# Exceptional sequences and rooted labeled forests

Kiyoshi Igusa<EMAIL_ADDRESS>and  Emre Sen<EMAIL_ADDRESS>
###### Abstract.

We give a representation-theoretic bijection between rooted labeled forests
with $n$ vertices and complete exceptional sequences for the quiver of type
$A_{n}$ with straight orientation, i.e., for each labeled rooted forest we
construct a unique exceptional sequence. The ascending and descending vertices
in the forest correspond to relatively projective and relatively injective
objects in the exceptional sequence. We conclude that every object in an
exceptional sequence is either relatively projective or relatively injective
or both.

We construct a natural action of the extended braid group on rooted labeled
forests and show that it agrees with the known action of the braid group on
complete exceptional sequences. We also describe the action of $\Delta$, the
Garside element of the braid group, on rooted labeled forest using
representation theory.

###### 2020 Mathematics Subject Classification:

16G20: 05C05

Supported by the Simons Foundation

## Introduction

For a quiver $Q$ of Dynkin type, an _exceptional sequence_ is a sequence of
indecomposable representations $(E_{1},\cdots,E_{k})$ of $Q$ over any field
such that

$\operatorname{Hom}(E_{j},E_{i})=0=\operatorname{Ext}(E_{j},E_{i})$

for all $i<j$. The exceptional sequence is called _complete_ if $k=n$, the
number of vertices of $Q$. Exceptional sequences first occurred as sequences
of line bundles over projective space [11], [22]. The number of exceptional
sequences for a quiver of any Dynkin type is well-known [24], [20], see also
[18].

In this paper we consider the quiver $A_{n}$ with straight orientation:

$1\to 2\to 3\to\cdots\to n.$

The number of exceptional sequences of type $A_{n}$ is $(n+1)^{n-1}$. This is
also the number of parking functions and rooted labeled forests. A bijection
between parking functions and complete exceptional sequences was given in [12]
and can be used to construct a bijection between complete exceptional
sequences and rooted labeled forests via Prüfer codes. We give a new proof of
this in Section 1 and an example in Section 3.3.

The purpose of this paper is to prove the two variable generating function for
exceptional sequences Theorem D. (Using variables $a,b$ with $c=1$.) We first
tried the known indirect bijection using parking functions and Prüfer codes
but that did not work to prove Theorem D. The bijection which works can be
given by combining [13], [19] with other bijections. (See Section 2.4.)

In Section 2 we construct a direct bijection between complete exceptional
sequences and rooted labeled forest giving a representation theoretic
interpretation of the relation between nodes. As we show in Section 3.3, our
bijection is different from the one coming from the bijection with parking
functions.

An indecomposable representation of a linear quiver of type $A_{n}$ is given
by its support which is a closed interval $[a,b]$ where $1\leq a\leq b\leq n$.
The representation with this support, denoted $M_{ab}$, has _top_ the simple
module at $a$ and _socle_ or _bottom_ the simple module at $b$.

######  Theorem A.

[12] (Theorem 1.3) For the linear $A_{n}$ quiver, a bijection between complete
exceptional sequences and parking functions is given by assigning to each
exceptional sequence $(E_{1},\cdots,E_{n})$ the sequence of tops
$(a_{1},\cdots,a_{n})$ where $E_{i}=M_{a_{i},b_{i}}$.

In Section 1 we will repeat the construction of the inverse function (from
parking functions to complete exceptional sequences) in order to compare the
braid group action given in [12] with the new action given in Section 3

######  Theorem B.

$(1)$ (Theorem 2.12) For each complete exceptional sequences
$(E_{1},\cdots,E_{n})$ for $A_{n}$ consider the partial ordering on $[n]$
given by $i<j$ if the support of $E_{i}$ is contained in the support of
$E_{j}$. The Hasse diagram is a rooted labeled forest. Conversely, any rooted
labeled forest is the Hasse diagram of a unique exceptional sequence.

$(2)$ (Theorem 2.20) Furthermore, $E_{i}$ is relatively projective if and only
if either the corresponding node $v_{i}$ is a root of a tree in the forest
(i.e., the support of $E_{i}$ is maximal) or $i<j$ for $E_{j}$ the smallest
object containing $E_{i}$ in its support. Dually, $E_{i}$ is relatively
injective if and only if either $v_{i}$ is a root of a tree in the forest or
$i>j$ for $E_{j}$ the smallest object containing $E_{i}$ in its support.

We observe in Section 2.4 that Theorem B(1) follows from [13] and [3].

######  Corollary C.

(Corollary 2.21) Every object in a complete exceptional sequence for linear
$A_{n}$ is either relatively projective or relatively injective and at least
one object is both.

Using the three variable generating function for rooted labeled forests from
[10] (which is a special case of an earlier result in [9]) we obtain the
following.

######  Theorem D.

(Theorem 2.29) We have a three variable generating function

$\sum a^{p}b^{q}c^{r}=c(a+(n-1)b+c)(2a+(n-2)b+c)\cdots((n-1)a+b+c)$

where the sum is over all complete exceptional sequences $E_{\ast}$ for a
linear quiver of type $A_{n}$ and the monomial $a^{p}b^{q}c^{r}$ is given by
letting

$p=$ number of relatively projective but not relatively injective objects in
$E_{\ast}$,

$q=$ number of relatively injective but not relatively projective objects in
$E_{\ast}$,

$r=$ number of objects in $E_{\ast}$ which are both relatively projective and
relatively injective.

In Section 3 we construct a natural action of the braid group $B_{n}$ on $n$
strands on the set of rooted labeled forests on $n$ vertices and we show in
Theorem 3.1 that this action corresponds to the known action of $B_{n}$ on
complete exceptional sequences.

In Section 4 we extend the braid group action on rooted labeled forests to the
“extended braid group” $\widetilde{B}_{n}=B_{n}\rtimes{\mathbb{Z}}_{2}$ by
adding an outer involution to the braid group which acts by inverting the
labels of the forest (Section 4.4). We also describe the action of the Garside
element $\Delta$ and show on forests how it transforms cluster tilting sets to
signed exceptional sequences (Theorem 4.15).

##  1\. Parking functions

A _parking function_ on $n$ is a function $f:[n]\to[n]$, where
$[n]=\\{1,2,\cdots,n\\}$ so that $f^{-1}[k]$ has at least $k$ elements for all
positive $k\leq n$. This is equivalent to the condition that there are at most
$p$ elements $i\in[n]$ with $f(i)\geq n-p+1$.

The name comes from an interpretation using $n$ cars parking in $n$ spaces
numbered $1$ through $n$. Assume that Car $i$ has preferred parking space
$a_{i}$. If that space is not available, Car $i$ will park in the next
available space. If $f(i)=a_{i}$ is a parking function, each car $i$ will find
a space $b_{i}\geq a_{i}$ and $g(i)=b_{i}$ will give a bijection
$g:[n]\to[n]$.

As an example, suppose $n=4$ and consider the parking function $(1,1,2,2)$.
Assume the cars come in reverse order. Then Car 4 will park in its preferred
spot $b_{4}=2$, Car 3 will go to the next spot $b_{3}=3$, Car 2 will park in
its favorite spot $b_{2}=1$. Finally, Car 1 will park in space $b_{1}=4$. The
result is an exceptional sequence:

$(M_{14},M_{11},M_{23},M_{22}).$

######  Lemma 1.1.

Let $(a_{1},a_{2},\cdots,a_{n})$ be a nondecreasing parking function on $n$.
Let $b_{1},\cdots,b_{n}$ be the permutation of $n$ given by the locations of
the cars with preferred positions $a_{i}$ assuming they park in reverse order
as outlined above. Then $M_{a_{1},b_{1}},\cdots,M_{a_{n},b_{n}}$ is an
exceptional sequence for linear $A_{n}$.

###### Proof.

For $i<j$ we have $\operatorname{Ext}(M_{a_{j},b_{j}},M_{a_{i},b_{i}})=0$
since $a_{i}\leq a_{j}$. It remains to show

$\operatorname{Hom}(M_{a_{j},b_{j}},M_{a_{i},b_{i}})=0$

for $i<j$. When $a_{i}=a_{j}$, we have $b_{i}>b_{j}$ since Car $j$ has parked
in space $b_{j}$ after finding spaces $a_{j}$ to $b_{j}-1$ occupied.
Therefore, Car $i$ will find that spaces $a_{i}$ to $b_{j}$ are taken and must
park in $b_{i}>b_{j}$. Then $M_{a_{j},b_{j}}$ is a quotient of
$M_{a_{i},b_{i}}$ and $\operatorname{Hom}(M_{a_{j},b_{j}},M_{a_{i},b_{i}})=0$.

When $a_{i}<a_{j}$, there is no morphism $M_{a_{j},b_{j}}\to M_{a_{i},b_{i}}$
since $b_{i}$ is not in the closed interval $[a_{j},b_{j}]$ since those spaces
are occupied by the time Car $i$ parks. Thus either

  1. (1)

$b_{i}<a_{j}$, making the supports of the two modules disjoint or

  2. (2)

$b_{i}>b_{j}$ in which case $M_{a_{j},b_{j}}$ is a subquotient of
$M_{a_{i},b_{i}}$ and there is no morphism.

Therefore $M_{a_{1},b_{1}},\cdots,M_{a_{n},b_{n}}$ is an exceptional sequence.
∎

######  Lemma 1.2.

Let $(E_{1},\cdots,E_{n})$ be an exceptional sequence with sequence of tops
$(a_{1},\cdots,a_{n})$. Then, for any $i$, there is another exceptional
sequence with $a_{i},a_{i+1}$ switched in the sequence of tops.

###### Proof.

If $a_{i}=a_{i+1}$ there is nothing to do. Otherwise, suppose $a_{i}<a_{i+1}$.
Then there is a unique object $E_{i+1}^{\prime}$ so that
$(E_{1},\cdots,E_{i+1}^{\prime},E_{i},E_{i+2},\cdots,E_{n})$ is an exceptional
sequence. But, the dimension vector of $E_{i+1}^{\prime}$ is congruent to
$\operatorname{\underline{dim}}E_{i+1}$ modulo
$\operatorname{\underline{dim}}E_{i}$. So, the top of $E_{i+1}^{\prime}$ is
equal to that of $E_{i+1}$, i.e,, the tops $a_{i},a_{i+1}$ have switched. The
case $a_{i}>a_{i+1}$ is similar. ∎

######  Theorem 1.3.

[12](Theorem A) The construction above gives a unique exceptional sequence
having any given parking function as sequence of tops.

###### Proof.

By Lemma 1.1, the construction works for any nondecreasing parking function.
By Lemma 1.2, we can permute the parking function. Since the two sets have the
same cardinality $(n+1)^{n-1}$, this construction gives a bijection. ∎

##  2\.  Rooted labeled forests

We will construct a 1-1 correspondence between complete exceptional sequences
for linear $A_{n}$ and rooted labeled forest. Then we derive some
representation theoretic consequences.

###  2.1. The Hasse diagram of an exceptional sequence

We consider the Hasse diagram of an exceptional sequence and derive some of
its properties. For example, we show that it is a rooted forest.

######  Definition 2.1.

A _rooted labeled forest_ is a forest with vertices numbered $1$ through $n$
and a root chosen for each component. By adding an extra vertex labeled 0 and
attaching it to each of the roots, such a structure is equivalent to a tree
with $n+1$ vertices labeled $0$ through $n$.

######  Remark 2.2.

The nodes of a rooted tree are partially ordered by paths to the root.
Conversely, the Hasse diagram of a finite poset is a rooted forest if it has
the property that each node has at most one parent. In that case, the maximal
elements are roots of disjoint trees.

Let $E_{\ast}=(E_{1},\cdots,E_{n})$ be a complete exceptional sequence for
$A_{n}$. (Unless otherwise noted, we always take the orientation $1\to
2\to\cdots\to n$.) We define a partial ordering on the objects $E_{i}$ by
$E_{i}\leq E_{j}$ if the support of $E_{i}$ is contained in the support of
$E_{j}$. In other words, $a_{j}\leq a_{i}\leq b_{i}\leq b_{j}$. We claim that
the Hasse diagram of this partial ordering is a rooted labeled forest. This
follow from the following lemma.

