcal{O}_{0}\otimes\chi_{2}$, and $\mathcal{O}_{0}\otimes V^{\vee}$ are in
the left orthogonal of the images of the positive-dimensional loci. In
particular, the sheaf $M\coloneqq\mathbb{C}[x,y]/(x^{2}+y^{2},x^{2}-y^{2})$ is
also in the left orthogonal, since as a representation

(4) $M\cong\mathbf{1}\oplus V^{\vee}\oplus\chi_{2}.$

######  Proposition 6.

The collection

(5)
$\left(\mathcal{O}_{0}\otimes\chi_{2},\,M,\,\mathcal{O}_{0}\otimes\chi_{2}\chi_{3},\,\mathcal{O}_{0}\otimes\chi_{2}\chi_{3}\chi_{4},\,\mathcal{O}_{0}\otimes
V,\,\mathcal{O}_{0}\right)$

is an exceptional collection in $\mathcal{D}[X/G]$.

###### Proof.

First observe that all objects are exceptional: the twists of
$\mathcal{O}_{0}$ by characters are immediate, the twist by $V$ follows from
Proposition 2, we check $M$ now. Its endomorphism algebra is computed by
resolving $M$ with a Koszul complex and computing the cohomology of

$M\xrightarrow{0}M\chi_{2}\chi_{3}\oplus
M\chi_{2}\chi_{3}\chi_{4}\xrightarrow{0}M\chi_{4}$

($\chi_{2}\chi_{3}$ and $\chi_{2}\chi_{3}\chi_{4}$ are the weights of
$x^{2}+y^{2}$, $x^{2}-y^{2}$ respectively), so by (4) it only has one
invariant summand in degree 0, i.e. $M$ is exceptional.

We now check the orthogonalities. All the conditions not involving $M$ follow
from Lemma 3, and all the ones involving $M$ are similar: we only compute
$\mathrm{Ext}^{*}(M,\mathcal{O}_{0}\otimes\chi_{2})^{G}=0$ here. Once again,
resolve $M$ with a Koszul complex and consider the cohomology of

$\mathcal{O}_{0}\chi_{2}\to\mathcal{O}_{0}\chi_{3}\oplus\mathcal{O}_{0}\chi_{3}\chi_{4}\to\mathcal{O}_{0}\chi_{2}\chi_{4},$

which has no invariant summands. ∎

For $i=1,...,4$, let $C_{i}$ denote the image of
$\Phi_{i}\colon\mathcal{D}(\bar{X}^{i})\to\mathcal{D}[X/G]$. For $i\geq 5$,
define $C_{i}$ to be the image of the twisted functors

$\displaystyle\Phi^{\prime}_{5}$ $\displaystyle=\Phi_{5}(-)\otimes\chi_{2}$
$\displaystyle\Phi^{\prime}_{8}$
$\displaystyle=\Phi_{8}(-)\otimes\chi_{2}\chi_{3}\chi_{4}$
$\displaystyle\Phi^{\prime}_{6}$ $\displaystyle=\Phi_{6}(-)\otimes M$
$\displaystyle\Phi^{\prime}_{9}$ $\displaystyle=\Phi_{9}(-)\otimes V$
$\displaystyle\Phi^{\prime}_{7}$
$\displaystyle=\Phi_{7}(-)\otimes\chi_{2}\chi_{3}$
$\displaystyle\Phi^{\prime}_{10}$ $\displaystyle=\Phi_{10}(-).$

######  Theorem 4.

There is a motivic semiorthogonal decomposition

(6) $\mathcal{D}[X/G]=\left\langle
C_{1},C_{2},C_{3},C_{4},C_{5},C_{6},C_{7},C_{8},C_{9},C_{10}\right\rangle.$

###### Proof.

For all $i=1,...,10$, the component $C_{i}$ is equivalent to
$\mathcal{D}(\bar{X}^{i})$. This is the content of Proposition 5 (the
generators of $C_{5},...,C_{10}$ are still exceptional objects by Proposition
6).

Proposition 6 also shows that $C_{5},...,C_{10}$ are semiorthogonal. Moreover,
it immediately follows from Lemma 2 that
$\mathrm{Ext}^{\ast}_{[X/G]}(C_{j},C_{i})=0$ for $i=1,2,3,4$ and $j=5,...,10$.
The first statement of Lemma 1 implies
$\mathrm{Ext}^{\ast}_{[X/G]}(C_{i},C_{1})=0$ for $i=2,3,4$. Now let $i,j$ be
distinct elements of $\left\\{2,3,4\right\\}$. Then, the Fourier-Mukai kernels
$Z^{i}$ and $Z^{j}$ only intersect at the origin, therefore the second
statement of Lemma 1 implies the vanishing
$\mathrm{Ext}^{\ast}_{[X/G]}(C_{i},C_{j})=0$.

