I have been interested in contact structures on supermanifolds. I though it would be useful, and fun to examine a low dimensional example to illustrate the definitions. Let \(M\) be a supermanifold. We will understand supermanifolds to be “manifolds” with commuting and anticommuting coordinates.
For manifolds there are several equivalent definitions. The one that is most suitable for generalisation to supermanifolds is the following:
Definition A differential one form \(\alpha \in \Omega^{1}(M)\) is said to be a contact form if
- \(\alpha\) is nowhere vanishing.
- \(d\alpha\) is nondegenerate on \(ker(\alpha)\)
This needs a little explaining. First we have to think about the grading here. Naturally, any one form decomposes into the sum of even and odd parts. To simplify things it makes sense to consider homogeneous structures, so we have even and odd differential forms. Due to the natural grading of differentials as fibre coordinates on antitangent bundle a Grassmann odd form will be known as an even contact structure and vice versa. The reason for this will become clearer later.
The definition of a nowhere vanishing one form is that there exists vector fields \(X \in Vect(M)\) such that \(i_{X}\alpha =1\). Again, via our examples this condition will be made more explicit.
The kernel of a one form is defined as the span of all the vector fields that annihilate the one form. Thus we have
\(ker(\alpha) = \{X \in Vect(M)| i_{X}( \alpha)=0 \}\).
The condition of nondegeneracy on \(d\alpha\) is that \(i_{X}(d \alpha)=0\) implies that \(X=0\). That is there are non-nonzero vector fields in the kernal of the contact form that annihilate the exterior derivative of the contact form.
On to simple examples. Consider the supermanifold \(R^{1|1}\) equipped with natural coordinates \((t, \tau)\). here \(t\) is an even or commuting coordinate and \(\tau\) is an odd or anticommuting coordinate.
I claim that the odd one form \(\alpha_{0} = dt + \tau d \tau\) is an even contact structure.
First due to our conventions, \(dt\) is odd and \(d \tau\) is even, so the above one form is homogeneous and odd.
Next we see that if we consider \(X = \frac{\partial}{\partial t}\) then the nowhere vanishing condition holds. Maybe more intuitively we see that considering when \(t = \tau =0\) the one form does not vanish.
The kernel is given by
\(ker(\alpha_{0}) = Span\left \{ \frac{\partial}{\partial \tau} {-} \tau \frac{\partial}{\partial t} \right \}\).
That is we have a single odd vector field as a basis of the kernel. That is we have one less even vector field as compared to the tangent bundle. Thus we have a codimension \((1|0)\) distribution.
Then \(d \alpha_{0}= d \tau d \tau\) , so it is clear that the nondegeneracy condition holds.
Thus I have proved my claim.
This is just about the simplest even contact structure you can have.
The odd partner to this is given by
\(\alpha_{1} = d \tau {-} \tau dt\)
This is clearly an even one form that is nowhere vanishing. The kernel is given by
\(ker(\alpha_{1}) = Span\left\{ \frac{\partial}{\partial t} {-} \tau \frac{\partial}{\partial \tau} \right\}\).
Thus we have a codimension \((0|1)\) distribution.
The nondegeneracy condition also follows directly.
For those who know a little contact geometry compare these with the standard contact structure on \(R^{3}\).
There is a lot more to say here, but it can wait. For those of you that cannot wait, see Grabowski’s preprint arXiv:1112.0759v2 [math.DG]. Odd contact structures are also discussed in my preprints arXiv:1111.4044 and arXiv:1101.1844.