Is there a quaternionic algebraic geometry ? - MathOverflow

Is there a quaternionic algebraic geometry ?

2011-01-13T14:46:50Z

Let $\mathbb{H}$ be the skew-field of quaternions. I'm aware of the

Theorem 1. A function $f:\mathbb{H}\to\mathbb{H}$ which is $\mathbb{H}$-differentiable on the left (i.e. the usual limit $h^{-1}\cdot (f(x+h)-f(x))$, for $h\to 0$, exists for every $x\in \mathbb{H}$) is a quaternionic affine function on the right (i.e. of the form $x\mapsto x\cdot \alpha + v$).

This means that there are no interesting smooth quaternionic funcions, hence no interesting "quaternionic-smooth manifolds" (which is not the same as the quaternionic-Kahler or hyperkahler structures you encounter in differential geometry and complex analytic geometry).

I think I can also recall the

Theorem 2. If a function $f:\mathbb{H}\to\mathbb{H}$ is locally $\mathbb{H}$-analytic (i.e. it can be locally developped in power series, for the suitable noncommutative notion of "power series"), than it corresponds to a real-analytic function $f:\mathbb{R}^4\to \mathbb{R}^4$, and any real-analytic funcion $f:\mathbb{R}^4\to \mathbb{R}^4$ can be obtained in this way.

That says that $\mathbb{H}$-analytic functions are essentially the same as quadruples of real-analytic functions of 4 variables. Hence there is no "quaternionic-analytic geometry" distinguishable from $4n$-dimentional real-analytic geometry. I think the same happens with quaternionic (noncommutative) polynomials: they're just 4-tuples of real polynomials in 4 variables.

But, is it reasonable that the zero locus on $\mathbb{H}^n$ of a "noncommutative polynomial" with $\mathbb{H}$-coefficients doesn't have any further mathematical structure than it's real-algebraic variety structure? It would be nice to be able to see things such as $\mathbb{HP}^1$ as a "quaternionic curve ", and to speak of a point " $\mathrm{Spec}(\mathbb{H})$ " (whatever it means) if it possible...

Is there a theory of "quaternionic algebraic geometry", maybe as a branch or particular case of some noncommutative (algebraic) geometry theory?

Of course, if such a theory has some sense, it cannot be the "obvious analog" of complex algebraic or analytic geometry, as theorems 1. and 2. above show.

Answer by Spinorbundle for Is there a quaternionic algebraic geometry ?

2011-01-13T16:04:25Z

The answer is yes!(at least if quaternionic holomorphic geometry counts) "Qauternionic holomorphic geometry" provides a very elegant description of surfaces in 3- and 4-dimensional space.

The first paper in this field is more or less

Franz Pedit and Ulrich Pinkall, Quaternionic Analysis on Riemann Surfaces and Differential Geometry, in: Proocedings of the international con- gress of mathematicians, Berlin 1998, II Documenta Mathematica, Journal der Deutschen Mathematiker-Vereinigung, Extra Volume ICM 1998, 389-400. Click me

A good introduction to "Quaternionic holomorphic geometry" is given by

Francis E. Burstall, Dirk Ferus, Katrin Leschke, Franz Pedit and Ulrich Pinkall, Conformal geometry of surfaces in S4 and quaternions, Lecture Notes in Mathematics 1772, Springer-Verlag, Berlin, 2002. http://arxiv.org/abs/math/0002075

and

Dirk Ferus, Katrin Leschke, Franz Pedit and Ulrich Pinkall, Quaternionic Holomorphic Geometry: Plucker Formula, Dirac Eigenvalue Esitmates and Energy Estimates of Harmonic 2-Tori, Inventiones Mathematicae 146 (2001),no. 3, 507-593 arxiv.org/abs/math/0012238v1

But just to clarify things: "Of course, if such a theory has some sense, it cannot be the "obvious analog" of complex algebraic or analytic geometry, as theorems 1. and 2. above show."

This isn't really true, because the quaternionic holomorphic geometry developed in the papers above, is a kind of generalization of complex geometry, i.e. if you look how a "quaternionic holomorphic structure" is defined, you see that it is a kind of $\overline{\partial}$-Operator with some extra data (a so called Hopf field).

If you need more references, there are plenty available (just type "quaternionic holomorphic geometry" into google)

Answer by Ben Webster for Is there a quaternionic algebraic geometry ?

2011-01-13T19:20:19Z

I actually have no idea how this relates to the "quaternionic geometry" mentioned above, but there is also hyperkähler geometry. A hyperkähler manifold is a real manifold such that every tangent space has an action of the quaternions, and there is a single metric which makes the manifold Kähler in the complex structure induced by $I,J$, or $K$. This sounds a little differential geometric at first, but one can get a purely algebraic object by turning $J$ and $K$ into their respective symplectic forms $\omega_J$ and $\omega_K$. The sum $\omega_J+i\omega_K$ is a holomorphic symplectic form for the complex structure $I$, and many examples of these come from purely algebro-geometric sources. See, for example, the survey of Kaledin.

Answer by Tony Pantev for Is there a quaternionic algebraic geometry ?

2011-01-14T02:44:14Z

Take a look at Dominic Joyce's paper "A theory of quaternionic algebra, with applications to hypercomplex geometry".

Answer by David Ben-Zvi for Is there a quaternionic algebraic geometry ?

2011-01-14T03:27:03Z

In a more analytic direction there is also a recent theory of "split quaternionic analysis" developed by Igor Frenkel and Matvei Libine, starting here with a survey here, with applications to representation theory and physics.

Answer by Roberto Frigerio for Is there a quaternionic algebraic geometry ?

2011-01-14T11:55:02Z

You may perhaps be also interested in the quite recent paper

Gentili, Graziano; Stoppato, Caterina Zeros of regular functions and polynomials of a quaternionic variable. Michigan Math. J. 56 (2008), no. 3, 655–667

(and other papers by the same authors).