It remains to show that the category $\mathcal{T}\coloneqq\left\langle
C_{1},...,C_{10}\right\rangle$ coincides with $\mathcal{D}[X/G]$. It suffices
to show that $\mathcal{T}$ contains the spanning class

$\Omega\coloneqq\left\\{\mathcal{O}_{p}\otimes\rho\;|\;p\in
X,\rho\in\mathrm{Irrep}(G_{p})\right\\}.$

In this case, in fact, every object $E\in\mathcal{T}^{\perp}$ satisfies
$\mathrm{Hom}(\omega,E)=0$ for all $\omega\in\Omega$, which implies $E=0$ by
definition of a spanning class, and therefore
$\mathcal{T}^{\perp}=\left\\{0\right\\}$.

If $p=0$, then $G_{p}\cong G$, and we recover all irreducible representations
as follows: from the exceptional collection (5) we get

$\chi_{2},\chi_{2}\chi_{3},\chi_{2}\chi_{3}\chi_{4},V,\mathbf{1},\mbox{ and
}V^{\vee}$

(the last one from the composition series of $M$, whose factors are
$(\chi_{2},V^{\vee},\mathbf{1})$). With these, and using Corollary 2, we
obtain $\chi_{3},\chi_{4},\chi_{3}\chi_{4}$, and $\chi_{2}\chi_{4}$.

If $p\neq 0$, then $G_{p}$ is either trivial or $\mu_{2}$, in which case $p\in
X^{i}$ for some $i=2,3,4$. If $G_{p}=0$, then $\Phi_{1}(\mathcal{O}_{\pi(p)})$
contains $\mathcal{O}_{p}$ as a direct summand (Proposition 4), hence
$\mathcal{O}_{p}\in\mathcal{T}$.

If $p$ has $\mu_{2}$ stabilizer, we apply Proposition 4: on the one hand, the
pull-back along $X^{i}\to\bar{X}^{i}$ contains $\mathcal{O}_{p}$ as a direct
summand. On the other hand, the pull-back $\pi^{*}(\mathcal{O}_{\pi(p)})$ is
non-reduced of length 2. Then, the non-trivial irreducible representation is
obtained as the kernel of the map

$\pi^{*}(\mathcal{O}_{\pi(p)})\to\mathcal{O}_{p}.\qed$

##  4\. Motivic Semiorthogonal Decomposition for Type (C)

###  4.1. Setup

We recall here the construction of Type (C) quotients, following [ALQ18] (see
Theorem 3). Let $\Lambda=\mathbb{Z}[i]$ be the Gaussian lattice and
$E\cong\mathbb{C}/\Lambda$ the corresponding elliptic curve with $\mu_{4}$
automorphism group. The action of $G\coloneqq G(4,2,2)$ on
$\mathbb{A}^{2}\simeq\mathbb{C}^{2}$ described in Section 3 preserves the
lattice $\Lambda^{\oplus 2}$ and descends to an action on $B\coloneqq E^{2}$.
As it turns out, the quotient $B/G$ is not smooth, but one considers another
Abelian surface $A\coloneqq E^{2}$ and an isogeny $\pi\colon B\to A$ defined
by the matrix

$\nu=\begin{pmatrix}1&-1\\\ 0&i-1\end{pmatrix}.$

Let $t_{0}$ denote the only non-trivial $\xi$-invariant element of $E$, then
the kernel of $\pi$ is

$\ker(\pi)=\Delta=\\{(t_{0},t_{0})\\},$

and we have $B/\Delta\cong A$. Moreover, $G$ acts trivially on $\Delta$, so we
have an action of $\Delta\times G$ on $B$. This action descends to an action
of $G$ on $A$ and induces a natural isomorphism $B/\Delta\rtimes G\equiv A/G$.
Explicitly, every element $g$ of $G$ acts on $A$ via the matrix
$\nu\lambda\nu^{-1}$ where $\lambda$ is the matrix expression of $g$ acting on
$\mathbb{A}^{2}$. In particular, the generators
$(-1,1,1),(-i,i,1),(1,1,\sigma)$ act on $A$ via the matrices:

$\alpha=\begin{pmatrix}-1&1+i\\\ 0&1\end{pmatrix},\
\beta=\begin{pmatrix}-i&i-1\\\ 0&i\end{pmatrix},\
\gamma=\begin{pmatrix}-1&0\\\ i-1&1\end{pmatrix}.$

###  4.2. Fixed loci and stabilizers

We use the numbering of Prop. 3 for the conjugacy classes and the centralizers
of $G$, and we list representatives of conjugacy classes and fixed loci. By
$E[n]$ we denote the set $\\{x\in E\mid nx=0\\}$ of $n$-torsion points. The
conjugacy classes are:

$\displaystyle D_{1}$ $\displaystyle=[\mathrm{Id}]$ $\displaystyle D_{2}$
$\displaystyle=[\gamma]$ $\displaystyle D_{3}$ $\displaystyle=[\alpha]$
$\displaystyle D_{4}$ $\displaystyle=[\beta\gamma]$ $\displaystyle D_{5}$
$\displaystyle=[-\mathrm{Id}]$ $\displaystyle D_{6}$
$\displaystyle=[\alpha\beta]$ $\displaystyle D_{7}$
$\displaystyle=[-\alpha\beta]$ $\displaystyle D_{8}$
$\displaystyle=[\alpha\beta\gamma]$ $\displaystyle D_{9}$
$\displaystyle=[\alpha\gamma]$ $\displaystyle D_{10}$ $\displaystyle=[\beta].$

