How to construct a complex projective variety with several classes of non-intersecting divisors? How to keep the answer concrete and simple, so that explicit calculations can be done? And the problem - there should be a complex Lie group acting on the whole construction and a factorisation problem to solve...

This is from the theory of integrable systems. Why do loop groups appear in relativistic integrable systems? One answer is that the 1+1 D Lorentz group (AKA the multiplcative reals) acts on it. Now take factorisations of loops. Solitons correspond to meromorphic loops, and the residual part gives (via Riemann-Hilbert factorisation) the dispersive solutions. It solves the sine Gordon equation, the principal chiral equation, and gives many soliton solutions to affine Toda. Space-time evolution is given by a reversal of order factorisation $a(x,t)\,\phi_0=\phi(x,t)b(x,t)$, where $\phi_0$ and $\phi(x,t)$ are meromorphic loops, and $a(x,t),b(x,t)$ are `analytic' loops (i.e. analytic where the mero loops have poles). Lots of dynamic information is encoded into the residues of the meromorphic loops. This is reasonably standard inverse scattering, as in Segal, Wilson, Novikov, Manakov, Pitaevskii, Zakharov, Fadeev etc.

The big problem is how to generalise this construction to higher dimensions. As far as I know (please correct me) there is only one construction of higher dimensional Lorentz invariant systems with soliton solutions - Richard Ward's mini-twistor construction. (Note: I am not talking about abstract ideas of integrability, I am talking about concrete methods of constructing solitons.)

The obvious thing to do is to replace the Riemann sphere with a dimension >1 complex variety on which the complexified Lorentz group acts (or an extension - $SL_2(\mathbb{C})$ would do nicely, or a conformal group). Specify some unitarity condition on a real submanifold preserved by the real Lorentz group. Take meromorphic matrix valued functions, with poles on some divisors, perform a reversal of order factorisation, and get solitons. Minor details, like just what differential equation the system actually solves, can be filled in later.

But life is not quite so simple. The algebraic geometry is more complicated than in one complex dimension, and likely line bundles and other complications occur. But the most immediate complication lies in getting examples to calculate with which have non-intersecting divisors. There are of course some such examples of non-intersecting divisors (the product of two Riemann shperes is such an example), but a problem is getting a reasonable number of examples to experiment with, and distinct classes of such divisors (up to group action), to try to get the factorisation problem to work. I shall sketch a simplified analogue below - adding line bundles may be necessary:

The "classical vacuum map" $a:S\to \mathrm{Ananytic}(U,M_N)$ maps space-time $S$ to analytic matrix valued functions on an open subset $U$ of the variety $X$. It is, in some sense, Lorentz invariant. The "meromorphic" (or suitable generalisation) functions $\phi:X\to M_n$ have divisors contained in $U$, and with suitable normalisation there is a unique reversal of order factorisation $a\phi_0=\phi b$ for mero $\phi_0,\phi$ and analytic $a,b$. (All details are up for revision - it is the well defined order reversal, and the underlying group action, that is vital.)

Please pardon any offense against etiquette, and give advice on such, I am new here. :)

  • $\begingroup$ The simplest construction is by blowing up several points in an algebraic variety; the exceptional divisors don't intersect. I have no idea if this gives something interesting for your purposes. $\endgroup$ – Angelo Dec 20 '12 at 12:48
  • $\begingroup$ This would give a family of varieties, graded by the number of blow-ups, and on which the group acting would act by shifting the blown-up points. The problem would be to construct some non-trivial factorisation on (matrix valued) sections of bundles which would have residues on the blown-up points. This may be possible, but I was really hoping for other answers which would involve the geometry of the construction to give a single manifold on which the group acted - this may be too much. thanks for the comment. $\endgroup$ – Edwin Beggs Dec 20 '12 at 13:34

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