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Paul and Akimi Kotta founded Mellow Monk's Green Teas in Livermore, California, in 2003. The green tea they sell comes from the rural Japanese town of Aso, where Akimi was born and raised. They lived in Aso for seven years; their daughter was born there.
All of Mellow Monk's growers are certified eco friendly under the Japanese government's Eco Farmer Program.
Mellow Monk® is a certified member of Green America's Green Business Network.
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text:
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redpajama_set_name: RedPajamaCommonCrawl


Below is a list of Girls Birthday Party Ideas.
See my other page All Kids Parties A-Z for a list of party ideas arranged by theme or age group .
A lot of party themes really are unisex these days. But it's a fact of life that some themes are going to be more attractive to girls and others preferred by boys. I realize it's hard not to think about kids party themes in terms of genders and traditional sterotypes, but I would suggest you try not to. It's old-fashioned and your party will be better for a more modern approach.
However the chances are that there probably aren't that many girls who are going to be interested in a Roman Gladiator party theme for example.
So below is a list of party themes that might be particularly suited to girls. That doesn't mean some of these themes aren't just a suitable for boys too. Likewise some themes listed on the Boys Birthday Party Ideas page might be just as suitable for girls. So I would suggest that you also check out the All Kids Parties A-Z page which contains themes for both sexes.
If you have some good girls birthday party ideas (or you've tried one of my ideas from the list above) share details with other party-ideas-by-a-pro.com visitors.
Use the form below to submit your girls birthday party ideas.


text:
Morning View är ett album av bandet Incubus som släpptes den 23 oktober 2001.

Låtlista
"Nice to Know You" – 4:43
"Circles" – 4:09
"Wish You Were Here" – 3:35
"Just a Phase" – 5:31
"11am" – 4:14
"Blood on the Ground" – 4:35
"Mexico" – 4:21
"Warning" – 4:41
"Echo" – 3:34
"Have You Ever" – 3:15
"Are You In?" – 4:24
"Under My Umbrella" – 3:32
"Aqueous Transmission" – 7:47

Musikalbum 2001
Album av Incubus

redpajama_set_name: RedPajamaWikipedia


Apartment type 4/6 Pax. Composed of two bedrooms, two bathrooms, living room with sofa bed for two people, kitchen. Sea view balcony. Central heating. Satellite TV. Safe. Wi-fi available on payment.
Apartment type 5 Pax. Composed of two bedrooms, one bathroom, separate kitchen, living room with sofa bed. Close to the beach, parking and main shopping street. WiFi on payment.
Apartment type 2/3 Pax. Composed of bathroom, kitchen, living room with sofa bed for two people.
Apartment type 2 Pax. Composed of bathroom, kitchen, living room with sofa bed for two people.
Apartment type 2/4 pax, in the centre of Lloret, near discos and pubs. Composed by a livingroom with american kitchen, 1 bedroom and bathroom.


As for the second piece, written by a contemporary composer, Lindemann calls it spectacular.
"The Rising Sun" has Japanese motifs, and while a new piece, is very accessible, Lindemann said.
As a child, Lindemann wanted to be a drummer. When he joined the school band at age 12, he was forced to either play trumpet or clarinet because at his school, only the best musicians could play drums. His audition to play the drums didn't go well.
Lindemann said it's always good to approach playing any instrument with a classical background.
Since he quickly learned the trumpet and became very good at it, some may think he was a prodigy.
So he went to the Juilliard School in New York City.
Lindemann committed to being a trumpet soloist when he was 17.
He competed in 40 solo competitions in eight years.
What he learned was to find his own voice.
That's when Lindemann's solo career began to take off. Now, he's considered to be one of the top performers of his generation.


text:
Mark Andrew prides himself on ability to bring people together
Our first profile in a series on the mayoral candidates
Sarah McKenzie / [email protected]
Updated: April 11, 2013 - 2:51 pm
Mayoral candidate Mark Andrew at a recent campaign event. Credit: By Kristin Lebben
Editor's note: The Journals will be profiling the self-declared candidates for mayor for the next several issues leading up to the DFL City Convention on June 15. We are kicking off our candidate profile series with a story on Mark Andrew. Stay tuned for future profiles on Jackie Cherryhomes, Don Samuels, Gary Schiff, Betsy Hodges, Jim Thomas and Cam Winton, an Independent candidate not seeking the DFL endorsement.
Mark Andrew believes he has a natural gift that is essential for a modern mayor.
"The single most important quality that I have is also the most important quality that a mayor has to have — I'm a natural uniter. I'm a collaborator. It's my gift," he said during a recent interview.
The next mayor of Minneapolis will have to be adept at bringing people together and working with different layers of government, whether it's the county, School Board or Park Board, he said.
Andrew, a resident of the Lynnhurst neighborhood, has a diverse background in public service and the private sector. Most recently he's been an environmental entrepreneur with GreenMark, a company he founded in 2007. GreenMark helps buildings and public venues become more environmentally sustainable.
One of Andrew's most significant accomplishments with GreenMark has been helping Target Field secure LEED certification and create a partnership with Pentair to install a rainwater recycling system at the ballpark.
Twins President Dave St. Peter had high praise for Andrew for his work on Target Field.
"He played a significant role in helping the Minnesota Twins understand the opportunities and the responsibilities that go with environmental stewardship, and I was personally very appreciative of having Mark's knowledge and talents and his vision that played an important role in making Target Field what we think is the greenest sports facility of its type in the country," he said.
Before launching GreenMark, Andrew was a senior vice president at Tunheim Partners, a strategic communications firm — a job that followed a lengthy tenure on the Hennepin County Board. He served as a county commissioner for 16 years — from 1982-1999 — and was elected chair of the board four times.
As for his greatest accomplishments while a county commissioner, he points to his work launching the county's recycling program, which was the biggest local recycling program in North America for over a decade. He also takes credit for spearheading the Midtown Greenway Corridor and playing a key role in negotiations for the Hiawatha Light Rail Transit and Central Corridor transit lines.
Andrew is a lifelong resident of Minneapolis and a graduate of Washburn High School and the University of Minnesota. He got his start in politics while in college when he become the co-founder and first president of MPIRG, a student-run environmental group.
Anita Tabb, a commissioner for the Park Board's 4th District and Andrew's campaign co-chair, agreed with the mayoral candidate that one of his best qualities is his ability to bring people together — an asset for a mayor working within Minneapolis' system of government.
"I love the fact that he has such a robust and diverse background — private sector, public sector. He's an entrepreneur," she said. "I love that Mark has a vision for the whole city. We can't be North Minneapolis versus South Minneapolis, or the east versus the west. We want to come together and be one whole city."
Andrew has three overarching goals for the city: make Minneapolis the greenest city in America; reduce the achievement gap between high performing and low performing students; and launch an aggressive campaign to accelerate development throughout the city.
As for the first goal, Andrew said he'd like to see the city make more investments in green space and other environmental amenities — infrastructure improvements that could attract new businesses to the city.
"We want to do this because it's smart business to be the greenest city in America," he said. "It will attract all kinds of positives."
Andrew's other top priority is reducing the achievement gap among Minneapolis students.
He said he has had education experts tell him the problem has gotten substantially worse in the last decade.
"Now we need to have the mayor be the convener and collaborator to attack that problem because the achievement gap — if it persists — will drive higher taxes in the city and in the future drive more government services," he said. "It's a prescription for disaster."
He said he would commission a blue ribbon task force made up of a wide group of community leaders to come up with a vision for tackling the achievement gap.
"It has to be the highest priority for the city," he said at mayoral candidate forum April 3 at Solomon's Porch in the Kingfield neighborhood.
Alberto Monserrate, chair of the Minneapolis School Board, is a supporter of Andrew's campaign.
"There are a number of good candidates this year for mayor, but I believe Andrew will be the best at forming the types of collaborative agreements between the city, school district, Park Board and county needed to eliminate the learning gap in Minneapolis," he said.
To accomplish the third goal, Andrew wants to see more high-density development in the city, particularly in North Minneapolis, along the riverfront and in the city's commercial nodes.
"I want to do somewhat higher density development as a way to more quickly drive the tax base and create jobs," he said. "It will generate more taxes and drive more employment, which is a critical need."
He cited statistics indicating Minneapolis has 28,000 fewer jobs today than in 2001. (Note: the Southwest Journal fact checked that figure and found it's based on an outdated report.)
Improving conditions in North Minneapolis to make it more attractive for development would be a special focus area. He'd also work to recruit new companies to the area.
"For every nickel of tax base we create on the North Side, we take pressure off the tax base on the South Side," he said.
Vikings stadium
As for the new Vikings stadium, Andrew said "any mayoral candidate who voted against the stadium was irresponsible."
"I will not preside over a hole in the ground. That will not happen on my watch," he said. "However, the funding mechanism for the stadium was fatally flawed from the start, and I was not a Johnny-come-lately on that issue. I've been critical of the financing plan throughout the legislative process."
Andrew said he is opposed to using gambling revenue to finance public projects.
Still, despite his reservations about the financial framework for the Vikings stadium, he said: "You can't be for jobs and against the stadium."
When asked to weigh in on what should be done about property taxes — a perennial hot topic in Minneapolis — Andrew said he has both short-and long-term strategies to keep taxes down.
If elected, Andrew said in the short term, he'd have his team do a thorough review of city programs. "My guess is that we're going to be able to find some regulatory functions that are redundant or unnecessary. … All of those savings are going to reduce property taxes."
The long-term strategy involves Andrew's economic development plans that call for more green amenities to attract more private investment in the city, which in turn would strengthen the tax base.
Andrew said his property taxes have doubled in recent years.
"They were already high, and here is the troubling realty: There is no game plan other than to let property taxes continue to rise," he said. "That's unacceptable."
At a glance: Mark Andrew
Neighborhood: Lynnhurst
Family: wife Connie and children Samantha and John
Resume snapshot: Founder of GreenMark, an environmental consulting firm. Before launching the company he worked at Tunheim Partners, a post he took after serving 16 years on the Hennepin County Board.
Education: bachelor's of arts degree, University of Minnesota-Twin Cities
Fun fact: Andrew has owned the Real S'mores and World's Greatest French Fries stand at the Minnesota State Fair since he was in college.
Website: www.markforminneapolis.com
Twitter: @MarkForMpls
Facebook: www.facebook.com/MarkForMinneapolis
Southwest light rail protest to highlight loss of trees
Council approves Southwest Light Rail Transit design


text:
\section{Introduction}

Recall that an Enriques surfaces $S$ is a smooth projective surface which can be represented
as the quotient of a K3 surface by an involution acting without fixed points.
Consecutively the canonical class $K_S$ of $S$ is a 2-torsion, $2K_S = 0$.
More than 10 years ago S.Zube~\cite{Zu} made the following nice observation

\begin{theorem}[\cite{Zu}]\label{zube}
Let $F_i^+$, $F_i^-$, $i =1,\dots,10$, be the multiple fibers of the $10$ elliptic pencils on $S$.
Then the line bundles $\{\CO_S(-F_i^+)\}_{i=1}^{10}$ form a completely orthogonal exceptional collection in $\D^b(S)$.
\end{theorem}

Since then it was an intriguing question to describe the orthogonal subcategory
\begin{equation}\label{cas}
\CA_S = {}^\perp\langle \{ \CO_S(-F_i^+) \}_{i=1}^{10} \rangle.
\end{equation}
The answer to this question is still not known. Our goal is to give a partial
answer for {\sf nodal Enriques surfaces}. Recall that Enriques surface $S$ is called {\sf nodal}
(another name is {\sf Reye congruence}), if the image of $S$ with respect to a morphism given
by a linear system $\frac13\sum F_i^+$ (called {\sf Fano model of $S$}) is contained in a quadric.
The main result of the present paper is a description of the category $\CA_S$
for nodal Enriques surfaces in other terms. Actually, we show that $\CA_S$ can be represented as a semiorthogonal
component of the derived category of a very special quartic double solid.

Recall that a quartic double solid is a double covering of $\PP^3$ ramified in a quartic surface $D \subset \PP^3$.
The bridge between quartic double solids and nodal Enriques surfaces is provided by special quartic surfaces,
the so-called {\sf quartic symmetroids}.