The corresponding fixed loci are:

$\displaystyle A^{1}$ $\displaystyle=A$ $\displaystyle A^{2}$
$\displaystyle=\\{(x,y)\in A\mid x=ix\\}\cong\\{e,t_{0}\\}\times E$
$\displaystyle A^{3}$ $\displaystyle=\\{(x,y)\in A\mid 2x=(1+i)y\\}\cong
E[2]\times E$ $\displaystyle A^{4}$ $\displaystyle=\\{(x,y)\in A\mid
x=iy\\}\cong E$ $\displaystyle A^{5}$ $\displaystyle=E[2]^{2}$ $\displaystyle
A^{6}$ $\displaystyle=...=A^{10}=\\{e,t_{0}\\}^{2}.$

We point out that the only points of $A$ with non-trivial stabilizers are:

Points | Stabilizer
---|---
$\\{e,t_{0}\\}^{2}$ | $G$
$E[2]^{2}\setminus\\{e,t_{0}\\}^{2}$ | $\mu_{2}\times\mu_{2}$
$(A^{2}\cup A^{3}\cup A^{4})\setminus E[2]^{2}$ |  $\mu_{2}$.

Finally, we compute the quotients $\bar{A}^{i}$. Observe that the group $G$
acts on $E[2]^{2}$ with 7 orbits. Four orbits are given by elements of
$\\{e,t_{0}\\}\times\\{e,t_{0}\\}$, which are fixed by $G$. The remaining
three orbits (let $a\coloneqq\frac{1}{2}$ and $b\coloneqq\frac{i}{2}$) are the
sets:

$\displaystyle\\{e,t_{0}\\}\times\\{a,b\\}$
$\displaystyle\\{a,b\\}\times\\{e,t_{0}\\}$ $\displaystyle\\{a,b\\}^{2}.$

This shows immediately that $\bar{A}^{5}=E[2]^{2}/G$ consists of the 7 orbits
above, and that the quotients

$\bar{A}^{6}=...=\bar{A}^{10}=\\{e,t_{0}\\}^{2}.$

We have $\bar{A}^{1}=\mathbb{P}^{2}$ by [ALQ18, Theorem 1.1]. It is left to
compute the quotients of the one dimensional loci. The computations are
similar for $i=2,3,4$. For example, we have

$\bar{A}^{3}=\\{(x,y)\in A\mid x=t_{0}y+E[2]\\}/\langle\alpha,\beta\rangle,$

and one sees that, for any $t\in E[2]$, we have

$\beta\begin{pmatrix}t_{0}y+t\\\ y\end{pmatrix}=\begin{pmatrix}t_{0}(iy)-it\\\
iy\end{pmatrix}$

In other words, $\beta$ acts by identifying two copies of $E$ (and by negation
on each of them) and by the order 4 automorphism on the other two copies:

$\bar{A}^{3}\simeq\mathbb{P}^{1}\cup\mathbb{P}^{1}\cup\mathbb{P}^{1}.$

Similarly, one sees $\bar{A}^{2}=\mathbb{P}^{1}\cup\mathbb{P}^{1}$ and
$\bar{A}^{4}=\mathbb{P}^{1}$.

###  4.3. The motivic semiorthogonal decomposition

We construct Fourier-Mukai functors as in Section 3.2, by taking the structure
sheaves of (equivariantized) graphs of the inclusions of the fixed loci in
$A^{i}\times A$. We denote by
$\Phi_{i}\colon\mathcal{D}(\bar{A}^{i})\to\mathcal{D}[A/G]$ the corresponding
Fourier-Mukai functors. The fixed loci $A^{i}$ with $i\geq 6$ only contain
points that are fixed by $G$. The action of $G$ locally at each of these
points coincides with that described in Section 3, so we can modify the
$\Phi_{i}$ by twisting

$\displaystyle\Phi^{\prime}_{6}$ $\displaystyle=\Phi_{6}(-)\otimes M$
$\displaystyle\Phi^{\prime}_{9}$ $\displaystyle=\Phi_{9}(-)\otimes V$
$\displaystyle\Phi^{\prime}_{7}$
$\displaystyle=\Phi_{7}(-)\otimes\chi_{2}\chi_{3}$
$\displaystyle\Phi^{\prime}_{10}$ $\displaystyle=\Phi_{10}(-),$
$\displaystyle\Phi^{\prime}_{8}$
$\displaystyle=\Phi_{8}(-)\otimes\chi_{2}\chi_{3}\chi_{4}$

where $M$ is the torsion sheaf supported at $\\{e,t_{0}\\}^{2}$ whose fiber at
every point is locally isomorphic to the module
$\mathbb{C}[x,y]/(x^{2}+y^{2},x^{2}-y^{2})$ (see Section 3.3).

