Timeline for $\mathcal{B}(\mathcal{H})$ as the reduced $C^*$-algebra of a groupoid
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Oct 25 at 9:07 | comment | added | Narutaka OZAWA | @David Gao: "Do you also know of a way around the normality issue?" It should be obvious, but not to me... | |
| Oct 24 at 3:52 | comment | added | David Gao | @NarutakaOZAWA Ah, never mind, I’ve figured out the normality part. As long as a faithful expectation exists, a faithful normal expectation exists. So the only remaining issue is how to prove $C_0(G^{(0)})$ is atomic whenever a faithful normal expectation exists. I’d still assume some modular automorphism argument works…? But my knowledge of modular theory is quite limited. | |
| Oct 23 at 5:57 | comment | added | David Gao | @NarutakaOZAWA Oh, that’s a great argument, thanks! Do you also know of a way around the normality issue? There certainly are faithful non-normal expectations, but it should always have a faithful normal part. The normal part is not automatically an expectation though. (And, for that matter, I’ve realized the argument I had in mind for when $E$ is normal is wrong. I’ve mistakenly assumed the modular automorphism group fixes $C_0(G^{(0)})$ pointwise, which it does not need to.) | |
| Oct 23 at 2:36 | comment | added | Narutaka OZAWA | @David Gao: If a $\mathrm{C}^\ast$-algebra $A\subset B(H)$ is the range of a faithful conditional expectation $E$, then it is a von Neumann algebra. Indeed, if $a$ is the SOT-limit of a bounded increasing net $(a_i)$ in $A_+$, then $E(a) - a \geq 0$, because $E(a)\geq E(a_i)=a_i$ for all $i$, which implies $E(a) = a$ by faithfulness of $E$. | |
| Oct 18 at 21:17 | comment | added | David Gao | @NarutakaOZAWA I can see that the faithfulness of $E$ implies $C_0(G^{(0)})$ is atomic, assuming $C_0(G^{(0)})$ is actually a von Neumann subalgebra of $B(H)$ and $E$ is normal, by applying modular theory. But I don’t see how one would show this without normality assumption? | |
| Oct 18 at 14:37 | comment | added | PKOA | We know $L^{\infty}(X, \mu)$ is a cartan subalgebra of $\mathcal{B}(L^{2}(X, \mu)$ as a von Neumann algebra. Is that the same reason $C(X)$ may not be cartan in $\mathcal{B}(L^{2}(X, \mu))$(embedded as multiplication operator) in $C^{\ast}$ sense? | |
| Oct 18 at 0:56 | comment | added | Narutaka OZAWA | A groupoid C*-algebra $\mathrm{C}^\ast_{\mathrm{r}}G$ has a faithful conditional expectation $E$ onto the Cartan subalgebra (in the $\mathrm{C}^*$-sense) $C_0(G^{(0)})$. I guess faithfulness of $E$ (restricted to $K(H)$) implies $C_0(G^{(0)})$ atomic and $C_0(G^{(0)})$ cannot be a Cartan subalgebra. I haven't checked the detail. (The norm-closure of those $T$ on $\ell_2\mathbb{N}$ such that $\sup_i |\{ j : |T_{i,j}|+|T_{j,i}|>0 \}|<\infty$ is an etale groupoid $\mathrm{C}^\ast$-algebra, but it is strictly smaller than $B(\ell_2\mathbb{N})$.) | |
| Oct 17 at 7:56 | comment | added | David Gao | @PKOA The issue is simply that the matrix units generate finite-dimensional matrix algebras as $C^\ast$-algebras, but only generate infinite-dimensional $B(H)$ as von Neumann algebras, and the latter is not enough to get a groupoid $C^\ast$-algebra structure. There’s also no other obvious choice, which is probably why this question is supposedly still open. | |
| Oct 17 at 7:53 | comment | added | PKOA | @Yemon choi, for finite dimensional case it does exist just take matrix units, so I was curious as $\mathcal{B}(\mathcal{H})$ has matrix units!! | |
| Oct 17 at 2:51 | comment | added | Yemon Choi | My apologies to the OP for not noticing that the error in the title was introduced by someone else. | |
| Oct 17 at 2:50 | comment | added | Yemon Choi | Well at least the question is consistent, but why do you think such a groupoid exists? And why have you combined the words in my comment when the combination you have chosen has some redundancy? | |
| Oct 17 at 2:50 | comment | added | LSpice | Apologies (to @YemonChoi and @PKOA) for "group" in the title—I tried to edit to a more informative title, and didn't notice I'd changed "groupoid" to "group". | |
| Oct 17 at 1:29 | comment | added | PKOA | Edited@Yemon Choi. Hope it is clear now. | |
| Oct 17 at 1:28 | history | edited | PKOA | CC BY-SA 4.0 |
added 32 characters in body; edited title
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| Oct 17 at 0:12 | comment | added | Yemon Choi | Your title says "group" but your question says "groupoid". Which one do you mean? Also, what assumptions are you putting on your group(oid) - discrete? locally compact? etale? etc | |
| Oct 16 at 22:52 | history | edited | LSpice | CC BY-SA 4.0 |
More informative title
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| Oct 16 at 18:55 | comment | added | David Gao | I believe it is not known what such a $G$ exists. See mathoverflow.net/questions/466749/… | |
| Oct 16 at 18:39 | history | edited | PKOA | CC BY-SA 4.0 |
added 1 character in body
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| Oct 16 at 18:39 | history | edited | PKOA | CC BY-SA 4.0 |
added 21 characters in body
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| S Oct 16 at 18:38 | review | First questions | |||
| Oct 16 at 20:14 | |||||
| S Oct 16 at 18:38 | history | asked | PKOA | CC BY-SA 4.0 |