Timeline for In which commutative algebras does any derivation possess a flow?
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| Sep 23, 2010 at 23:04 | comment | added | Fiktor | Yes, I'm interested only in the Hamiltonian derivations, and yes, not all the derivations are Hamiltonian derivations. But I don't know, how can it help me except for the example 1' in my question, where I should use only the derivations, generated by the Hamiltonians with compact level sets. | |
| Sep 23, 2010 at 19:08 | comment | added | Greg Muller | Ah, interesting. But then you don't necessarily care about all derivations, so much as you care about those coming from the Poisson bracket, for a Poisson algebra. This can reduce the complexity quite a bit; notice that an irrationally sloped vector field on R^2/Z^2 does have a corresponding flow, but is not the flow corresponding to any Hamiltonian function. I suspect there are stronger results for when all Poisson derivations exponentiate. | |
| Sep 23, 2010 at 19:00 | comment | added | Fiktor | @Greg Muller I care about this kind of algebras too. I can explain, why. I'm reading a course about algebraic approach to Hamiltonian mechanics. All definitions are given there for general commutative (and supercommutative) algebras. The classical Hamiltonian mechanics appears here as the case $A=C^\infty(M)$ and I'm interested in other useful and/or interesting cases. The problem in the question is the most important problem with many of them. Although finite-dimensional algebras are not so important for the goal of the course, they are good for simple examples and exercises. | |
| Sep 23, 2010 at 18:26 | history | edited | Greg Muller | CC BY-SA 2.5 |
added 331 characters in body; added 8 characters in body; added 17 characters in body
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| Sep 23, 2010 at 18:24 | comment | added | Greg Muller | Fine, `no geometric information' was needlessly severe. I meant that it wasn't a ring of real-valued functions on some space, which seemed to be the motivation behind to examples given; particularly the choice to use R instead of C. | |
| Sep 23, 2010 at 16:44 | comment | added | Theo Johnson-Freyd | Also, I object to the idea that $\mathbb R[x]/x^n$ "contain[s] no geometric information". It is precisely the ring of $\mathbb R$-valued functions on an order-$n$ formal neighborhood of a point in a line. In particular, the "Taylor series" map $\mathcal C^\infty(\mathbb R) \to \mathbb R[x]/x^n$ is the restriction of a smooth function to a very small formal neighborhood. The restriction remembers precisely the first $n$ derivatives of the function: this is definitely "geometric" information. | |
| Sep 23, 2010 at 16:41 | comment | added | Theo Johnson-Freyd | Note that $\frac\partial{\partial x}$ is not a derivation of $\mathbb R[x]/x^n$. It would be the derivation that sends $x\mapsto1$ if it exists, and this determines the derivation, but then $0=x^n\mapsto nx^{n-1}\neq0$ by the Leibniz rule. On the other hand, as fiktor points out, $x\frac\partial{\partial x}$, which is the derivation extending $x\mapsto x$, does exists: it acts on polynomials by $x^k\mapsto kx^k$. (Under $x=e^z$, $x\frac\partial{\partial x}=\frac\partial{\partial z}$.) All the derivations of $\mathbb R[x]/x^n$ are of the form $p(x)\frac\partial{\partial x}$ for $p(0)=0$. | |
| Sep 23, 2010 at 14:33 | comment | added | Fiktor | "$\xi^n=0$" is wrong. Consider $\xi=xQ(x)\frac{\partial}{\partial x}$. It is well defined. In order to show this, we should ensure that $[\xi P]$ does not depend on the polynomial $P$ from equivalence class $[P]\in\mathbb{R}/x^n$. We can check this for $[P]=0$. In this case P is multiple of $x^n$ then $\xi P$ is also multiple of $x^n$, so $[\xi P]=0$. If, for example, $Q(x)=1$, then $\xi^n(x)=x\neq 0$. However from your approach I see, that all derivations in finite-dimensional algebras possess a flow, because for matrices $\xi$ the power serie you wrote converges. | |
| Sep 23, 2010 at 14:09 | history | answered | Greg Muller | CC BY-SA 2.5 |