We then modify $\Phi_{5}$ by only twisting by a character on points with
stabilizer $G$, and defining for $X\in\mathcal{D}(\bar{A}^{5})$:

$\Phi_{5}^{\prime}(X)\coloneqq\begin{cases}\Phi_{5}(X)\otimes\chi_{2}\ \
&\mbox{ if }\ \ \mathrm{Supp}(X)\subset\\{e,t_{0}\\}^{2}\\\ \Phi_{5}(X)\ \
&\mbox{ if }\ \
\mathrm{Supp}(X)\subset\bar{A}^{5}\setminus\\{e,t_{0}\\}^{2}\end{cases}$

and extending additively to $\mathcal{D}(\bar{A}^{5})$. Then we have:

######  Theorem 5.

In the case of a smooth quotient $A/G$ of type (C), the functors
$\Phi_{1},...,\Phi_{4},\Phi_{5}^{\prime},...,\Phi_{10}^{\prime}$ give rise to
a motivic semiorthogonal decomposition

(7)
$\mathcal{D}[A/G]=\left\langle\mathcal{D}(\bar{A}^{1}),...,\mathcal{D}(\bar{A}^{10})\right\rangle.$

###### Proof of Theorem 5.

Everything can be checked locally around each point $p\in A$. If the
stabilizer of $p$ is not $\mu_{2}\times\mu_{2}$, (7) restricts to the
semiorthogonal decomposition (6), and the statement follows from Theorem 4.

On the other hand, suppose $p$ has a $\mu_{2}\times\mu_{2}$ stabilizer: the
local model around $p$ is $\mathbb{A}^{2}$ where each copy of $\mu_{2}$ acts
by negating a coordinate, and it has a motivic semiorthogonal decomposition

(8)
$\mathcal{D}[\mathbb{A}^{2}/\mu_{2}\times\mu_{2}]=\langle\mathcal{D}(\bar{\mathbb{A}^{2}}),\mathcal{D}(\bar{\mathbb{A}^{1}}),\mathcal{D}(\bar{\mathbb{A}^{1}}),\mathcal{D}(\mathrm{pt})\rangle.$

When restricted to $p$, the categories in (7) yield the semiorthogonal
decomposition (8). For example, let $p=(e,a)$. We have

$p\in A^{1}\cap A^{2}\cap\alpha_{*}A^{2}\cap A^{5}.$

Then we have local isomorphisms

$\displaystyle\Phi_{1}(\mathcal{O}_{p})$
$\displaystyle\simeq\mathbb{C}[u,v]/(u^{2},v^{2})$
$\displaystyle\Phi_{2}(\mathcal{O}_{p})$
$\displaystyle\simeq\mathbb{C}[u,v]/(u^{2},uv,v^{2})$
$\displaystyle\Phi_{5}(\mathcal{O}_{p})$
$\displaystyle\simeq\mathbb{C}[u,v]/(u,v),$

which satisfy the correct orthogonalities by (8) and generate
$\mathcal{D}[A/G]$ at $p$. ∎

######  Corollary 3.

The category $\mathcal{D}[A/G]$ admits a full exceptional collection.

###### Proof.

In fact, each of the pieces $\mathcal{D}(\bar{A}^{i})$ appearing in (7) admits
a full exceptional collection: the $\bar{A}^{i}$ are computed in Section 4.2
and are unions of projective spaces and points. ∎

##  5\. Motivic semiorthogonal decompositions for $\mathcal{D}[A/G]$

We will break the proof up into three parts. In §5.1, we prove the case of
zero-dimensional fixed loci where $G$ acts by group automorphism. In §5.2, we
prove the case of positive dimensional fixed loci where $G$ acts by group
automorphism. In §5.3, we use the results of §5.1 and §5.2 to prove the case
where $G$ is an arbitrary finite group of automorphisms, i.e. not necessarily
fixing the identity.

###  5.1. Zero dimensional fixed loci

Suppose $G$ acts on $A$ by group automorphisms. We assume $G$ acts irreducibly
on $T_{e}(A)$ and hence $\dim(A^{G})=0$.

######  Theorem 6.

Suppose $G$ is a finite group of automorphisms of an Abelian variety $A$ that
fix the identiy. If $G$ acts irreducibly on $T_{e}(A)$, then
$\mathcal{D}[A/G]$ has a motivic semiorthogonal decomposition.

###### Proof.

The case $G\cong C^{n}\rtimes S_{n}$ with $C\neq 1$ has already been done in
[PVdB19].

We now consider the case $G\cong S_{n+1}$ and $A\subset X=E^{n+1}$ is the
divisor given by $x_{1}+\cdots+x_{n+1}=0$. We apply [LP17, Thm 1.2.1]. Let
$W(\lambda)=C(\lambda)/(\lambda)$. Then $W(\lambda)$ acts on $X_{\lambda}$ and
we set $X_{\lambda}^{fr}$ to be the free locus. Then we need to show that
$A\cap X_{\lambda}^{fr}$ is dense in $A\cap X_{\lambda}$.