A quartic symmetroid is a quartic surface in $\PP^3 = \PP(W)$ which is a discriminant
locus of a family of quadrics in another $\PP^3 = \PP(V)$ parameterized by $\PP(W)$,
such that the quadrics in the family have no common points in $\PP(V)$.
The quartic symmetroid $D \subset \PP(W)$ parameterizes degenerate quadrics
in the family. It has 10 singular points corresponding to the quadrics of corank 2.
It turns out that the resolution of singularities $D'$ of $D$ has a natural
involution $\iota$ without fixed points and the quotient $S := D'/\iota$ is a nodal
Enriques surface.

On the other hand, one can consider the double covering $X \to \PP(W)$ ramified in the symmetroid~$D$.
As it was shown by F.Cossec~\cite{Co} such double coverings are exactly the Artin--Mumford conic bundles~\cite{AM}.
The main result of the present paper is a representation of the subcategory $\CA_S \subset \D^b(S)$
as a semiorthogonal component in $\D^b(X^+)$, the derived category of a small resolution $X^+$ of singularities of $X$.
More precisely we show that there is a semiorthogonal decomposition
$$
\D^b(X^+) = \langle \CA_S, \CO_{X^+}(-h), \CO_{X^+} \rangle,
$$
where $h$ stands for the positive generator of the Picard group of $\PP(W)$.

There is a subtle point in the above result. Namely, in general a minimal resolution of singularities
$X^+ \to X$ is not algebraic, in fact it is a Moishezon variety. However, since each Moishezon variety
is an algebraic space, its derived category of coherent sheaves (which is by definition is the subcategory
of the unbounded derived category of $\CO$-modules with bounded and coherent cohomology) is well-defined.
Moreover, it has all the properties the usual derived categories of coherent sheaves have.

On the other hand, if one wants to avoid nonalgebraic varieties, one can consider the blowup $X'$ of $X$
which is again a resolution of singularities, not small. Then we conjecture that one can realize
the whole derived category $\D^b(S)$ of the Enriques surface $S$ inside $\D^b(X')$.
More precisely, we conjecture that there is a semiorthogonal decomposition
$$
\D^b(X') = \langle \D^b(S), \CO_{X'}(-h), \{ \CO_{X'}(-e_i) \}_{i=1}^{10}, \CO_{X'} \rangle,
$$
where $e_i$ are the classes of the exceptional divisors of the blowup $X' \to X$.
We are going to return to this conjecture in future.

The paper is organized as follows.
In Section~\ref{s-pre} we remind some notions and constructions, such as
semiorthogonal decompositions, mutation functors and Serre functors.
In Section~\ref{s-qds} we discuss the derived categories of quartic double solids.
In Section~\ref{s-symm} we recall the quartic symmetroids and state the main result of the paper.
In Section~\ref{s-proof} we give the proof of the main result.

{\bf Acknowledgement.}
The first author would like to thank A.Bondal for helpful discussions.
The second author would like to thank L.Katzarkov and D.Orlov for helpful discussions
and is very grateful to I.Dolgachev for sharing many interesting facts about Enriques surfaces.

\section{Preliminaries}\label{s-pre}

The base field is the field $\C$ of complex numbers.

If $X$ is a Moishezon variety we denote by $\D^b(X)$ the full subcategory
of the unbounded derived category of $\CO_X$-modules with bounded and coherent cohomology.
This category is triangulated. For any morphism $f:X \to Y$ of Moishezon varieties we denote
by $f_*:\D^b(X) \to \D^b(Y)$ and by $f^*:\D^b(Y) \to \D^b(X)$ the {\em derived}\/
push-forward and pull-back functors (in first case we need $f$ to be proper,
and in the second to have finite $\Tor$-dimension, these assumptions ensure the functors
to preserve both boundedness and coherence). Similarly, $\otimes$
stands for the {\em derived}\/ tensor product. If $F \in \D^b(X)$
the derived tensor product functor $F\otimes -:\D^b(X) \to \D^b(X)$ will be denoted by $\TT_F$.

For a proper morphism of finite $\Tor$-dimension $f:X \to Y$ we will also use the right adjoint $f^!$
of the push-forward functor, which is given by the formula
$$
f^!(F) \cong f^*(F)\otimes\omega_{X/Y}[\dim X - \dim Y],
$$
where $\omega_{X/Y}$ is the relative canonical line bundle.

\subsection{Semiorthogonal decompositions}

Let $\CT$ be a triangulated category.

\begin{definition}[\cite{BK,BO95}]
A {\sf semiorthogonal decomposition}\/ of a triangulated category $\CT$ is a sequence of
full triangulated subcategories $\CA_1,\dots,\CA_n$ in $\CT$ such that
$\Hom_{\CT}(\CA_i,\CA_j) = 0$ for $i > j$
and for every object $T \in \CT$ there exists a chain of morphisms
$0 = T_n \to T_{n-1} \to \dots \to T_1 \to T_0 = T$ such that
the cone of the morphism $T_k \to T_{k-1}$ is contained in $\CA_k$
for each $k=1,2,\dots,n$.
\end{definition}

We will write $\CT = \langle \CA_1,\CA_2,\dots,\CA_n \rangle$ for a semiorthogonal
decomposition of a triangulated category $\CT$ with components $\CA_1,\CA_2,\dots,\CA_n$.

An important property of a triangulated subcategory $\CA \subset \CT$ ensuring that
it can be extended to a semiorthogonal decomposition is admissibility.

\begin{definition}[\cite{BK,B}]
A full triangulated subcategory $\CA$ of a triangulated category $\CT$ is called
{\sf admissible}\/ if for the inclusion functor $i:\CA \to \CT$ there is
a right adjoint $i^!:\CT \to \CA$, and a left adjoint $i^*:\CT \to \CA$ functors.
\end{definition}

\begin{lemma}[\cite{BK,B}]\label{sos_sod}
$(i)$ If\/ $\CA_1,\dots,\CA_n$ is a semiorthogonal sequence
of admissible subcategories in a triangulated category $\CT$\/
{\rm(}i.e. $\Hom_{\CT}(\CA_i,\CA_j) = 0$ for $i > j${\rm)}\/ then
$$
\lan\CA_1,\dots,\CA_k,
{}^\perp\lan\CA_1,\dots,\CA_k\ran \cap \lan\CA_{k+1},\dots,\CA_n\ran{}^\perp,
\CA_{k+1},\dots,\CA_n\ran
$$
is a semiorthogonal decomposition.

$(ii)$ If\/ $\D^b(X) = \lan \CA_1,\CA_2,\dots,\CA_n \ran$ is a semiorthogonal decomposition
of the derived category of a smooth projective variety $X$ then each subcategory $\CA_i \subset \D^b(X)$
is admissible.
\end{lemma}

Actually the second part of the Lemma holds for any {\em saturated}\/ (see~\cite{BK}) triangulated category.

\begin{definition}[\cite{B}]
An object $F \in \CT$ is called {\em exceptional}\/ if $\Hom(F,F)=\C$
and $\Ext^p(F,F)=0$ for all $p\ne 0$. A collection of exceptional
objects $(F_1,\dots,F_m)$ is called {\em exceptional}\/ if
$\Ext^p(F_l,F_k)=0$ for all $l > k$ and all $p\in\ZZ$.
\end{definition}

Assume that $\CT$ is {\em $\Ext$-finite}\/ (which means that for any objects $G,G' \in \CT$
the graded vector space $\Ext^\bullet(G,G') := \oplus_{t\in\ZZ} \Hom(G,G'[t])$ is finite-dimensional).

\begin{lemma}[\cite{B}]\label{eo}
The subcategory $\lan F \ran$ of $\D^b(X)$ generated by an exceptional object $F$ is
admissible and is equivalent to the derived category of vector spaces $\D^b(\C)$.
\end{lemma}

As a consequence of~\ref{sos_sod} and of~\ref{eo} one obtains the following

\begin{corollary}[\cite{BO95}]\label{sodgen}
If $\CT$ is an $\Ext$-finite triangulated category then any exceptional collection $F_1,\dots,F_m$ in $\CT$
induces a semiorthogonal decomposition
$$
\CT = \langle \CA , F_1, \dots, F_m \rangle
$$
where $\CA = \langle F_1, \dots, F_m \rangle^\perp = \{F \in \CT\ |\ \text{$\Ext^\bullet(F_k,F) = 0$ for all $1 \le k \le m$}\}$.
\end{corollary}

\subsection{Serre functors}

\begin{definition}[\cite{BK},\cite{BO01}]
Let $\CT$ be a triangulated category.
A covariant additive functor $\SS:\CT\to\CT$ is a {\sf Serre functor}\/
if it is a category equivalence and for all objects $F,G\in\CT$ there are
given bi-functorial isomorphisms $\Hom(F,G)\to\Hom(G,\SS(F))^\vee$.
\end{definition}

If a Serre functor exists then it is unique up to a canonical
functorial isomorphism. If $X$ is a smooth projective variety
then $\SS(F):=F\otimes\omega_X[\dim X]$ is a Serre functor
in $\D^b(X)$.

\begin{lemma}[\cite{B}]\label{mut_funct}
If $\CT$ admits a Serre functor $\SS$ and $\CA\subset\CT$ is right admissible
then $\CA$ admits a Serre functor $\SS_\CA = i^!\circ\SS\circ i$,
where $i:\CA \to \CT$ is the inclusion functor.
\end{lemma}
\begin{proof}
If $A,A'\in\CA$ then
$\Hom_\CA(A,i^!\SS i A') \cong
\Hom_\CT(i A,\SS i A') \cong
\Hom_\CT(i A',i A)^\vee \cong
\Hom_\CA(A',A)^\vee$.
\end{proof}

\subsection{Mutations}

If a triangulated category $\CT$ has a semiorthogonal decomposition then
usually it has quite a lot of them. More precisely, there are two groups
acting on the set of semiorthogonal decompositions --- the group of
autoequivalences of $\CT$, and a certain braid group. The action of the braid
group is given by the so-called mutations.

Roughly speaking, the mutated decomposition is obtained by dropping one of the components
of the decomposition and then extending the obtained semiorthogonal collection
by inserting new component at some other place as in Lemma~\ref{sos_sod}.
More precisely, the basic two operations are defined as follows.

\begin{lemma}[\cite{B}]\label{mutfun}
Assume that $\CA \subset \CT$ is an admissible subcategory, so that we have
two semiorthogonal decompositions $\CT = \lan \CA^\perp, \CA \ran$
and $\CT = \lan \CA, {}^\perp\CA \ran$.
Then there are functors $\LL_\CA,\RR_\CA: \CT \to \CT$
vanishing on~$\CA$ and inducing mutually inverse equivalences
${}^\perp\CA \to \CA^\perp$ and $\CA^\perp \to {}^\perp\CA$ respectively.
\end{lemma}
\begin{proof}
Let $i:\CA \to \CT$ be the embedding functor. For any $F \in \CT$ we define
$$
\LL_\CA(F) = \Cone (i i^! F \to F),
\qquad
\RR_\CA(F) = \Cone (F \to i i^* F)[-1].
$$
Note that the cones in this triangles are functorial due to the semiorthogonality.
All the properties are verified directly.
\end{proof}

The functors $\LL_\CA$ and $\RR_\CA$ are known as the {\sf left}\/ and the {\sf right mutation functors}.

\begin{remark}
If $\CA$ is generated by an exceptional object $E$ we can use explicit formulas for
the adjoint functors $i^!$, $i^*$ of the embedding functor $i:\CA \to \CT$. Thus we obtain
the following distinguished triangles
\begin{equation}\label{excmut}
\Ext^\bullet(E,F)\otimes E \to F \to \LL_E(F),
\qquad
\RR_E(F) \to F \to \Ext^\bullet(F,E)^\vee\otimes E.
\end{equation}
\end{remark}

It is easy to deduce from Lemma~\ref{mutfun} the following

\begin{corollary}[\cite{B}]
Assume that $\CT = \lan \CA_1,\CA_2,\dots,\CA_n \ran$ is a semiorthogonal decomposition with
all components being admissible. Then for each $1 \le k \le n-1$ there is a semiorthogonal decomposition
$$
\CT = \lan \CA_1,\dots,\CA_{k-1},
\LL_{\CA_k}(\CA_{k+1}),
\CA_k,\CA_{k+2},\dots,\CA_n \ran
$$
and for each $2 \le k \le n$ there is a semiorthogonal decomposition
$$
\CT = \lan \CA_1,\dots,\CA_{k-2},\CA_k,
\RR_{\CA_k}(\CA_{k-1}),
\CA_{k+1},\dots,\CA_n \ran
$$
\end{corollary}