######  Lemma 2.3.

Given two objects $E_{i}=M_{a_{i},b_{i}}$ and $E_{j}=M_{a_{j},b_{j}}$, the
intervals $[a_{i},b_{i}]$, $[a_{j},b_{j}]$ are “noncrossing”, i.e. either
these closed interval are disjoint or one is contained in the other.

###### Proof.

Given two objects whose supports cross, say, $M_{ab}$, $M_{cd}$ where $a<c\leq
b<d$, we have

$\operatorname{Hom}(M_{cd},M_{ab})\neq 0,\operatorname{Ext}(M_{ab},M_{cd})\neq
0.$

So, both pairs $(M_{ab},M_{cd})$ and $(M_{cd},M_{ab})$ are not exceptional.
Therefore, they cannot occur in an exceptional sequence in either order. ∎

######  Lemma 2.4.

The set of elements above $E_{i}$ form a linearly ordered set.

###### Proof.

Suppose $E_{i}\leq E_{j},E_{k}$. Then the supports of $E_{j},E_{k}$ meet and
therefore, by Lemma 2.3, one is contained in the other. So, they are linearly
ordered. ∎

######  Proposition 2.5.

The Hasse diagram of the object $E_{i}$ in an exceptional sequence form a
rooted labeled forest. Moreover, the roots are the maximal elements.

###### Proof.

By Lemma 2.4 a node cannot have two parents. Therefore, each connected
component of the Hasse diagram is a tree with root at the top. ∎

######  Example 2.6.

Consider the exceptional sequence for $A_{7}$:

$(S_{4},S_{5},M_{15},M_{67},S_{6},M_{12},S_{1})$

The corresponding Hasse diagram is:

---
$\textstyle{E_{3}=M_{15}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{4}=M_{67}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{1}=S_{4}}$$\textstyle{E_{2}=S_{5}}$$\textstyle{E_{6}=M_{12}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{5}=S_{6}}$$\textstyle{E_{7}=S_{1}}$

We note that the leaves are simple modules. More generally, we have the
following.

######  Proposition 2.7.

The length of the object $E_{i}=M_{a_{i},b_{i}}$ (given by $b_{i}-a_{i}+1$) is
the number of objects $E_{j}$ which are $\leq E_{i}$ in the partial ordering.
In particular, the leaves (the minimal elements) have length 1, i.e., they are
simple modules.

Given a rooted labeled tree, define the _weight_ of any node $v_{i}$ to be the
number of nodes which are $\leq v_{i}$. The statement of the proposition is
that the weight of $v_{i}$ as a node in the Hasse diagram is equal to the
length of the corresponding module $E_{i}$.

###### Proof.

We use the fact that the dimension vectors
$\operatorname{\underline{dim}}E_{i}$ form a vector space basis for
${\mathbb{R}}^{n}$. Suppose that $E_{i}=M_{ab}$ with length $k=b-a+1$. Take
the linear map $\pi:{\mathbb{R}}^{n}\twoheadrightarrow{\mathbb{R}}^{k}$ given
by projection onto coordinates $a$ through $b$. Then
$\operatorname{\underline{dim}}E_{i}$ maps to the vector $(1,1,\cdots,1)$. For
any $E_{j}$ whose support is not contained in the interval $[a,b]$ we have by
Lemma 2.3 that the support of $E_{j}$ is either disjoint from the interval
$[a,b]$ or it contains that interval. In either case,
$\pi(\operatorname{\underline{dim}}E_{j})$ is a scalar multiple of
$\operatorname{\underline{dim}}E_{i}$. Therefore, the image of $\pi$ is
spanned by $\pi(\operatorname{\underline{dim}}E_{j})$ for those $E_{j}$ with
support in $[a,b]$. So, there must be at least $k$ of these. However, there
are at most $k$ linearly independent vectors with support in $[a,b]$.
Therefore, the number of $E_{j}$ with support in $[a,b]$ is exactly $k$, the
length of $E_{i}$. So, the length of $E_{i}$ is the weight of $v_{i}$ as
claimed. ∎

Proposition 2.5 gives a mapping

$\left\\{\text{complete exceptional sequence for
$A_{n}$}\right\\}\xrightarrow{H}\left\\{\text{rooted labeled forests on
$n$}\right\\}$

We claim that this map is a bijection. Since the two sets have the same
cardinality it will suffice to show that the map is surjective. So, for every
rooted labeled forest on $n$ we will construct the unique exceptional sequence
having that forest as Hasse diagram. We will not prove uniqueness directly.
Uniqueness follows from the counting argument.

###  2.2. The exceptional sequence of a rooted labeled forests

We give an explicit construction of the necessarily unique complete
exceptional sequence for any rooted labeled forest. The construction will be
recursive.

Given $i\in[n]$, let $v_{i}$ be the node labeled $i$ in the forest $F$. Let
$T_{i}$ be the tree having root $v_{i}$, i.e., $T_{i}$ consists of all nodes
$v_{j}\leq v_{i}$ in $F$. Then $v_{j}\leq v_{i}$ if and only if $T_{j}\subset
T_{i}$.

First, consider the connected case, i.e., $F$ is a tree. Let $v_{r}$ be the
root of $F$, i.e., $F=T_{r}$. Let $v_{i_{1}},v_{i_{2}},\cdots,v_{i_{k}}$ be
the children of the root $v_{r}$. Without loss of generality, we may choose
the labels so that

$i_{1}<i_{2}<\cdots<i_{p}<r<i_{p+1}<\cdots<i_{k}$

where $0\leq p\leq k$. Recall that the weight $w(v_{i})$ is the size of
$T_{i}$. In particular, the weight of $v_{r}$ is $n$ and

$n=1+\sum w(v_{i_{j}})$

We assign the module $E_{r}:=M_{1n}$ to $v_{r}$. Since $M_{1n}$ is uniserial,
it has a filtration with $k+1$ subquotients, one of length 1 and the others of
length $w(v_{i_{j}})$. Denote the filtration:

$0\subset A_{p}\subset A_{p-1}\subset\cdots\subset A_{2}\subset A_{1}\subset
B_{k}\subset\cdots\subset B_{p+1}\subset B_{p}=M_{1n}$

We may chose these so that $A_{x}/A_{x-1}$ has length $w(v_{i_{x}})$ when
$x\leq p$ and $B_{y-1}/B_{y}$ has length $w(v_{i_{y}})$ for $y\geq p+1$. (This
makes $B_{k}/A_{1}$ simple, of length 1.) Then we let

$E_{i_{x}}=A_{x}/A_{x-1},\qquad E_{i_{y}}=B_{y-1}/B_{y}.$

######  Proposition 2.8.

$(E_{i_{1}},\cdots,E_{i_{p}},E_{r},E_{i_{p+1}},\cdots,E_{i_{k}})$ is an
exceptional sequence.

###### Proof.

It suffices to show that $(X,Y)$ is an exceptional pair for any subsequence
$X,Y$ of length 2.

Case 1: $X=E_{r}$. Then $Y=E_{i_{y}}$ for $y>p$ which has no morphism to
$E_{r}$ since its support does not contain $n$ by construction.
$\operatorname{Ext}(Y,E_{r})=0$ since $E_{r}$ is injective.

Case 2: $Y=E_{r}$, Then $X=E_{i_{x}}$ for $x\leq p$. These cannot be quotients
of $Y$. Hence there is no morphism $Y\to X$ and
$\operatorname{Ext}(E_{r},X)=0$ since $E_{r}$ is projective.

For the remaining pairs, there is no morphism since they have disjoint
supports.

Case 3: $X=E_{i_{x}}$ where $x\leq p$ and $Y=E_{i_{y}}$ where $y\geq p+1$.
Then the supports of $X,Y$ are disjoint and not consecutive. Therefore there
is no morphism and no extension.

In the remaining cases, $X,Y$ are on the same side of $E_{r}$. So, their
supports are disjoint and $\operatorname{Ext}(Y,X)=0$ since, in any extension,
$Y$ would be the submodule.∎

We come to the recursion step. Take any subtree of $F$ with root $v_{s}$.
Suppose the corresponding module has been constructed, say $E_{s}=M_{ab}$ with
length $b-a+1=w(v_{s})$. In particular $E_{s}$ will be simple when $v_{s}$ is
a leaf. When $v_{s}$ is not a leaf, we repeat the construction above for the
children $v_{j_{z}}$ of $v_{s}$. This will assign modules $E_{j_{z}}$ of
length equal to the weight $w(v_{j_{z}})$ having supports which are closed
disjoint subintervals of $[a,b]$ with union $[a,b]$ minus one point which we
call the “gap”. (In the construction above, the gap is the support of the
simple module $B_{k}/A_{1}$.) Continuing in this way we construct all modules
$E_{t}$ in the case when $F$ is connected.

Now consider the case when $F$ is a disjoint union of trees
$T_{1},\cdots,T_{k}$ with roots $v_{j_{1}},\cdots,v_{j_{k}}$ where
$j_{1}<j_{2}<\cdots<j_{k}$. Then the weights $w(v_{j_{z}})$ add up to $n$. Let
$E_{j_{1}},\cdots,E_{j_{k}}$ be the modules with disjoint supports of length
$w(v_{j_{z}})$ which are consecutive. For instance $E_{j_{1}}=M_{1x}$ where
$x=w(v_{j_{1}})$, $E_{j_{2}}=M_{x+1,y}$ where $y=x+w(v_{j_{2}})$, etc.

######  Proposition 2.9.

$(E_{j_{1}},\cdots,E_{j_{k}})$ is an exceptional sequence.

###### Proof.

There are no morphisms between these objects since their supports are
disjoint. For any $x<y$, $\operatorname{Ext}(E_{j_{y}},E_{j_{x}})=0$ since in
any extension, $E_{j_{y}}$ will be the submodule. ∎

######  Remark 2.10.

We say that a sequence of modules $X_{1},\cdots,X_{k}$ have _consecutive
supports_ if $X_{i}=M_{a_{i},b_{i}}$ where each $a_{i}=b_{i-1}+1$. For
example, the modules in Proposition 2.9 have consecutive supports. Modules
with consecutive support form an exceptional sequence.

We list the basic properties of the modules $E_{i}$ corresponding to the nodes
$v_{i}$ under the construction above.

######  Proposition 2.11.

Given any rooted labeled forest $F$, let $E_{i}$ be the module corresponding
to node $v_{i}$ by the above recursive construction. Then

  1. (1)

Each $E_{i}$ has length equal to $w(v_{i})$.

  2. (2)

The support of $E_{i}$ is contained in the support of $E_{j}$ if and only if
$v_{i}\leq v_{j}$.

  3. (3)

The supports of $E_{i},E_{j}$ are disjoint if and only if $v_{i},v_{j}$ are
not comparable.

  4. (4)

If $v_{i_{1}},\cdots,v_{i_{k}}$ are the children of $v_{r}$ and
$i_{1}<i_{2}<\cdots<i_{p}<r<i_{p+1}<\cdots<i_{k}$ then

    1. (a)

$E_{i_{x}}$ is a submodule of $E_{r}$ if and only if $x=p$, i.e., if
$v_{i_{x}}$ is the child of $v_{r}$ with the largest label less than $r$.
(None of the $E_{i_{x}}$ are submodules of $E_{r}$ when $p=0$.)

    2. (b)

$E_{i_{y}}$ is a quotient of $E_{r}$ if and only if $y=p+1$, i.e., if
$v_{i_{y}}$ is the child of $v_{r}$ with the smallest label greater than $r$.
(None of the $E_{i_{y}}$ are quotients of $E_{r}$ when $p=k$.)

    3. (c)

$E_{i_{1}},\cdots,E_{i_{p}}$ have consecutive supports.