To that end, write $\lambda=(1^{r_{1}})(2^{r_{2}})\cdots(k^{r_{k}})$. Then we
can identify $W_{\lambda}\cong S_{r_{1}}\times S_{r_{2}}\times\cdots\times
S_{r_{k}}$. We can then identify

$X_{\lambda}\cong E^{r_{1}}\times E^{r_{2}}\times\cdots E^{r_{k}}$

then we can identify $A\cap X_{\lambda}$ as the set of points $e\in X$ such
that

$e_{1}+\cdots+e_{r_{1}}+2e_{r_{1}+1}+\cdots+2e_{r_{1}+r_{2}}+\cdots+ke_{r_{1}+\cdots+r_{k-1}+1}+\cdots+ke_{r_{1}+\cdots+r_{k}}=0.$

Consequently $A\cap X_{\lambda}^{fr}$ is the subset of $A\cap X_{\lambda}$
such that

$e_{1}\neq\cdots\neq e_{r_{1}},e_{r_{1}+1}\neq\cdots\neq
e_{r_{1}+r_{2}},\ldots,e_{r_{1}+\cdots+r_{k-1}+1}\neq\cdots\neq
e_{r_{1}+\cdots+r_{k}}.$

Solving for $e_{r_{1}+\cdots+r_{k}}$ we see that this is the open set

$\\{e\in X_{\lambda}\mid e_{r_{1}+\cdots+r_{k}}\in-e_{1}-\cdots-
e_{r_{1}}-2e_{r_{1}+1}-\cdots-2e_{r_{1}+r_{2}}-\cdots-
ke_{r_{1}+\cdots+r_{k}-1}+E[k]\\}$

where $E[k]$ is the $k$-torsion of $E$. Since any non-free point can be moved
while staying in this set, it is open on each connected component of $A\cap
X_{\lambda}^{fr}$. Thus it is dense. We conclude the MSOD conjecture holds for
$[A/S_{n+1}]$.

Lastly, if $A$ is a surface of Type (C), this is Theorem 5. ∎

###  5.2. Positive dimensional fixed loci

Suppose now that $\dim(A^{G})>0$. Let $A_{0}\subset A^{G}$ be the connected
component containing the identity and $P_{G}$ a complementary Prym subvariety.
Set $\Delta=P_{G}\cap A_{0}$, then there is a short exact sequence of Abelian
varieties

$1\to\Delta\to P_{G}\times A_{0}\to A\to 1$

and an isomorphism

$P_{G}\times A_{0}/\Delta\cong A$

where $\Delta$ acts freely on $A_{0}$. Set $\tilde{A}=P_{G}\times A_{0}$ for
convenience. Note, the $\Delta$-equivariant derived category of $\tilde{A}$ is
equivalent to the derived category of $A$:

$\mathcal{D}(\tilde{A})^{\Delta}\cong\mathcal{D}(A)$

via the pushforward map

Moreover, since $\Delta\subset A^{G}$, we see that $\Delta$ and $G$ commute
with eachother inside $\mathrm{Aut}(P_{G}\times A_{0})$. So we can consider

$[\tilde{A}/G\times\Delta]$

and the corresponding derived equivalence:

$\mathcal{D}[\tilde{A}/G]^{\Delta}\cong\mathcal{D}[\tilde{A}/G\times\Delta]\cong\mathcal{D}[A/G].$

Note, the action is diagonal as $\Delta$ acts on both $P_{G}$ and $A_{0}$. The
quotient stack $[\tilde{A}/G]$ possesses a motivic semiorthogonal
decomposition. Consider the functors defining it:

$\Phi_{\lambda}\colon\mathcal{D}(P_{G}^{\lambda}/C_{G}(\lambda)\times
A_{0})\hookrightarrow\mathcal{D}[\tilde{A}/G].$

The functor $\Phi_{\lambda}$ corresponds to a $G$-equivariant kernel
$\mathcal{K}_{\lambda}$ on $\tilde{A}^{\lambda}\times\tilde{A}$, where $G$
acts on the right.

######  Theorem 7.

For each $\lambda\in G/\sim$, the functors $\Phi_{\lambda}$ are fully-faithful
and there exists a total order on $G/\sim$ such that the functors give a
motivic semiorthogonal decomposition of $\mathcal{D}[A/G]$

###### Proof.

We show the functors defining motivic semiorthogonal decompositions for
$\mathcal{D}[\tilde{A}/G]$ descend to a semiorthogonal decomposition of
$\Delta$-equivariant derived categories by showing that they are invariant
under the action of $\Delta$. To that end, notice that there are isomorphisms

$\tilde{A}^{\lambda}/C_{G}(\lambda)\times\Delta\cong A/C_{G}(\lambda)$

which give derived equivalences

$\mathcal{D}(\tilde{A}^{\lambda}/C_{G}(\lambda))^{\Delta}\cong\mathcal{D}(A/C_{G}(\lambda))$

and

$\mathcal{D}[\tilde{A}/G]^{\Delta}\cong\mathcal{D}[A/G].$

Since $A/G$ is smooth, so is $P_{G}/G$. If $G$ does not act irreducible on
$T_{e}(P_{G})$, then as before the motivic decompositions will decompose as
well. So we can assume that $[P_{G}/G]$ is of types A, B, or C.