There are two cases when the action of the mutation functors is particularly simple.

\begin{lemma}\label{perpmut}
Assume that $\CT = \lan \CA_1,\CA_2,\dots,\CA_n \ran$ is a semiorthogonal decomposition with
all components being admissible. Assume also that the components $\CA_k$ and $\CA_{k+1}$
are completely orthogonal, i.e. $\Hom(\CA_k,\CA_{k+1}) = 0$ as well as $\Hom(\CA_{k+1},\CA_k) = 0$. Then
$$
\LL_{\CA_k}(\CA_{k+1}) = \CA_{k+1}
\qquad\text{and}\qquad
\RR_{\CA_{k+1}}(\CA_k) = \CA_k,
$$
so that both the left mutation of $\CA_{k+1}$ through $\CA_k$ and the right mutation of $\CA_k$ through $\CA_{k+1}$
boil down to just a permutation and
$$
\CT = \lan \CA_1,\dots,\CA_{k-1},\CA_{k+1},\CA_k,\CA_{k+2},\dots,\CA_n \ran
$$
is the resulting semiorthogonal decomposition of $\CT$.
\end{lemma}

\begin{lemma}\label{longmut}
Let $X$ be a smooth projective algebraic variety and $\CA \subset \D^b(X)$ an admissible subcategory.
Then
$$
\LL_{\CA^\perp}(\CA) = \CA\otimes\omega_X
\qquad\text{and}\qquad
\RR_{{}^\perp\CA}(\CA) = \CA\otimes\omega_X^{-1}.
$$
\end{lemma}

An analogue of this Lemma holds for any triangulated category which has a Serre functor (see~\cite{BK}).
In this case tensoring by the canonical class should be replaced by the action of the Serre functor.
Whenever such mutation is performed we will say that
the component $\CA$ of a semiorthogonal decomposition $\D^b(X) = \langle \CA^\perp,\CA \rangle$
is {\sf translated to the left}
(respectively the component $\CA$ of a semiorthogonal decomposition $\D^b(X) = \langle \CA,{}^\perp\CA \rangle$
is {\sf translated to the right}).

We will also need the following evident observation.

\begin{lemma}\label{tensmut}
Let $\Phi$ be an autoequivalence of $\CT$. Then
$$
\Phi\circ\RR_\CA \cong \RR_{\Phi(\CA)}\circ\Phi,
\qquad
\Phi\circ\LL_\CA \cong \LL_{\Phi(\CA)}\circ\Phi.
$$
In particular, if $L$ is a line bundle on $X$
and $E$ is an exceptional object in $\D^b(X)$ then
$$
\TT_L\circ\RR_E \cong \RR_{E\otimes L}\circ\TT_L,
\qquad
\TT_L\circ\LL_E \cong \LL_{E\otimes L}\circ\TT_L,
$$
where $\TT_L:\D^b(X) \to \D^b(X)$ is the functor of tensor product by $L$.
\end{lemma}

\section{Derived category of a quartic double solid}\label{s-qds}

In this section we discuss the structure of the derived categories of quartic double solids
(smooth and nodal) and their resolutions of singularities.

\subsection{Smooth double covering}

We start with the following general result.

\begin{lemma}\label{pbec}
Let $f:X \to Y$ be a $2$-fold covering ramified in a divisor $D \subset Y$.
Assume that for some $H \in \Pic Y$ we have $D \sim 2H \sim -K_Y$.
Assume that $\CE_1,\dots,\CE_n$ is an exceptional collection in $\D^b(Y)$ such that
the collection
\begin{equation}\label{col1}
\CE_1(-H),\dots,\CE_n(-H),\CE_1,\dots,\CE_n
\end{equation}
is also exceptional. Then the collection
$$
f^*\CE_1,\dots,f^*\CE_n \in \D^b(X)
$$
is exceptional in $\D^b(X)$.
\end{lemma}
\begin{proof}
Note that $X = \Spec_Y(\CO_Y \oplus \CO_Y(-H))$, hence $f_*\CO_X \cong \CO_Y \oplus \CO_Y(-H)$.
Now we have
$$
\Ext^\bullet(f^*\CE_i,f^*\CE_j) =
\Ext^\bullet(\CE_i,f_*f^*\CE_j) =
\Ext^\bullet(\CE_i,\CE_j\otimes (\CO_Y \oplus \CO_Y(-H))) =
\Ext^\bullet(\CE_i,\CE_j) \oplus \Ext^\bullet(\CE_i,\CE_j(-H))).
$$
Note that the second summand vanishes for all $i,j$ and the first summand vanishes
for $i > j$ since the collection~\eqref{col1} is exceptional.
Moreover, for $i = j$ we obtain exceptionality of $f^*\CE_i$.
\end{proof}

Now once we have constructed an exceptional collection in $\D^b(X)$, we can extend
it to a semiorthogonal decomposition. It turns out that the additional component has
a very interesting property.

\begin{proposition}\label{serre}
Assume that the conditions of Lemma~{\rm\ref{pbec}} are satisfied.
Denote the orthogonal to the exceptional collection of the Lemma by
$\CA_X = \langle f^*\CE_1, \dots, f^*\CE_n \rangle^\perp$.
Then
$$
\D^b(X) = \langle \CA_X, f^*\CE_1, \dots, f^*\CE_n \rangle
$$
is a semiorthogonal decomposition. Moreover,
if collection~\eqref{col1} is full in $\D^b(Y)$ then
the Serre functor $\SS_{\CA_X}$ of the category $\CA_X$ is given by
$$
\SS_{\CA_X} \cong \tau[\dim X - 1],
$$
where $\tau$ is the involution of the double covering $f:X \to Y$.
\end{proposition}
\begin{proof}
The first part follows from Corollary~\ref{sodgen} and Lemma~\ref{pbec}.
So, the only thing we have to compute is the Serre functor. For this
we have to check that for any $\CF_1,\CF_2 \in \CA_X$ we have
$$
\Hom(\CF_1,\tau\CF_2[\dim X - 1])^\vee \cong \Hom(\CF_2,\CF_1),
$$
a bifunctorial isomorphism.
For this we start with the Serre duality on $X$. Note that $K_X = -H$, hence
$$
\Hom(\CF_1,\tau\CF_2[\dim X - 1])^\vee \cong \Hom(\tau\CF_2,\CF_1(-H)[1]) \cong \Hom(\CF_2,\tau\CF_1(-H)[1]).
$$
On the other hand, it is easy to see that for each $\CF_1$ we have a canonical distinguished triangle
$$
f^*f_*\CF_1 \to \CF_1 \to \tau\CF_1(-H)[1].
$$
So, to prove the claim it suffices to check that $\Hom(\CF_2,f^*f_*\CF_1) = 0$ for all $\CF_1,\CF_2 \in \CA_X$.
But since $K_{X/Y} = \CO(H)$ we have $f^! \cong f^*\circ\TT_{\CO(H)}$, so
$$
\Hom(\CF_2,f^*f_*\CF_1) \cong
\Hom(\CF_2(H),f^!f_*\CF_1) \cong
\Hom(f_*\CF_2(H),f_*\CF_1).
$$
By definition of $\CA_X$ we have $\Hom(\CE_i,f_*\CF_j) = 0$ for all $i,j$, hence
$$
f_*\CF_1 \in \langle \CE_1,\dots,\CE_n \rangle^\perp = \langle \CE_1(-H),\dots,\CE_n(-H) \rangle
$$
because~\eqref{col1} is full. By the same reason $f_*\CF_2(H) \in \langle \CE_1,\dots,\CE_n \rangle$.
The latter two subcategories are semiorthogonal, hence the $\Hom$ in question vanishes.
\end{proof}

\begin{remark}
This behavior of the Serre functor of $\CA_X$ has a nice generalization, see~\cite{K09}.
One can take $X$ to be a double covering of (or a hypersurface in) arbitrary smooth projective variety $Y$
which has a rectangular Lefschetz decomposition. Then analogously defined subcategory $\CA_X$ of $\D^b(X)$
is a fractional Calabi--Yau category.
\end{remark}

Now we apply this result to a quartic double solid. Recall that a quartic double solid
is a double covering of $\PP^3$ ramified in a quartic. Denote by $h$ the positive generator
of $\Pic\PP^3$ as well as its pullback to the double covering. The standard exceptional collection
$\CO(-3h),\CO(-2h),\CO(-h),\CO$ on $\PP^3$ has the form~\eqref{col1} for $H = 2h$, hence we obtain the following

\begin{corollary}
Let $D \subset \PP^3$ be a quartic surface and $f:X \to \PP^3$ --- the associated double covering.
Then $(\CO_X(-h),\CO_X)$ is an exceptional pair and we have a semiorthogonal decomposition
$$
\D^b(X) = \langle \CA_X, \CO_X(-h), \CO_X \rangle,
$$
where $\CA_X = \langle \CO_X(-h), \CO_X \rangle^\perp$. Moreover, $\SS_{\CA_X} \cong \tau[2]$.
\end{corollary}

Note that the behavior of the Serre functor of $\CA_X$ is analogous to that of an Enriques surface.
We are going to show that this is not just a coincidence. But first of all we are going to extend
the construction of subcategory $\CA_X$ to nodal quartic double solids.

\subsection{Nodal case, big resolution}

Now let $D \subset \PP^3$ be a quartic surface with a finite number $N$ of ordinary double points $y_1,\dots,y_N \in \PP^3$.
Then the corresponding double covering $f:X \to \PP^3$ also has $N$ ordinary double points, $x_1 = f^{-1}(y_1)$, \dots, $x_N = f^{-1}(y_N)$.
Therefore, the category $\CA_X$ in this case is singular. If we want to replace it by a smooth category, we have two possibilities.
The first, is to replace the singular threefold $X$ by its blowup in points $x_1,\dots,x_N$,
$$
X' = \Bl_{x_1,\dots,x_N} X.
$$
It is easy to see that $X'$ can also be represented as a double covering.

\begin{lemma}\label{xprime}
Let
$$
Y = \Bl_{y_1,\dots,y_N}\PP^3
$$
be the blowup of $\PP^3$ in points $y_1,\dots,y_N$ and $D' \subset Y$ the proper preimage of $D$.
Then $X'$ is the double covering of $Y$ ramified in $D'$.
\end{lemma}
\begin{proof}
Evident.
\end{proof}

Denote the morphism $X' \to Y$ also by $f$. Denote the exceptional divisors of $Y \to \PP^3$ by $E_1,\dots,E_N$,
and the exceptional divisors of $X' \to X$ by $Q_1,\dots,Q_N$. Each $E_i$ is isomorphic to $\PP^2$ and each $Q_i$
is isomorphic to $\PP^1\times\PP^1$ (since $x_i$ is an ordinary double point). Moreover, $f$ maps $Q_i$ to $E_i$
and is a double covering ramified in a conic.
$$
\xymatrix{
\bigsqcup\limits_{i=1}^N \PP^1\times\PP^1 \ar@{=}[r] &
\bigsqcup\limits_{i=1}^N Q_i \ar[d]_{2:1} \ar@{=}[r] &
X' \ar[rr]^{\Bl_{x_1,\dots,x_N}} \ar[d]_{2:1}^f &&
X \ar[d]_{2:1}^f \\
\bigsqcup\limits_{i=1}^N \PP^2 \ar@{=}[r] &
\bigsqcup\limits_{i=1}^N E_i \ar[r] &
Y \ar[rr]_{\Bl_{y_1,\dots,y_N}} &&
\PP^3
}
$$
The naive replacement of the category $\CA_X$ would be the orthogonal to $\langle \CO_{X'}(-h), \CO_{X'} \rangle$ in $\D^b(X')$.
However, there is a better replacement. Note that the ramification divisor $D'$ lies in the linear system $4h - 2\sum e_i$,
where $e_i$ is the class of the exceptional divisor of $Y$ over the point $y_i \in \PP^3$. Denote one half of it by $H$:
\begin{equation}\label{defH}
H = 2h - \sum e_i.
\end{equation}
Note that
\begin{equation}\label{ky}
K_Y = -2H = 2\sum e_i - 4h.
\end{equation}

\begin{proposition}
Consider the collection of line bundles
$$
(\CE_1,\dots,\CE_{N+2}) = (\CO_X(-h),\{\CO_X(-e_i)\}_{i=1}^{N},\CO_X).
$$
Then the collection $(\CE_1(-H),\dots,\CE_{N+2}(-H),\CE_1,\dots,\CE_{N+2})$ is a full exceptional collection in $D^b(Y)$.
\end{proposition}
\begin{proof}
We start with a standard exceptional collection of a blowup (see~\cite{Or})
$$
\D^b(Y) = \langle \CO_Y(-3h),\CO_Y(-2h),\CO_Y(-h),\CO_Y,\{\CO_{E_i}\}_{i=1}^{N},\{\CO_{E_i}(1)\}_{i=1}^{N} \rangle
$$
and perform a sequence of mutations.

