morphisms of path algebras induced from morphisms of quivers


Let Q=(Q0,Q1,s,t), Q=(Q0,Q1,s,t) be quivers and let F:QQ be a morphism of quivers.

PropositionPlanetmathPlanetmath 1. If w=(α1,,αn) is a path in Q, then

F(w)=(F1(α1),,F1(αn))

is a path in Q.

Proof. Indeed, for any i=1,,n-1 we calculate

t(F1(αi))=F0(t(αi))=F0(s(αi+1))=t(F1(αi+1)),

which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Proposition 2. Let w,u be paths in Q. If w is compatible (http://planetmath.org/PathAlgebraOfAQuiver) with u then F(w) is compatible (http://planetmath.org/PathAlgebraOfAQuiver) with F(u). The inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath implicationMathworldPlanetmath holds if and only if F0 is an injective function.

Proof. Assume that we have the following presentationsMathworldPlanetmathPlanetmath:

w=(w1,,wn);
u=(u1,,un).

If t(wn)=s(u1), then

t(F1(wn))=F0(t(wn))=F0(s(u1))=s(F1(u1))

which shows the first part of the thesis.

For the second part note, that if F0 is injective, then the above equalities can be reversed to obtain that t(wn)=s(u1).

On the other hand assume that F0 is not injective, i.e. F0(a)=F0(b) for some distinct vertices a,bQ0. Then for stationary paths ea and eb we have that

t(F1(ea))=F0(t(ea))=F0(a)=F0(b)=F0(s(eb))=s(F1(eb))

so paths (F1(ea)) and (F1(eb)) are compatible (http://planetmath.org/PathAlgebraOfAQuiver), although (ea), (eb) are not.

Definition. Let k be a field. The linear map

F¯:kQkQ

defined on a basis of kQ by

F¯(w)=F(w)

is said to be induced from F.

Proposition 3. The linear map F¯:kQkQ induced from F:QQ is a homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of algebras if and only if F0 is injective.

Proof. Indeed, we will show that F¯ preservers multiplicationPlanetmathPlanetmath of compatible paths (http://planetmath.org/PathAlgebraOfAQuiver). If

w=(w1,,wn);
u=(u1,,um)

are compatible paths (http://planetmath.org/PathAlgebraOfAQuiver) in Q, then

F¯(wu)=F¯((w1,,wn,u1,,um))=(F1(w1),,F1(wn),F1(u1),F1(um))=F¯(w)F¯(u),

which completes this part.

Now assume that w, u are paths, which are not compatible (http://planetmath.org/PathAlgebraOfAQuiver). If F0 is injective, then by proposition 2 F(w) and F(u) are also not compatible (http://planetmath.org/PathAlgebraOfAQuiver) and thus

F¯(wu)=F¯(0)=0=F¯(w)F¯(u).

On the other hand, if F0 is not injective, then there are paths w, u which are not compatible (http://planetmath.org/PathAlgebraOfAQuiver), but F(w), F(u) are. Assume, that F¯ is a homomorphism of algebras. Then

0=F¯(0)=F¯(wu)=F¯(w)F¯(u)0

because of the compatibility (http://planetmath.org/PathAlgebraOfAQuiver). The contradictionMathworldPlanetmathPlanetmath shows that F¯ is not a homomorphism of algebras. This completes the proof.

Title morphisms of path algebras induced from morphisms of quivers
Canonical name MorphismsOfPathAlgebrasInducedFromMorphismsOfQuivers
Date of creation 2013-03-22 19:17:03
Last modified on 2013-03-22 19:17:03
Owner joking (16130)
Last modified by joking (16130)
Numerical id 6
Author joking (16130)
Entry type Definition
Classification msc 14L24