aarjaneiro/planetmath huggingface dataset:
morphisms between quivers

morphisms between quivers

Recall that a quadruple

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

is a quiver, if

Q

0

is a set (whose

elements

are called vertices),

Q

1

is also a set (whose elements are called arrows) and

s

,

t

:

Q

1

→

Q

0

are

functions

which take each arrow to its source and target respectively.

Definition

.

A

morphism

from a quiver

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

to a quiver

Q

′

=

(

Q

0

′

,

Q

1

′

,

s

′

,

t

′

)

is a pair

F

=

(

F

0

,

F

1

)

such that

F

0

:

Q

0

→

Q

0

′

,

F

1

:

Q

1

→

Q

1

′

are functions which

satisfy

s

′

⁢

(

F

1

⁢

(

α

)

)

=

F

0

⁢

(

s

⁢

(

α

)

)

;

t

′

⁢

(

F

1

⁢

(

α

)

)

=

F

0

⁢

(

t

⁢

(

α

)

)

.

In this case we write

F

:

Q

→

Q

′

. In other words

F

:

Q

→

Q

′

is a morphism of quivers, if for an arrow

\xymatrix

⁢

x

⁢

\ar

⁢

[

r

]

α

⁢

&

⁢

y

in

Q

the following

\xymatrix

⁢

F

0

⁢

(

x

)

⁢

\ar

⁢

[

r

]

F

1

⁢

(

α

)

⁢

&

⁢

F

0

⁢

(

y

)

is an arrow in

Q

′

.

If

F

:

Q

→

Q

′

and

G

:

Q

′

→

Q

′′

are

morphisms between quivers

, then we have the

composition

G

∘

F

:

Q

→

Q

′′

defined by

G

∘

F

=

(

G

0

∘

F

0

,

G

1

∘

F

1

)

.

It can be easily checked, that

G

∘

F

is again a morphism between quivers.

The class of all quivers, all morphisms between together with the composition is a category. In particular we have a notion of

isomorphism

. It can be shown, that two quivers

Q

,

Q

′

are isomorphic if and only if there exists a morphism of quivers

F

:

Q

→

Q

′

such that both

F

0

and

F

1

are

bijections

.

For example quivers

\xymatrix

⁢

Q

:

1

⁢

\ar

⁢

[

r

]

⁢

&

⁢

2

⁢

&

⁢

&

⁢

&

⁢

Q

′

:

1

⁢

&

⁢

2

⁢

\ar

⁢

[

l

]

are isomorphic, although not equal.

Title

morphisms between quivers

Canonical

name

MorphismsBetweenQuivers

Date of creation

2013-03-22 19:16:57

Last modified on

2013-03-22 19:16:57

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:04 2018 by

LaTeXML
path algebra of a disconnected quiver

path algebra of a disconnected quiver

Let

Q

be a disconnected quiver, i.e.

Q

can be written as a

disjoint union

of two quivers

Q

′

and

Q

′′

(which means that there is no path starting in

Q

′

and ending in

Q

′′

and vice versa) and let

k

be an arbitrary field.

Proposition

.

The path

algebra

k

⁢

Q

is

isomorphic

to the

product

of path algebras

k

⁢

Q

′

×

k

⁢

Q

′′

.

Proof.

If

w

is a path in

Q

, then

w

belongs either to

Q

′

or

Q

′′

. Define linear map

T

:

k

⁢

Q

→

k

⁢

Q

′

×

k

⁢

Q

′′

by

T

⁢

(

w

)

=

(

w

,

0

)

if

w

∈

Q

′

or

T

⁢

(

w

)

=

(

0

,

w

)

if

w

∈

Q

′′

and extend it linearly to entire

k

⁢

Q

. We will show that

T

is an

isomorphism

of algebras.

If

w

,

w

′

are paths in

Q

, then since

Q

′

and

Q

′′

are

disjoint

, then each of them entirely lies in

Q

′

or

Q

′′

. Now since

Q

′

and

Q

′′

don’t have common vertices it follows that

w

⋅

w

′

=

w

′

⋅

w

=

0

. Without loss of generality we may assume, that

w

is in

Q

′

and

w

′

is in

Q

′′

. Then we have

T

⁢

(

w

⋅

w

′

)

=

T

⁢

(

0

)

=

(

0

,

0

)

=

(

w

,

0

)

⋅

(

0

,

w

′

)

=

T

⁢

(

w

)

⋅

T

⁢

(

w

′

)

.