    4. (d)

$E_{i_{p+1}},\cdots,E_{i_{k}}$ have consecutive supports.

###### Proof.

Properties (1) and (4) follow from the construction. To prove (2) and (3) we
note that, for any $i,j$ there are only two possibilities: either one is
$\leq$ the other or they are not comparable. In the first case, suppose
$v_{i}<v_{j}$. Then there is a sequence
$v_{i}=v_{k_{0}}<v_{k_{1}}<v_{k_{2}}<\cdots<v_{j}$ where each $v_{k_{x}}$ is a
child of $v_{k_{x+1}}$ in that case,
$\operatorname{supp}E_{k_{x}}\subset\operatorname{supp}E_{k_{x+1}}$ by
construction. So, $\operatorname{supp}E_{i}\subset\operatorname{supp}E_{j}$.
In the second case, either $v_{i},v_{j}$ lie in disjoint trees with distinct
roots $v_{k},v_{\ell}$ in which case $E_{k}$ and $E_{\ell}$ have disjoint
supports by Proposition 2.9. Otherwise, $v_{i},v_{j}$ lie in the same tree.
Let $v_{k}$ be the smallest node $\geq v_{i},v_{j}$. Then $v_{i},v_{j}$ are
$\leq$ two children of $v_{k}$ where the corresponding modules have disjoint
supports. So, $E_{i},E_{j}$ have disjoint supports. ∎

######  Theorem 2.12.

(Theorem B(1)) Given any rooted labeled forest $F$, let $E_{i}$ be the module
corresponding to node $v_{i}$ by the above recursive construction. Then

$(E_{1},E_{2},\cdots,E_{n})$

is a complete exceptional sequence.

To prove this theorem we consider the three cases when $E_{i}$ is a submodule
of $E_{j}$, $E_{i}$ is a quotient module of $E_{j}$ and $E_{i},E_{j}$ extend
each other.

######  Lemma 2.13.

If $E_{i}$ is a submodule of $E_{j}$ and $i\neq j$ then $v_{i}<v_{j}$ and the
nodes $v_{k_{1}},\cdots,v_{k_{t}}$ connecting $v_{i}$ to $v_{j}$ in $F$ have
increasing labels: $i<k_{1}<k_{2}<\cdots<k_{t}<j$.

###### Proof.

Since $v_{i}<v_{k_{t}}<v_{j}$,
$\operatorname{supp}E_{i}\subset\operatorname{supp}E_{k_{t}}\subset\operatorname{supp}E_{j}$.
Since $E_{i}$ contains the socle of $E_{j}$, so does $E_{k_{t}}$. So,
$E_{k_{t}}$ is a submodule of $E_{j}$ and $v_{k_{t}}$ is a child of $v_{j}$.
It follows that the label must be smaller by the construction: $k_{t}<j$. By
the same argument $k_{p}<k_{p+1}$ for all $p$ and $i<k_{1}$. ∎

######  Lemma 2.14.

If $E_{i}$ is a quotient module of $E_{j}$ and $i\neq j$ then $v_{i}<v_{j}$
and the nodes $v_{k_{1}},\cdots,v_{k_{t}}$ connecting $v_{i}$ to $v_{j}$ in
$F$ have decreasing labels: $i>k_{1}>k_{2}>\cdots>k_{t}>j$.

###### Proof.

This statement is dual to Lemma 2.13 and has an analogous proof. ∎

######  Lemma 2.15.

$\operatorname{Ext}(E_{i},E_{j})\neq 0$ if and only if $E_{i},E_{j}$ have
(disjoint) consecutive supports and, in that case, $E_{i}$ is a submodule of
some $E_{k}$ and $E_{j}$ is a quotient module of some $E_{\ell}$ where
$v_{k},v_{\ell}$ are sibling and

$i\leq k<\ell\leq j.$

###### Proof.

Let $\operatorname{Ext}(E_{i},E_{j})\neq 0$. Then either the supports are
consecutive or they overlap. But the second case is excluded by Proposition
2.11 (2), (3). The converse is clear.

Suppose the supports of $E_{i},E_{j}$ are consecutive, say
$\operatorname{supp}E_{i}=[a,b]$, $\operatorname{supp}E_{j}=[b+1,c]$. Let
$v_{k}$ be maximal in $F$ so that $v_{i}\leq v_{k}$ and $v_{j}\not\leq v_{k}$.
Then $E_{k}$ contains the support of $E_{i}$ and is disjoint from the support
of $E_{j}$. So, $b\in\operatorname{supp}E_{k}$ and
$b+1\notin\operatorname{supp}E_{k}$ making $E_{i}$ a submodule of $E_{k}$. By
Lemma 2.13, $i\leq k$.

Similarly, let $v_{\ell}$ be maximal in $F$ so that $v_{j}\leq v_{\ell}$ and
$v_{i}\not\leq v_{\ell}$. Then $\ell\leq j$ by Lemma 2.14. By maximality of
$v_{k},v_{\ell}$, they must be sibling (or roots of disjoint trees). In either
case, $k<\ell$ since $E_{k},E_{\ell}$ have consecutive supports. (This follows
from either Proposition 2.11(4) or Proposition 2.9.) This proves the lemma. ∎

###### Proof of Theorem 2.12.

These lemmas imply that $(E_{1},\cdots,E_{n})$ is a complete exceptional
sequence. Indeed, if $i<j$ we have $\operatorname{Hom}(E_{j},E_{i})=0$ since
$E_{j}$ is not a submodule of $E_{i}$ by Lemma 2.13 and $E_{i}$ is not a
quotient module of $E_{j}$ by Lemma 2.14. Also,
$\operatorname{Ext}(E_{j},E_{i})=0$ by Lemma 2.15. ∎

######  Example 2.16.

It is easy to convert a rooted labeled forest into a complete exceptional
sequence. Start with the following.

132456789

By Remark 2.10, the modules $E_{3},E_{5}$ corresponding to the roots
$v_{3},v_{5}$ have consecutive supports. By Proposition 2.11 (1), these
supports have lengths 3,6. So, $E_{3}=M_{13}$ and $E_{5}=M_{49}$. Since $1<3$,
$E_{1}$ is the length 2 submodule of $E_{3}$. So, $E_{1}=M_{23}$. By
Proposition 2.11(4), $E_{6}$ has support at $4$ followed by a gap, then
$E_{2},E_{4}$ have consecutive supports in the interval $[6,9]$. So,
$E_{6}=S_{4}$, $E_{2}=S_{6}$ and $E_{4}=M_{79}$. The children of $v_{4}$ are
leaves with larger labels, so the corresponding modules are simples of
injective type: $E_{7}=S_{7}$, $E_{9}=S_{8}$. The complete exceptional
sequence is

$E_{\ast}=(M_{23},S_{6},M_{13},M_{79},M_{49},S_{4},S_{7},S_{2},S_{8})$

This same example is done using chord diagrams following [13] in Figure 1.

###  2.3. Relatively projective and injective objects

######  Definition 2.17.

An object $E_{i}$ in a complete exceptional sequence $(E_{1},\cdots,E_{n})$ is
called _relatively projective_ if it is a projective object of the full
subcategory of $mod\text{-}\Lambda$ of all modules $Y$ so that

$\operatorname{Hom}_{\Lambda}(E_{j},Y)=0=\operatorname{Ext}_{\Lambda}(E_{j},Y)$

for all $j>i$. This is called the _right perpendicular_ category of
$Z=E_{i+1}\oplus\cdots\oplus E_{n}$ and is denoted $Z^{\perp}$.

Similarly, an object $E_{i}$ in a complete exceptional sequence
$(E_{1},\cdots,E_{n})$ is called _relatively injective_ if it is an injective
object of the full subcategory of $mod\text{-}\Lambda$ of all modules $X$ so
that

$\operatorname{Hom}_{\Lambda}(X,E_{j})=0=\operatorname{Ext}_{\Lambda}(X,E_{j})$

for all $j<i$. This is called the _left perpendicular_ category of
$Z=E_{1}\oplus\cdots\oplus E_{i-1}$ and is denoted $\,{}^{\perp}Z$.

######  Proposition 2.18.

If any two of the objects in a short exact sequence $0\to A\to B\to C\to 0$
lie in $Z^{\perp}$ (or $\,{}^{\perp}Z$) then so does the third.

###### Proof.

This follows from the $6$-term exact sequence:

$\begin{split}0\to&\operatorname{Hom}(Z,A)\to\operatorname{Hom}(Z,B)\to\operatorname{Hom}(Z,C)\\\
&\to\operatorname{Ext}(Z,A)\to\operatorname{Ext}(Z,B)\to\operatorname{Ext}(Z,C)\to
0.\end{split}$

An analogous argument works for $\,{}^{\perp}Z$. ∎

Since $X\in Y^{\perp}$ if and only if $Y\in\,^{\perp}X$, we get the following.

######  Corollary 2.19.

For any short exact sequence $0\to A\to B\to C\to 0$, $(A\oplus
B)^{\perp}=A^{\perp}\cap B^{\perp}$ is contained in $C^{\perp}$ and similarly,
$\,{}^{\perp}(A\oplus B)\subseteq\,^{\perp}C$.∎

######  Theorem 2.20.

(Theorem B(2)) Let $E_{\ast}=(E_{1},\cdots,E_{n})$ be a complete exceptional
sequence for a linear quiver of type $A_{n}$. Let $F$ be the corresponding
rooted labeled forest. For each $E_{i}$ in $E_{\ast}$, let $v_{i}$ be the
corresponding node of $F$. Then,

  1. (1)

$E_{i}$ is relatively projective and relatively injective if and only if
$v_{i}$ is a root of $F$.

  2. (2)

For $v_{i}$ not a root of $F$, let $v_{j}$ be the parent of $v_{i}$. Then,
$E_{i}$ is relatively projective, resp. injective in $E_{\ast}$ if and only if
$i<j$, resp $i>j$.

######  Corollary 2.21.

(Corollary C) Every object in a complete exceptional sequence of type $A_{n}$
with straight orientation is either relatively projective or relatively
injective and at least one object is both.∎

To prove Theorem 2.20, let $v_{i_{1}},\cdots,v_{i_{k}}$ be the children of
$v_{j}$ with

$i_{1}<i_{2}<\cdots<i_{p}<j<i_{p+1}<\cdots<i_{k}.$

Then we have by construction that $E_{j}=M_{ab}$ and
$E_{i_{1}},\cdots,E_{i_{p}}$ is a sequence of modules with consecutive support
in the interval $[a+1,b]=\operatorname{supp}rM_{ab}$ and so that $E_{i_{p}}$
is a submodule of $M_{ab}$. In more details, we have
$E_{i_{s}}=M_{a_{s},b_{s}}$ where $a<b_{1}<b_{2}<\cdots<b_{p}=b$ and
$a_{s}=b_{s-1}+1$ for $1<s\leq p$.

######  Lemma 2.22.

The objects $E_{i_{y}}$ for $y>p$ are not relatively projective in $E_{\ast}$.

###### Proof.

Let $W$ be the direct sum of the objects $E_{t}$ which come after $E_{i_{y}}$,
i.e., $t>i_{y}$. We will show that $E_{t}$ is not a projective object of
$W^{\perp}$.

Claim: For any $p<x\leq y$, $W^{\perp}$ contains a module $M_{cb}$ with socle
$S_{b}$ which maps onto $E_{i_{x}}$.

Since $E_{i_{x}}$ is a proper quotient of $M_{cb}$, $E_{i_{x}}$ cannot be
projective in $W^{\perp}$ for $p<x\leq y$. When $x=y$ this proves the lemma.
So, it suffices to prove this claim.

