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Let $(M,g)$ be a Riemannian manifold. Then there is the well-known Sasaki metric that makes $(TM,\hat{g})$ a Riemannian manifold. In a similar way, one can construct a Sasaki metric $\bar{g}$ on the cotangent bundle. My question is, is the map $g\colon (TM,\hat{g}) \to (T^*M,\bar{g})$ an isometry?

For completeness, the Sasaki metrics are given as follows (not 100% sure about the cotangent one). Let $X,Y$ be vector fields on $M$ and $\alpha,\beta$ be one-forms on $M$, let $\pi$ be the projection from both bundles onto $M$ and let superscripts $h$ and $v$ denote horizontal and vertical lifts. Then these Sasaki metrics are uniquely defined by the following conditions: $$ \hat{g}(X^h,Y^h) = \hat{g}(X^v,Y^v) = g(X,Y) \circ \pi, \quad \hat{g}(X^h,Y^v) = 0 $$ and $$ \bar{g}(X^h,Y^h) = g(X,Y) \circ \pi, \quad \bar{g}(\alpha^v,\beta^v) = g^{-1}(\alpha,\beta) \circ \pi, \quad \bar{g}(X^h,\alpha^v) = 0. $$

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The answer is yes, and you can see this by realizing that the vertical (resp. the horizontal) subbundle of $TTM$ is mapped to the vertical (resp. the horizontal) subbundle of $TT^*M.$ For the vertical part, this follows from the fact that $g\colon TM\to T^*M$ is a bundle homomorphism, and for the horizontal bundle this follows from the fact that $g\colon TM\to T^*M$ is parallel (with respect to the natural connections defined by $g$ which are used to define the horizontal bundles).

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  • $\begingroup$ Thanks for the clear answer! Now I also found an index error in my coordinate calculation, which now agrees on this. Btw. there's a small typo in your answer: it should be $TT^*M$ instead of $T^*TM$, but edits need to be more than 6 characters, so I left it. $\endgroup$
    – Jaap Eldering
    Commented Sep 25, 2015 at 14:15

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