*(this question is about a particular aspect of [a previous question](https://mathoverflow.net/q/229825/22606), which was not duly stressed)*


Let $(M,g)$ a Riemannian $n$-dimensional manifold, and let
$$
\widetilde{M}:=\mathbb{P}T^\*M
$$
be the $(2n-1)$-dimensional manifold of tangent hyperplanes to $M$.



> 
> QUESTION: Is $\widetilde{M}$ a sub-Riemannian manifold in a canonical way?
> 
> 
> 


**Remarks & my own partial answer.**


Obviously, $\widetilde{M}$ is a contact manifold. Let $H\in T^\*M$ be a tangent hyperplane, and denote by $\tau:\widetilde{M}\to M$ the canonical projection: then
$$
C\_H:= \tau\_\ast^{-1}(H)\subset T\_H\widetilde{M}
$$
is the contact plane at $H$.


By definition, $C\_H$ fits into the exact sequence
$$
0\to (H^\*\otimes H^\perp)\to C\_H\stackrel{\tau^\*}{\to} H\to 0\, ,
$$
where $H^\*\otimes H^\perp$ is the vertical tangent space to $\widetilde{M}$ at $H$. Now both the endpoints are equipped with a metric, essentially obtained from the restricted metric $g|\_H$ and its dual (the "twisting line" $H^\perp$ can be normalised and hence neglected), so I figured that I can claim that $C\_H$ possesses a unique metric which makes both arrows metric-preserving (when $TM$ is trivial, my reasoning certainly works).


If the answer to my question is positive, I would appreciate a link to the existing literature on the subject.



Is $\mathbb{P}T^*M$ a sub-Riemannian manifold if $M$ is Riemannian?
I also believe that the answer to your question is yes, but I think that things are more complicated than indicated in the answer by @Ben\_McKay and in your partial solution. The point is that while you have an exact sequence of the form claimed in your answer, requiring that the two arrows are metric does not pin down an inner product on the middle term. Essentially, the inner products which have this property are parametrized by all splittings of the sequence. (A metric gives a splitting via mapping the quotient back to the orthocomplement of the subspace and a splitting gives a metric by declaring the subspace and the image of the split to be orthogonal to each other.) You can use the Levi-Civita connection to split the corresponding sequence on the level of $TT^\*M$ and then prove that this descends to the projectivization. 