Step 1. Translate the block $\{\CO_{E_i}(1)\}_{i=1}^{N}$ to the left.
Since $(\omega_Y)_{|E_i} \cong \CO_Y(2e_i - 4h)_{|E_i} = \CO_{E_i}(-2)$,
it follows from Lemma~\ref{longmut}
that we will obtain
$$
\D^b(Y) = \langle \{\CO_{E_i}(-1)\}_{i=1}^{N},\CO_Y(-3h),\CO_Y(-2h),\CO_Y(-h),\CO_Y,\{\CO_{E_i}\}_{i=1}^{N} \rangle.
$$

Step 2. Apply the left mutation through the block $\{\CO_{E_i}(-1)\}_{i=1}^{N}$ to $\CO_Y(-3h)$ and $\CO_Y(-2h)$.
Note that
$$
\Ext^t(\CO_{E_i}(-1),\CO_Y(-kh)) =
\Ext^{3-t}(\CO_Y(-kh),\CO_{E_i}(-3))^\vee =
H^{3-t}(E_i,\CO_{E_i}(-3))^\vee =
H^{t-1}(E_i,\CO_{E_i})
$$
(the first equality is the Serre duality on $Y$, in the second we use $\CO_Y(h)_{|E_i} \cong \CO_{E_i}$,
and the third is the Serre duality on $E_i$).
Thus we have one $\Ext^1$ for each $i$. Therefore it is clear that the mutation is given
by the following exact sequence
$$
0 \to \CO_Y(-kh) \to \CO_Y(-kh + \sum_{i=1}^{N} e_i) \to \oplus_{i=1}^{N} \CO_{E_i}(-1) \to 0,
$$
for both $k = 2$ and $k = 3$, that is $\LL_{\langle \{\CO_{E_i}(-1) \}_{i=1}^N \rangle} \CO_Y(-kh) = \CO_Y(-kh + \sum_{i=1}^N e_i)$.
Thus, we obtain
$$
\D^b(Y) = \langle \CO_Y(-3h + \sum_{i=1}^{N} e_i),\CO_Y(-2h + \sum_{i=1}^{N} e_i),\{\CO_{E_i}(-1)\}_{i=1}^{N},\CO_Y(-h),\CO_Y,\{\CO_{E_i}\}_{i=1}^{N} \rangle.
$$

Step 3. Apply the left mutation through $\CO_Y$ to the block $\{\CO_{E_i}\}_{i=1}^{N}$,
and simultaneously the left mutation through $\CO_Y(-2h + \sum_{i=1}^{N} e_i)$
to the block $\{\CO_{E_i}(-1)\}_{i=1}^{N}$.
Note that
$$
\Ext^\bullet(\CO_Y,\CO_{E_i}) = H^\bullet(E_i,\CO_{E_i}),
$$
so we have one $\Hom$ for each $i$. It follows that the mutation of $E_i$
is given by the following exact sequence
$$
0 \to \CO_Y(-e_i) \to \CO_Y \to \CO_{E_i} \to 0,
$$
that is $\LL_{\CO_Y}\CO_{E_i} = \CO_Y(-e_i)$. Hence
$$
\LL_{\CO_Y(-2h + \sum_{i=1}^N e_i)} \CO_{E_i}(-1) = \LL_{\CO_Y(-H)} \CO_{E_i}(-H) = \CO_Y(-H-e_i)
$$
by Lemma~\ref{tensmut}.
So, finally we obtain
\begin{equation}\label{coly}
\D^b(Y) = \langle \CO_Y(-h-H),\{\CO_Y(-e_i-H)\}_{i=1}^{N},\CO_Y(-H),\CO_Y(-h),\{\CO_Y(-e_i)\}_{i=1}^{N},\CO_Y \rangle
\end{equation}
which proves the Proposition.
\end{proof}

Applying Lemma~\ref{pbec} and Proposition~\ref{serre} we obtain the following

\begin{corollary}\label{caxp}
There is a semiorthogonal decomposition
\begin{equation}\label{sodxp}
\D^b(X') = \langle \CA_{X'}, \CO_{X'}(-h),\{\CO_{X'}(-e_i)\}_{i=1}^{N},\CO_{X'} \rangle,
\end{equation}
where $\CA_{X'} = \langle \CO_{X'}(-h),\{\CO_{X'}(-e_i)\}_{i=1}^{N},\CO_{X'} \rangle^\perp$.
Moreover, $\SS_{\CA_{X'}} \cong \tau[2]$.
\end{corollary}

\subsection{Nodal case, small resolution}

Another alternative is to replace $X$ by its {\em small resolution of singularities}.
Note that an analytic neighborhood of an ordinary double point $x_i \in X$ is analytically isomorphic
to an analytic neighborhood of the vertex of the 3-dimensional quadratic cone. In particular, it has a resolution
of singularities (actually two resolutions!) with the exceptional locus being $\PP^1$. Choosing
such resolution for each $x_i$ and gluing all of them together, we obtain a smooth analytic manifold $X^+$
with a map $\sigma:X^+ \to X$. This manifold does not admit an algebraic structure in general,
however it is always a Moishezon variety, in particular an algebraic space.
Therefore one can speak about its derived category of coherent sheaves.

Note also that since the construction of the minimal resolution $X^+$ involves a choice
of one of two local resolutions for each of the singular points $x_1,\dots,x_N$, there are
$2^N$ different minimal resolutions. Each of them has a semiorthogonal decomposition
analogous to that of $X$.

\begin{proposition}\label{dbxplus}
There is a semiorthogonal decomposition
$$
\D^b(X^+) = \langle \CA_{X^+}, \CO_{X^+}(-h), \CO_{X^+} \rangle,
$$
where $\CA_{X^+} = \langle \CO_{X^+}(-h), \CO_{X^+} \rangle^\perp$.
\end{proposition}
\begin{proof}
Note that $\sigma_*\CO_{X^+} \cong \CO_X$ since ordinary double point is a rational singularity, hence
$$
\Ext^\bullet(\sigma^*\CE,\sigma^*\CE') \cong
\Ext^\bullet(\CE,\sigma_*\sigma^*\CE') \cong
\Ext^\bullet(\CE,\CE'\otimes\sigma_*\CO_{X^+}) \cong
\Ext^\bullet(\CE,\CE').
$$
This shows that $(\CO_{X^+}(-h),\CO_{X^+})$ is an exceptional pair in $\D^b(X^+)$.
Hence it gives the required decomposition.
\end{proof}

\begin{remark}
A natural question to ask is whether the Serre functor of $\CA_{X^+}$ can be written as an involution composed with the shift by 2.
It turns out that the answer is negative. Indeed, since $K_{X^+} = -2h$, we have
$$
\SS_{\CA_{X^+}}^{-1} \cong
\LL_{\langle \CO_{X^+}(-h),\CO_{X^+} \rangle}\circ \SS^{-1}_{\D^b(X^+)} \cong
\LL_{\langle \CO_{X^+}(-h),\CO_{X^+} \rangle}\circ \TT_{\CO_{X^+}(2h)}[-3].
$$
It is easy to see that applied to $\sigma^*\CF$ this functor gives $\sigma^*\tau\CF[-2]$.
On the other hand, if $C$ is an exceptional curve of $X^+$ then
$$
\SS_{\CA_{X^+}}(\CO_{C}(-1)) \cong \CO_{C}(-1)[-3].
$$
The shifts are different, hence the behavior of $\CA_{X^+}$ is more complicated in this case.
\end{remark}

\begin{remark}
If we have a family $X_t$ with smooth generic member and $X_0$ being a nodal double solid,
then one can consider $\CA_{X_0^+}$ as a {\em smooth degeneration} of $\CA_{X_t}$. And the category
$\CA_{X'_0}$ can be thought of as a (categorical) blowup of the category $\CA_{X'_0}$.
\end{remark}

\begin{lemma}\label{dbxp}
There is a fully faithful embedding $\CA_{X^+} \to \CA_{X'}$ which extends
to a semiorthogonal decomposition $\CA_{X'} = \langle \CA_{X^+}, \{ \CO_{Q_i}(-1,0) \}_{i=1}^N \rangle$.
\end{lemma}
\begin{proof}
Note that the blow up of the minimal resolution $X^+$ in the union of the exceptional curves
$C_1 = \sigma^{-1}(x_1)$, \dots, $C_N = \sigma^{-1}(x_N)$ coincides with the blowup $X'$ of $X$.
Denote the exceptional divisors by $Q_1$, \dots, $Q_N$.
Note that $Q_i \cong C_i \times \PP^1$.

The standard semiorthogonal decomposition of the blowup in this case gives
$$
\D^b(X') = \langle \D^b(X^+), \{ \CO_{Q_i}(-1,0), \CO_{Q_i} \}_{i=1}^N \rangle.
$$
Replacing $\D^b(X^+)$ by its decomposition we obtain
$$
\D^b(X') = \langle \CA_{X^+}, \CO_{X'}(-h), \CO_{X'}, \{ \CO_{Q_i}(-1,0), \CO_{Q_i} \}_{i=1}^N \rangle.
$$
Now we note that the sheaves $\CO_{Q_i}(-1,0)$ are completely orthogonal to $\CO_{X'}(-h)$ and $\CO_{X'}$.
Indeed, since the latter are pullbacks via $\sigma':X' \to X$, it suffices to observe that
$\sigma'_*\CO_{Q_i}(-1,0) = 0$, since $\sigma'$ contracts each of $Q_i$ to a point and the sheaf
$\CO_{Q_i}(-1,0)$ on $Q_i$ is acyclic. Therefore we can move all $\CO_{Q_i}(-1,0)$ to the left.
We obtain a semiorthogonal decomposition
$$
\D^b(X') = \langle \CA_{X^+}, \{ \CO_{Q_i}(-1,0) \}_{i=1}^N, \CO_{X'}(-h), \CO_{X'}, \{ \CO_{Q_i} \}_{i=1}^N \rangle.
$$
Finally, mutating $\CO_{Q_i}$ through $\CO_{X'}$ we obtain
$$
\D^b(X') = \langle \CA_{X^+}, \{ \CO_{Q_i}(-1,0) \}_{i=1}^N, \CO_{X'}(-h), \{ \CO_{X'}(-e_i) \}_{i=1}^N, \CO_{X'} \rangle.
$$
Comparing it with~\eqref{sodxp} we deduce the claim.
\end{proof}

\section{Quartic symmetroids and special double solids}\label{s-symm}

In this section we discuss the quartic double solids associated with quartic symmetroids and state the main result of the paper.
The following construction is very classical. For more details see e.g.~\cite{Co}.

