Timeline for Pairwise orthogonality for partitions of unity in a *-algebra
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 30 at 6:43 | vote | accept | JP McCarthy | ||
| Jan 29 at 19:54 | answer | added | Tobias Fritz | timeline score: 7 | |
| Jan 29 at 19:43 | comment | added | David Gao | I’m thinking it might be possible to directly prove, using perhaps a combinatorial argument, in the free $\ast$-algebra generated by four self-adjoint idempotents $p_1, \cdots, p_4$ with relation $p_1 + \cdots + p_4 = 1$, you have $p_1p_2 \neq 0$. The algebra in question is the free product of four copies of $\mathbb{C}^2$, mod the kernel generated by $p_1 + \cdots + p_4 - 1$. It’s just a matter of showing $p_1p_2$ is not in the kernel - feels certainly doable, no? | |
| Jan 29 at 13:21 | comment | added | Tobias Fritz | @DavidGao: whoops, right, I indeed forgot to include the most important relation :) | |
| Jan 29 at 11:20 | comment | added | David Gao | @TobiasFritz I think you also need an additional relation stating $p_1 + \cdots + p_4 = 1$? But yeah, all the relations are preserved under reversal of words, so reversing words indeed defines an involution on it. I wonder why the authors didn’t notice this. | |
| Jan 29 at 9:21 | comment | added | Tobias Fritz | Well, all I'm saying is that having an example as claimed in the paper would be enough to answer the question. To add further to the mystery: the orthogonality holds automatically for all $N$ in every finite-dimensional algebra (without involution). This is because it holds in matrix algebras, and every finite-dimensional algebra embeds into a matrix algebra. | |
| Jan 29 at 8:55 | comment | added | JP McCarthy | @TobiasFritz I am not sure. I guess my question could include... does anyone know Radjavi's example? | |
| Jan 29 at 8:29 | comment | added | Tobias Fritz | Of course not, but the with that presentation clearly does (you just reverse all words), no? | |
| Jan 29 at 8:11 | comment | added | JP McCarthy | @TobiasFritz I don't think an algebra necessarily admits an involution. | |
| Jan 29 at 8:04 | comment | added | Tobias Fritz | The counterexample attributed in that paper to Heydar Radjavi should at least give enough a somehwat non-explicit counterexampe: it implies the desired non-orthogonality in the algebra with presentation $\langle p_1, \dots, p_4 \mid p_i^2 = p_i \rangle$, and that's enough because this algebra is clearly a $*$-algebra if you make the generators self-adjoint. So it seems to me that the authors could immediately have resolved Problem 5.2 in the negative. Am I missing something? | |
| Jan 29 at 7:11 | comment | added | JP McCarthy | @TobiasFritz IIRC person $X$ is a coauthor on that paper. I wonder if the people I asked confused [17] with an answer to Problem 5.2? Or maybe person $X$ found a counterexample since 2017. Thanks for comment. | |
| Jan 29 at 6:55 | comment | added | Tobias Fritz | This appears as Problem 5.2 in Algebras, Synchronous Games and Chromatic Numbers of Graphs. A few paragraphs before that, it is noted that you will at least need 4 projections for a potential counterexample. The paper is from 2017, and I don't know if this has been resolved since then. | |
| Jan 29 at 6:31 | history | asked | JP McCarthy | CC BY-SA 4.0 |