If both lie in the same

component

, for example in

Q

′

, then

T

⁢

(

w

⋅

w

′

)

=

(

w

⋅

w

′

,

0

)

=

(

w

,

0

)

⋅

(

w

′

,

0

)

=

T

⁢

(

w

)

⋅

T

⁢

(

w

′

)

.

Since

T

preservers multiplication on paths, then

T

preserves multiplication and thus

T

is an algebra homomorphism.

Obviously by definition

T

is 1-1.

It remains to show, that

T

is onto. Assume that

(

a

,

b

)

∈

k

⁢

Q

′

⊕

k

⁢

Q

′′

. Then we can write

(

a

,

b

)

=

∑

i

,

j

λ

i

,

j

⁢

(

v

i

,

w

j

)

=

∑

i

,

j

λ

i

,

j

⁢

(

v

i

,

0

)

+

∑

i

,

j

λ

i

,

j

⁢

(

0

,

w

j

)

,

where

v

i

are paths in

Q

′

and

w

j

are paths in

Q

′′

. It can be easily checked, that

T

⁢

(

∑

i

,

j

λ

i

,

j

⁢

(

v

i

+

w

j

)

)

=

(

a

,

b

)

.

Here we consider all

v

i

and

w

j

as paths in

Q

.

Thus

T

is an isomorphism, which

completes

the proof.

□

Title

path algebra of a disconnected quiver

Canonical name

PathAlgebraOfADisconnectedQuiver

Date of creation

2013-03-22 19:16:25

Last modified on

2013-03-22 19:16:25

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Theorem

Classification

msc 14L24

Generated on Fri Feb 9 18:00:09 2018 by

LaTeXML
path algebra of a quiver

path algebra of a quiver

Let

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

be a quiver, i.e.

Q

0

is a set of vertices,

Q

1

is a set of arrows,

s

:

Q

1

→

Q

0

is a source

function

and

t

:

Q

1

→

Q

0

is a target function.

Recall that a

path

of length

l

⩾

1

from

x

to

y

in

Q

is a

sequence

of arrows

(

a

1

,

…

,

a

l

)

such that

s

⁢

(

a

1

)

=

x

;

t

⁢

(

a

l

)

=

y

;

t

⁢

(

a

i

)

=

s

⁢

(

a

i

+

1

)

for any

i

=

1

,

2

,

…

,

l

-

1

,

l

.

Also we allow paths of length

0

, i.e.

stationary

paths.

If

a

=

(

a

1

,

…

,

a

l

)

and

b

=

(

b

1

,

…

,

b

k

)

are two paths such that

t

⁢

(

a

l

)

=

s

⁢

(

b

1

)

then we say that

a

and

b

are

compatibile

and in this case we can form another path from

a

and

b

, namely

a

∘

b

=

(

a

1

,

…

,

a

l

,

b

1

,

…

,

b

k

)

.

Note, that the length of

a

∘

b

is a sum of lengths of

a

and

b

. Also a path

a

=

(

a

1

,

…

,

a

l

)

of

positive

length is called a

cycle

if

t

⁢

(

a

l

)

=

s

⁢

(

a

1

)

. In this case we can compose

a

with itself to produce new path.

Also if

a

is a path from

x

to

y

and

e

x

,

e

y

are stationary paths in

x

and

y

respectively, then we define

a

∘

e

y

=

a

and

e

x

∘

a

=

a

.

Let

k

⁢

Q

be a

vector space

with a basis consisting of all paths (including stationary paths). For paths

a

and

b

define

multiplication

as follows:

If

a

and

b

are

compatible

, then put

a

⁢

b

=

a

∘

b

and put

a

⁢

b

=

0

otherwise. This

operation

extendes bilinearly to entire

k

⁢

Q

and it can be easily checked that

k

⁢

Q

becomes an

associative algebra

in this manner called the

path

algebra

of

Q

over

k

.

Title

path algebra of a quiver

Canonical

name

PathAlgebraOfAQuiver

Date of creation

2013-03-22 19:16:19

Last modified on

2013-03-22 19:16:19

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:12 2018 by

LaTeXML
predecessors and succesors in quivers

predecessors and succesors in quivers

Let

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

be a quiver, i.e.

Q

0

is a set of vertices,

Q

1

is a set of arrows and

s

,

t

:

Q

1

→

Q

0

are

functions

called source and target respectively. Recall, that

ω

=

(

α

1

,

…

,

α

n

)

is a path in

Q

, if each

α

i

∈

Q

1

and

t

⁢

(

α

i

)

=

s

⁢

(

α

i

+

1

)

for all

i

=

1

,

…

,

n

-

1

. The length of

ω

is defined as

n

.

Definition

.

If

a

,

b

∈

Q

0

are vertices such that there exists a path

ω

=

(

α

1

,

…

,

α

n

)

with

s

⁢

(

α

1

)

=

a

and

t

⁢

(

α

n

)

=

b

, then

a

is said to be a

predecessor

of

b

and

b

is said to be a

successor

of

a

. Additionally if there is such path of length

1

, i.e. there exists an arrow from

a

to

b

, then

a

is a

direct predecessor

of

b

and

b

is a

direct succesor

of

a

.

For a given vertex

a

∈

Q

0

we define the following sets:

a

-

=

{

b

∈

Q

0

|

b

⁢

is a direct predecessor of

⁢

a

}

;

a

+

=

{

b

∈

Q

0

|

b

⁢

is a direct successor of

⁢

a

}

.

The

elements

in

a

-

∪

a

+

are called

neighbours

of

a

.

Example.

Consider the following quiver:

\xymatrix

⁢

&

⁢

&

⁢

&

⁢

3

⁢

0

⁢

\ar

⁢

[

r

]

⁢

&

⁢

1

⁢

\ar

⁢

[

r

]

⁢

&

⁢

2

⁢

\ar

⁢

[

r

⁢

u

]

⁢

\ar

⁢

[

r

⁢

d

]

⁢

&

⁢

&

⁢

&

⁢

4

Then

2

-

=

{

1

}

;

2

+

=

{

3

,

4

}

;

and

1

,

3

,

4

are all neighbours of

2

. Also

0

is a predecessor of

2

, but not direct.

Title

predecessors and succesors in quivers

Canonical

name

PredecessorsAndSuccesorsInQuivers

Date of creation

2013-03-22 19:17:47

Last modified on

2013-03-22 19:17:47

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:14 2018 by

LaTeXML
properties of admissible ideals

properties of admissible ideals

Let

Q

be a quiver,

k

a field and

I

an

admissible

ideal (see parent object) in the path

algebra

k

⁢

Q

. The following

propositions

and proofs are taken from

[

1

]

.

Proposition 1.

If

Q

is finite, then

k

⁢

Q

/

I

is finite dimensional algebra.

Proof.

Let

R

Q

be the arrow ideal in

k

⁢

Q

. Since

R

Q

m

⊆

I

for some

m

, then we have a

surjective

algebra homomorphism

k

⁢

Q

/

R

Q

m

→

k

⁢

Q

/

I

. Thus, it is enough to show, that

k

⁢

Q

/

R

Q

m

is finite dimensional. But since

Q

is a finite quiver, then there is finitely many paths of length at most

m

. It is easy to see, that these paths form a basis of

k

⁢

Q

/

R

Q

m

as

vector space

over

k

. This

completes

the proof.

□

Proposition 2.

If

Q

is finite, then

I

is a

finitely generated

ideal.

Proof.

Consider the

short exact sequence

\xymatrix

⁢

0

⁢

\ar

⁢

[

r

]

⁢

&

⁢

R

Q

m

⁢

\ar

⁢

[

r

]

⁢

&

⁢

I

⁢

\ar

⁢

[

r

]

⁢

&

⁢

I

/

R

Q

m

⁢

\ar

⁢

[

r

]

⁢

&

⁢

0

of

k

⁢

Q

modules. It is well known that in such

sequences

the middle term is finitely generated if the end terms are. Of course

R

Q

m

is finitely generated, because

Q

is finite so there is finite number of paths of length

m

.

