most recent 30 from http://mathoverflow.net 2013-05-18T23:41:42Z http://mathoverflow.net/feeds/question/210 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/210/questions-about-quivers streklin 2009-10-09T00:14:55Z 2010-03-04T03:50:01Z

<p>Hi,</p> <p>The definition I have for a Path Algega of a quiver Q is that it is the algebra whose basis is formed by the oriented paths in Q, including the trivial ones. Apparently multiplication is given by concatenation of paths, and those that can't be concatenated are considered zero. That part I think I understand, but I am wondering if there is a non-formal interpretation of the addition operation on two elements of a path algebra someone could provide?</p>

http://mathoverflow.net/questions/210/questions-about-quivers/231#231 Ben Webster 2009-10-09T17:21:34Z 2009-10-09T17:21:34Z

<p>The short answer is no. You just have to think of them as formal sums, in the same way that you can only think of elements of a group algebra as formal sums.</p> <p>What you can do is think of the path category of a quiver, which is the category whose objects are elements are vertices of the quiver, and whose morphisms are paths in the quiver. A representation of the path algebra as an algebra is essentially just a functor from this path category to vectors spaces.</p>

http://mathoverflow.net/questions/210/questions-about-quivers/17037#17037 Helene Tyler 2010-03-04T01:24:07Z 2010-03-04T01:24:07Z

<p>I realize that this question is four months old, but I'm new to the site.</p> <p>Another thing you can do is use the relationship between path algebras and other algebras to breathe some life into the formal construction. For example, it is known that any elementary hereditary algebra over a field is isomorphic to a path algebra.</p>

http://mathoverflow.net/questions/210/questions-about-quivers/17051#17051 Pavel Etingof 2010-03-04T03:50:01Z 2010-03-04T03:50:01Z

<p>Maybe it is useful to add that a path algebra is a special case of a tensor algebra of a bimodule. Namely, the path algebra of a finite quiver $Q$ is the tensor algebra <code>$T_RV$</code> of a finite dimensional bimodule <code>$V=V_Q$</code> over a split semisimple finite dimensional commutative algebra <code>$R=\oplus_{i\in I}k$</code> over a field $k$. More precisely, if <code>$M_{ij}$</code> are the simple (1-dimensional) $R$-bimodules, then <code>$V_Q=\oplus n_{ij}(Q)M_{ij}$</code>, where <code>$n_{ij}(Q)$</code> is the number of edges in $Q$ going from $i$ to $j$. </p>