In each of the irreducible cases, the kernels defining Fourier-Mukai functors
are supported on:

$Z_{\lambda}=\bigcup_{g\in G}g\cdot\Gamma$

with its reduced structure. In particular, the image of the Fourier-Mukai
functor $\Phi_{\lambda}(\mathcal{D}(\bar{A}^{\lambda}))$ is supported on

$\cup_{\mu\sim\lambda}A^{\mu}.$

Since the action of $\Delta$ preserves the support, i.e.
$\mu(\delta(x))=\delta(x)$ for all $x\in X^{\mu}$ and $\delta\in\Delta$, each
of the images of $\Phi_{\lambda}(\mathcal{D}(\bar{A}^{\lambda}))$ is
$\Delta$-invariant. Thus the motivic semiorthogonal decompositions we have
constructed for $\mathcal{D}[\tilde{A}/G]$ descend to a semiorthogonal
decomposition for $\mathcal{D}[A/G]$, [AE21, Corollary 3.18]. Finally, the
$\Delta$-equivariant derived category of
$\mathcal{D}(P_{G}^{\lambda}/C_{G}(\lambda)\times A_{0})$ is equivalent to
$\mathcal{D}[A^{\lambda}/C_{G}(\lambda)]$ as
$[(P_{G}^{\lambda}/C_{G}(\lambda)\times A_{0})/\Delta]\cong
A^{\lambda}/C_{G}(\lambda)$. ∎

###  5.3. Translations

Suppose now that $G$ does not fix the origin. That is, $G$ possesses a normal
subgroup of translations. Let $T\subset G$ be the subgroup generated by
translations and $H$ be the subgroup that fixes the origin. Notice that $T$
acts freely and $G$ admits a semi-direct product decomposition:

$G=T\rtimes H.$

Let $A_{T}=A/T$, then the action of $H$ on $A$ descends to an action on
$A_{T}$. Given by $h(\bar{a})=\overline{h(a)}$ where $a$ is any preimage of
$\bar{a}$ under the quotient map $\pi\colon A\to A_{T}$. This is well-defined
as $T$ is a normal subgroup. Our goal will be to construct a motivic
semiorthogonal decomposition of $\mathcal{D}[A/G]$ by using the one we know
exists for $\mathcal{D}[A_{T}/H]$.

In fact, we can prove a more general statement. Suppose a finite group
$G=K\rtimes H$ acts effectively on a smooth quasi-projective variety $X$ with
$K$ acting freely so that $X_{K}=X/K$ is also a smooth quasi-projective
variety. Then $H$ acts on $X_{K}$.

######  Lemma 4.

Pick conjugacy class representatives $(k_{1},h),\ldots,(k_{r},h)$ so that
$(k,h)$ is conjugate to one of $(k_{i},h)$. Then there is an isomorphism of
varieties

$\coprod_{i=1}^{r}X^{(k_{i},h)}/C_{G}((k_{i},h))\to X_{K}^{h}/C_{H}(h).$

###### Proof.

Since all varieties involved are quasi-projective and normal, it is enough to
check that it is a bijection on closed points. For surjectivity, we take
$\bar{x}\in X_{K}^{h}$ and show there exists $k\in K,h\in H$ and $x\in X$ so
that $(k,h)(x)=x$ and $\bar{x}$ is conjugate to the image of $x$ under the
quotient $X\to X_{K}$. Pick any lift $x$ of $\bar{x}$. Since
$h(\bar{x})=\bar{x}$, $h(x)$ is in the same orbit as $x$ under the $K$ action.
That is, there exists $k\in K$ so that $k(h(x))=x$. There exists
$(k^{\prime},h^{\prime})$ so that conjugating $(k,h)$ by
$(k^{\prime},h^{\prime})$ is one of the $(k_{i},h)$ classes. It follows that
$x$ is conjugate to the image of $k_{i}(h(x))$ under the action of $H$. This
proves surjectivity.

Since the action of $K$ is free, if $k\neq k^{\prime}$, then $X^{(k,h)}\cap
X^{(k^{\prime},h)}=\emptyset$. Indeed, any $x$ in the intersection would force
$k^{-1}k^{\prime}h(x)=h(x)$ and hence $k^{-1}k^{\prime}$ stabilizes a point.
But only the identity stabilizes a point. Hence, we have $X^{(k_{i},h)}\cap
X^{(k_{j},h)}=\emptyset$ for $i\neq j$.

Now if $x_{i}\in X^{(k_{i},h)}$ and $x_{j}\in X^{(k_{j},h)}$, then the images
of $x_{i}$ and $x_{j}$ are not identified by an element in the centralizer of
$C_{H}(h)$. Indeed, suppose they were, i.e. that there exists $h^{\prime}\in
H$ with $h^{\prime}h=h^{\prime}h$ and $h^{\prime}(\bar{x}_{i})=\bar{x}_{j}$.
Then $h^{\prime}(x_{i})\in X^{(k_{j},h)}$. So

$x_{i}=(1,h^{\prime})^{-1}(k_{j},h)(h^{\prime}(x_{i}))=(k_{j}^{h^{\prime}},h)(x_{i}).$

Hence, $(k_{j},h)$ is conjugate to $(k_{i},h)$ but this cannot happen if
$i\neq j$. It follows that disjoint components have disjoint images.

