The goal of this post is to work through the following theorem:
The manifold of 1-jets of functions (with its natural contact structure) is contact diffeomorphic to
, the manifold of co-oriented contact elements in
.
Recall: the natural contact structure of the space of 1-jets is given by
, where
.
A point in , the space of 1-jets of
, is given by:
, where
.
Since is the gradient of
at q, it means that
gives us the direction of fastest ascent. This direction lives in the tangent space of
at
. This tangent space is the space orthogonal to
, i.e.
. Hence,
.
It follows from the structure given in that the natural contact structure in
is given by:
.
What is the tautological contact structure on ?
We denote a point in to be
, where
and
.
The tautological contact structure is the field of hyperplanes in such that the projection of each hyperplane to
gives us a co-oriented contact element; in this case, it would be
(
and
is a co-oriented hyperplane in
centered at
).
corresponds uniquely to an oriented 1-form
, which identifies uniquely with a ray emanating from the origin,
. We can think of
as one of the two rays in the space orthogonal to
.
Thus, we can see that each co-oriented contact element, corresponds to unique a point
.
Since can be thought of as the space of vectors orthogonal to
emanating from
, it follows that:
The corresponding hyperplane in the tautological contact structure is the pre-image of this space in under the natural projection
. This hyperplane is the set of vectors,
emanating from
such that
, i.e.
Hence, the tautological contact structure in is given by
.
Claim:
given by
is a contact diffeomorphism.
Proof:
Since ,
, and
, the map is well-defined and that
. Hence,
is bijective. It is easy to see that
is smooth. Hence, it is a diffeomorphism.
We now show that maps the natural contact structure in
into the tautological contact structure
.
It suffices to show that given at
, the corresponding
at
(i.e. the image of
) satisfies
.
Recall, and
.
Thus, , and so,
(*)
Since , it follows that
. We also know that
. Hence,
(*) = .
Since , it follows that
, and so,
. The above gives us,
.
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