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created Noether’s theorem – no-nonsense version
In my vague memory, Tamarkin was explaining to us in his Paris lectures 2004 that in the derived version, I mean in the setup like in Beilinson-Drinfeld with coisson algebras etc. the direction is reverse: one has the conservation laws as more basic and the symmetries of the system always appear as coming from them; the converse does not exist always. Is something of the sort in Costello’s work ?
In noncommutative geometry, the question of Noether theorem is so far very unsatifactory.
In my vague memory, Tamarkin was explaining to us in his Paris lectures 2004 that in the derived version, I mean in the setup like in Beilinson-Drinfeld with coisson algebras etc. […]
I need to better learn all this. I’ll create now a floating TOC on this subject complex. What should we call it? “Variational calculus”?
In noncommutative geometry, there are many toy field theories developed, the things go quite well. But the extension of the usual correspondence between the consevation laws and symmetries, well, there has been very little success there. For one of the paper discussing the difficulty and proposing a work-around see
The “dynamical” aspect here is about some half-satisfactory work-around. Even the authors are not happy with it, but this is what we so far have. For a follow-up of that approach see
This is a potentially interesting intersection between the n-Forum and John’s new Azimuth forum.
I had never thought much about Navier-Stokes, but saw some relations to some earlier work I’ve done on noncommutative geometry. I’m pretty sure this comment presents an original result where the Navier-Stokes equation is shown to be a zero curvature condition in noncommutative geometry.
In a nutshell, beginning with a $G L(1, \mathbb{R})$-valued noncommutative 1-form connection $A$ in $(3+1)$-dimensions, the vanishing of the curvature
$F = d A + A A$results in the Navier-Stokes equations. This demonstrates the Navier-Stokes as something that might curiously be described as a noncommutative abelian gauge field theory. Abelian because the Lie group is abelian. Noncommutative because 0-forms and 1-forms do not commute.
Thank you Eric. I do not understand, but will in due time. The zero curvature equations in soliton theory usually show the integrability of the system. I know that there are some noncommutative generalization, but I am surprised to hear that for the equations on commutative spaces there is an interpretation with noncommutative differential calculus. But Dimakis and Mueller-Hoisen are known mathematical physicists so it makes sense to look at what they wrote. On the other hand, I am interested in some other integrability aspects in noncommutative geometry, namely about the underlying geometry of the noncommutative analogues of Painleve transcendents studied recently by Roubtsov, Retakh etc. In the commutative case, there is interesting algebraic geometry behind the scenes.
Hi Zoran,
In the paper by Dimakis and Mueller-Hoissen on soliton equations from zero curvature condition in noncommutative geometry, they make a minor technical error with the Burgers equation. I explain/correct the error and provide some calculations that might help understand the corresponding calculations for Navier-Stokes here:
@Zoran #7:
I had a look at the first paper you referred to:
and the first thing that strikes me is the Moyal $\star$-product. I bump against this once every few years, but never spent the time to fully grok it.
In particular, the relation
$[\hat{x}^\mu,\hat{x}^\nu] = i\theta^{\mu\nu}$is eerily similar to the relation I’m accustomed to, i.e.
$[d x^\mu,x^\nu] = \theta^{\mu\nu} d t.$It would be interesting if existing work done with the Moyal $\star$-product can be mapped to noncommutative geometry of commutative algebras (that I’m accustomed to) and vice versa.
Edit: A potential step for the mapping is possibly to introduce a new product
$d f\cdot d g = [d f,g].$The product rule ensures this product is commutative and we have
$d f\cdot d g = \theta^{\mu\nu} \left(\partial_\mu f\right)\left(\partial_\nu g\right) d t.$I know it is not correct, but I’ll write the following suggestion in case it sparks some ideas:
$\left(f\star g\right) d t = m\left[e^{d\otimes d}(f\otimes g)\right]$where
$m(f\otimes g) = f g d t$and
$m(d f\otimes d g) = d f\cdot d g.$Or something…
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