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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?

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up vote 7 down vote accepted

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.

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.

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Either mathoverflow is acting weird, or my home computer is. I posted this question last night, but now it shows up as unknown(google) so I am afraid I can't actually mark this as the answer. ... feel free to imagine you have 15 more points if you like - and thanks. – streklin Oct 9 '09 at 17:56
@streklin: it should be better now; I merged the two users and added the other OpenID as an alternate on your account. The problem is that Google is issuing different OpenIDs on different occasions for some reason. If it happens again, flag the post for moderator attention. – Anton Geraschenko Oct 9 '09 at 19:29
@Anton - cool - thanks for the help. Good work so far by the way – streklin Oct 9 '09 at 19:35

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 $T_RV$ of a finite dimensional bimodule $V=V_Q$ over a split semisimple finite dimensional commutative algebra $R=\oplus_{i\in I}k$ over a field $k$. More precisely, if $M_{ij}$ are the simple (1-dimensional) $R$-bimodules, then $V_Q=\oplus n_{ij}(Q)M_{ij}$, where $n_{ij}(Q)$ is the number of edges in $Q$ going from $i$ to $j$.

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But the (tensor) product of elements of $T_RV$ is not the product in the path algebra, or is it? – Vít Tuček Aug 4 '13 at 11:11

I realize that this question is four months old, but I'm new to the site.

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.

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Elementary==basic, in some circles, and the field has to be algebraically closed (as $\mathbb C$ is not a path $\mathbb R$-algebra, for example) – Mariano Suárez-Alvarez Mar 4 '10 at 2:30
By the way, you shouldn't feel any compunction about "resurrecting" an old question. It's one of the great abilities of the site. – Ben Webster Mar 4 '10 at 2:41

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