On the other hand

I

/

R

Q

m

is an ideal in

k

⁢

Q

/

R

Q

m

, which is finite dimensional by proposition 1. Thus

I

/

R

Q

m

is a finite dimensional vector space over

k

. But then it is finitely generated

k

⁢

Q

module (see this entry (

http://planetmath.org/FiniteDimensionalModulesOverAlgebra

) for more details), which completes the proof.

□

Proposition 3.

If

Q

is finite, then there exists a

finite set

of

relations

(

http://planetmath.org/RelationsInQuiver

)

{

ρ

1

,

…

,

ρ

m

}

such that

I

is generated by them.

Proof.

By proposition 2 there is a finite set of

generators

{

a

1

,

…

,

a

n

}

of

I

. Generally the don’t have to be relations. On the other hand, if

e

x

denotes the

stationary

path in

x

∈

Q

0

, then it can be easily checked, that every element of the form

e

x

⋅

a

i

⋅

e

y

is either zero or a relation. Also, note that

a

i

=

∑

x

,

y

∈

Q

0

e

x

⋅

a

i

⋅

e

y

.

Since

Q

is finite, then this completes the proof.

□

References

1

I. Assem, D. Simson, A. SkowroÃÆski,

Elements of the

Representation Theory

of Associative Algebras, vol 1.

, Cambridge University Press 2006, 2007

Title

properties of admissible ideals

Canonical name

PropertiesOfAdmissibleIdeals

Date of creation

2013-03-22 19:16:48

Last modified on

2013-03-22 19:16:48

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Theorem

Classification

msc 14L24

Generated on Fri Feb 9 18:00:16 2018 by

LaTeXML
quiver

quiver

A quiver

Q

is a directed

multigraph

. So it consists of a set

I

of vertices and a set

A

of arrows between vertices. Each arrow

a

∈

A

is uniquely characterized by its start vertex and its end vertex in

I

.

Title

quiver

Canonical

name

Quiver

Date of creation

2013-03-22 15:34:19

Last modified on

2013-03-22 15:34:19

Owner

mathangel (11460)

Last modified by

mathangel (11460)

Numerical id

7

Author

mathangel (11460)

Entry type

Definition

Classification

msc 14L24

Synonym

Köcher

Related topic

Multigraph

Defines

quiver

Generated on Fri Feb 9 18:00:18 2018 by

LaTeXML
quiver representations and representation morphisms

quiver representations and representation morphisms

Let

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

be a quiver, i.e.

Q

0

is a set of vertices,

Q

1

is a set of arrows and

s

,

t

:

Q

1

→

Q

0

are functions such that

s

maps each arrow to its source and

t

maps each arrow to its target.

A

representation

𝕍

of

Q

over a field

k

is a family of

vector spaces

{

V

i

}

i

∈

Q

0

over

k

together with a family of

k

-

linear maps

{

f

a

:

V

s

⁢

(

a

)

→

V

t

⁢

(

a

)

}

a

∈

Q

1

.

A

morphism

F

:

𝕍

→

𝕎

between representations

𝕍

=

(

V

i

,

g

a

)

and

𝕎

=

(

W

i

,

h

a

)

is a family of

k

-linear maps

{

F

i

:

V

i

→

W

i

}

i

∈

Q

0

such that for each arrow

a

∈

Q

1

the following

relation

holds:

F

t

⁢

(

a

)

∘

g

a

=

h

a

∘

F

s

⁢

(

a

)

.

Obviously we can compose

morphisms of representations

and in this the case class of all representations and representation morphisms together with the standard composition is a category. This category is

abelian

.

It can be shown that for each finite quiver

Q

(i.e. with

Q

0

finite) and field

k

there exists an

algebra

A

over

k

such that the category of representations of

Q

is equivalent to the category of modules over

A

.

A representation

𝕍

of

Q

is called

trivial

iff

V

i

=

0

for each vertex

i

∈

Q

0

.

