most recent 30 from http://mathoverflow.net 2013-05-21T14:53:20Z http://mathoverflow.net/feeds/question/93187 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93187/ring-completion-of-kq Amied 2012-04-05T04:26:46Z 2012-04-05T13:01:10Z

<p>Hello,</p> <p>Let $Q$ be a finite quiver, let $M$ denote the arrow ideal and let $kQ$ denote the path algebra. Endow $kQ$ with the $M$-adic topology. Now let $\mathcal{A}$ be the set of all formal series ${\sum_{\gamma} a_{\gamma} \gamma , a_{\gamma} \in K}$ where $\gamma$ is a path. Then $kQ$ is naturally a $k$-algebra (same operations we use for a group ring). Define $t$ as follows:</p> <p>$t(\sum a_{\gamma} \gamma)= n$ if $a_{p}$ is non-zero for at least one path p of length $n$ and $a_{q}=0$ for all paths of length smaller that $n$. This induces a metric $(a,b) \mapsto 2^{-t(a-b)}$, then $\mathcal{A}$ becomes a topological algebra with this metric. </p> <p>Question: why is $\mathcal{A}$ isomorphic (as a topological algebra) to the completion of $kQ$ endowed with the $M$-adic topology?</p>

http://mathoverflow.net/questions/93187/ring-completion-of-kq/93211#93211 Benjamin Steinberg 2012-04-05T13:01:10Z 2012-04-05T13:01:10Z

<p>The metric you have given induces the M-adic uniformity on kQ. A ball in the metric around 0 contains all linear combinations of paths whose shortest term with non-zero coefficient is a path of length greater than or equal to n for a certain n which can be computed from the radius of the ball. So $M^n$ is contained in this ball. On the other hand $M^n$ is a ball around 0 by choosing the radius appropriately. The isomorphism follows from uniqueness of completions. </p>