Timeline for The $K_0$ mapping of an automorphism induced by a derivation
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2023 at 2:14 | history | edited | Sanae Kochiya | CC BY-SA 4.0 |
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| Jan 26, 2023 at 2:13 | comment | added | Sanae Kochiya | @JamieGabe Thank you for your corrections. I will edit my post but look forward to seeing more info about $e^{\delta}$. | |
| Jan 25, 2023 at 13:38 | comment | added | Jamie Gabe | The norm estimate I wrote above was absolute rubbish. Whoops! It should had been $\| e^{\epsilon \delta} - \mathrm{id}_{\mathfrak A}\| \leq e^{\epsilon \| \delta\|} - 1$ (which can be easily verified using the power series). | |
| Jan 25, 2023 at 10:14 | comment | added | Jamie Gabe | You probably want to define $\delta^0 = \mathrm{id}_{\mathfrak A}$ otherwise $e^\delta$ is not linear (for instance, $e^\delta(0) = 1$). With this modification, $[0,1] \ni t\mapsto e^{t\delta}$ does indeed give a homotopy (even in the norm topology, and not just in the point-norm topology) from $e^\delta$ to $\mathrm{id}_{\mathfrak A}$ (it is norm continuous since $\| e^{\epsilon \delta} \| \leq \epsilon e^{\| \delta\|}$). So $K_0(e^\delta) = \mathrm{id}_{K_0(\mathfrak A)}$. | |
| Jan 24, 2023 at 18:33 | history | asked | Sanae Kochiya | CC BY-SA 4.0 |