A representation

𝕍

of

Q

is called

locally

finite-dimensional

iff

dim

k

⁢

V

i

<

∞

for each vertex

i

∈

Q

0

and

finite-dimensional

iff

𝕍

is locally finite-dimensional and

V

i

=

0

for almost all vertices

i

∈

Q

0

.

Title

quiver representations and representation morphisms

Canonical name

QuiverRepresentationsAndRepresentationMorphisms

Date of creation

2013-03-22 19:16:15

Last modified on

2013-03-22 19:16:15

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:20 2018 by

LaTeXML
quotient quiver

quotient quiver

Let

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

be a quiver.

Definition.

An

equivalence relation

on

Q

is a pair

∼

=

(

∼

0

,

∼

1

)

such that

∼

0

is an equivalence relation on

Q

0

,

∼

1

is an equivalence relation on

Q

1

and if

α

∼

1

β

for some arrows

α

,

β

∈

Q

1

, then

s

⁢

(

α

)

∼

0

s

⁢

(

β

)

⁢

and

⁢

t

⁢

(

α

)

∼

1

t

⁢

(

β

)

.

If

∼

is an equivalence relation on

Q

, then

(

Q

0

/

∼

0

,

Q

1

/

∼

1

,

s

′

,

t

′

)

is a quiver, where

s

′

⁢

(

[

α

]

)

=

[

s

⁢

(

α

)

]

t

′

⁢

(

[

α

]

)

=

[

t

⁢

(

α

)

]

.

This quiver is called

the quotient quiver

of

Q

by

∼

and is denoted by

Q

/

∼

.

It can be easily seen, that if

Q

is a quiver and

∼

is an equivalence relation on

Q

, then

π

:

Q

→

Q

/

∼

given by

π

=

(

π

0

,

π

1

)

, where

π

0

and

π

1

are

quotient maps

is a

morphism

of quivers. It will be called

the quotient morphism

.

Example.

Consider the following quiver

\xymatrix

⁢

&

⁢

2

⁢

\ar

⁢

[

d

⁢

r

]

c

⁢

&

⁢

1

⁢

\ar

⁢

[

u

⁢

r

]

a

⁢

\ar

⁢

[

d

⁢

r

]

b

⁢

&

⁢

&

⁢

3

⁢

&

⁢

4

⁢

\ar

⁢

[

u

⁢

r

]

d

⁢

&

If we take

∼

by putting

2

∼

0

4

and

a

∼

1

b

,

c

∼

1

d

, then the corresponding quotient quiver is

isomorphic

to

\xymatrix

⁢

1

⁢

\ar

⁢

[

r

]

⁢

&

⁢

2

⁢

\ar

⁢

[

r

]

⁢

&

⁢

3

Title

quotient quiver

Canonical name

QuotientQuiver

Date of creation

2013-03-22 19:17:22

Last modified on

2013-03-22 19:17:22

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:22 2018 by

LaTeXML
relations in quiver

relations in quiver

Let

Q

be a quiver and

k

a field.

Definition.

A

relation

in

Q

is a

linear combination

(over

k

) of paths of length at least

2

such that all paths have the same source and target. Thus a relation is an element of the path

algebra

k

⁢

Q

of the form

ρ

=

∑

i

=

1

m

λ

i

⋅

w

i

such that there exist

x

,

y

∈

Q

0

with

s

⁢

(

w

i

)

=

x

and

t

⁢

(

w

i

)

=

y

for all

i

, all

w

i

are of length at least

2

and not all

λ

i

are zero.

If a relation

ρ

is of the form

ρ

=

w

for some path

w

, then it is called a

zero relation

and if

ρ

=

w

1

-

w

2

for some paths

w

1

,

w

2

, then

ρ

is called a

commutativity

relation

.

Title

relations in quiver

Canonical name

RelationsInQuiver

Date of creation

2013-03-22 19:16:45

Last modified on

2013-03-22 19:16:45

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:24 2018 by

LaTeXML
representations of a bound quiver

representations of a bound quiver

Let

(

Q

,

I

)

be a bound quiver (

http://planetmath.org/AdmissibleIdealsBoundQuiverAndItsAlgebra

) over a field

k

.

