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I started reading A tour of subriemannian geometry, their geodesics and applications by Montgomery. There's one thing I don't quite understand, which probably boils down to linear algebra, but I'm just not seeing it.

Let $(Q, \mathcal{H}, \langle \cdot,\cdot\rangle)$ be a sub-Riemannian manifold. A cometric on $Q$ is a twice-contravariant symmetric tensor field. It is equivalent to giving a bundle morphism $\beta: T^*Q \to TQ$ such that $\beta$ corresponds to the adjoint $\beta^*$ under the natural isomorphism $T^{**}Q \cong TQ$. Fine. Then he says that a sub-Riemannian structure uniquely determines a cometric subject to the conditions

(i) ${\rm Im}(\beta) = \mathcal{H}$;

(ii) $p(v) = \langle \beta(p),v\rangle$ for all $p \in T_x^*Q$, $v\in \mathcal{H}_x$, $x \in Q$.

Here's what I understand: given $p \in T_x^*Q$, since we require $\beta(p)$ to be horizontal, it suffices to know what $\langle \beta(p),v\rangle$, is for all $v \in \mathcal{H}_x$, as $\langle\cdot,\cdot\rangle$ is non-degenerate, so requiring such product to equal $p(v)$ defines $\beta$, makes it as smooth as $\mathcal{H}$, takes care of (ii), and already gives us ${\rm Im}(\beta) \subseteq \mathcal{H}$.

I can't see why actual equality ${\rm Im}(\beta) = \mathcal{H}$ should hold. Namely, I don't see a completely natural way to produce $p \in T_x^*Q$ such that $\beta(p)=w$, where $w \in \mathcal{H}_x$ was arbitrary. Obviously condition (ii) defines $p$ on $\mathcal{H}_x$, but how to extend it? One can take an auxiliary Riemannian metric on $Q$ and let $p$ annihilate $\mathcal{H}_x^\perp$, but then uniqueness of the cometric is not clear and it feels like cheating.

What am I missing?

P.S.: I was sure this forum had a sub-Riemannian geometry tag but I'm on mobile and can't find it. Whatever.

Ivo Terek
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$\newcommand{\H}{\mathcal{H}}$ If $\H^\perp \subset T^*$ is the annihilator of $\H$, then $$ \H^* \simeq T^*/\H^\perp.$$ The inner product on $\H$ is a symmetric tensor $g \in S^2\H^*$. If $\pi: T^* \rightarrow \H^*$ is the quotient map, then the cometric is $\pi^*g$. In other words, $$g(v,v) = g(\pi(v),\pi(v)).$$ Associated with $g$ is a linear isomorphism $\phi: \H^* \rightarrow \H$. This can also be pulled back to a map $\phi\circ\pi: T^* \rightarrow \H$, which has maximal rank.

Deane
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    This is great and simple. Thank you! – Ivo Terek May 08 '21 at 22:37
  • Hello, sorry for the late reply. I wonder if you could expand on the statement that $\phi \circ \pi$ has maximal rank and on the quotient map $\pi$? Thank you!! – user57 May 12 '23 at 15:33
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    @user57, I suggest that you start by working everything out using bases. Start with a basis of $\mathcal{H}$, extend it to a basis of $T$, and use the dual basis of $T^*$. – Deane May 12 '23 at 18:31