# admissible ideals,, bound quiver and its algebra

Assume, that $Q$ is a quiver and $k$ is a field. Let $kQ$ be the associated path algebra. Denote by $R_{Q}$ the two-sided ideal in $kQ$ generated by all paths of length $1$, i.e. all arrows. This ideal is known as the arrow ideal.

It is easy to see, that for any $m\geqslant 1$ we have that $R_{Q}^{m}$ is a two-sided ideal generated by all paths of length $m$. Note, that we have the following chain of ideals:

$R_{Q}^{2}\supseteq R_{Q}^{3}\supseteq R_{Q}^{4}\supseteq\cdots$

Definition. A two-sided ideal $I$ in $kQ$ is said to be admissible if there exists $m\geqslant 2$ such that

$R_{Q}^{m}\subseteq I\subseteq R_{Q}^{2}.$

If $I$ is an admissible ideal in $kQ$, then the pair $(Q,I)$ is said to be a bound quiver and the quotient algebra $kQ/I$ is called bound quiver algebra.

The idea behind this is to treat some paths in a quiver as equivalent. For example consider the following quiver

$\xymatrix{&2\ar[dr]^{{b}}&\\ 1\ar[rr]^{{c}}\ar[dr]_{{e}}\ar[ur]^{{a}}&&3\\ &4\ar[ur]_{{f}}}$

Then the ideal generated by $ab-c$ is not admissible ($ab-c\not\in R^{2}_{Q}$) but an ideal generated by $ab-ef$ is. We can see that this means that ,,walking” from $1$ to $3$ directly and through $2$ is not the same, but walking in the same number of steps is.

Note, that in our case there is no path of length greater then $2$. In particular, for any $m>2$ we have $R_{Q}^{m}=0$.

More generally, it can be easily checked, that if $Q$ is a finite quiver without oriented cycles, then there exists $m\in\mathbb{N}$ such that $R_{Q}^{m}=0$

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Reference
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## Mathematics Subject Classification

14L24 Geometric invariant theory

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## Info

Owner: joking
Added: 2011-02-18 - 22:58
Author(s): joking

## Versions

(v6) by joking 2013-03-22
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