Now if $x,y\in X^{(k_{i},h)}$ are such that $\bar{x}=\bar{y}$. Then, since $K$
acts freely, there exists a unique $k\in K$ such that $k(x)=y$. We need to see
that $(k,1)$ centralizes $(k_{i},h)$. Conjugation of $(k_{i},h)$ by $(k,1)$
fixes $x$:

$(k_{i},h)^{(k,1)}(x)=(k,1)^{-1}(k_{i},h)(k,1)(x)=(k,1)^{-1}(h_{i},k)(y)=(k,1)^{-1}(y)=x.$

We conclude that the fixed locus of $(k_{i},h)^{(k,1)}$ and $(k_{i},h)$
intersect. But this only happens if $(k_{i},h)^{(k,1)}=(k_{i},h)$ and hence
$(k,1)$ centralizes $(k_{i},h)$. ∎

The following is immediate.

######  Corollary 4.

Let $(k_{1},h),\ldots,(k_{r},h)$ be as in Lemma 4. Then

$\mathcal{D}(X_{K}^{h}/C_{H}(h))\cong\bigoplus_{i=1}^{r}\mathcal{D}(X^{(k_{i},h)}/C_{G}(k_{i},h)).$

######  Lemma 5.

There is an equivalence of categories:

$\mathfrak{coh}(X)^{G}\cong\mathfrak{coh}(X_{K})^{H}$

induced by the quotient map $\pi\colon X\to X_{K}$. This functor can also be
described by taking $T$-invariants. In particular, this equivalence of Abelian
categories induces a derived equivalence:

$\mathcal{D}[X/G]\cong\mathcal{D}[X_{K}/H].$

###### Proof.

This is straightforward. If $\mathcal{F}$ is a sheaf on $X$, then a
$G$-equivariant structure is the data of a $K$-equivariant structure with a
compatible $H$-equivariant structure. The data of a $K$-equivariant structure
is the data of a pullback of a sheaf $\bar{\mathcal{F}}$ on $X_{K}$. The data
of a compatible $H$-equivariant structure is the data of an $H$-equivariant
structure on $\bar{\mathcal{F}}$. The details are left to the reader. ∎

######  Theorem 8.

Suppose $G=K\rtimes H$ acts on a smooth quasi-projective variety $X$ so that
$K$ acts freely. Consider the induces action of $H$ on the quotient
$X_{K}=X/H$ and suppose $\mathcal{D}[X_{K}/H]$ admits a motivic semiorthogonal
decomposition of the form

$\mathcal{D}[X_{K}/H]=\langle\mathcal{D}(\bar{X}_{K}^{h_{1}}),\ldots,\mathcal{D}(\bar{X}_{K}^{h_{t}})\rangle.$

Then $\mathcal{D}[X/G]$ admits a motivic semiorthogonal decomposition.

###### Proof.

The equivalence of Lemma 5 gives a derived equivalence

$\mathcal{D}[X/G]\cong\mathcal{D}[X_{K}/H].$

The motivic semiorthogonal decomposition of $\mathcal{D}[X_{K}/H]$ directly
induces one of $\mathcal{D}[X/G]$. The pieces $\mathcal{D}(X^{h}/C_{H}(h))$ of
the motivic semiorthogonal decomposition of $\mathcal{D}[X_{K}/H]$ decompose
into $\mathcal{D}(X^{(k_{i},h)}/C_{G}(k_{i},h))$. Since these are pairwise
completely orthogonal, any total order suffices. Thus we get a semiorthogonal
decomposition

$\mathcal{D}[X/G]=\langle\bigoplus_{i=1}^{r_{1}}\mathcal{D}(X^{(k_{i},h_{1})}/C_{G}((k_{i},h_{1}))),\ldots,\bigoplus_{i=1}^{r_{t}}\mathcal{D}(X^{(k_{i},h_{t})}/C_{G}((k_{i},h_{t}))\rangle.$

Finally $\mathcal{D}(X/G)$ linearity follows from
$\mathcal{D}(X_{K}/H)$-linearity using the canonical isomorphism $X/G\cong
X_{K}/H$. ∎

Combining Theorem 8 with our earlier work gives motivic semiorthogonal
decompositions for all Abelian varieties with smooth quotients.

######  Theorem 9.

Let $G=T\rtimes H$ be a finite group of automorphisms of an Abelian variety
$A$ such that $A/G$ is smooth. Then there exists a motivic semiorthogonal
decomposition for $\mathcal{D}[A/G]$.

##  6\. Examples

###  6.1. Curves

Quotients of an elliptic curve $E$ by a finite group of group automorphisms
are weighted projective lines in the sense of Geigle and Lenzing [GL87]. The
group $G$ is $\mu_{n}$ for $n=2,3,4$, or $6$. All quotients have
$\mathbb{P}^{1}$ as a coarse space. There are 4 $\mu_{2}$ stabilizers for the
generic case $n=2$ and 3 stabilizers for the other cases, of orders $(3,3,3)$,
$(4,4,2)$, and $(6,3,2)$ respectively. From the point of view of
representation theory, the categories $\mathfrak{coh}([E/G])$ are the module
categories over a canonical algebra in the sense of Ringel [Rin90]. In this
context, the motivic semiorthogonal decompositions of Theorem 1 recover the
exceptional collections constructed in [Mel95].

