path algebra of a disconnected quiver
Let be a disconnected quiver, i.e. can be written as a disjoint union of two quivers and (which means that there is no path starting in and ending in and vice versa) and let be an arbitrary field.
Proof. If is a path in , then belongs either to or . Define linear map
by if or if and extend it linearly to entire . We will show that is an isomorphism of algebras.
If are paths in , then since and are disjoint, then each of them entirely lies in or . Now since and don’t have common vertices it follows that . Without loss of generality we may assume, that is in and is in . Then we have
If both lie in the same component, for example in , then
Since preservers multiplication on paths, then preserves multiplication and thus is an algebra homomorphism.
Obviously by definition is 1-1.
It remains to show, that is onto. Assume that . Then we can write
where are paths in and are paths in . It can be easily checked, that
Here we consider all and as paths in .
Thus is an isomorphism, which completes the proof.
|Title||path algebra of a disconnected quiver|
|Date of creation||2013-03-22 19:16:25|
|Last modified on||2013-03-22 19:16:25|
|Last modified by||joking (16130)|