The proof of the claim is by induction on $x$. For $x=p+1$, $E_{i_{p+1}}$ is a
quotient of $E_{j}=M_{ab}$ which lies in $W^{\perp}$ since it is to the left
of $E_{i_{y}}$ in $E_{\ast}$. We take $c=a$ in that case. Suppose the claim
holds for $x<y$. Then, by induction, there is a module $M_{cb}\in W^{\perp}$
which maps onto $E_{i_{x}}$ which also lies in $W^{\perp}$. Therefore the
kernel $M_{db}$ lies in the abelian category $W^{\perp}$. But $M_{db}$ maps
onto $E_{i_{x+1}}$ since $E_{i_{x}},E_{i_{x+1}}$ have consecutive supports.
So, the claim holds for $x+1$. This proves the claim and the lemma. ∎

######  Lemma 2.23.

$rM_{ab}$ is a projective object in the perpendicular category
$M_{ab}^{\perp}$.

###### Proof.

When $b=n$, $rM_{ab}$ is projective and lies in $M_{ab}^{\perp}$.

For $b<n$, $rM_{ab}$ is a quotient of $\tau M_{ab}=M_{a+1,b+1}$. So,
$\operatorname{Hom}(rM_{ab},\tau M_{ab})=0$ and
$\operatorname{Hom}(M_{ab},rM_{ab})=0$. So, $rM_{ab}\in M_{ab}^{\perp}$.

To show that $rM_{ab}$ is projective in $M_{ab}^{\perp}$, suppose not. Then
there is another object $X\in M_{ab}^{\perp}$ and an epimorphism
$X\twoheadrightarrow rM_{ab}$. But then $X=M_{a+1,c}$ for some $c>b$. Then we
have an epimorphism $X\twoheadrightarrow\tau M_{ab}$ contradicting the
assumption that $X\in M_{ab}^{\perp}$. So, $rM_{ab}$ is projective in
$M_{ab}^{\perp}$. ∎

Lemma 2.23 is the case $s=p$ of the following lemma:

######  Lemma 2.24.

$M_{a+1,b_{s}}$ is a projective object in the perpendicular category
$W^{\perp}$ where $W=M_{ab}\oplus E_{i_{s+1}}\oplus\cdots\oplus E_{i_{p}}$.

###### Proof.

For $b_{s}=b$, $M_{a+1,b_{s}}=rM_{ab}$ and the previous lemma applies.

For $b_{s}<b$, $M_{a+1,b_{s}}$ is a quotient of $\tau M_{ab}=M_{a+1,b+1}$. So,
$M_{a+1,b_{s}}\in M_{ab}^{\perp}$. Also, $M_{a+1,b_{s}}\in E_{i_{t}}^{\perp}$
for $s<t\leq p$ since the support of $E_{i_{t}}$ lies in $(b_{s},n]$. So,
$M_{a+1,b_{s}}\in W^{\perp}$

The module $M_{a,b_{s}}$ is an iterated quotient of components of $W$ since
$M_{a,b_{p-1}}=M_{ab}/E_{i_{p}}$ and $M_{a,b_{s}}=M_{a,b_{s+1}}/E_{i_{s+1}}$.
Therefore, $W^{\perp}\subset M_{a,b_{s}}^{\perp}$ by Corollary 2.19.
$M_{a+1,b_{s}}$ is projective in $M_{a,b_{s}}^{\perp}$ by the previous lemma
and therefore it is projective in $W^{\perp}$. ∎

######  Theorem 2.25.

Let $v_{i}$ be a child of $v_{j}$ in the rooted labeled forest $F$. Then the
object $E_{j}$ in the corresponding complete exceptional sequence
$E_{\ast}=(E_{1},\cdots,E_{n})$ is relatively projective if and only if $i<j$.

###### Proof.

If $i>j$ then, by Lemma 2.22, $E_{i}$ is not relatively projective. So,
suppose $i<j$. Then, in the notation above, $i=i_{s}$ for some $s\leq p$.

Let $W$ be the direct sum of all objects $E_{k}$ in the exceptional sequence
with $k>i$. This includes the summands of $W_{0}=E_{j}\oplus
E_{i_{s+1}}\oplus\cdots\oplus E_{i_{p}}$. Thus $W^{\perp}\subset
W_{0}^{\perp}$. By Lemma 2.24, $W_{0}^{\perp}$ contains as a projective object
the module $M_{a+1,b_{s}}$ which has $E_{i_{s}}=M_{a_{s},b_{s}}$ as a
submodule. Since $W_{0}^{\perp}$ is a hereditary abelian category, this makes
$E_{i_{s}}$ a projective object of $W_{0}^{\perp}$. So, $E_{i_{s}}$ is a
projective object of $W^{\perp}\subset W_{0}^{\perp}$. ∎

We have the following dual statement with an analogous argument.

######  Theorem 2.26.

Let $v_{i}$ be a child of $v_{j}$ in the rooted labeled forest $F$. Then the
object $E_{j}$ in the corresponding complete exceptional sequence
$E_{\ast}=(E_{1},\cdots,E_{n})$ is relatively injective if and only if $i>j$.∎

By the discussion above, the only $E_{j}$ which have a chance to be both
relatively projective and relatively injective are ones corresponding to the
roots of the components of $F$. Let $v_{j_{1}},v_{j_{2}},\cdots,v_{j_{k}}$ be
the roots of $F$ with $j_{1}<j_{2}<\cdots<j_{k}$. Then the modules
$E_{j_{1}},\cdots,E_{j_{k}}$ have consecutive supports with union $[1,n]$.

######  Theorem 2.27.

The objects $E_{j_{1}},\cdots,E_{j_{k}}$ corresponding to the roots of the
forest $F$ are relatively projective and relatively injective in the complete
exceptional sequence $E_{\ast}$.

###### Proof.

First, $E_{j_{k}}$ is projective since it is a submodule of the projective
module $M_{1n}$. Then, as in the proof of Lemma 2.24, the quotient
$M_{1n}/E_{j_{k}}$ is a projective object in the category $E_{j_{k}}^{\perp}$.
Since $E_{j_{k-1}}$ is a submodule of $M_{1n}/E_{j_{k}}$, it also becomes
projective in $E_{j_{k}}^{\perp}$ and therefore projective in $W^{\perp}$
where $W$ is the direct sum of all $E_{t}$ for $t>j_{k-1}$. Repeating this
process, we see that each $E_{j_{t}}$ is relatively projective in $E_{\ast}$.

The dual arguments show that each $E_{j_{t}}$ is relatively injective in
$E_{\ast}$. ∎

Theorem 2.20 is given by combining Theorems 2.25, 2.26 and 2.27.

###  2.4. Chord diagrams and rooted labeled trees

First, we note that a rooted forest with $n$ labeled vertices is equivalent to
a rooted tree with $n+1$ vertices and $n$ labeled edges. (Just move the edge
labels to the endpoint furthest from the root, delete the root of the tree,
then declare the adjacent vertices to be roots of the resulting forest.)

A bijection between exceptional sequences on quiver of type $A_{n}$ (with any
orientation) and rooted trees wth $n$ labeled edges is given as follows.
(Thanks to Hugh Thomas for explaining this to the first author many years
ago.)

12345678901$\ast$53972684$\ast$123456789 Figure 1. The exceptional sequence
$(M_{23},S_{6},M_{13},M_{79},M_{49},S_{4},S_{7},S_{2},S_{8})$ from Example
2.16 is drawn as the sequence of chords $(24,67,14,70,40,45,78,23,89)$ in red.
The corresponding tree with root is in the region containing the arc $0$ to
$1$ is indicated in blue. Deleting the root gives the rooted labeled forest in
Example 2.16.

Goulden and Yon [13] (and earlier [19]) showed that rooted trees with $n$
labeled edges are in bijection with factorizations of the $n+1$ cycle
$(012\cdots n)$ as a product of $n$ transpositions $\tau_{i}=(a_{i},b_{i})$.
The bijection is given by drawing chords (labeled with$i$) between points
$a_{i}$ and $b_{i}$ in a circle as indicated in Figure 1. The dual graph of
these chords is the tree. The root is the region adjacent to the arc from $0$
to $1$. Brady and Watt [3] showed that such factorizations of the $n+1$ cycle
are in bijection with exceptional sequences where
$(\tau_{1},\cdots,\tau_{n})=((a_{1},b_{1}),\cdots,(a_{n},b_{n}))$ corresponds
to the exceptional sequence $(M_{a_{1},b_{1}-1},\cdots,M_{a_{n},b_{n}-1})$
where the indices are taken modulo $n+1$.

It is easy to see why the bijection between exceptional sequences and rooted
labeled forests (or trees), as illustrated in Figure 1 is the same as the
bijection given in Theorem 2.12. Any chords in the construction, say $14$,
separates the circle in two parts. One part contains the integers in the
interval $[1,4]$. The other part contains the arc $01$ and therefore the root
of the tree. So, any other chord $(a,b)$ will be further away from the root if
and only if $[a,b]\subsetneq[1,4]$, or, equivalently, $[a,b)\subsetneq[1,4)$.
Therefore, the rooted forest is the Hasse diagram of the set of corresponding
intervals $[a_{i},b_{i})$ which are the supports of the objects in the
exceptional sequence.

The correspondence between exceptional sequences and factorizations of the
Coxeter element of the Weyl group (the $n+1$ cycle in the symmetric group for
$A_{n}$) has been generalized to any quiver in [15]. The correspondence with
chord diagrams is also known for any orientation of $A_{n}$ [8]. See also [17]
for the relation to noncrossing partitions.

###  2.5. Generating function for exceptional sequences

The following theorem and its proof come from [10]. This is a special case of
a result from [9].

######  Theorem 2.28.

For each $n\geq 1$, let $P_{n}(a,b,c)$ be the three variable polynomial for
rooted labeled forest given by $P_{n}(a,b,c)=\sum a^{p}b^{q}c^{r}$ where the
sum is over all rooted labeled forests, $r$ is the number of roots of the
forest, $p$ is the number of nodes $v_{i}$ whose parent $v_{j}$ has label
$j>i$, $q$ is the number of nodes $v_{i}$ with parent $v_{j}$ so that $j<i$.
Then

$P_{n}(a,b,c)=c(a+(n-1)b+c)(2a+(n-2)b+c)\cdots((n-1)a+b+c).$

###### Proof.

We associate a weight to each node of a forest. The weight will be $a,b,c$
depending on which type of node it is. Then the monomial $a^{p}b^{q}c^{r}$ for
a forest is the product of the weights of its nodes.

The proof is by induction on $n$. For $n=1$, $P_{1}(a,b,c)=c$ since there is
only one forest with one node which has weight $c$.

For $n\geq 2$, we write $P_{n}(a,b,c)$ as the sum of two generating function

$P_{n}(a,b,c)=P_{n}(a,b,c)_{1}+P_{n}(a,b,c)_{2}$

the first is for forest in which $v_{1}$ is a root. We remove the root
$v_{1}$, decrease the labels of the other nodes by 1 and keep the weights of
the children of $v_{1}$ to be $b$. This gives a rooted labeled forest with
$n-1$ nodes in which the roots have weight either $b$ or $c$. Since a node
with weight $c$ was removed we have:

$P_{n}(a,b,c)_{1}=cP_{n-1}(a,b,b+c).$

For the remaining forests, we remove the node $v_{1}$ and make the children of
$v_{1}$ roots, but keeping the weight $b$. We obtain a forest with roots of
weight either $b$ or $c$. To reassemble the original forest, we need to choose
one node to be the parent of $v_{1}$, we need to make the root of the tree in
which this node lives to have weight $c$. The other $b$ weighted roots become
the children of $v_{1}$. So,

$P_{n}(a,b,c)_{2}=a(n-1)\frac{c}{b+c}P_{n-1}(a,b,b+c).$

A straightforward calculation proves the result. ∎

Let $P_{A_{n}}(a,b,c)=\sum\lambda_{pqr}a^{p}b^{q}c^{r}$ where $\lambda_{pqr}$
is the number of complete exceptional sequences for linear $A_{n}$ having

  1. (1)

$p$ objects $E_{i}$ which are relatively projective but not relatively
injective,

  2. (2)

$q$ objects which are relatively injective but not relatively projective

  3. (3)

$r$ objects which are both relatively projective and relatively injective.

