Timeline for The tensor product of two path algebras
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2012 at 10:53 | vote | accept | Aimin Xu | ||
| May 9, 2012 at 21:13 | comment | added | Benjamin Steinberg | Hmm, my picture of the Hasse diagram of a diamond didn't draw well in the comments. | |
| May 9, 2012 at 21:12 | comment | added | Benjamin Steinberg | I think he means take the Cartesian product of the quivers and then put in commutativity relations. For instance, if you take the quiver 1->2 and take the tensor product of this path algebra with itself you get the incidence algebra of the diamond 1 / \ a b \ / 0 The Hasse diagram is the quiver of this algebra, but you have the relation that the two paths to the top are the same. | |
| May 9, 2012 at 12:51 | comment | added | Andreas Blass | How does that work? Specifically, which element in the tensor product would be the image of the element $(a,b)$ of the Cartesian product? The "obvious" choice, $a\otimes b$ won't work, because addition in the Cartesian product ($(a,b)+(c,d)=(a+c,b+d)$) doesn't match addition in the tensor product, so you won't have a homomorphism. (Multiplication by scalars from $k$ doesn't match either unless $k$ is the two-element field.) | |
| May 9, 2012 at 12:04 | history | answered | Tore Forbregd | CC BY-SA 3.0 |