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Let $A$ be a Banach or a $C^*$ algebra. We consider the differential equation $$(*)\;\;\;\;Z'=Z^2-Z$$ on $A$.

Obviously the singularities of this systems are just the idempotents of the algebra.

It can be easily shown that the group of invertible elements is invariant under this flow. (Edit:According to the answer of Robert Israel we get that the space of Left zero divisors is flow invariant too.) Furthermore the group of invertible elements does not contain any periodic orbit(Except the trivial case of singularity $Z_0=1$ but strictly speaking a singular point can not be regarded as a periodic orbit). Furthermore non of the following algebras can have a periodic orbit of the above systems:

1) The Matrix algebra

2)$C^*_{\text{red}} F_1=C^*_{\text{red}} \mathbb{Z}=C(S^1)$

Our questions:

1)Is there a Banach or $C^*$ algebra $A$ for which the system $(*)$ has a periodic orbit?

2)In the literature, are there some researches devoted to Kaplansky or Kadison Kaplansky conjecture via dynamical consideration of the equation $(*)$? As we see in this post, the three key elements of the Kaplansky conjecture is meaningfully involved with the dynamical interpretation of $Z'=Z^2 -Z$.These $3$ concepts are "Invertibles", "zero divisors" and "idempotent".

Proof of the fact that the group of invertible elements of a $C^*$ algebra $A$ is invariant under flow of $(*)$:

The group of invertible elements of $A$ is denoted by $G(A)$.Let $Z(t)$ be a solution of $(*)$ with $Z(0)=Z_0\in G(A)$. For some $t_0>0$, let $Z(t)\in G(A),\; \forall t\in [0,t_0)$ but $Z(t_0)$ is non invertible. Note that $W(t)=Z(t)^{-1}$ is a solution of $$(**)\;\; W'=W-I$$ Obviousely this vector field $(**)$ is a complet vector field, i.e. all soltions has maximal interval of definitions equal to $(-\infty, +\infty)$. In particular $W(t)$ is defined at $t_0$ hence $W(t)$ is bounded around $t_0$. This situation contradicts to the following lemma which is proved in Functional Analysis by W. Rudin.(10.17 lemma page 256).

Lemma: Let $Z_n$ be a sequence of invertible elements of a Banach algebra which converges to a non invertible element then the sequence $W_n=Z_n^{-1}$ is an unbounded sequence.

Remark: Please see the comment conversations to the following link as some suggestions for consideration of dynamical methods in the idempotent problem.

Smooth derivations of a Banach space

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The solution $Z(t)$ of your differential equation with $Z(0) = Z_0$ satisfies $$ Z(t) (e^t + (1-e^t) Z_0) = Z_0 $$ In order for this to be periodic with period $p$, you'd need $(1-e^p) Z_0 (1-Z_0) = 0 $. $1-e^p = 0$ (for real $p$) only if $p=0$, while if $Z_0 (1-Z_0) = 0$ we have a fixed point.

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  • $\begingroup$ 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 $\frac{-e^t}{1-e^t}$. $\endgroup$ – Ali Taghavi Nov 26 '19 at 14:50
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    $\begingroup$ 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 $t = \log(2)$. $\endgroup$ – Robert Israel Nov 26 '19 at 17:05
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    $\begingroup$ 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 $t \in (a,b)$. $\endgroup$ – Robert Israel Nov 26 '19 at 17:13
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    $\begingroup$ 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. $\endgroup$ – Robert Israel Nov 26 '19 at 17:16
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    $\begingroup$ 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 $Q(t) = 0$. $\endgroup$ – Robert Israel Dec 5 '19 at 0:52

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