######  Theorem 2.29.

(Theorem D)

$P_{A_{n}}(a,b,c)=P_{n}(a,b,c)=c\prod_{j=1}^{n-1}(ja+(n-j)b+c).$

###### Proof.

This follows from Theorem 2.28 and the properties of the bijection between
complete exceptional sequences and rooted labeled forests given by Theorem
2.20. ∎

######  Remark 2.30.

Theorem 2.29 implies that there is only one exceptional sequence in which all
objects are both relatively projective and relatively injective. This is never
true for other orientations of $A_{n}$ since such an exceptional sequence is
given by a sequence of simple modules in which the support of each object is a
source in the complement of the supports of the ones on its left and there are
many such sequences for nonlinear $A_{n}$.

##  3\. Braid group action on rooted labeled forests

It is well-known that the braid group acts transitively on the set of complete
exceptional sequences for any Dynkin quiver, in particular, for type $A_{n}$
[6], [21]. In this section we construct an natural action of the braid group
on rooted labeled forests and we show, in Theorem 3.1, that this action
corresponds to the action on complete exceptional sequences under the
bijection given in the previous section. We also show that this action of the
braid group is different from the one given in [12] since we are using a
different bijection between the two sets.

The _braid group_ $B_{n}$ on $n$ strands is given by generators and relations
as follows. The generators are $\sigma_{i}$ for $1\leq i<n$ with two
relations:

  1. (1)

$\sigma_{i},\sigma_{j}$ commute if $|j-i|\geq 2$

  2. (2)

$\sigma_{i}\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1}$.

Given a rooted labeled forest $F$, two node $v_{i},v_{j}$ are said to be
_close_ if one of the following holds.

  1. (1)

$v_{i},v_{j}$ are roots of distinct components of $F$.

  2. (2)

$v_{i},v_{j}$ are sibling, i.e., they have the same parent.

  3. (3)

One of $v_{i},v_{j}$ is the parent of the other.

All remaining cases are not close.

To simplify notation, we add a root $v_{0}$, which we call the _master root_ ,
at the top of the forest giving a rooted tree $F_{+}$. Then Case 1 is a
special case of Case 2: Roots in $F$ are sibling in $F_{+}$.

###  3.1. Braid group action

The braid group $B_{n}$ acts on the set of rooted labeled forests on $n$ as
follows. We give the action of $\sigma_{i}$ for $1\leq i<n$. The partial
ordering on $[n]$ with Hasse diagram $F$ and the one with Hasse diagram
$\sigma_{i}(F)$ are the same for nodes other than $v_{i},v_{i+1}$. We keep the
same notation for these nodes in $\sigma_{i}(F)$. But, we denote by
$v_{i}^{\prime},v_{i+1}^{\prime}$ the nodes of $\sigma_{i}(F)$ with labels
$i,i+1$. These are given as follows.

Case 0: If nodes $v_{i},v_{j}$ are not close then $\sigma_{i}(F)$ is given by
switching the labels $i,i+1$, i.e., $\sigma_{i}(F)=F$ with nodes relabeled:
$v_{i}^{\prime}=v_{i+1},v_{i+1}^{\prime}=v_{i}$.

Case 1: Suppose $v_{i}$ is the parent of $v_{i+1}$, $v_{k}$ is the parent of
$v_{i}$ in $F_{+}$, $X$ is the set of other children of $v_{i}$ and $Y$ is the
set of children of $v_{i+1}$. Then $\sigma_{i}(F)_{+}$ is given by removing
node $v_{i+1}$, relabeling $v_{i}$ to $v_{i+1}^{\prime}$, then adding a new
node $v_{i}^{\prime}$ and making it a child of $v_{i+1}^{\prime}$ and a parent
of all elements of $X$. Thus $Y\cup\\{v_{i}^{\prime}\\}$ is the set of
children of the new node $v_{i+1}^{\prime}$ and $v_{k}$ is the parent of
$v_{i+1}^{\prime}$ in $\sigma_{i}(F)_{+}$.

Case 2: Suppose $v_{i}$ is a child of $v_{i+1}$, $v_{k}$ is the parent of
$v_{i+1}$ in $F_{+}$, $Y$ is the set of other children of $v_{i+1}$ and $X$ is
the set of children of $v_{i}$. Then $\sigma_{i}(F)_{+}$ is given by removing
node $v_{i+1}$, relabeling $v_{i}$ to $v_{i+1}^{\prime}$, then adding a new
node $v_{i}^{\prime}$ which will be the parent of the elements of $Y$. The
nodes $v_{i}^{\prime},v_{i+1}^{\prime}$ will be sibling in
$\sigma_{i}(F)_{+}$, with parent $v_{k}$.

Case 3: Suppose $v_{i},v_{i+1}$ are sibling with parent $v_{k}$ in $F_{+}$ and
$Y,X$ are the sets of children of $v_{i},v_{i+1}$, resp. Then
$\sigma_{i}(F)_{+}$ is given by removing node $v_{i+1}$, relabeling $v_{i}$ to
$v_{i+1}^{\prime}$, then adding a new node $v_{i}^{\prime}$ with parent
$v_{k}$ and children $v_{i+1}^{\prime}$ and the elements of $X$.

Note that $\sigma_{i}$ takes examples from Cases 1,2,3 to examples from Cases
2,3,1 respectively. Furthermore, $\sigma_{i}^{3}$ is the identity map on these
three cases. See Figure 2

|  |
---|---|---
$\textstyle{v_{k}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v_{k}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v_{k}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma_{i}}$$\textstyle{v_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma_{i}}$$\textstyle{v_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma_{i}}$$\textstyle{(\text{Case
1})}$$\textstyle{X}$$\textstyle{Y}$$\textstyle{X}$$\textstyle{Y}$$\textstyle{X}$$\textstyle{Y}$Case
1Case 2Case 3 Figure 2. Cases 1,2,3 of the braid move $\sigma_{i}$ are
indicated above. $\sigma_{i}^{3}$ is the identity in these cases. In all
cases, $v_{k}$ is the smallest node in $F_{+}$ which is above both $v_{i}$ and
$v_{i+1}$ and $Z$ is the set of other children of $v_{k}$.

######  Theorem 3.1.

The action of $\sigma_{i}$ on the set of rooted labeled forests described
above corresponds to the action of $\sigma_{i}$ on exceptional sequences under
the bijection given in the previous section.

######  Corollary 3.2.

The action of $\sigma_{i}$ on the set of rooted labeled forests satisfies the
braid relations and therefore gives an action of the braid group.∎

The following property of $\sigma_{i}$, which follows easily from its
description, will be needed in the next section.

######  Proposition 3.3.

The vertex $v_{i+1}$ is a root of the forest $F$ if and only if
$v_{i}^{\prime}$ is a root of $\sigma_{i}F$.∎

###  3.2. Proof of Theorem 3.1

######  Lemma 3.4.

Let $F$ be a rooted labeled forest with associated exceptional sequence
$E_{\ast}=(E_{1},\cdots,E_{n})$. Then for any $i<n$ we have the following.

  1. (1)

$E_{i}$ is a submodule of $E_{i+1}$ if and only if $v_{i}$ is a child of
$v_{i+1}$ in $F$.

  2. (2)

$E_{i+1}$ is a quotient of $E_{i}$ if and only if $v_{i+1}$ is a child of
$v_{i}$.

  3. (3)

$E_{i},E_{i+1}$ have disjoint consecutive supports if and only if
$v_{i},v_{i+1}$ are sibling in $F_{+}$.

  4. (4)

$E_{i},E_{i+1}$ are $\operatorname{Hom}$ and $\operatorname{Ext}$ orthogonal
if and only if none of the above hold, i.e., if and only if $v_{i},v_{i+1}$
are not close.

###### Proof of Theorem 3.1.

The braid move $\sigma_{i}$ acts on $E_{\ast}$ by moving $E_{i}$ to position
$i+1$ and inserting a uniquely determined new object $E_{i}^{\prime}$ in
position $i$ to give a new exceptional sequence

$\sigma_{i}(E_{\ast})=(E_{1},\cdots,E_{i-1},E_{i}^{\prime},E_{i},E_{i+2},\cdots,E_{n}).$

Case a: If $E_{i},E_{i+1}$ are $\operatorname{Hom}$ and $\operatorname{Ext}$
orthogonal then they commute and $E_{i}^{\prime}=E_{i+1}$.

Case b: Suppose that $E_{i}$ is a submodule of $E_{i+1}$. Then we know that
$E_{i}^{\prime}$ is the quotient module $E_{i+1}/E_{i}$. By Lemma 2.13,
$v_{i}$ must be a child of $v_{i+1}$ since there are no $k$ with $i<k<i+1$.
Let $Y$ denote the other children of $v_{i+1}$. These have disjoint supports
whose union is the support of $E_{i+1}/E_{i}$ minus one point. Therefore, $Y$
is the set of children of $v_{i}^{\prime}$ the node in $\sigma_{i}(F)$
corresponding to $E_{i}^{\prime}$. $E_{i}$ has not changes except for its
label which is now $i+1$. So, it has the same set of children. This is Case 2
in subsection 3.1 and $\sigma_{i}(F)$ is the Hasse diagram of
$\sigma_{i}(E_{\ast})$.

Case c: Suppose that $E_{i+1}$ is a quotient of $E_{i}$, say $E_{i+1}=M_{ab}$
and $E_{i}=M_{ac}$. Then $E_{i}^{\prime}=M_{b+1,c}$ is the kernel of the
epimorphism $E_{i}\twoheadrightarrow E_{i+1}$. By Lemma 2.14, $v_{i+1}$ must
be a child of $v_{i}$. If $X$ is the set of the other children of $v_{i}$ in
$F$ then the corresponding modules must have support in $[a,c]$ but disjoint
from $[a,b]$. So, their supports are in
$\operatorname{supp}E_{i}^{\prime}=[b+1,c]$. So, $X$ is the set of children of
the new $v_{i}^{\prime}$ in $\sigma_{i}(F)$. Since $v_{i+1}$ in $F$ has been
removed, its children $Y$ become children of the new
$v_{i+1}^{\prime}\in\sigma_{i}(F)$.

Case d: Suppose that $v_{i},v_{i+1}$ are sibling in $F_{+}$ with sets of
children $Y,X$, resp. Then $E_{i},E_{i+1}$ must have disjoint consecutive
supports by construction (Proposition 2.11), say
$\operatorname{supp}E_{i}=[a,b]$, $\operatorname{supp}E_{i+1}=[b+1,c]$. Since
$E_{i}$ extends $E_{i+1}$, the new exceptional pair $(E_{i}^{\prime},E_{i})$
must be $(M_{ac},M_{ab})$. The children of the new $v_{i}^{\prime}$ with
module $M_{ac}$ are the new $v_{i+1}^{\prime}$ with module $M_{ab}$ and $X$.
The children of $v_{i+1}^{\prime}$ with module $M_{ab}$ are $Y$.

This concludes the proof of Theorem 3.1. See Figure 3 for an example. ∎

Figure 3 illustrates the sequence of five rooted labeled forests corresponding
to the five complete exceptional sequences for $A_{10}$:

$\begin{array}[]{rl}&(M_{99},M_{1,10},M_{46},M_{16},M_{12},\
M_{79},M_{77},M_{11},M_{44},M_{55})\\\
\xrightarrow{\sigma_{3}}&(M_{99},M_{1,10},M_{13},M_{46},M_{12},\
M_{79},M_{77},M_{11},M_{44},M_{55})\\\
\xrightarrow{\sigma_{2}}&(M_{99},M_{4,10},M_{1,10},M_{46},M_{12},\
M_{79},M_{77},M_{11},M_{44},M_{55})\\\
\xrightarrow{\sigma_{5}}&(M_{99},M_{4,10},M_{1,10},M_{46},M_{79},\
M_{12},M_{77},M_{11},M_{44},M_{55})\\\
\xrightarrow{\sigma_{4}}&(M_{99},M_{4,10},M_{1,10},M_{49},M_{46},\
M_{12},M_{77},M_{11},M_{44},M_{55})\end{array}$

16347852910$\sigma_{3}$16437852910$\sigma_{2}$16527943108$\sigma_{5}$15627943108$\sigma_{4}$16479523108
Figure 3. Braid moves $\sigma_{5},\sigma_{3},\sigma_{2},\sigma_{4}$
illustrated above are examples of Cases a,b,c,d in the proof of Theorem 3.1.

