# The sum of trivial paths for a finite quiver is 1?

let $Q=(E_0, E_1)$ be a quiver and let $P_Q$ be a path algebra of $Q$. Let $p_i$ be the trivial path associated to each vertex $i$ in $E_0$.

Then why is $\sum_{i\in E_0} p_i=1$ for a finite quiver?

Are the $p_i$'s viewed as self-loops (arrows whose source and target is $i$, i.e., $\mathrm{End}(V_i)$ where $V_i$ is a vector space assigned to $i\in E_0$) or are the $p_i$'s viewed as "no arrows" (i.e., $p_i$ is the zero endomorphism from $V_i$ to itself)?

-

The $p_i$s are self-loops, and special ones, since they're required to compose trivially. That is, if you look at the quiver as a category, they're the identity morphisms. Then in any representation they're mapped to the identity automorphism of each vector space.
As to your question about their sum: take an arbitrary element $\pi=\sum c_j \pi_j,$ where the $\pi_j$ are arbitrary paths in $Q$ and $c_j$ constants from whatever you're taking $P_Q$ over.
Now in considering the product $(\sum p_i) \pi,$ the claim is that each $c_j\pi_j$ is annihilated by all but one of the $p_i,$ and sent to itself by the last. This is because the paths can't concatenate if the end point of $\pi_j$ isn't the start (and end) point of $p_i$, in which case the product is defined as 0, while the $p_i$ that does match the end point of $\pi_j$ is required to composed trivially with it. So each component path in $\pi$ comes back out of the product once, and only once.
Thanks Kevin! $\:\:\:$ – math-visitor Aug 2 '12 at 19:38