Assume $V$ and $W$ are vector spaces, both of dimension $4$, and choose a generic embedding
$$
s:W \subset S^2V^\vee.
$$
Alternatively, $s$ can be considered as a tensor in $W^\vee \otimes S^2V^\vee$.
The latter space can be considered as the space of global sections of various natural vector bundles.
Considering zero loci of these sections we obtain some natural varieties associated with $s$.
For example, considering the line bundle $\CO(1,2)$ on $\PP(W)\times\PP(V)$ we obtain
a family of quadrics
$$
Q \subset \PP(W)\times\PP(V).
$$
The degeneration locus of this family $D \subset \PP(W)$ is the zero locus of $\det s \in S^4W^\vee \otimes (\det V^\vee)^2$,
hence is a quartic surface. Quartic surfaces obtained in this way are called {\sf quartic symmetroids}.
If $s$ is generic then $D$ has 10 ordinary double points. The blowup $D'$ of $D$ can be realized
as the zero locus of $s$ considered as a global section of the vector bundle $W^\vee\otimes\CO(1,1)$
on $\PP(V)\times\PP(V)$ (the map $D' \to D$ takes a point $(v,v') \in D'$ to the unique $w \in \PP(W)$
such that quadric $Q_w$ is singular at $v$). On the other hand, the transposition of factors
$\iota:\PP(V)\to\PP(V)$ acts on K3-surface $D'$ without fixed points, hence
$$
S = D'/\iota
$$
is an Enriques surface.

\begin{remark}
Enriques surfaces associated in this way with quartic symmetroids are known
as {\sf nodal Enriques surfaces} (the name is slightly misleading, since
they are smooth), or {\sf Reye congruences}. The family of nodal Enriques
surfaces is known to be a divisor in the family of all Enriques surfaces.
\end{remark}

Our goal is the following

\begin{conjecture}
Let $X'$ be the blowup of the nodal quartic double solid $X$ associated with the quartic symmetroid $D \subset \PP^3$
and $S = D'/\iota$ --- the corresponding Enriques surface.
Then
$$
\CA_{X'} \cong \D^b(S),
$$
where subcategory $\CA_{X'} \subset \D^b(X')$ is defined in Corollary~{\rm\ref{caxp}}.
\end{conjecture}

Instead of proving this conjecture we concentrate on proving the following
slightly simpler result. Note that both categories $\CA_{X'}$ and $\D^b(S)$
have a completely orthogonal exceptional collections of length~10:
for $\CA_{X'}$ it is given by the sheaves $\CO_{Q_i}(-1,0)$ (see Lemma~\ref{dbxp}),
and for $\D^b(S)$ it is given by the line bundles $\CO_S(-F_i^+)$ (see Theorem~\ref{zube}).
The orthogonal complements of these collections are the categories
$\CA_{X^+}$ and $\CA_S$ respectively.
The main result of the paper is the following

\begin{theorem}\label{cxpcs}
Let $X^+$ be a small resolution of singularities of the nodal quartic double solid $X$ associated with the quartic symmetroid $D \subset \PP^3$
and $S$ the corresponding Enriques surface.
Then
$$
\CA_{X^+} \cong \CA_S,
$$
where subcategories $\CA_{X^+} \subset \D^b(X^+)$ and $\CA_S \subset \D^b(S)$ are defined in
Proposition~{\rm\ref{dbxplus}} and~\eqref{cas} respectively.
\end{theorem}

\begin{remark}
Note that there are $2^{10}$ ways to choose one multiple fiber
in each of 10 elliptic pencils on~$S$, precisely the same number
as the number of small resolutions of singularities of $X$.
We will show below that there is a bijection between these sets
and the equivalence of Theorem~\ref{cxpcs} is for proved compatible choices.
\end{remark}

\begin{remark}
Note that different small resolutions of $X$ are related by flops.
These flops induce equivalences of the corresponding subcategories $\CA_{X^+}$.
On the other hand, different choices of multiple fibers in elliptic pencils on $S$
lead to subcategories $\CA_S$ related by appropriate mutation functors.
\end{remark}

The proof of the Theorem is given in the next section.

\section{The relative scheme of lines}\label{s-proof}

The main idea of the proof is to analyze carefully the derived category of
the relative scheme of lines $M$ on the family $Q \to \PP^3$ of quadrics associated with $s$.
One one hand, we show that $M$ is the blowup of $\Gr(2,V)$ in the Enriques surface $S$,
whereof one realizes $\D^b(S)$ as a component of $\D^b(M)$. On the other hand,
there is a general description of the relative scheme of lines~\cite{K10},
one component of which is the derived category of $X^+$. Finally, a sequence
of mutations allows to transform one of these decompositions of $\D^b(M)$
to the other and establishes the required equivalence.

\subsection{Relation to the Enriques surface}

First of all we will need another description of the Enriques surface associated with the quartic symmetroid.
For this consider the product of the Grassmannians $\Gr(2,V) \times \Gr(2,W)$. Denote the tautological subbundles
on those as $\CU$ and $\CU_W$ respectively. Then $s$ can be thought of as a global section of the vector
bundle $S^2\CU^\vee \boxtimes \CU_W$. Consider its zero locus $\TS \subset \Gr(2,V)\times\Gr(W)$.
Set-theoretically, $\TS$ is the set of all pairs $(L,K)$ of twodimensional subspaces $L \subset V$,
$K \subset W$ such that the line $\PP(L)$ is contained in all the quadrics in the pencil $\PP(K) \subset \PP(W)$.
It is known (see e.g.~\cite{Be}, Chapter VIII) that $\TS$ is isomorphic to the Enriques
surface associated with~$s$ (the map $D' \to \TS$ takes a point $(v,v') \in D' \subset \PP(V)\times \PP(V)$
to the point $(L,K) \in \Gr(2,V)\times\Gr(2,W)$, where $L$ is the linear span of $v$ and $v'$
and $K$ consists of all $w \in \PP(W)$ such that $Q_w$ contains both $v$ and~$v'$) and that
the projection onto $\Gr(2,V)$ gives an embedding of $S = \TS$ into $\Gr(2,V)$.

Below we will need a resolution of the structure sheaf of $S$ on $\Gr(2,V)$.

\begin{lemma}
The structure sheaf of $S$ in $\Gr(2,V)$ has the following resolution
\begin{equation}\label{os}
0 \to S^2\CU(-3g) \to W^\vee\otimes\CO_{\Gr(2,V)}(-3g) \to \CO_{\Gr(2,V)} \to \CO_S \to 0.
\end{equation}
\end{lemma}
\begin{proof}
By definition of $\TS$ we have the following Koszul resolution for $\CO_{\tilde{S}}$:
$$
\begin{array}{rcl}
0 \to \CO(-6,-3) \to
S^2\CU(-4)\boxtimes\CU_W(-2) \to \\
{S^2\CU(-3)\boxtimes S^2\CU_W(-1)} & \oplus &  (S^4\CU(-2) \oplus \CO(-4))\boxtimes\CO(-2) \to \\
{\CO(-3)\boxtimes S^3\CU_W} & \oplus & (S^4\CU(-1) \oplus S^2\CU(-2))\boxtimes \CU_W(-1) \to \\
S^2\CU(-1)\boxtimes S^2\CU_W & \oplus & (S^4\CU \oplus \CO(-2)) \boxtimes \CO(-1) \to \\
&& \qquad\qquad\qquad\qquad\qquad S^2\CU\boxtimes\CU_W \to
{\CO} \to \CO_{\tilde{S}} \to 0.
\end{array}
$$
Since the projection to $\Gr(2,V)$ identifies $\tilde{S}$ with $S$,
we can take the pushforward of the above resolution to get a resolution of $\CO_S$.
Note that the bundles $\CU_W$, $\CO(-1)$, $S^2\CU_W$, $\CU_W(-1)$, $\CO(-2)$, $\CU_W(-2)$ and $\CO(-3)$ on $\Gr(2,W)$
are acyclic, the nontrivial pushforward will only come from the terms
$S^2\CU(-3)\boxtimes S^2\CU_W(-1)$, $\CO(-3)\boxtimes S^3\CU_W$, and $\CO$.
Since
$$
H^2(\Gr(2,W),S^2\CU_W(-1)) = \C,\qquad
H^2(\Gr(2,W),S^3\CU_W) = W^\vee,\qquad
H^0(\Gr(2,W),\CO) = \C,
$$
and the other cohomology vanishes, we obtain the desired resolution.
\end{proof}

Now consider the relative scheme of lines $M$ for the family of quadrics $Q \to \PP(W)$.
By definition $M$ is a subscheme in $\Gr(2,V)\times \PP(W)$, consisting of all pairs $(L,w)$,
where $L$ is a 2-dimensional subspace of $V$ and $w \in \PP(W)$, such that the line $\PP(L)$
lies on the quadric $Q_w \subset \PP(V)$. Thus, it is the zero locus of a section
of the vector bundle $S^2\CU^\vee\otimes\CO_{\PP(W)}(1)$ on $\Gr(2,V)\times\PP(W)$,
given by $s$. Since $s$ is generic the section is regular.
Denoting the positive generator of $\Pic(\Gr(2,V))$ by $\CO(g)$
we obtain the following

\begin{lemma}
The structure sheaf of $M$ in $\Gr(2,V)\times\PP(W)$ has the following resolution
\begin{equation}\label{om}
0 \to \CO(-3g-3h) \to S^2\CU(-g-2h) \to S^2\CU(-h) \to \CO \to \CO_M \to 0.
\end{equation}
\end{lemma}
\begin{proof}
This is just the Koszul resolution.
\end{proof}

Now we are ready to show that map $\pi:M \to \Gr(2,V)$ is the blowup of $S \subset \Gr(2,V)$.

\begin{lemma}
The map $\pi:M \to \Gr(2,V)$ is the blowup of $S \subset \Gr(2,V)$.
Moreover, if $e$ is the class of the exceptional divisor then $h = 3g - e$.
\end{lemma}
\begin{proof}
First, let us describe the fibers of $\pi$. It is clear that the fiber over a point of $\Gr(2,V)$
corresponding to a subspace $L \subset V$ is the linear subspace of $\PP(W)$ consisting of quadrics,
containing the line $\PP(L)$. This is a condition of codimension 3, hence the generic fiber is a point.
Moreover, the fiber is nontrivial if there is a pencil of quadrics containing $\PP(L)$, that is if $L$
is in the image of the map $S = \TS \to \Gr(2,V)$. This shows that $\pi$ is the blowup of an ideal
supported at $S$. It remains to check that this is the ideal of~$S$.
For this consider the sheaf $\pi_*(\CO_M(h))$. Since $h$ is very ample over $\Gr(2,V)$
this sheaf is a twist of the ideal in question.
To compute $\pi_*\CO_M(h))$ we use~\eqref{om}. It gives
$$
0 \to S^2\CU \to W^\vee\otimes\CO \to \pi_*\CO_M(h) \to 0.
$$
Comparing with~\eqref{os} we see that $\pi_*\CO_M(h) \cong I_S(3g)$.
Thus $M$ is the blowup of $S$. Moreover, it follows also that $h = 3g - e$.
\end{proof}

Denote the exceptional divisor of $\pi$ by $E$, and the restriction of $\pi$ to $E$ by $p$.
Let also $i:E \to M$ be the embedding. Using~\cite{Or} we obtain the following

\begin{corollary}\label{dbmpi}
We have the following semiorthogonal decomposition
\begin{equation}\label{som0}
\D^b(M) = \langle \pi^*(\D^b(\Gr(2,V))), i_*p^*(\D^b(S)) \rangle.
\end{equation}
\end{corollary}

Since $\D^b(\Gr(2,V))$ is generated by an exceptional collection,
we see that $\D^b(M)$ consists of several exceptional objects and the category $\CA_S$.
We will continue using this line in subsection~\ref{ss-mut}, but now
we will look at $M$ from the other point of view.

\subsection{Relation to the double solid}

So far we considered $M$ as a blowup of $\Gr(2,V)$.
Now consider the projection $\rho:M \to \PP(W)$.
By definition of $M$ it is the relative scheme of lines
for the family of quadrics $Q \to \PP(W)$. Therefore the generic fiber of $M$
is a disjoint union of two conics; while over the divisor $D$ of degenerate quadrics
the fiber is a single conic, and over 10 points corresponding to quadrics of corank~2
the fiber is a union of two planes intersecting in a point. Note that the Stein
factorization for the map $\rho:M \to \PP(W)$ is
$$
\xymatrix{M \ar[rr]^\mu \ar[dr]_\rho && X \ar[dl]^f \\ & \PP(W)},
$$
where $f:X \to \PP(W)$ is the double covering ramified in $D$, that is $X$ is the special
double solid we are interested in.

