Timeline for On differential equation $Z'=Z^2-Z$ on a $C^*$ algebra
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 9, 2019 at 13:25 | history | bounty ended | Ali Taghavi | ||
| Dec 6, 2019 at 10:15 | comment | added | Ali Taghavi | So every orbit is a commutative set. | |
| Dec 5, 2019 at 20:58 | comment | added | Robert Israel | In the case of matrix algebra, $Z(t)$ has the same generalized eigenspaces as $Z_0$, with eigenvalue $\lambda$ of $Z_0$ becoming $\lambda/(e^t + (1-e^t)\lambda)$ for $Z(t)$ (the solution ceases to exist when the denominator becomes $0$). So yes, the rank is fixed along trajectories. | |
| Dec 5, 2019 at 15:53 | comment | added | Ali Taghavi | Very beautifull argument.!Now I understand your firmula. Just another question: In the mateix algebra do you think that rank is fixed along trajectories. Or in case of operator algebra, Do you think that "trace class operatir or Hilbert Schmidt operatores are flow invariant? | |
| Dec 5, 2019 at 0:56 | comment | added | Robert Israel | Together with the similar formula $(e^t + (1-e^t) Z_0) Z(t) = Z_0$, it does prove that the set of invertible elements is invariant under the flow. | |
| Dec 5, 2019 at 0:52 | comment | added | Robert Israel | The formula does not depend on anything being invertible. It just comes from differentiating $Q(t) = Z(t)(e^t +(1-e^t) Z_0) - Z_0$, substituting $Z'(t)$ from the differential equation, and noting that the result is $Z(t) Q(t)$. Since $Q(0)=0$ and the local uniqueness theorem applies to the differential equation $Q'(t) = Z(t) Q(t)$, we get $Q(t) = 0$. | |
| Dec 4, 2019 at 20:14 | comment | added | Ali Taghavi | Is the formula you provided valid in the case that the group of invertible elements is not dense on our algebra? Does you formula give an alternative proof of invariance of invertible elements under flow? | |
| Nov 26, 2019 at 17:16 | comment | added | Robert Israel | If $Z_0$ is invertible but $e^t + (1-e^t) Z_0$ is not invertible, it tells you that $Z(t)$ does not exist. | |
| Nov 26, 2019 at 17:13 | comment | added | Robert Israel | So your statement "the group of invertible elements is flow invariant" really means: if $Z_0$ is invertible and there is a solution $Z(t)$ on an interval $(a,b)$ containing $0$ with $Z(0)=Z_0$, then $Z(t)$ is invertible for all $t \in (a,b)$. | |
| Nov 26, 2019 at 17:05 | comment | added | Robert Israel | The problem is that the solution may not exist globally. Thus for the equation on the real numbers with $Z(0) = 2$, the solution is $Z(t) = 2/(2 - e^t)$ which blows up at $t = \log(2)$. | |
| Nov 26, 2019 at 14:50 | comment | added | Ali Taghavi | Thank you very much for your answer! Am I mistaken to think that the formulation of solution you wrote is in conflict with the fact that the group of invariant elements is flow invariant? Let $Z_0$ be invertible but has a real spectrum as $\frac{-e^t}{1-e^t}$. | |
| Nov 25, 2019 at 19:11 | history | edited | Robert Israel | CC BY-SA 4.0 |
added 57 characters in body
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| Nov 25, 2019 at 18:20 | history | answered | Robert Israel | CC BY-SA 4.0 |