Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.
6
votes
0
answers
95
views
Condition of indecomposability in the definition of wild representation type
In Tame and wild matrix problems, Drozd defines an algebra $A$ to be wild if there is an $A-k\langle x,y \rangle$-bimodule $M$, such that ${M \otimes}-$ reflects isomorphisms.
Usually I've seen the de …
6
votes
1
answer
445
views
Tame-Wild dichotomy; why can't tame algebras be wild?
I would like to understand the Tame-Wild dichotomy, and in particular why an algebra cannot be tame and (semi-)wild at the same time. I've looked in the papers by Drozd and Crawley-Boevey [D80, CB88]. …
6
votes
3
answers
365
views
Is there a finite dimensional algebra with left finitistic dimension different from its righ...
Let $\Lambda$ be finite dimensional algebra over a field $k$. The (left) finitistic dimension of a finite dimensional algebra is defined as
$$\operatorname{findim}(\Lambda)=\sup\{\operatorname{pd}M | …
3
votes
1
answer
185
views
Explicit proof that algebra is derived wild
Following the terminology of
Drozd, Yuriy A., Derived tame and derived wild algebras, Algebra Discrete Math. 2004, No. 1, 57-74 (2004). ZBL1067.16028.
let $A$ and $R$ be algebras over a field $k$. A s …
2
votes
Is there a finite dimensional algebra with left finitistic dimension different from its righ...
Let $\Lambda$ be the path algebra of the quiver
with relations $(a^2, ac, ba, cbc)$. Then I claim $\operatorname{findim}(\Lambda) \geq 1$, while $\operatorname{findim}(\Lambda^{op})=0$.
The projectiv …
1
vote
Accepted
Explicit proof that algebra is derived wild
A few such examples are constructed in
Bekkert, Viktor; Drozd, Yuriy; Futorny, Vyacheslav, Derived tame local and two-point algebras, J. Algebra 322, No. 7, 2433-2448 (2009). ZBL1191.16017.