The derived category of a relative scheme of lines for a family of 2-dimensional quadrics
was described in~\cite{K10}. To state the answer obtained therein we need some notation to be introduced.
First of all, for each point $y_i$ corresponding to a quadric of corank 2 choose one of two planes
in the fiber of $M$ over $y_i$. Denote it by $\Sigma_i^+$, and the other by $\Sigma_i^-$.
Further, consider the sheaf $\CB_0$ of even parts of Clifford algebras on $\PP(W)$
associated with the family of quadrics $Q \to \PP(W)$ (see~\cite{K08a} for details).
Let $\D^b(\PP(W),\CB_0)$ denote the derived category of sheaves of $\CB_0$-modules on $\PP(W)$.

\begin{theorem}\label{dbm}
For each choice of $10$ planes $\Sigma^+_i$ there exists a minimal resolution $X^+$ of $X$
and a semiorthogonal decomposition
$$
\D^b(M) = \langle \D^b(X^+), \D^b(\PP(W),\CB_0), \{ \CO_{\Sigma_i^+} \}_{i=1}^{10} \rangle.
$$
\end{theorem}

We will also need a description of the embedding functors for the first two components of this decomposition.
First of all, the functor $\D^b(X^+) \to \D^b(M)$ can be constructed as follows.
One can check (see~\cite{K10}) that the normal bundle to any of the planes $\Sigma_i^+$ is $\CO(-1) \oplus \CO(-1)$.
Hence one can perform a flip $M \dashrightarrow M^+$ in all these planes. The obtained Moishezon variety $M^+$
has a structure of a $\PP^1$-fibration $\mu_+:M^+ \to X^+$  over $X^+$. In particular, the pullback functor $\mu_+^*$
gives an embedding $\D^b(X^+) \to \D^b(M^+)$. Further, the flip $M \dashrightarrow M^+$ can be represented as
a composition of a blowup $\xi:\TM \to M$ and a blowdown $\xi_+:\TM \to M^+$. The composition
of functors $\xi_*\xi_+^*:\D^b(M^+) \to \D^b(M)$ is also fully faithful by a result of Bondal and Orlov (see~\cite{BO95}).
Composing we will obtain an embedding $\Phi_0 = \xi_*\xi_+^*\mu_+^*:\D^b(X^+) \to \D^b(M)$. What we will need further is the following observation

\begin{lemma}\label{phi0}
Let $f_+:X_+ \to \PP(W)$ be the composition of the contraction $X^+ \to X$ and the double covering $X \to \PP(W)$.
For any object $\CF \in \D^b(\PP(W))$ one has
$$
\Phi_0(f_+^*\CF) \cong \rho_+^*\CF.
$$
Moreover, the functor $\Phi_0$ is $\PP(W)$-linear, that is
$$
\Phi_0(\CG\otimes f_+^*\CF) \cong \Phi_0(\CG)\otimes\rho_+^*\CF.
$$
\end{lemma}
\begin{proof}
Follows immediately from the projection formula.
\end{proof}

Now consider the second component.
Note that besides the sheaf of even parts of Clifford algebras $\CB_0$ one has also
the sheaf $\CB_1$ of odd parts of Clifford algebras on $\PP(W)$.
As sheaves on $\PP(W)$ they have the following structure
$$
\CB_0 = \CO \oplus \Lambda^2V(-h) \oplus \Lambda^4V(-2h),\qquad
\CB_1 = V\otimes\CO \oplus \Lambda^3V(-h).
$$
This pair of sheaves naturally extends to a sequence of sheaves $\CB_k$ of $\CB_0$-modules
defined by
$$
\CB_{2k} = \CB_0(kh),
\qquad
\CB_{2k+1} = \CB_1(kh).
$$
For each $k \in \ZZ$ consider the morphism $\CU\boxtimes\CB_{k-1} \to \CB_k$ of sheaves of $\CB_0$-modules
on $\Gr(2,V)\times\PP(W)$ induced by the embedding $\CU\subset V\otimes\CO$ and the action $V\otimes\CB_k \to \CB_{k+1}$.
Let $\alpha:M \to \Gr(2,V) \times \PP(W)$ be the embedding.
Another result of~\cite{K10} is the following

\begin{theorem}[\cite{K10}]\label{phi}
There are isomorphisms
$\Coker(\CU\boxtimes\CB_{k-1} \to \CB_k) \cong \alpha_*\CS_k$,
where the sheaves $\CS_k$ are defined as
\begin{equation}\label{c2k}
0 \to \CO(kh) \to \CS_{2k} \to \CO(g + (k-1)h) \to 0,\qquad
\CS_{2k+1} = (V/\CU)(kh),
\end{equation}
and the extension is nontrivial. Moreover, we have
\begin{equation}\label{rhosk}
\rho_*(\CS_k(-g)) = 0
\end{equation}
for all $k \in \ZZ$.
Finally, the functor $\Phi_1:\D^b(\PP(W),\CB_0) \to \D^b(M)$, $\CF \mapsto \CS_0\otimes_{\CB_0} \rho^*\CF$ is fully faithful
and
$$
\Phi_1(\CB_k) \cong \CS_k.
$$
\end{theorem}

On the other hand, it follows from~\cite{K08a}, Theorem~5.5 that $\D^b(\PP(W),\CB_0)$ is generated by an exceptional collection
$\CB_{k-3},\CB_{k-2},\CB_{k-1},\CB_k$, where $k$ can be taken arbitrarily. Indeed there might be an additional component,
the derived category of the intersection of all quadrics in $\PP(V)$ parameterized by $W$. But since $s$ is generic this
is the intersection of 4 generic quadrics in $\PP^3$, and so is empty. Combining this result with the Theorem above
we obtain the following

\begin{corollary}\label{phi1}
The subcategory $\Phi_1(\D^b(\PP(W),\CB_0)) \subset \D^b(M)$ is generated by an exceptional collection
$$
\langle \CS_{k-3},\CS_{k-2},\CS_{k-1},\CS_k \rangle,
$$
where $k$ can be taken arbitrarily.
\end{corollary}

Combining the above results we can write down the following semiorthogonal decomposition
\begin{equation}\label{dbmpi1}
\D^b(M) = \langle \Phi_0(\D^b(X^+)), \CS_{-1}, \CS_0, \CS_1, \CS_2,  \{ \CO_{\Sigma_i^+} \}_{i=1}^{10} \rangle.
\end{equation}

\subsection{The mutations}\label{ss-mut}

Now our strategy is the following. We start with decomposition~\eqref{som0} of $\D^b(M)$
and then apply a sequence of mutations to get decomposition~\eqref{dbmpi1}.
As a result we will obtain a relation between $\CA_{X^+}$ and $\D^b(S)$.
More precisely, instead of the first component of~\eqref{som0}
we substitute the following exceptional collection on $\Gr(2,V)$:
$$
\D^b(\Gr(2,V)) = \langle \CO(-2g), \CO(-g), V/\CU(-g), \CO, V/\CU, \CO(g) \rangle.
$$
Plugging it into~\eqref{som0} and denoting
\begin{equation}\label{psi0}
\Psi_0 = i_*p^*: \D^b(S) \to \D^b(M),
\end{equation}
we obtain a semiorthogonal decomposition
\begin{equation}\label{som1}
\D^b(M) = \langle \CO(-2g), \CO(-g), V/\CU(-g), \CO, V/\CU, \CO(g), \Psi_0(\D^b(S)) \rangle.
\end{equation}

Now we perform a sequence of mutations.

Step 1. Apply the left mutation through $V/\CU$ and $\CO(g)$ to $\Psi_0(\D^b(S))$. We obtain
\begin{equation}\label{som2}
\D^b(M) = \langle \CO(-2g), \CO(-g), V/\CU(-g), \CO, \Psi_1(\D^b(S)), V/\CU, \CO(g) \rangle,
\end{equation}
where
\begin{equation}\label{psi1}
\Psi_1 = \LL_{\langle V/\CU, \CO(g)\rangle}\circ i_*p^*: \D^b(S) \to \D^b(M).
\end{equation}

Step 2. Translate bundles $V/\CU$ and $\CO(g)$ to the left. By Lemma~\ref{longmut} we obtain
\begin{equation}\label{som3}
\D^b(M) = \langle V/\CU(-g-h), \CO(-h), \CO(-2g), \CO(-g), V/\CU(-g), \CO, \Psi_1(\D^b(S)) \rangle
\end{equation}
since $K_M = - 4g + e = - g - h$.

Step 3. Apply the left mutation through $V/\CU(-g-h)$ and $\CO(-h)$ to $\CO(-2g)$.

\begin{lemma}
We have $\Ext^\bullet(\CO(-h),\CO(-2g)) = \Ext^\bullet(V/\CU(-g-h),\CO(-2g)) = 0$.
\end{lemma}
\begin{proof}
We have to compute
$$
H^\bullet(M,\CO(h-2g)) = H^\bullet(M,\CO(g-e)),
\qquad\text{and}\qquad
H^\bullet(M,\CU^\perp(h-g)) = H^\bullet(M,\CU^\perp(2g-e)),
$$
where $\CU^\perp := (V/\CU)^\vee$.
Since $\pi_*(\CO(-e)) = I_S$, the projection formula implies that this is the same as
$H^\bullet(\Gr(2,V),I_S(g))$ and $H^\bullet(\Gr(2,M),I_S\otimes\CU^\perp(2g))$ respectively.
Using~\eqref{os} as a resolution of $I_S$ and Borel--Bott--Weil on $\Gr(2,V)$
we obtain the required vanishing.
\end{proof}

Because of the orthogonality proved above and by Lemma~\ref{perpmut} we obtain
\begin{equation}\label{som4}
\D^b(M) = \langle \CO(-2g), V/\CU(-g-h), \CO(-h), \CO(-g), V/\CU(-g), \CO, \Psi_1(\D^b(S)) \rangle.
\end{equation}

Step 4. Apply the left mutation through $\CO(-h)$ to $\CO(-g)$.

\begin{lemma}
We have $\Ext^\bullet(\CO(-h),\CO(-g)) = \C[-1]$.
\end{lemma}
\begin{proof}
We have to compute $H^\bullet(M,\CO(h-g)) = H^\bullet(M,\CO(2g-e))$.
Since $\pi_*(\CO(-e)) = I_S$, the projection formula implies that this is the same as
$H^\bullet(\Gr(2,V),I_S(2g))$. By Borel--Bott--Weil we have $H^\bullet(\Gr(2,V),\CO(-g)) = 0$
and $H^\bullet(\Gr(2,V),S^2\CU(-g)) = \C[-2]$. Using~\eqref{os} as a resolution of $I_S$ we
deduce the claim.
\end{proof}

It follows that $\LL_{\CO(-h)}(\CO(-g))$ is the unique extension of $\CO(-h)$ by $\CO(-g)$.
By Theorem~\ref{phi} it is isomorphic to $\CS_0(-g)$. Thus we obtain
\begin{equation}\label{som5}
\D^b(M) = \langle \CO(-2g), V/\CU(-g-h), \CS_0(-g), \CO(-h), V/\CU(-g), \CO, \Psi_1(\D^b(S)) \rangle.
\end{equation}

Step 5. Apply the right mutation through $V/\CU(-g)$ to $\CO(-h)$.

\begin{lemma}
We have $\Ext^\bullet(\CO(-h),V/\CU(-g)) = 0$.
\end{lemma}
\begin{proof}
We have to compute $H^\bullet(M,V/\CU(h-g)) = H^\bullet(M,V/\CU(2g-e))$.
Since $\pi_*(\CO(-e)) = I_S$, the projection formula implies that this is the same as
$H^\bullet(\Gr(2,M),I_S\otimes V/\CU(2g))$.
Using~\eqref{os} as a resolution of $I_S$ and Borel--Bott--Weil on $\Gr(2,V)$
we obtain the required vanishing.
\end{proof}

Because of the orthogonality proved above and by Lemma~\ref{perpmut} we obtain
\begin{equation}\label{som6}
\D^b(M) = \langle \CO(-2g), V/\CU(-g-h), \CS_0(-g), V/\CU(-g), \CO(-h), \CO, \Psi_1(\D^b(S)) \rangle.
\end{equation}

Step 6. Translate bundle $\CO(-2g)$ to the right. By Lemma~\ref{longmut} we obtain
\begin{equation}\label{som7}
\D^b(M) = \langle V/\CU(-g-h), \CS_0(-g), V/\CU(-g), \CO(-h), \CO, \Psi_1(\D^b(S)), \CO(h-g) \rangle.
\end{equation}

Step 7. Apply the right mutation through $\CO(h-g)$ to $\Psi_1(D^b(S))$. We obtain
\begin{equation}\label{som8}
\D^b(M) = \langle V/\CU(-g-h), \CS_0(-g), V/\CU(-g), \CO(-h), \CO, \CO(h-g), \Psi_2(\D^b(S)) \rangle,
\end{equation}
where
\begin{equation}\label{psi2}
\Psi_2 = \RR_{\CO(h-g)}\circ \LL_{\langle V/\CU, \CO(g)\rangle}\circ i_*p^*: \D^b(S) \to \D^b(M).
\end{equation}

Step 8. Apply the left mutation through $\CO$ to $\CO(h-g)$. Analogously to step 4 we have $\LL_\CO(\CO(h-g)) = \CS_0(h-g) = \CS_2(-g)$.
Thus we obtain
\begin{equation}\label{som9}
\D^b(M) = \langle V/\CU(-g-h), \CS_0(-g), V/\CU(-g), \CO(-h), \CS_2(-g), \CO, \Psi_2(\D^b(S)) \rangle.
\end{equation}

Step 9. Finally, apply the left mutation through $\CO(-h)$ to $\CS_2(-g)$.

\begin{lemma}
We have $\Ext^\bullet(\CO(-h),\CS_2(-g)) = 0$.
\end{lemma}
\begin{proof}
Follows immediately from~\eqref{rhosk}.
\end{proof}

Because of the orthogonality and by Lemma~\ref{perpmut} we obtain
$$
\D^b(M) = \langle V/\CU(-g-h), \CS_0(-g), V/\CU(-g), \CS_2(-g), \CO(-h), \CO, \Psi_2(\D^b(S)) \rangle.
$$
Finally, rewriting $V/\CU = \CS_1$, $V/\CU(-h) = \CS_{-1}$ (see Theorem~\ref{phi}) we see that we have obtained
\begin{equation}\label{som10}
\D^b(M) = \langle \CS_{-1}(-g), \CS_0(-g), \CS_1(-g), \CS_2(-g), \CO(-h), \CO, \Psi_2(\D^b(S)) \rangle.
\end{equation}

\subsection{Multiple elliptic fibers}

Now we have to take into account the structure of $\D^b(S)$, namely the semiorthogonal decomposition
\begin{equation}\label{dbs}
\D^b(S) = \langle \{ \CO_S(-F_i^+) \}_{i=1}^{10}, \CA_S \rangle,
\end{equation}
where $F_i^\pm$ are the multiple fibers of the 10 elliptic pencils on $S$ and $\CA_S = {}^\perp\langle \{ \CO_S(-F_i^+) \}_{i=1}^{10} \rangle$.
It turns out that the functor $\Psi_2$ relates the multiple elliptic fibers $F_i^\pm$
to the planes $\Sigma^\pm_i$ of $M$.