###  3.3. Comparison with the braid group action on parking functions

There is a well-known bijection between parking functions and rooted labeled
forests using Prüfer codes which we will review briefly here. We will use this
bijection to compare the braid group action on parking functions given in [12]
(using Theorem 1.3) to the one we give in the previous section using rooted
labeled forests.

We take as an example, the parking function $(1,1,1,1)$ on $A_{4}$ and the
action of $\sigma_{1}$ then $\sigma_{2}$ then $\sigma_{3}$ using the top
function (Theorem 1.3):

$(1,1,1,1)\xrightarrow{\sigma_{1}}(4,1,1,1)\xrightarrow{\sigma_{2}}(4,3,1,1)\xrightarrow{\sigma_{3}}(4,3,2,1).$

The corresponding Prüfer codes are:

$(0,0,0)\to(2,0,0)\to(4,3,0)\to(4,4,4)$

The _Prüfer code_ corresponding to a parking function
$(a_{1},a_{2},\cdots,a_{n})$ is given by

$(a_{2}-a_{1},a_{3}-a_{2},\cdots,a_{n}-a_{n-1})$

where these numbers are taken modulo $n+1$.

The complete exceptional sequence corresponding to the parking function
$(1,1,1,1)$ is the sequence of injective modules
$(M_{14},M_{13},M_{12},M_{11})$. The sequence of braid moves from above with
corresponding rooted labeled forests is given by:

$\begin{array}[]{cccc}\text{Exceptional
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The Prüfer code corresponding to a rooted labeled forest is given by
$(p_{1},\cdots,p_{n-1})$ where $p_{1}$ is the label of the unique node
adjacent to the leaf of the forest with the largest label ($p_{1}=0$ if this
maximal leaf is a root). When that leaf is removed from the forest, the label
of the unique node adjacent to the leaf with the largest label is $p_{2}$ and
so on. The corresponding parking functions are given in the fourth column of
the table above.

We see that, using the standard bijection between rooted labeled forests and
parking functions, the braid group actions are different. Using the bijection
through exceptional sequences, the braid group actions are the same by
definition (since the action comes from the action on exceptional sequences).

###  3.4. Examples: $A_{2}$ and $A_{3}$

For $n=2$, there are 3 rooted labeled forests and $\sigma_{1}$ permutes them
in a 3-cycle:

12$\sigma_{1}$$\sigma_{1}$$\sigma_{1}$122112 Figure 4. The three rooted binary
forests on $n=2$ are cyclically permuted by $\sigma_{1}$.

For $n=3$ there are 16 rooted labeled forests. Figure 5 illustrates 8 of them.
Figure 6 illustrates how the rooted labeled forest can be visualized when the
exceptional sequence is embedded in the Auslander-Reiten quiver of the path
algebra.

$\sigma_{1}$213123$\sigma_{2}$$\sigma_{1}$231123213$\sigma_{2}$$\sigma_{1}$213$\sigma_{2}$321$\sigma_{1}$312$\sigma_{2}$213
Figure 5. The action of $\sigma_{1}$ and $\sigma_{2}$ is illustrated in the
case $n=3$. 123$\sigma_{1}$213$\sigma_{2}$321 Figure 6. The last three forests
in Figure 5 are embedded in the Auslander-Reiten sequence of $A_{3}$.

##  4\. Action the Garside element

The action of the fundamental braid $\delta$ on forests has an easy
description. However, the action of the Garside element $\Delta$ is more
difficult. This is an important element of the braid group for many reasons
[7]. For example, $\Delta^{2}=\delta^{n}$ generates the center of the braid
group [5] and the action of $\Delta$ on exceptional sequences converts
clusters to $c$-vectors. (This a comment made without proof in [4].) We also
use $\Delta$ to describe an action of the extended braid group
$\widetilde{B}_{n}\rtimes{\mathbb{Z}}_{2}$ on the set of rooted labeled
forests.

To do these detailed computations, we need more notation about exceptional
sequences. First, the _Euler pairing_ for $\Lambda$-modules $X,Y$, also called
the _Euler characteristic_ of $\Lambda$ [1] is given by

$\left<\operatorname{\underline{dim}}X,\operatorname{\underline{dim}}Y\right>_{\Lambda}:=\dim\operatorname{Hom}_{\Lambda}(X,Y)-\dim\operatorname{Ext}_{\Lambda}(X,Y).$

We will use the shorthand notation:

$\chi_{\Lambda}(X,Y):=\left<\operatorname{\underline{dim}}X,\operatorname{\underline{dim}}Y\right>_{\Lambda}.$

###  4.1. The fundamental braid $\delta$

This is the braid group element

$\delta=\delta_{n}=\sigma_{1}\sigma_{2}\cdots\sigma_{n-1}.$

This pushed the last ($n$th) strand underneath the first $n-1$ strands to put
it on the far left. (See Figure 7.) The action of $\delta$ on an exceptional
sequence is

$\delta(E_{1},E_{2},\cdots,E_{n})=(E_{n}^{\prime},E_{1},E_{2},\cdots,E_{n-1}).$

######  Remark 4.1.

Each object in a complete exceptional sequence is uniquely determined up to
isomorphicm by its position and the other objects. In this case we must have

  1. (1)

$E_{n}^{\prime}=\tau E_{n}$, the Auslander-Reiten translation of $E_{n}$ if
$E_{n}$ is not projective since, in that case,

$\chi_{\Lambda}(E_{n},X)=-\chi_{\Lambda}(X,\tau E_{n}).$

  2. (2)

When $E_{n}=P_{i}$ is projective with simple top $S_{i}$ then

$\chi_{\Lambda}(P_{i},X)=\dim X_{i}=\chi_{\Lambda}(X,I_{i})$

where $I_{i}$ is the injective envelope of $S_{i}$. So, $E_{n}^{\prime}=I_{i}$
in this case.

(2) can also be seen combinatorially using Proposition 3.3 which implies that
$v_{n}$ is a root of $F$ if and only if $v_{1}^{\prime}$ is a root of
$F^{\prime}=\delta F$. But this is equivalent to $E_{n}^{\prime}$, the first
object of $\delta E_{\ast}$ being injective. We use the shorthand notation

$\tau_{\Lambda}^{\ast}(X):=\begin{cases}\tau X&\text{if $X$ is not
projective}\\\ I_{i}&\text{if $X=P_{i}$}\end{cases}$

$E_{1}$$E_{2}$$\cdots$$E_{n-1}$$E_{n}$$E_{n}^{\prime}$$E_{1}$$E_{2}$$\cdots$$E_{n-1}$
Figure 7. The braid $\delta$ pulls the last strand to the first position
underneath the other strands. On exceptional sequences, it moves the last
object $E_{n}$ to the first position and changes it to a uniquely determined
$E_{n}^{\prime}$

Since the formula for $\delta$ depends on whether $E_{n}$ is projective, we
need a combinatorial characterization of vertices corresponding to the
projective (and injective) objects.

######  Lemma 4.2.

The projective objects $P_{1}\subset P_{2}\subset\cdots\subset P_{p}$ of a
complete exceptional sequence correspond to vertices
$v_{j_{1}},v_{j_{2}},\cdots,v_{j_{p}}$ of the forest $F$ given as follows.

  1. (1)

$v_{j_{p}}$ is the last root of $F$, i.e., the other roots have smaller
labels.

  2. (2)

$v_{j_{i}}$ is the last relatively projective child of $v_{j_{i+1}}$, i.e.,
$j_{i}$ is maximal among labels of children of $v_{j_{i+1}}$ so that
$j_{i}<j_{i+1}$.

######  Lemma 4.3.

The injective objects $I_{1}\twoheadrightarrow
I_{2}\twoheadrightarrow\cdots\twoheadrightarrow I_{q}$ of a complete
exceptional sequence correspond to vertices
$v_{k_{1}},v_{k_{2}},\cdots,v_{k_{q}}$ of the forest $F$ given as follows.

  1. (1)

$v_{k_{1}}$ is the first root of $F$, i.e., the other roots have larger
labels.

  2. (2)

$v_{k_{i+1}}$ is the first relatively injective child of $v_{k_{i}}$, i.e.,
$k_{i+1}$ is minimal among labels of children of $v_{k_{i}}$ with
$k_{i+1}>k_{i}$.

We call the vertices $v_{j}$ satisfying Lemma 4.2 _projective vertices_ and
those satisfying Lemma 4.3 the _injective vertices_ of the rooted labeled
forest $F$. These lemmas imply the following.

######  Lemma 4.4.

In a complete exceptional sequence for linear $A_{n}$, $E_{n}$ is a projective
module if and only if $v_{n}$ is a root of $F$. Similarly, $E_{1}$ is
injective if and only if $v_{1}$ is a root.

######  Theorem 4.5.

Let $F,\delta F$ be the rooted labeled forests of
$E_{\ast}=(E_{1},\cdots,E_{n})$ and $\delta
E_{\ast}=(\tau^{\ast}E_{n},E_{1},\cdots,E_{n-1})$ with labeled vertices
$v_{i}$ and $v_{i}^{\prime}$ respectively.

  1. (1)

If $v_{n}$ is not a root of $F$, then $\delta F$ is equal to $F$ with the
labels cyclically permuted so that $v_{1}^{\prime}=v_{n}$ and
$v_{i}^{\prime}=v_{i-1}$ for $1<i\leq n$.

  2. (2)

If $v_{n}$ is a root of $F$, then $\delta F$ is obtained by cyclically
permuting the labels of $F$, then exchanging $v_{1}^{\prime}=v_{n}$ with the
master root $v_{0}$.

In both cases $\delta F_{+}\cong F_{+}$.

###### Proof.

If $v_{n}$ is not a root of $F$ then, by Lemma 4.4, $E_{n}$ is not projective.
So, $E_{n}^{\prime}=\tau E_{n}$ which has the same length as $E_{n}$. Since
the other objects are the same as before, all objects have the same length as
before. So, the support containment relations cannot change. So, $\delta F$
must be the same forest as $F$ with labels cyclically permuted.

If $v_{n}$ is a root of $F$, then $E_{n}$ is projective. Every other object in
the sequence has support contained either in the support of $E_{n}$ or the
support of the corresponding injective object
$E_{n}^{\prime}=\tau_{\Lambda}^{\ast}E_{n}$. Therefore, the new root
$v_{1}^{\prime}$ of $\delta F$ covers exactly those vertices of $F$ which were
not descendants of $v_{n}$. This is equivalent to saying that $v_{n}$ has been
exchanged with the master root and relabeled $v_{1}^{\prime}$. ∎

213$\delta$$\delta$321123123$\delta$ Figure 8. Theorem 4.5 in the case $n=3$
is illustrated (from Figure 5).

######  Corollary 4.6.

Let $F,\delta^{n}F$ be the rooted labeled forests of
$E_{\ast}=(E_{1},\cdots,E_{n})$ and
$\delta^{n}E_{\ast}=(\tau^{\ast}E_{1},\cdots,\tau^{\ast}E_{n})$. Let
$v_{j_{1}},\cdots,v_{j_{p}}$ be the projective vertices of $F$. Then
$\delta^{n}F_{+}\cong F_{+}$ with projective vertices together with the master
root $v_{0}$ cyclically permuted and all other vertices remaining the same.
Thus $v_{j_{p}}^{\prime}=v_{0}$, $v_{0}^{\prime}=v_{j_{1}}$,
$v_{j_{i}}^{\prime}=v_{j_{i+1}}$ for $1\leq i<p$ and $v_{k}^{\prime}=v_{k}$
for all other values of $k$. In particular,
$v_{j_{1}}^{\prime},\cdots,v_{j_{p}}^{\prime}$ are the injective vertices of
$\delta^{n}F$ and the roots of $\delta^{n}F$ consist of $v_{j_{1}}^{\prime}$
and the children of $v_{j_{1}}$ in $F$. (See Figure 9.)

