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Let's say a quiver $Q$ is covered by cycles if each of it’s arrows can be included in an oriented cycle.

It's easy to prove that if a path-algebra with relations $KQ/I$ (where $I$ is an admissible ideal) is selfinjective then $Q$ is covered by cycles. The following question occured to me: is the opposite true? Is it true that if a quiver $Q$ is covered by cycles then there exist an admissible ideal $I\triangleleft KQ$ such that the algebra $KQ/I$ is selfinjective?

I found an answer on my own. It's true indeed. But the proof is quite difficult. Maybe this result is already known?

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Benjamin- if there are no arrows, than any condition that begins "each arrow..." is vacuously true. –  Ben Webster Nov 27 '11 at 22:05
    
Sorry, I thought it said vertices. –  Benjamin Steinberg Nov 27 '11 at 22:38
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1 Answer

up vote 7 down vote accepted

One way to prove a finite-dimensional algebra is self-injective is to find an isomorphism of modules $A\cong A^*$, that is a non-degenerate bilinear form such that $(ab,c)=(a,bc)$. Of course, such a form is uniquely determined by the linear map $t(a)=(a,1)$ (you can recover the form by $(a,b)=t(ab)$).

Thus, for any such linear map on an algebra, we have a canonical quotient on which it is non-degenerate (you can check that the kernel of $(-,-)$ is a 2-sided ideal), which is self-injective if it is finite dimensional.

Of course, if we consider a path algebra $kQ$, and consider a linear map that kills all paths of length $\geq N$, this will obviously have finite dimensional self-injective quotient.

Thus, we need to see that if $Q$ is covered by cycles, we can find such a form whose kernel is an admissible ideal. We can do this in a very silly way: pick any set of cycles $C_i$ which cover the graph and pick $a_i$ non-zero elements of $k$. Let $t$ be the linear map which sends any path that traces out $C_i$ (with any starting point) to $a_i$ (you also need to add a term which gives non-zero values on all isolated vertices). The kernel of this contains all paths longer than all $C_i$, so it is finite dimensional. On the other hand, for any non-zero linear combination of idempotents (length 0 paths) and length 1 paths, one can always take pairing against a path either tracing one of the cycles starting at one of the vertices (if idempotents appear) or completing one of the cycles (if length 1 paths appear). Thus, none of these lie in the kernel, and we are done.

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Thank you ! ! ! –  Sergey Nov 28 '11 at 14:42
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