Let

𝕍

be a

representation

of

Q

over

k

composed by

{

f

⁢

(

q

)

}

q

∈

Q

1

a family of linear maps. If

w

=

(

α

1

,

…

,

α

n

)

is a path in

Q

, then we have the evaluation map

f

w

=

f

⁢

(

α

n

)

∘

f

⁢

(

α

n

-

1

)

∘

⋯

∘

f

⁢

(

α

2

)

∘

f

⁢

(

α

1

)

.

For stationary paths we define

f

e

x

:

V

x

→

V

x

by

f

e

x

=

0

. Also, note that if

ρ

is a

relation

(

http://planetmath.org/RelationsInQuiver

) in

Q

, then

ρ

=

∑

i

=

1

m

λ

i

⋅

w

i

where all

w

i

’s have the same source and target. Thus it makes sense to talk about evaluation in

ρ

, i.e.

f

ρ

=

∑

i

=

1

n

λ

i

⋅

f

w

i

.

In particular

f

ρ

:

V

s

⁢

(

w

i

)

→

V

t

(

w

i

)

is a linear map.

Recall that the ideal

I

is generated by relations (see this entry (

http://planetmath.org/PropertiesOfAdmissibleIdeals

))

{

ρ

1

,

…

,

ρ

n

}

.

Definition.

A representation

𝕍

of

Q

over

k

with linear mappings

{

f

⁢

(

q

)

}

q

∈

Q

1

is said to be

bound by

I

if

f

ρ

i

=

0

for every

i

=

1

,

…

,

n

.

It can be easily checked, that this definition does not depend on the choice of (relation)

generators

of

I

.

The

full subcategory

of the

category

of all representations which is composed of all representations bound by

I

is denoted by

REP

⁢

(

Q

,

I

)

. It can be easily seen, that it is

abelian

.

Title

representations of a bound quiver

Canonical name

RepresentationsOfABoundQuiver

Date of creation

2013-03-22 19:16:51

Last modified on

2013-03-22 19:16:51

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

6

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:26 2018 by

LaTeXML
subquiver and image of a quiver

subquiver and image of a quiver

Let

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

be a quiver.

Definition.

A quiver

Q

′

=

(

Q

0

′

,

Q

1

′

,

s

′

,

t

′

)

is said to be a

subquiver

of

Q

, if

Q

0

′

⊆

Q

0

,

Q

1

′

⊆

Q

1

are such that if

α

∈

Q

1

′

, then

s

⁢

(

α

)

,

t

⁢

(

α

)

∈

Q

0

′

. Furthermore

s

′

⁢

(

α

)

=

s

⁢

(

α

)

,

t

′

⁢

(

α

)

=

t

⁢

(

α

)

.

In this case we write

Q

′

⊆

Q

.

A subquiver

Q

′

⊆

Q

is called

full

if for any

x

,

y

∈

Q

0

′

and any

α

∈

Q

1

such that

s

⁢

(

α

)

=

x

and

t

⁢

(

α

)

=

y

we have that

α

∈

Q

1

′

. In other words a subquiver is full if it ,,inherits” all arrows between points.

If

Q

′

is a subquiver of

Q

, then the

mapping

i

=

(

i

0

,

i

1

)

where both

i

0

,

i

1

are

inclusions

is a

morphism

of quivers. In this case

i

is called

the inclusion morphism

.

If

F

:

Q

→

Q

′

is any morphism of quivers

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

and

Q

′

=

(

Q

0

′

,

Q

1

′

,

s

′

,

t

′

)

, then the quadruple

Im

⁢

(

F

)

=

(

Im

⁢

(

F

0

)

,

Im

⁢

(

F

1

)

,

s

′′

,

t

′′

)

where

s

′′

,

t

′′

are the

restrictions

of

s

′

,

t

′

to

Im

⁢

(

F

1

)

is called

the image of

F

. It can be easily shown, that

Im

⁢

(

F

)

is a subquiver of

Q

′

.