###### Proof.

The first part follows from Theorem 4.5. The last part follows from Lemma 4.3
since $j_{1}$ must be smaller than the labels of the other roots of
$\delta^{n}F$. These other roots in $\delta^{n}F$ were the children of
$v_{j_{1}}$ in $F$ which must have larger index by Lemma 4.2.2.∎

$\textstyle{v_{j_{p}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{p+1}}$$\textstyle{X_{1}}$$\textstyle{v_{j_{1}}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{p}}$$\textstyle{X_{2}}$$\textstyle{v_{j_{2}}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F:}$$\textstyle{v_{j_{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta^{n}=\Delta^{2}}$$\textstyle{\delta^{n}F:}$$\textstyle{X_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v_{j_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}}$$\textstyle{v_{j_{p}}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}}$$\textstyle{X_{p+1}}$
Figure 9. Illustrating Corollary 4.6, the projective vertices of $F$ become
the injective vertices of $\delta^{n}F$. The children of the smallest
projective vertex $v_{j_{1}}$ in $F$ have larger indices and become the other
roots of $\delta^{n}F$ with the injective vertex $v_{j_{1}}^{\prime}$ being
the first root of $\delta^{n}F=\Delta^{2}F$.

###  4.2. Garside element

The _Garside element_ $\Delta$ of the braid group $B_{n}$ is given by

$\Delta=\delta_{n-1}\delta_{n-2}\cdots\delta_{2}\delta_{1}$

where, for each $k<n$, $\delta_{k}$ is given by

$\delta_{k}=\sigma_{1}\sigma_{2}\cdots\sigma_{k}.$

In particular $\delta_{n-1}=\delta$ is the fundamental braid. The commutator
relation:

$\Delta\sigma_{i}\Delta^{-1}=\sigma_{n-i}$

implies that $\Delta^{2}$ lies in the center of the braid group $B_{n}$. In
fact, $\Delta^{2}$ generates the center of $B_{n}$ by [5]. It is also an easy
observation that $\Delta^{2}=\delta^{n}$. So, the action of the central
element $\Delta^{2}$ is given by Corollary 4.6 and Figure 9.

Figure 10. The Garside braid $\Delta$ for $n=5$. Each strand goes over the
ones on its right and under the ones on its left.
$\Delta$213$\Delta$$\Delta$123321123 Figure 11. The action of
$\Delta=\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2}$ is
shown in the case $n=3$ using Figure 5. Proposition 4.11 explains the
combinatorics of $\Delta$.

The Garside braid is shown in Figure 10. Examples of the action of $\Delta$ on
exceptional sequences for $A_{3}$ were computed in Figure 5 and are summarized
in Figure 11 below. The use of the Garside element in cluster theory is
explained in the next subsection. Here we discuss the action of $\Delta$ on
forests.

We use the notation

$\Delta(E_{1},\cdots,E_{n})=(E_{1}^{\prime},\cdots,E_{n}^{\prime}).$

The first thing to notice is that $E_{n}^{\prime}=E_{1}$ since the first
strand goes over the other strands in each step of the braid $\Delta$.
Similarly, every strand goes over the ones on its right and under the ones on
its left.

Let $\mathcal{C}_{k}=Z^{\perp}$, the right perpendicular category of
$Z=E_{k+1}\oplus\cdots\oplus E_{n}$. This is the category of all
$\Lambda$-modules $X$ so that $\chi_{\Lambda}(E_{j},X)=0$ for all $j>k$.

######  Lemma 4.7.

The sequence $(E_{1},\cdots,E_{k})$ is a complete exceptional sequence for the
category ${\mathcal{C}}_{k}$.

###### Proof.

The objects $E_{1},\cdots,E_{k}$ lie in ${\mathcal{C}}_{k}$ by definition of
an exceptional sequence. To show it is complete, suppose we could add an
object of ${\mathcal{C}}_{k}$ to this sequence. Then, that object could also
be added to the complete exceptional sequence $(E_{1},\cdots,E_{n})$ to the
left of $E_{k+1}$ contradicting its maximality. ∎

We recall our main theorem (Theorem 2.20) which says that $v_{k}$ is ascending
in $F$ (either a root or the child of some $v_{j}$ for $j>k$) if and only if
$E_{k}$ is relatively projective which means $E_{k}$ is a projective object of
the category ${\mathcal{C}}_{k}$.

######  Lemma 4.8.

The object $E_{n-k+1}^{\prime}$ in $\Delta E_{\ast}$ is the unique
indecomposable object of ${\mathcal{C}}_{k}$ having the property that
$(E_{n-k+1}^{\prime},E_{1},\cdots,E_{k-1})$ is a complete exceptional sequence
for ${\mathcal{C}}_{k}$. Consequently,

  1. (1)

If $E_{k}$ is not relatively projective, then
$E_{n-k+1}^{\prime}=\tau_{k}E_{k}$ where $\tau_{k}$ is Auslander-Reiten
translation in ${\mathcal{C}}_{k}$. In that case,
$\chi_{\Lambda}(E_{k},E_{n-k+1}^{\prime})=-1$.

  2. (2)

If $E_{k}$ is relatively projective, then $E_{n-k+1}^{\prime}$ is the
injective envelope in ${\mathcal{C}}_{k}$ of the relatively simple
${\mathcal{C}}_{k}$-top of $E_{k}$. In this case,
$\chi_{\Lambda}(E_{k},E_{n-k+1}^{\prime})=+1$.

In particular, $E_{1}^{\prime}=\tau^{\ast}E_{n}$ (just as in $\delta
E_{\ast}$).

###### Proof.

The braid move $\delta_{k-2}\cdots\delta_{1}$ on $E_{\ast}$ produces the
exceptional sequence

$(E_{n-k+2}^{\prime},\cdots,E_{n}^{\prime},E_{k},E_{k+1},\cdots,E_{n})$

which implies that $(E_{n-k+2}^{\prime},\cdots,E_{n}^{\prime})$ is a complete
exceptional sequence for ${\mathcal{C}}_{k-1}$. Applying the braid move
$\delta_{k-1}$ produces the exceptional sequence

$(E_{n-k+1}^{\prime},E_{n-k+2}^{\prime},\cdots,E_{n}^{\prime},E_{k+1},\cdots,E_{n})$

which implies that $E_{n-k+1}^{\prime}$ is the unique object of
${\mathcal{C}}_{k}$ in the right perpendicular category of
${\mathcal{C}}_{k-1}$. Equivalently,
$(E_{n-k+1}^{\prime},E_{1},\cdots,E_{k-1})$ is a complete exceptional sequence
for ${\mathcal{C}}_{k}$ as claimed. The rest follows by analogy with Theorem
4.5. ∎

This lemma immediately implies the following.

######  Lemma 4.9.

Let $\Delta E_{\ast}=(E_{1}^{\prime},\cdots,E_{n}^{\prime})$. Then $E_{k}$ is
relatively projective in $E_{\ast}$ if and only if $E_{n-k+1}^{\prime}$ is
relatively injective in $\Delta E_{\ast}$.

In the proof of Lemma 4.8 we saw that

$\delta_{k-1}^{-1}(E_{n-k+1}^{\prime},E_{n-k+2}^{\prime},\cdots,E_{n}^{\prime})=(E_{n-k+2}^{\prime},\cdots,E_{n}^{\prime},E_{k}).$

If we apply this braid move to the last $k$ objects in $\Delta E_{\ast}$ we
obtain the following.

######  Lemma 4.10.

With the notation
$\Delta(E_{1},\cdots,E_{n})=(E_{1}^{\prime},\cdots,E_{n}^{\prime})$ we have
that

$(E_{1}^{\prime},\cdots,E_{n-k}^{\prime},E_{n-k+2}^{\prime},\cdots,E_{n}^{\prime},E_{k})$

is a complete exceptional sequence. In particular,

$\chi_{\Lambda}(E_{k},E^{\prime}_{j})=0$

for $j\neq n-k+1$.

######  Proposition 4.11.

Let $F^{\prime}=\Delta F$ for a rooted labeled forest $F$ with $n$ vertices.
Then

  1. (1)

$v_{j_{1}},\cdots,v_{j_{p}}$ are the projective vertices of $F$ if and only if
$v_{n-j_{1}+1}^{\prime},\cdots,v_{n-j_{p}+1}^{\prime}$ are the roots of
$\Delta F$.

  2. (2)

$v_{k_{1}},\cdots,v_{k_{r}}$ are the roots of $F$ if and only if
$v_{n-k_{1}+1}^{\prime},\cdots,v_{n-k_{r}+1}^{\prime}$ are the injective
vertices of $\Delta F$.

  3. (3)

If $E_{i_{1}}$ is the smallest projective in $E_{\ast}$ and
$E_{i_{2}},\cdots,E_{i_{s}}$ are its children, then
$E_{n-i_{1}+1},\cdots,E_{n-i_{s}+1}$ are the projective objects of $\Delta
E_{\ast}$.

  4. (4)

If $E_{\ell_{1}},\cdots,E_{\ell_{q}}$ are the injective objects of $E_{\ast}$
with $\ell_{1}<\ell_{2}<\cdots<\ell_{q}$, then $E_{n-\ell_{1}+1}^{\prime}$ is
the last (and smallest) injective object of $\Delta E_{\ast}$ and
$E_{n-\ell_{2}+1}^{\prime},\cdots,E_{n-\ell_{q}+1}^{\prime}$ are its children.

###### Proof.

Since $\Delta^{2}=\delta^{n}$, it follows from Corollary 4.6 that (1) and (2)
are equivalent. We prove only (1) since (3) and (4) are not needed in this
paper.

Statement (1) says that $v_{k}$ is a projective vertex of $F$ if and only if
$v_{n-k+1}^{\prime}$ is a root of $F^{\prime}=\Delta F$. We prove this by
induction on $n-k$. For $k=n$, $E_{n}$ is projective if and only if $v_{n}$ is
a root of $F$ if any only if the first vertex $v_{1}^{\prime}$ of $\Delta F$
is a root (See Remark 4.1). So, the statement holds for $k=n$.

Now suppose the statement holds for $k+1$. Then, we observe that the braid
move $\sigma_{k}$ moves $E_{k}$ to the $k+1$st position in
$\sigma_{k}E_{\ast}$:

$\sigma_{k}(E_{1},\cdots,E_{k},\cdots,E_{n})=(E_{1},\cdots,E_{k-1},E_{k+1}^{\ast},E_{k},E_{k+2},\cdots,E_{n}).$

Therefore, by induction on $n-k$ we have that $E_{k}$ is a projective module
if and only if the $(n-k)$th vertex of $\Delta\sigma_{k}F=\sigma_{n-k}\Delta
F$ is a root. By Proposition 3.3, this is equivalent to the $(n-k+1)$st vertex
of $\Delta F$ being a root. Thus the statement holds for $k$.

This proves (1) in the proposition. ∎

######  Corollary 4.12.

$\Delta E_{\ast}$ has a simple injective object if and only if $E_{\ast}$ has
only one injective objects. Similarly, $E_{\ast}$ has a simple projective
object if and only if $\Delta E_{\ast}$ has only one projective object.

4$F:$516372$\Delta$$\Delta^{-1}F:$3412567$\Delta$$\Delta F:$7346512 Figure 12.
Illustrating Proposition 4.11: The projective vertices $v_{1},v_{2},v_{3}$ of
$F$ become the roots $v_{7},v_{6},v_{5}$ resp. of $\Delta F$. The only root
$v_{3}$ of $F$ becomes the only injective vertex $v_{5}$ of $\Delta F$. The
injective vertices $v_{3},v_{7}$ of $F$ correspond to the last injective
$v_{5}$ in $\Delta F$ and its child $v_{1}$. The first projective in $F$ and
its children become the projective vertices of $\Delta F$. For the placement
of $v_{2}$ in $\Delta F$ we go to $\Delta^{-1}F$ then apply
$\delta^{7}=\Delta^{2}$.

