Timeline for Pairwise orthogonality for partitions of unity in a *-algebra
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 3 at 22:21 | history | edited | Tobias Fritz | CC BY-SA 4.0 |
typo
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| Jan 31 at 8:12 | comment | added | JP McCarthy | They take also not take bounded but all real valued functions. | |
| Jan 31 at 7:20 | comment | added | JP McCarthy | Ah yes, I am stupid. Two of those maps have a $-1-x$ and they are not operators on that Hilbert space... and if we change the Hilbert space appropriately then they are not bounded. | |
| Jan 31 at 7:15 | comment | added | Tobias Fritz | Good point, then I'm confused as well... Perhaps let us know if you figure it out? | |
| Jan 31 at 6:38 | comment | added | JP McCarthy | .... but the same paper claims you cannot sum four non-trivial idempotents to zero in a Banach algebra... I am actually giving a talk in a small seminar today we might tease this out there. | |
| Jan 31 at 5:20 | comment | added | Tobias Fritz | It looks like they do, right? So you can even get a non-orthogonal partition of unity (consisting of non-self-adjoint idempotents) in a C*-algebra. | |
| Jan 30 at 19:29 | comment | added | JP McCarthy | The latest (surely stupid) confusion: why don't the $P,Q,R,S$ define bounded operators on $L^2([0,1])$... | |
| Jan 30 at 7:16 | comment | added | JP McCarthy | OK I understand that the $\pi$ is just an algebra homomorphism. | |
| Jan 30 at 7:05 | comment | added | Tobias Fritz | Uhm not exactly, I'm not trying to construct an involution on $A$, but only on the universal algebra $B$ (where the involution is your $\tau$, provided that your opp also involves complex conjugation in order to make the involution conjugate linear). Like you, I'm also skeptical about $A$ having an involution. Having an involution on $B$ is enough because the generating projections in $B$ are non-orthogonal, by virtue of the $\pi(p_i)$ being non-orthogonal in $A$. | |
| Jan 30 at 6:51 | comment | added | JP McCarthy | Great, thank you. So, if I don't want to get down in the weeds in $A=*\text{-}\operatorname{alg}\langle P,Q,R,S,I\rangle$ and $A^{\text{opp}}$, but still be fussy, with $B$ the universal *-algebra in this question, and by the universal property I have *-homomorphisms $\pi:B\to A$ and $\tau:B\to B^{\text{opp}}$, and I can define an involution on $A$ by: $\pi(f)^*=\pi(\tau(f))$... which is just exactly what you are saying. | |
| Jan 30 at 6:43 | vote | accept | JP McCarthy | ||
| Jan 30 at 6:42 | comment | added | Tobias Fritz | Yes. the involution amounts to choosing a homomorphism to its (complex conjugate) opposite algebra, and this follows from an application of the universal property defined by the presentation. You cannot always do this, for example for $\langle a, b \mid a b = 1\rangle$ you clearly do not get a homomorphism to the opposite algebra by mapping each generator to itself. As mentioned in the answer, it works in the case at hand because the relations only involve palindromes. | |
| Jan 30 at 6:38 | comment | added | JP McCarthy | by the universal property. And that feels OK. But surely in that case we should be able to put the *-structure on $\operatorname{alg}\langle P,Q,R,S,I\rangle\subseteq L(\ell^{\infty}(\mathbb{R}))$ and just have an explicit counterexample in *-$\operatorname{alg}\langle P,Q,R,S,I\rangle\subseteq L(\ell^{\infty}(\mathbb{R}))$? It must be the case that for algebras that do not admit a * that the opposite algebra is really very different. | |
| Jan 30 at 6:35 | comment | added | JP McCarthy | Example 4.2 gives four projections $P,Q,R,S$ in $L(\ell^\infty(\mathbb{R}))$. I am happy with the unital algebra generated by these, say $A_0=\operatorname{alg}\langle P,Q,R,S,I\rangle$. I am tentative about the *-operation here, and in the two universal algebras above and in the comments. Essentially, I see ye choosing as the involution a homomorphism $A_0\to A_0^{\text{opp}}$. And my concern is that it feels like you can always do this... which seems wrong. Perhaps the point I am missing is that if $A$ is universal and $A^{\text{op}}$ satisfies same relations then there is a homomorphism... | |
| Jan 30 at 6:07 | history | edited | Tobias Fritz | CC BY-SA 4.0 |
notation consistent with OP, typo
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| Jan 29 at 19:54 | history | answered | Tobias Fritz | CC BY-SA 4.0 |