Let $y_i \in \PP(w)$ be one of 10 nodes of the symmetroid $D$.
Then as we know the quadric $Q_{y_i}$ is a union of two planes,
and $\rho^{-1}(y_i) = \Sigma_i^+ \cup \Sigma_i^-$ is also a union of two planes.

\begin{lemma}
The exceptional divisor $E$ of the blowup $\pi:M \to \Gr(2,V)$ intersects $\Sigma_i^\pm$ along an elliptic curve.
Its image $F_i^\pm = \pi(\Sigma_i^\pm \cap E) \subset S$ is a multiple elliptic fiber of $S$.
\end{lemma}
\begin{proof}
Note that $e = 3g - h$, hence $\CO(e)_{|\Sigma_i^\pm} = \CO(3)$.
On the other hand, $\Sigma_i^\pm$ is not contained in $E$, since otherwise
we would have $\PP^2 = \pi(\Sigma_i^\pm) \subset S$ which is impossible.
Hence the intersection is a plane cubic. But the only plane cubic curves
on $S$ are the multiple elliptic fibers.
\end{proof}

\begin{proposition}\label{FS}
Let $F_i^\pm = \pi(E \cap \Sigma^\pm_i)$ be a multiple elliptic fiber of $S$. Then
$$
\Psi_2(\CO_S(2g - F_i^\pm)) \cong \CO_{\Sigma_i^\pm}(-1).
$$
\end{proposition}
\begin{proof}
To unburden the notation we will write $\Sigma$ instead of $\Sigma_i^\pm$ and $F$ instead of $F_i^\pm$.
Now consider several exact sequences. First, note that $\pi(\Sigma) = \Gr(2,3) \subset \Gr(2,V)$ is the zero locus
of a regular section of a vector bundle $\CU^\vee$; hence we have the following Koszul resolution on $\Gr(2,V)$
$$
0 \to \CO(-g) \to \CU \to \CO \to \CO_{\pi(\Sigma)} \to 0.
$$
Identifying $\CU = \CU^\vee(-g)$ and pulling back to $M$ we obtain
\begin{equation}\label{ppsk}
0 \to \CO(-g) \to \CU^\vee(-g) \to \CO \to \CO_{\pi^{-1}(\pi(\Sigma))} \to 0.
\end{equation}
On the other hand, the preimage $\pi^{-1}(\pi(\Sigma))$ clearly has two components,
one is $\Sigma$ itself, and the other is $p^{-1}(\pi(\Sigma) \cap S) = p^{-1}(F)$.
The components intersect along the cubic curve $\Sigma \cap p^{-1}(F)$.
So, the structure sheaf of $\pi^{-1}(\pi(\Sigma))$ fits into the following exact sequence
\begin{equation}\label{ppsu}
0 \to \CO_{\Sigma}(-3) \to \CO_{\pi^{-1}(\pi(\Sigma))} \to i_*p^*\CO_{F} \to 0.
\end{equation}
Also we have the following Koszul resolutions
\begin{equation}\label{ipf}
0 \to i_*p^*\CO_S(-F) \to i_*p^*\CO_S \to i_*p^*\CO_{F} \to 0
\end{equation}
and
\begin{equation}\label{ips}
0 \to \CO(-e) \to \CO \to i_*p^*\CO_S \to 0.
\end{equation}
Combining~\eqref{ppsk}, \eqref{ppsu}, \eqref{ipf}, and~\eqref{ips} and twisting everything by $\CO(2g)$ we obtain a bicomplex
$$
\xymatrix{
&&& \CO(2g-e) \ar[r] \ar[d] & \CO_{\Sigma}(-1) \ar[d] \ar[r] & 0\\
0 \ar[r] & \CO(g) \ar[r] \ar[d] & \CU^\vee(g) \ar[r] \ar[d] & \CO(2g) \ar[r] \ar[d] & \CO_{\pi^{-1}(\pi(\Sigma))}(2g) \ar[r] \ar[d] & 0 \\
& 0 \ar[r] & i_*p^*\CO_S(2g-F) \ar[r] & i_*p^*\CO_S(2g) \ar[r] & i_*p^*\CO_{F}(2g) \ar[r] & 0
}
$$
Here the middle row is~\eqref{ppsk}, the bottom row is~\eqref{ipf},
the third column is~\eqref{ips}, and the fourth column is~\eqref{ppsu},
everything twisted by $\CO(g)$ as it was already mentioned.

Since the bottom two rows of the bicomplex are exact, the total complex is quasiisomorphic to the top row,
which can be rewritten as
\begin{equation}\label{rpi}
0 \to \CO(h-g) \to \CO_{\Sigma}(-1) \to 0
\end{equation}
(the map here is just the restriction
$\CO(h-g) \to \CO(h-g)_{|\Sigma} \cong \CO(2g-e)_{|\Sigma} \cong \CO_{\Sigma}(-1)$).
On the other hand, since the right two columns are exact, the total complex is quasiisomorphic to
\begin{equation}\label{lf}
0 \to \CO(g) \to \CU^\vee(g) \to i_*p^*\CO_S(2g-F) \to 0.
\end{equation}
Combining these observations we conclude that complexes~\eqref{rpi} and~\eqref{lf} are quasiisomorphic.

Now we claim that complex~\eqref{rpi} is quasiisomorphic to $\LL_{\CO(h-g)}(\CO_{\Sigma}(-1))$
and complex~\eqref{lf} is quasiisomorphic to $\LL_{\langle\CO(g),\CU^\vee(g)\rangle}(i_*p^*\CO_S(2g-F))$.
Indeed, the first is clear since
$$
\Ext^\bullet(\CO(h-g),\CO_{\Sigma}(-1)) \cong
H^\bullet(\CO(g-h)\otimes\CO_{\Sigma}(-1)) \cong
H^\bullet(\Sigma,\CO) \cong \C.
$$
For the second it suffices to check that~\eqref{lf}, or equivalently~\eqref{rpi}, is orthogonal to $\CO(g)$ and $\CU^\vee(g)$.
For this we check that both terms of~\eqref{rpi} are orthogonal to $\CO(g)$ and $\CU^\vee(g)$.
Indeed, for the second term, $\CO_\Sigma(-1)$, we have
$$
\begin{array}{l}
\Ext^\bullet(\CO(g),\CO_{\Sigma}(-1)) \cong H^\bullet(\Sigma,\CO(-2)) = 0,\\
\Ext^\bullet(\CU^\vee(g),\CO_{\Sigma}(-1)) \cong H^\bullet(\Sigma,\CU(-2)) = 0.
\end{array}
$$
Further, for the first term, $\CO(h-g)$, we have
$$
\begin{array}{l}
\Ext^\bullet(\CO(g),\CO(h-g)) = H^\bullet(M,\CO(h-2g)) = H^\bullet(M,\CO(g-e)) = H^\bullet(\Gr(2,V),I_S(g)),\\
\Ext^\bullet(\CU^\vee(g),\CO(h-g)) = H^\bullet(M,\CU(h-2g)) = H^\bullet(M,\CU(g-e)) = H^\bullet(\Gr(2,V),I_S\otimes \CU(g)).
\end{array}
$$
Using resolution~\eqref{os} and Borel--Bott--Weil Theorem we conclude that these cohomology vanish.

Thus we have checked that
\begin{equation}\label{llll}
\LL_{\CO(h-g)}(\CO_{\Sigma}(-1)) \cong \LL_{\langle\CO(g),\CU^\vee(g)\rangle}(i_*p^*\CO_S(2g-F)).
\end{equation}
On the other hand, recall that by Lemma~\ref{mutfun} the mutation functors $\LL_{\CO(h-g)}$ and $\RR_{\CO(h-g)}$
are mutually inverse equivalences between ${}^\perp\CO(h-g)$ and $\CO(h-g)^\perp$.
So, let us check that $\CO_{\Sigma}(-1) \in {}^\perp\CO(h-g)$,
i.e. that $\Ext^\bullet(\CO_{\Sigma}(-1),\CO(h-g)) = 0$.
Indeed, by Serre duality we have
$$
\Ext^t(\CO_{\Sigma}(-1),\CO(h-g))^\vee =
\Ext^{4-t}(\CO(2h),\CO_{\Sigma}(-1)) = H^{4-t}(\Sigma,\CO_\Sigma(-1)) = 0.
$$
Thus~\eqref{llll} implies that
$$
\RR_{\CO(h-g)}\LL_{\langle\CO(g),\CU^\vee(g)\rangle}(i_*p^*\CO_S(2g-F)) \cong \CO_\Sigma(-1).
$$
Finally note that $\langle V/\CU, \CO(g) \rangle = \langle \CU^\perp(g), \CO(g) \rangle = \langle \CO(g), \CU^\vee(g) \rangle$,
hence $\LL_{\langle V/\CU, \CO(g) \rangle} = \LL_{\langle \CO(g), \CU^\vee(g) \rangle}$.
In view of this, the above isomorphism proves the Proposition.
\end{proof}

\subsection{The proof of the Main Theorem}

Now we are in a position to prove the Main Theorem.

Indeed, let us start with a semiorthogonal decomposition~\eqref{som10}.
Replacing $\D^b(S)$ by its semiorthogonal decomposition~\eqref{dbs}, twisting by $\CO_S(2g)$, and taking into account Proposition~\ref{FS} we obtain
the following semiorthogonal decomposition
$$
\D^b(M) = \langle \CS_{-1}(-g), \CS_0(-g), \CS_1(-g), \CS_2(-g), \CO(-h), \CO, \{ \CO_{\Sigma^+_i}(-1) \}_{i=1}^{10}, \Psi_2(\CA_S\otimes\CO_S(2g)) \rangle.
$$
Let us twist it by $\CO_M(g)$. As a result we obtain
\begin{equation}\label{som11}
\D^b(M) = \langle \CS_{-1}, \CS_0, \CS_1, \CS_2, \CO(g-h), \CO(g), \{ \CO_{\Sigma^+_i} \}_{i=1}^{10}, \Psi_3(\CA_S) \rangle,
\end{equation}
where
$\Psi_3 = \TT_{\CO(g)}\circ\Psi_2\circ\TT_{\CO(2g)}$.
It remains to make several additional mutations.