###  4.3. Clusters and signed exceptional sequences

The Garside element $\Delta$ takes a “cluster tilting set” to the
corresponding “signed exceptional sequence” [16]. This is a comment made in
[4] using different terminology (before the introduction of signed exceptional
sequences).

######  Definition 4.13.

We define a _cluster tilting set_ for a Dynkin quiver with $n$ vertices to be
a set of $n$ nonisomorphic modules and “shifted projective modules”
$T_{1},\cdots,T_{n}$ (each $T_{i}$ is either an indecomposable module $M_{i}$
or a shifted indecomposable projective module $P_{k}[1]$) having the property
that $\operatorname{Hom}_{{\mathcal{D}}^{b}}(T_{i},T_{j}[1])=0$ in the bounded
derived category ${\mathcal{D}}^{b}={{\mathcal{D}}^{b}(mod\text{-}\Lambda)}$
for all $i,j$. In other words, $\operatorname{Ext}_{\Lambda}(M_{i},M_{j})=0$
for all $i,j$ and $\operatorname{Hom}_{\Lambda}(P_{k},M_{i})=0$ for all $k,i$.

For example, in $A_{3}$, $S_{1},S_{3},P_{2}[1]$ is a cluster tilting set since
$S_{1},S_{3}$ do not extend each other
($\operatorname{Ext}_{\Lambda}(S_{1},S_{3})=\operatorname{Ext}_{\Lambda}(S_{3},S_{1})=0$)
and $\operatorname{Hom}_{\Lambda}(P_{2},S_{1}\oplus S_{3})=0$.

######  Definition 4.14.

By a _signed exceptional sequence_ we mean an exceptional sequence
$E_{\ast}=(E_{1},\cdots,E_{n})$ together with a sequence $\varepsilon_{\ast}$
of signs $\varepsilon_{i}\in\\{+1,-1\\}$ so that $\varepsilon_{i}=-1$ implies
$E_{i}$ is relatively projective. We write the sequence as
$(X_{1},\cdots,X_{n})$ where

$X_{i}=\begin{cases}E_{i}[1]&\text{if }\varepsilon_{i}=-1\\\
E_{i}&\text{otherwise}.\end{cases}$

For example, in $A_{3}$, $(S_{1}[1],S_{3},P_{2})$ is a signed exceptional
sequence since $S_{1}$ is relatively projective. (The first object in an
exceptional sequence is always relatively projective.)

######  Theorem 4.15.

Any cluster tilting set can be arranged into a signed exceptional sequence
$E_{\ast}=(E_{1},\cdots,E_{n})$ with signs
$\varepsilon_{1},\cdots,\varepsilon_{n}$. Furthermore,

$\Delta E_{\ast}=(E^{\prime}_{1},\cdots,E^{\prime}_{n})\text{ with signs
}\varepsilon_{1}^{\prime},\cdots,\varepsilon_{n}^{\prime}$

is a signed exceptional sequence with the property that

(4.1)
$\epsilon_{k}\epsilon_{j}^{\prime}\chi_{\Lambda}(E_{k},E_{j}^{\prime})=\delta_{kj}.$

######  Remark 4.16.

It is shown in [4] that this holds for any hereditary algebra. Here we prove
it in the case of linear $A_{n}$ using rooted labeled forests. Also, Equation
(4.1) implies that
$-\varepsilon_{j}\operatorname{\underline{dim}}E_{j}^{\prime}$ are the
“$c$-vectors” of the “cluster” $\\{\operatorname{\underline{dim}}E_{k}\\}$ if
we take the “initial seed” to be the shifted projectives [14].

###### Proof.

It is an observation originally due to Schofield [23] that a support tilting
set can be arranged in an exceptional sequence. In the case of finite type,
this is very easy to see: take the left-to-right order of the objects as they
appear in the Auslander-Reiten quiver. The shifted projective objects can be
placed after these in increasing order of size. Since all negative (shifted)
objects are projective, the result is a signed exceptional sequence
$E_{\ast},\varepsilon_{\ast}$.

To see that the exceptional sequence $\Delta E_{\ast}$ with signs
$\varepsilon_{\ast}$ is a signed exceptional sequence we examine which objects
have negative signs.

  1. (1)

If $E_{k}$ is a projective module then $E_{k}$ and $E^{\prime}_{n-k+1}$ have
the same sign: $\varepsilon_{k}=\varepsilon^{\prime}_{n-k+1}$. Since
$v_{n-k+1}$ is a root of the forest $\Delta F$, $E^{\prime}_{n-k+1}$ is
relatively projective and either sign is allowed.

  2. (2)

If $E_{k}$ is any relative projective object which is not projective then it
is positive. So, $\varepsilon^{\prime}_{n-k+1}=\varepsilon_{k}=+1$.

  3. (3)

In all other cases, $E_{k}$ is relatively injective but not relatively
projective. So, $E_{k}$ is positive and
$\varepsilon^{\prime}_{n-k+1}=-\varepsilon_{k}=-1$. This is OK since
$E^{\prime}_{n-k+1}$ is relatively projective (but not relatively injective)
in $\Delta E_{\ast}$.

The conclusion is that all objects in $\Delta E_{\ast}$ which are relatively
projective but not relatively injective have a negative sign and those which
are relatively injective but not relatively projective have positive sign. The
roots might have either sign. So, $\Delta E_{\ast}$ with signs
$\varepsilon_{\ast}^{\prime}$ is a signed exceptional sequence.

By Lemma 4.8 we have
$\chi_{\Lambda}(E_{k},E^{\prime}_{n-k+1})=\varepsilon_{k}\varepsilon^{\prime}_{n-k+1}$.
By Lemma 4.10, $\chi_{\Lambda}(E_{k},E^{\prime}_{j})=0$ when $j\neq{n-k+1}$.
Equation (4.1) follows. ∎

######  Example 4.17.

None of the forests in Figure 12 are examples of cluster tilting sets. In the
first and second forest, there is a vertex with two relatively injective
children. These extend each other. The last $\Delta F$ has three roots. The
smaller two extend each other. (We could shift the projective module $E_{7}$.)
In Figure 11, the second and last forests are not cluster tilting for the same
reasons. The second has two relatively injective children of the root and the
last has three roots which is not allowed. The first forest is cluster tilting
if we shift the projective module $E_{3}$. This gives
$(S_{2},I_{2},S_{3}[1])$. The corresponding signed exceptional sequence is
given by the second forest with its root shifted: $(P_{1}[1],S_{1},S_{2})$. To
illustrate Equation (4.1) note that $\operatorname{Hom}(I_{2},S_{1})=K$, but
$\operatorname{Hom}(I_{2},S_{2})=0=\operatorname{Ext}(I_{2},S_{2})$ and
$\operatorname{Hom}(I_{2},P_{1})=0=\operatorname{Ext}(I_{2},P_{1})$.

In [16] it is shown that signed exceptional sequences are in 1-1
correspondence with ordered cluster tilting sets. Thus the number of signed
exceptional sequences of type $A_{n}$ is $n!$ times the Catalan number
$C_{n+1}=\frac{1}{n+2}\binom{2n+2}{n+1}$. This can also be seen from the
generating function. Since the relative projectives can have any sign, the
number of signed exceptional sequences is

$p_{n}(2,1,2)=2(n-1+2(2))(n-2+2(4))\cdots(1+2(n))=2\frac{(2n+1)!}{(n+2)!}=n!C_{n+1}.$

###  4.4. Action of extended braid group

The _extended braid group_ is

$\widetilde{B}_{n}:=B_{n}\rtimes_{C}{\mathbb{Z}}/2$

using the outer involution $C$ given by $\sigma_{i}\mapsto\sigma_{i}^{-1}$. We
refer to $C$ as _complex conjugation_. Equivalently, we use $D=\Delta C$ in
$\widetilde{B}_{n}$ conjugation by which gives

$D\sigma_{i}D=\sigma_{n-i}^{-1}.$

We call this _duality_ since it corresponds to $D:mod$-$\Lambda\to
mod$-$\Lambda^{op}$ given by

$DM=\operatorname{Hom}_{K}(M,K).$

This reverses the order of an exceptional sequence:

$D(E_{1},\cdots,E_{n})=(DE_{n},\cdots,DE_{1})$

since $\chi_{\Lambda}(X,Y)=\chi_{\Lambda^{op}}(DY,DX)$. The operation $D$
gives an involution on the set of complete exceptional sequences of the linear
$A_{n}$ quiver since this quiver is isomorphic to its opposite quiver. The
modules $DE_{i}$ have the same support inclusion relations as the $E_{i}$.
Therefore, duality $D$ acts on the corresponding rooted labeled forests by
reversing the order of vertices: $DF\cong F$ with vertex labels
$v_{i}^{\prime}=v_{n-i+1}$.

Complex conjugation $C=D\Delta$ has a cleaner description than the Garside
element:

######  Proposition 4.18.

Let $F^{\prime}=CF=D\Delta F$. Then $F=D\Delta F^{\prime}$ and

  1. (1)

$v_{j_{1}},\cdots,v_{j_{p}}$ are the projective vertices of $F$ if and only if
$v_{j_{1}}^{\prime},\cdots,v_{j_{p}}^{\prime}$ are the roots of $F^{\prime}$.

  2. (2)

$v_{k_{1}},\cdots,v_{k_{q}}$ are the injective vertices of $F$ with
$k_{1}<k_{2}<\cdots<k_{q}$ if and only if $v_{k_{1}}^{\prime}$ is the first
projective vertex of $F^{\prime}$ and
$v_{k_{2}}^{\prime},\cdots,v_{k_{q}}^{\prime}$ are its children.

  3. (3)

Dually, $v_{\ell_{q}}$ is the last injective vertex of $F$ and
$v_{\ell_{1}},\cdots,v_{\ell_{q-1}}$ are its children if and only if
$v_{\ell_{1}}^{\prime\prime},\cdots,v_{\ell_{q}}^{\prime\prime}$ are the
projective vertices of $F^{\prime\prime}=D\Delta^{-1}F$.

###  4.5. Comments

We note that the results of this paper partially answer a question in [25]
raised by the second author. The question was: “Is it possible to interpret
completed exceptional sequences of linear Nakayama algebras via certain rooted
labeled forests?” But another question is raised: In [25] a bijection is
constructed between rooted labeled trees of height at most 1 and complete
exceptional sequences for $A_{n}$ with radical square zero. How is this
related to the bijection constructed in the present paper?

There is also the question of what happens with other orientations of the
$A_{n}$ quiver where our main results do not hold for $A_{n}$ by Remark 2.30.
There is a bijection between exceptional sequences of $A_{n}$ (or any other
Dynkin quiver) with different orientations given by APR-tilting [2] and one
can ask how this affects the bijection with parking functions and rooted
labeled forest. We note that the bijection using [3], [13] works for any
orientation of $A_{n}$. However, it is not clear how this affects the
distibution of relatively injective and projective objects.

##  5\. Acknowledgements

The authors are grateful to Gordana Todorov for all her help and support. We
also thank Olivier Bernardi for the speculation that the generating function
for rooted labeled forests may be related to generating functions for
exceptional sequences of type $A_{n}$. We thank Shujian Chen for discussions
about chord diagrams and exceptional sequences. Also, Hugh Thomas explained
the bijection from [13] to the first author many years ago.

The main results of this paper were presented as conjectures at the 2021
Workshop on Cluster Algebras and Related Topics held at the Morningside Center
of Mathematics of the Chinese Academy of Sciences in Beijing held in August,
2021. The first author would like to thank the organizers of this conference
for the opportunity to present this material.

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