Title

subquiver and image of a quiver

Canonical name

SubquiverAndImageOfAQuiver

Date of creation

2013-03-22 19:17:19

Last modified on

2013-03-22 19:17:19

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:28 2018 by

LaTeXML
translation quiver

translation quiver

Let

Q

=

(

Q

0

,

Q

1

,

s

,

t

)

be a

locally finite quiver

without loops. Recall that a loop is an arrow

α

such that

s

⁢

(

α

)

=

t

⁢

(

α

)

. Let

X

,

Y

⊆

Q

0

.

Definition 1.

A pair

(

Q

,

τ

)

is said to be a

translation quiver

iff the following holds:

1.

τ

:

X

→

Y

is a bijection;

2.

If

x

∈

X

and

y

∈

x

-

is a direct predecessor of

x

, then the number of arrows from

y

to

x

is equal to the number of arrows from

τ

⁢

(

x

)

to

y

.

If

(

Q

,

τ

)

is a translation quiver then we will say that

τ

⁢

(

x

)

exists if

x

∈

X

and

τ

⁢

(

x

)

does not exist (or it is not defined) if

x

∉

X

.

Definition 2.

If

(

Q

,

τ

)

is a translation quiver, then a pair

(

Q

′

,

τ

′

)

is called a

translation

subquiver

if it is a translation quiver,

Q

′

is a full subquiver (

http://planetmath.org/SubquiverAndImageOfAQuiver

) of

Q

and

τ

′

⁢

(

x

)

=

τ

⁢

(

x

)

whenever

x

is a vertex in

Q

′

such that

τ

⁢

(

x

)

exists and belongs to

Q

′

.

Example.

Let

Q

be the following quiver:

\xymatrix

⁢

1

⁢

\ar

⁢

[

r

⁢

d

]

⁢

&

⁢

&

⁢

2

⁢

\ar

⁢

[

r

⁢

d

]

⁢

&

⁢

&

⁢

3

⁢

\ar

⁢

[

r

⁢

d

]

⁢

&

⁢

&

⁢

4

⁢

&

⁢

5

⁢

\ar

⁢

[

r

⁢

d

]

⁢

\ar

⁢

[

r

⁢

u

]

⁢

&

⁢

&

⁢

6

⁢

\ar

⁢

[

r

]

⁢

\ar

⁢

[

r

⁢

u

]

⁢

&

⁢

7

⁢

\ar

⁢

[

r

]

⁢

&

⁢

8

⁢

\ar

⁢

[

r

⁢

u

]

⁢

&

⁢

&

⁢

&

⁢

9

⁢

\ar

⁢

[

r

⁢

u

]

If we put

X

=

{

2

,

3

,

4

,

6

,

8

}

,

Y

=

{

1

,

2

,

3

,

5

,

6

}

and

τ

⁢

(

2

)

=

1

;

τ

⁢

(

3

)

=

2

;

τ

⁢

(

4

)

=

3

;

τ

⁢

(

6

)

=

5

;

τ

⁢

(

8

)

=

6

;

then the pair

(

Q

,

τ

)

is a translation quiver and

\xymatrix

⁢

&

⁢

2

⁢

\ar

⁢

[

r

⁢

d

]

⁢

&

⁢

5

⁢

\ar

⁢

[

r

⁢

u

]

⁢

\ar

⁢

[

r

⁢

d

]

⁢

&

⁢

&

⁢

6

⁢

&

⁢

9

⁢

\ar

⁢

[

r

⁢

u

]

⁢

&

is its translation subquiver, where

τ

′

⁢

(

6

)

=

5

.

Remark.

It is common to write translation quivers as in example. This means that

Q

is ,,oriented” to the right and in rows we have vertices such that ,,jumping” two places to the left gives us

τ

of this vertex. Note that in the example the vertex

7

is not written in the same row as

9

because

τ

⁢

(

7

)

is not

9

(indeed,

τ

⁢

(

7

)

is not defined).

Title

translation quiver

Canonical name

TranslationQuiver

Date of creation

2013-03-22 19:17:53

Last modified on

2013-03-22 19:17:53

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Definition

Classification

msc 14L24

Generated on Fri Feb 9 18:00:31 2018 by

LaTeXML