4
$\begingroup$

Given a representation-finite (connected) quiver algebra $A$ and a module $M$.

Is there a good way to test whether the set $\{ [N] \mid N \in \mathrm{add}(M) \}$ generates $K_0(\mbox{mod-}A)$?

Can this be done using the GAP-package QPA?

$\endgroup$
4
$\begingroup$

Suppose that $M = \oplus_{i=1}^n M_i$ with $M_i$ being indecomposable and assume that $M_i \not\simeq M_j$ for $i\neq j$ (that is, $M$ is basic) over a finite dimensional algebra $A$. Define a matrix $K$ with rows equal to the dimension vector of $M_i$ for $i = 1,2,\ldots,n$. Then $M$ generates Grothendieck group of $A$ if and only if the Smith normal form of $K$ over $\mathbb{Z}$ is of the form $\begin{bmatrix} I \\ \hline O\end{bmatrix}$ where $I$ is an identity matrix. So we could do the following in QPA:

gap> A := NakayamaAlgebra(Rationals, [3,4,3,3]); 
<Rationals[<quiver with 4 vertices and 4 arrows>]/
<two-sided ideal in <Rationals[<quiver with 4 vertices and 4 arrows>]>, (3 generators)>>
gap> P := IndecProjectiveModules(A);             
[ <[ 1, 1, 1, 0 ]>, <[ 1, 1, 1, 1 ]>, <[ 1, 0, 1, 1 ]>, <[ 1, 1, 0, 1 ]> ]
gap> I := IndecInjectiveModules(A);              
[ <[ 1, 1, 1, 1 ]>, <[ 1, 1, 0, 1 ]>, <[ 1, 1, 1, 0 ]>, <[ 0, 1, 1, 1 ]> ]
gap> M := Concatenation(P,I);                   
[ <[ 1, 1, 1, 0 ]>, <[ 1, 1, 1, 1 ]>, <[ 1, 0, 1, 1 ]>, <[ 1, 1, 0, 1 ]>, <[ 1, 1, 1, 1 ]>, 
  <[ 1, 1, 0, 1 ]>, <[ 1, 1, 1, 0 ]>, <[ 0, 1, 1, 1 ]> ]
gap> K := List( M, m -> DimensionVector( m ) ); 
[ [ 1, 1, 1, 0 ], [ 1, 1, 1, 1 ], [ 1, 0, 1, 1 ], [ 1, 1, 0, 1 ], [ 1, 1, 1, 1 ], [ 1, 1, 0, 1 ], 
  [ 1, 1, 1, 0 ], [ 0, 1, 1, 1 ] ]
gap> SNK := SmithNormalFormIntegerMat( K );
[ [ 1, 0, 0, 0 ], [ 0, 1, 0, 0 ], [ 0, 0, 1, 0 ], [ 0, 0, 0, 1 ], [ 0, 0, 0, 0 ], [ 0, 0, 0, 0 ], 
  [ 0, 0, 0, 0 ], [ 0, 0, 0, 0 ] ]
gap> Display(SNK);
[ [  1,  0,  0,  0 ],
  [  0,  1,  0,  0 ],
  [  0,  0,  1,  0 ],
  [  0,  0,  0,  1 ],
  [  0,  0,  0,  0 ],
  [  0,  0,  0,  0 ],
  [  0,  0,  0,  0 ],
  [  0,  0,  0,  0 ] ]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.