Timeline for Non-isomorphic Heisenberg groups over rings
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2016 at 17:40 | vote | accept | user35603 | ||
| Dec 2, 2016 at 17:38 | comment | added | user35603 | @YCor thank you for valuable comments. | |
| Dec 2, 2016 at 17:37 | comment | added | user35603 | Thanks, @BenjaminSteinberg I did not know about that paper. I posted the question and then went offline... so I missed the interesting discussion. | |
| Dec 2, 2016 at 3:24 | comment | added | Benjamin Steinberg | Great. We agree. | |
| Dec 2, 2016 at 3:22 | comment | added | YCor | OK thanks. So this one $\to\leftarrow$ yields basis $(u,v,w,a,b)$ with nonzero products $u^2=u$, $v^2=v$, $w^2=w$, $ua=a=av$, $wb=b=bv$ (the unit is $u+v+w$). That it's not isomorphic to its opposite (over a field) can also be seen naively: while right multiplication by $a+b$ has rank 2 as linear operator, the left multiplication on the whole algebra by any element of the Jacobson radical (with basis $(a,b)$) has rank $\le 1$. | |
| Dec 2, 2016 at 3:05 | comment | added | Benjamin Steinberg | The path algebra has basis all directed paths in the quiver including empty paths at each vertex. So it is 5 dimensional. | |
| Dec 2, 2016 at 2:59 | comment | added | YCor | I'm confused then about the definition. For this "$\to\leftarrow$" quiver, isn't the path algebra given by the basis $(1,a,b)$ with $ab=ba=0$ (thus commutative)? because there is no pair of composable edges... | |
| Dec 2, 2016 at 2:46 | comment | added | Benjamin Steinberg | Simple projective left modules correspond to sources and simple projective right modules to sinks since empty paths give a complete set of primitive idempotents and no two give isomorphic simples. | |
| Dec 2, 2016 at 2:43 | comment | added | Benjamin Steinberg | @YCor, Any such quiver works. By Gabriel's theorem two isomorphic basic finite dimensional algebras have isomorphic quivers. If you take two edges with the same head but different tails over any field it has two simple projective left modules but only one simple projective right module so it is not isomorphic to its opposite | |
| Dec 2, 2016 at 2:32 | comment | added | YCor | Not any such quiver works, if I'm correct (e.g. the connected quiver with 3 vertices, 2 edges with the same head doesn't). However some quiver with 4 vertices and 3 arrows works, with the path algebra $A$ with basis $(1,a,b,z,az,bz)$ and the obvious products ($1$ unit, all products 0 except those involving 1 and $a.z=az,b.z=az$). Then the left nil-center (not sure of terminology) $\{x:\forall y:xy=0\}$ is 3-dimensional for $A$ (basis $(az,bz,z)$) and 4-dimensional for $A^{\mathrm{op}}$ (basis $(az,bz,a,b)$), so $A$ and $A^{\mathrm{op}}$ are not isomorphic. | |
| Dec 2, 2016 at 2:23 | comment | added | Benjamin Steinberg | I added that the answer is positive even for infinite commutative rings. | |
| Dec 2, 2016 at 2:22 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
added 37 characters in body
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| Dec 2, 2016 at 2:17 | comment | added | Benjamin Steinberg | Take an acyclic connected quiver that is not isomorphic to the quiver obtained by reversing its arrows and then take the path algebra over a finite field. Such guys are indecomposable and not self dual and finite. | |
| Dec 2, 2016 at 2:12 | comment | added | YCor | I did it anyway :) btw you probably have in mind some finite ring not anti-isomorphic to itself? | |
| Dec 2, 2016 at 2:11 | history | edited | YCor | CC BY-SA 3.0 |
Clarified answer
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| Dec 2, 2016 at 2:07 | comment | added | Benjamin Steinberg | Probably not worth editing. Also finite is not important. | |
| Dec 1, 2016 at 23:35 | comment | added | YCor | Yes I know: this was no criticism but rather an excuse, since editing the tags, I sort of hid the only hint that the question was only in the commutative case. But of course it's great your answer both: just I would rather formulate it now saying that the answer is positive but not its extension to the non-commutative case. | |
| Dec 1, 2016 at 23:02 | comment | added | Benjamin Steinberg | @YCor, I never read tags. Anyway I give the answer in that case too. | |
| Dec 1, 2016 at 23:00 | comment | added | YCor | Actually before I fixed the tags, the only "ring" tag was "commutative rings", so the question was quite clearly for commutative rings (which should have been more explicit). I added the tag "ra-..." because Lie algebras are lurking behind... | |
| Dec 1, 2016 at 22:42 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |