4
$\begingroup$

The horizontal distribution in the Heisenberg group is the kernel of the standard contact form: $$ \alpha = dt + 2 \sum_{j=1}^n (x_j \, dy_j - y_j \, dx_j). $$

Question. Can one describe horizontal distribution in any Carnot group in terms of kernels of some $1$-forms?

I believe the answer should be in the positive and I am looking for references for explicit constructions of such forms so that in the case of the Heisenberg group these constructions would give the contact form.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
4
$\begingroup$

The distribution on any Carnot group is left invariant, so pick any set of covectors at the origin that annihilates the distribution at that point, then left translate it to give a global set of one forms whose kernel is the distribution.

On the Heisenberg group this procedures gives a scalar multiple of the standard contact form.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank, you that was very simple. $\endgroup$
    – Piotr Hajlasz
    Commented Feb 10, 2019 at 23:13
  • $\begingroup$ At the origin this are just linear forms on the tangent space. You get (left-invariant) differential forms once you translate. $\endgroup$
    – YCor
    Commented Feb 11, 2019 at 0:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .