Timeline for On equation $e^{xy-yx}=e^xe^ye^{-x}e^{-y}$ in $C^*$ algebras
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 15, 2019 at 4:21 | vote | accept | Ali Taghavi | ||
| Oct 14, 2019 at 18:07 | comment | added | YCor | @user44191 you're right, thanks for the correction. | |
| Oct 14, 2019 at 17:58 | comment | added | user44191 | @YCor I think you made a sign error when switching $x$ and $z$ in the inner bracket? $[x,[y,z]]+[y,[x,z]]=0$ becomes $[x,[y,z]]−[y,[z,x]]=0$, or $[x,[y,z]]=[y,[z,x]]$. You get equality when cycling, and sign change when switching any 2 of the 3, if I'm not mistaken. Which leads to using Jacobi and needing $3$ to not be a $0$-divisor, but $2$ is free. | |
| Oct 14, 2019 at 9:18 | answer | added | YCor | timeline score: 14 | |
| Oct 14, 2019 at 9:00 | comment | added | YCor | ... oh, but the latter is already solved here :mathoverflow.net/a/299943/14094. I'll post an answer. | |
| Oct 14, 2019 at 8:45 | comment | added | YCor | Conversely BCH for a unital Banach algebra satisfying $[x,[x,y]]=0$ identically writes as $\exp(x)\exp(y)=\exp(x+y+\frac12[xy])$ and this gives the commutator identity. So a unital Banach algebra satisfies the given commutator identity iff its underlying Lie algebra is 2-step nilpotent. What remains is to determine whether a $C^*$-algebra whose underlying Lie algebra is 2-step nilpotent is necessarily commutative. | |
| Oct 14, 2019 at 7:47 | comment | added | YCor | @user44191 polarization gives $[x,[y,z]]+[y,[x,z]]=0$ (for all $x,y,z$). This can be rewritten as $[y,[z,x]]=-[x,[y,z]]$, and iterating 3 times we get $[x,[y,z]]=-[y,[z,x]]=[z,[x,y]]=-[x,[y,z]]$. Hence if $2$ is invertible (the question is over $\mathbf{C}$) we get $[x,[y,z]]=0$ without using Jacobi. | |
| Oct 14, 2019 at 4:04 | comment | added | user44191 | @YCor I think it may be worth noting that the Jacobi identity is needed, that is, that polarization doesn't quite give the result directly. The derivation is straightforward, but not immediate. | |
| Oct 13, 2019 at 21:12 | comment | added | YCor | Taylor at order 3 $[+O(\|x\|^4+\|y\|^4)]$ yields that a unital Banach algebra with this identity has to be 2-step nilpotent (that is $[x,y]$) is central for all $x,y$); indeed it says that $[x,[x,y]]$ is zero, which by polarization gives the result. | |
| Oct 13, 2019 at 21:05 | comment | added | user44191 | I would expect that the Baker-Campbell-Hausdorff formula could be used to say that $A$ is commutative; failing that, that for every $X, Y$ we have that $[X, Y]$ commutes with $X$ and $Y$. Have you tried it? | |
| Oct 13, 2019 at 20:51 | history | edited | YCor | CC BY-SA 4.0 |
fixed typos
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| Oct 13, 2019 at 19:40 | comment | added | Yemon Choi | [deleted over hasty attempt at an answer, I need to revisit this] | |
| Oct 13, 2019 at 18:55 | vote | accept | Ali Taghavi | ||
| Oct 13, 2019 at 19:08 | |||||
| Oct 13, 2019 at 18:52 | comment | added | Ali Taghavi | @YemonChoi I thank you too for your revision. | |
| Oct 13, 2019 at 18:46 | comment | added | Yemon Choi | Thanks for the clarification. I have reworded the title in case anyone makes the same error as I did | |
| Oct 13, 2019 at 18:39 | history | edited | Yemon Choi | CC BY-SA 4.0 |
improved title and wording to avoid condusion
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| Oct 13, 2019 at 18:36 | comment | added | Ali Taghavi | @YemonChoi motivation is added | |
| Oct 13, 2019 at 18:36 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 184 characters in body
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| Oct 13, 2019 at 18:33 | comment | added | Ali Taghavi | @YemonChoi the left commutator is an an algebra commutator but the right one is a group commutator. So it holds for all commutative algebra. The equation in the title is misleading please read the equation in the body of the post. | |
| Oct 13, 2019 at 18:18 | history | asked | Ali Taghavi | CC BY-SA 4.0 |