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Hello,

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:

$t(\sum a_{\gamma} \gamma)= n$ if there exists $a_{p}$ such that theis non-zero for at least one path $p$ hasp of length $n$ and $0$ otherwise$a_{q}=0$ 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.

Question: why is $\mathcal{A}$ isomorphic (as a topological algebra) to the completion of $kQ$ endowed with the $M$-adic topology?

Hello,

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:

$t(\sum a_{\gamma} \gamma)= n$ if there exists $a_{p}$ such that the path $p$ has length $n$ and $0$ otherwise. This induces a metric $(a,b) \mapsto 2^{-t(a-b)}$, then $\mathcal{A}$ becomes a topological algebra with this metric.

Question: why is $\mathcal{A}$ isomorphic (as a topological algebra) to the completion of $kQ$ endowed with the $M$-adic topology?

Hello,

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:

$t(\sum a_{\gamma} \gamma)= n$ 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.

Question: why is $\mathcal{A}$ isomorphic (as a topological algebra) to the completion of $kQ$ endowed with the $M$-adic topology?

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Amied
  • 65
  • 4

Hello,

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:

$t(\sum a_{\gamma} \gamma)= n$ if there exists $a_{p}$ such that the length of the path $p$ is equal tohas length $s$$n$ and $0$ otherwise. This induces a metric $(a,b) \mapsto 2^{-t(a-b)}$, then $\mathcal{A}$ becomes a topological algebra with this metric.

Question: why is $\mathcal{A}$ isomorphic (as a topological algebra) to the completion of $kQ$ endowed with the $M$-adic topology?

Hello,

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:

$t(\sum a_{\gamma} \gamma)= n$ if there exists $a_{p}$ such that the length of the path $p$ is equal to $s$ and $0$ otherwise. This induces a metric $(a,b) \mapsto 2^{-t(a-b)}$, then $\mathcal{A}$ becomes a topological algebra with this metric.

Question: why is $\mathcal{A}$ isomorphic (as a topological algebra) to the completion of $kQ$ endowed with the $M$-adic topology?

Hello,

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:

$t(\sum a_{\gamma} \gamma)= n$ if there exists $a_{p}$ such that the path $p$ has length $n$ and $0$ otherwise. This induces a metric $(a,b) \mapsto 2^{-t(a-b)}$, then $\mathcal{A}$ becomes a topological algebra with this metric.

Question: why is $\mathcal{A}$ isomorphic (as a topological algebra) to the completion of $kQ$ endowed with the $M$-adic topology?

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Amied
  • 65
  • 4

Ring completion of $kQ$

Hello,

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:

$t(\sum a_{\gamma} \gamma)= n$ if there exists $a_{p}$ such that the length of the path $p$ is equal to $s$ and $0$ otherwise. This induces a metric $(a,b) \mapsto 2^{-t(a-b)}$, then $\mathcal{A}$ becomes a topological algebra with this metric.

Question: why is $\mathcal{A}$ isomorphic (as a topological algebra) to the completion of $kQ$ endowed with the $M$-adic topology?