Step 9.
Apply the left mutation through $\CO(g-h)$ and $\CO(g)$ to the block $\{ \CO_{\Sigma_i^+} \}_{i=1}^{10}$.
The sheaves $\CO_{\Sigma^+_i}$ are completely orthogonal to $\CO(g)$ and $\CO(g-h)$, since
$$
\Ext^\bullet(\CO(g-th),\CO_{\Sigma^+_i}) = H^\bullet(\Sigma_i^+,\CO_{\Sigma_i^+}(-1)) = 0.
$$
Therefore, by Lemma~\ref{perpmut} we obtain
\begin{equation}\label{som12}
\D^b(M) = \langle \CS_{-1}, \CS_0, \CS_1, \CS_2, \{ \CO_{\Sigma^+_i} \}_{i=1}^{10}, \CO(g-h), \CO(g), \Psi_3(\CA_S) \rangle.
\end{equation}

Step 10.
Translate $\CO(g-h)$, $\CO(g)$ and $\Psi_3(\CA_S)$ to the left. By Lemma~\ref{longmut} we obtain
\begin{equation}\label{som13}
\D^b(M) = \langle \CO(-2h), \CO(-h), \Psi_4(\CA_S), \CS_{-1}, \CS_0, \CS_1, \CS_2, \{ \CO_{\Sigma^+_i} \}_{i=1}^{10} \rangle,
\end{equation}
where
$\Psi_4 = \TT(-g-h)\circ\Psi_3 = \TT_{\CO(-h)}\circ\Psi_2\circ\TT_{\CO(2g)}$.

Comparing this decomposition with Theorem~\ref{dbm} we conclude that
$$
\Phi_0(\D^b(X_+)) = \langle \CO_M(-2h), \CO_M(-h), \Psi_4(\CA_S) \rangle.
$$
By Lemma~\ref{phi0} the functor $\Phi_0$ commutes with the twist by $\CO(-h)$, hence we can rewrite
$$
\Phi_0(\D^b(X_+)) = \langle \CO_M(-h), \CO_M, \Psi_5(\CA_S) \rangle,
$$
where
$\Psi_5 = \TT_{\CO(h)}\circ\Psi_4 = \Psi_2\circ\TT_{\CO(2g)}$.
Since again by Lemma~\ref{phi0} we have $\CO_M(-th) = \Phi_0(\CO_{X^+}(-th))$,
and $\Phi_0$ is fully faithful, we conclude that
\begin{equation}
\Psi_5(\CA_S) = \Phi_0\circ\TT_{\CO(2h)}(\CA_{X^+}).
\end{equation}

redpajama_set_name: RedPajamaArXiv


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I've been GMing for a couple of sessions now and I'm still struggling at making encounters or situations more difficult. I will describe dangerous situations and encounters in fiction but the mechanics make it too easy for my players to succeed.
For instance no matter how big I describe the Troll, if the fighter of the party charges him and swing his sword, roll+2 for Hack and slash, he'll easily hit the troll. Almost 40% chance to hit without consequences.
One thing you can do to make a big boss dangerous is just make normal Hack & Slash useless. Imagine they are fighting a storm giant or something the like, and they normal Hack & Slash, you can just say "ok, you are just chipping his toenails, that is not going to work."
Force them to be creative: climb the giant (defy danger), try to out maneuver it, use terrain to get higher, make it kneel, whatever. It's a giant, it makes sense a normal attack on his foot with a sword is not going to work (even if it works in other games).
— The troll comes at you with its huge club, what do you do?
And now is when you decide a hard or a soft move. You could do damage, or you could just put the player in a harder situation (you are not fast enough and the club grazes you taking you out of balance, your weapon goes flying and you are now lying on the floor, while the troll prepares to step on you!).
The narrative at hand is a perfect reason to make things difficult for the player. You should always ask questions like crazy, ie. ask for justification from the player about how he does what he does. He may want to hack and slash at the troll but he doesn't decide on what move his intended action corresponds to. Feel free to challenge his intent by clarifying details and then asking.
— I hack and slash at the troll!
— Cool, how do you do it?
— I just swing my sword to chop his head off!
— Well, he's swinging a huge tree trunk around, how do you intend to get close enough to bring your steel toohpick in striking range?
— Well, I try to look for an opening and dive in.
— Looks like discern realities, please roll+wis.
— That's an 8, I get to ask one question, ok, what should I be on the lookout for?
— His swing has a semi-random looking pattern but you can guess it out.
— So I can just run in with perfect timing to avoid his swing.
— Cool, that looks like defy danger with roll+int to outsmart the troll.
— What? Not dex? Why?
— Because you can see that even with speed and reflexes, you'd eventually get hit by the trunk unless you figure out his rythym.
— Well, here goes… a 4, ouch!
— There's no way you're getting in by timing it, sooner or later that trunk will push your skull into your ribcage unless you keep your distance. That's what you have been doing instinctively anyway without realizing that you have been cornered, with nowhere to retreat anymore. The troll seizes the opportunity to deliver its skull-to-ribcage promise by swinging the trunk straight down on you. What do you do?
— Whoa, I duck and tumble inside his guard.
— 9 this time. What now?
— Well, you're obviously not going to pull that off with a sword in your hand. Either let go of it or feel the impact of the trunk on your back, your call.
— I need the HP, so ok, I let go of the sword and tumble.
— Cool. You're now inside the troll's guard without your sword. The troll grins to show you his rotting teeth, drops the trunk which just falls on your sword, and lunges to grab you by your throat. What do you do?
Here, the GM diverts the player's action so that he's on the defensive rather than the offensive he intends. The player had to go through three potential points of failure and he still couldn't take a real swing at the troll.
That, in my opinion, is what makes this a difficult adversary. Not how improbable your chance of succeeding is, but how many other things you have to go through before you can take a shot at your quarry.
Dungeon World puts difficult after the roll, not before.
There are two misunderstandings combining when someone wants to start fiddling with the mechanics to make Dungeon World harder (or easier).
The roll is to determine success or failure.
Mechanical difficulty is assigned ahead of the roll based on the situation.
DM: "Jumping the chasm is a Hard check. Roll to see if you succeed."
Success: "You leap the chasm and land safely on the other side."
Failure: "You fall!" (rolling) "You suffer 23 damage from the fall. You're lying in a shallow stream at the bottom of the chasm."
This is not how Dungeon World works.
The roll is to determine whether the GM gets to make a move as well.
Difficulty is determined after the roll, but only by the GM's choice of narration.
DM: "Jumping the chasm is dangerous, how are you going to defy that danger?"
Paladin: "I'm going to take a running start and put my all my strength into it."
DM: "That sounds like Defying Danger by Powering Through, roll plus strength."
Hit: "You leap the chasm and land safely on the other side."
Partial hit: "You go to leap the chasm and then pull up short. You can feel that you're just not going to make it with your panoply of plate. You can remove your amour and leave it on this side and make the leap no problem. What do you do?"
Miss: "You leap the chasm, but only just. You slam into the far edge and barely manage to grab on to stop yourself from plummeting. Ranger, you can tell that Paladin is stuck, unable to pull herself up due to the weight of her plate armour. She's slipping, what do you do?"
First, there is no success or failure, just hits and misses. Both hits and misses need to be translated through the rules and some more choices first, before you can judge subjectively whether the outcome was a success or a failure.
Hits of either kind always become fundamentally successes, but only after you follow the move's rules, and we don't know about misses until the GM makes their move. The success of a partial hit can carry consequences that can hurt enough that it's still a success, but a Pyrrhic one. We always follow the rules of the moves to find out what happens exactly first, we don't just say "yay, you did what you were trying to."
For misses, we need another step of transmutation before we learn about success or failure: the GM move. Most GM moves are some kind of consequence that is tangential to what the PC was trying for: losing things, getting split up, being caught in a bad spot, noticing a temptation, etc. None of those moves say "you fail", so when the GM makes those moves, she can say that you succeeded at your main intent anyway. The only time the GM has to say you failed at your intent is when she picks a GM move that has in-game effects incompatible with whatever the PC was attempting to achieve. If the GM is making a soft move, succeeding at the original task is one of the ways of making a GM move softer. If the GM is making a hard move, then failure is likely part of the package. Success and failure, then, depend on what the in-game effect of the GM's move is. The miss isn't the failure, it's the consequences of the move the GM chooses.
Second, mechanical difficulty isn't determined before the roll. Instead, the situation has already been set up using soft (or sometimes hard) GM moves in response to the players declared what they're doing, saying, and looking at, and whatever else is happening at the same time. That sets the in-game difficulty of the situation, but the difficulty only gets applied to the task at hand after the roll, and only on a miss or if the move's rules says the GM gets to add something unfortunate. On a success, the difficulty simply didn't matter.
Defy Danger seems like an exception to this because it often happens before other moves, but it's not. The GM can't just call for Defy Danger on a whim, it has to be because the PC is doing something that puts them in danger—so it triggers independently of the GM's wishes, based on the fair judgement of the situation instead, just like every other move. It's the choice to do something that exposes them to danger that increases the difficulty, not the GM deciding it should be harder and assigning a mechanical modifier.
When a player declares their PC is attempting something that matches a move without exposing themself to danger first, the group just does what the move says. There is no difficulty involved at first. The roll is made, and a hit means they did what they were trying to do, easy as pie. Only when the roll gives the GM an opportunity to make a move, or to say something as part of the move's instructions, does the GM bring difficulty into the picture after the fact. If the roll "should" have been harder in some other system, now is the time to show just how hard it was by using a hard move to show the hard consequences of the risks taken. If the roll "should" have been easy, now is the time to show that it was easy by making a soft, kind move that might even include success, despite the miss.
A quick example might help demonstrate the difference: Imagine the Fighter is Hack & Slashing against an orc, and misses. The GM chooses a move that follows (a GM principle), choosing to have the Fighter slip and go to one knee as the orc grins and winds up for a strong overhand chop (soft move: show signs). Now imagine the Fighter is Hack & Slashing against a roper (a toothy tentacle monster), and misses. The GM chooses a move that follows, choosing to have the Fighter grabbed and pinned by three wrapping tentacles (hard move: put them in a spot). That's much worse! Fighting the roper is harder than fighting an orc, but it's harder because it's more dangerous when the Fighter messes up, not because of plusses or minuses.
This actually goes for hits too: even when you roll a hit, what you were trying to do... maybe that actually makes your life harder, and you did that all by yourself without the GM's help. Again, it's the situation caused by the result of the chain of moves that makes things harder or easier after the roll. That's not something that could be modelled by a bonus or penalty to the roll anyway, but it neatly flows from the non-bonus-based difficulty rules of Dungeon World.
You could make things mechanically more difficult by making custom moves, but I would not start there. Instead, narrate in the fiction greater consequences, via the GM's show an approaching/looming threat moves and follow up with hard moves, tougher monsters, and more damaging challenges.
For a boss with giant sword, lead up to the fact that he's dangerous: share that he's killed twenty knights in one-on-one combat, show carnage on the way to him, have him be encrusted with gore. The 16 HP Dragon is an example of what can be done. You could also make him mechanically tougher, of course.
I wouldn't make the lockpicking roll harder, but I would just ramp up the consequences of failure—again, appropriately telegraphed so that when they lose a finger or hand when they botch (6-) the roll they aren't completely surprised.
Can't I try to dodge the attack?
Why do I get damage automatically and why does it pass my armor?
Explain when you, as GM, are allowed to make hard moves like this. If they ask why their armor doesn't help, ask back: Do you think it's going to help? You got hit by a tree and landed on a wall. Do you think it would help you if you fell down the stairs? Remember the GM principle "Ask questions and use the answers", but also talk to your players so that they understand how the game works.
The simple answer is use hard moves.
When you have a chance to make a move, you get to choose whether it is a soft move, announcing danger; or a hard move, delivering on the promise of that danger.
So if you think fights are too easy, use more hard moves.
My experience is that players like it when they succeed, but bleed. That means they get through just by the skin of their teeth - they face dangers that do them harm and make tough choices and live with the consequences - but they achieve their goals. Maybe with a price they hadn't anticipated, but they succeed just the same. They will deny it, they will tell you that they like the succeed part just by itself, but they are wrong or they are lying. They like encounters that balance on the razor's edge - and why not? If the encounter was too much for them, they'd have been smarter to find a way around it. If it was a cakewalk, why even bother playing it out? It's the instances where it can go either way that are worth playing.
One of the great things about Dungeon World is that it allows you to dynamically adjust the difficulty of encounters instead of having to set everything up exactly beforehand. If your PCs are bowling over the goblins, they're going to succeed, but if you need to make them bleed before they do, use the horde tag to make them call for backup from their kin! And use organized to make sure they fight like intelligent creatures fighting for their lives - they'll volley with arrows and divide-and-conquer!
I highly recommend reading The Dungeon World Guide - it provides lots of great examples specifically on combat and hard-vs-soft moves.
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