###  6.2. Surfaces

In this section, we explicitly compute the motivic semiorthogonal
decompositions constructed here and in [PVdB19] when $A$ is a surface and the
action of $G$ fixes $e$ and acts irreducibly on $T_{e}(A)$. Such an action is
one of three types by Theorem 3. We have already written out Type (C) in
Section 4. We begin with Type (A):

######  Example 2.

Let $A\cong E^{2}$ and $G=\mu_{n}^{2}\rtimes S_{2}$ for $n=1,2,3,4,6$. The
generic case, $n=1$, has coarse quotient $\mathrm{Sym}^{2}(E)$ and the non-
trivial element fixes the diagonal. Hence,

$\mathcal{D}[A/G]=\langle\mathcal{D}(\mathrm{Sym}^{2}(E)),\mathcal{D}(E)\rangle.$

In all other cases, $[A/G]$ admits a full exceptional collection. The coarse
quotient is $\mathrm{Sym}^{2}(\mathbb{P}^{1})\simeq\mathbb{P}^{2}$.

If $n=2$, there are 5 conjugacy classes, whose fixed loci and centralizers are
as follows:

Representative $g$ | $A^{g}$ | $C(g)$ | $\bar{A}^{g}$
---|---|---|---
$(1,1,1)$ | $A$ | $G$ | $\mathbb{P}^{2}$
$(-1,-1,1)$ | $E[2]\times E[2]$ | $G$ | $\mathrm{pt}$
$(-1,1,1)$ | $E[2]\times E$ | $\langle(-1,1,1),(-1,-1,1)\rangle$ | $E[2]\times\mathbb{P}^{1}$
$(1,1,\sigma)$ | $E$ | $\langle(1,1,\sigma),(-1,-1,1)\rangle$ | $\mathbb{P}^{1}$
$(1,-1,\sigma)$ | $E[2]$ | $\langle(1,-1,\sigma)\rangle$ | $E[2]$

Thus we have

$\mathcal{D}[A/G]=\langle\mathcal{D}(\mathbb{P}^{2}),\underbrace{\mathcal{D}(\mathbb{P}^{1}),...,\mathcal{D}(\mathbb{P}^{1})}_{5\mbox{
copies}},\underbrace{\mathcal{D}(\mathrm{pt}),...,\mathcal{D}(\mathrm{pt})}_{8\mbox{
copies}}\rangle.$

A similar computation (omitted) computes the MSOD in the other cases. For
$n=3$ there are 9 conjugacy classes. The corresponding MSOD has one two-
dimensional component, 3 one-dimensional components, and 5 zero-dimensional
components:

$\mathcal{D}[A/G]=\langle\mathcal{D}(\mathbb{P}^{2}),\underbrace{\mathcal{D}(\mathbb{P}^{1}),...,\mathcal{D}(\mathbb{P}^{1})}_{7\mbox{
copies}},\underbrace{\mathcal{D}(\mathrm{pt}),...,\mathcal{D}(\mathrm{pt})}_{21\mbox{
copies}}\rangle.$

For $n=4$ there are 14 conjugacy classes. The corresponding MSOD has one two-
dimensional component, 4 one-dimensional components, and 9 zero-dimensional
components:

$\mathcal{D}[A/G]=\langle\mathcal{D}(\mathbb{P}^{2}),\underbrace{\mathcal{D}(\mathbb{P}^{1}),...,\mathcal{D}(\mathbb{P}^{1})}_{6\mbox{
copies}},\underbrace{\mathcal{D}(\mathrm{pt}),...,\mathcal{D}(\mathrm{pt})}_{33\mbox{
copies}}\rangle.$

For $n=6$ there are 27 conjugacy classes. The corresponding MSOD has one two-
dimensional component, 6 one-dimensional components, and 20 zero-dimensional
components:

$\mathcal{D}[A/G]=\langle\mathcal{D}(\mathbb{P}^{2}),\underbrace{\mathcal{D}(\mathbb{P}^{1}),...,\mathcal{D}(\mathbb{P}^{1})}_{\mbox{7
copies}},\underbrace{\mathcal{D}(\mathrm{pt}),...,\mathcal{D}(\mathrm{pt})}_{\mbox{26
copies}}\rangle.$

For type (B):

######  Example 3.

Consider $[E^{2}/S_{3}]$. The coarse quotient is $\mathbb{P}^{2}$. The group
$S_{3}$ acts on $T_{e}(E^{2})$ by

$(1,2)=\begin{pmatrix}0&1\\\ 1&0\\\ \end{pmatrix},\
(1,2,3)=\begin{pmatrix}0&-1\\\ 1&-1\end{pmatrix}.$

The conjugacy class corresponding to $(1,2)$ fixes the diagonal $\Delta\subset
E\times E$. The conjugacy class corresponding to $(1,2,3)$ fixes the diagonal
copy of the three torsion subgroup $E[3]$. The centralizers act trivially on
the fixed loci. By Theorem 6, there is a motivic semiorthogonal decomposition
of the form

$\mathcal{D}[E^{2}/S_{3}]=\langle\mathcal{D}(\mathbb{P}^{2}),\mathcal{D}(E),\mathcal{D}(E[3])\rangle.$

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