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I have always thought of Riemannian metrics as being an inner product assigned to each tangent space. That is, if $M$ is a manifold, then at any point $p \in M$, $$g_p: T_pM \times T_pM \rightarrow \mathbb{R}.$$ However I am following a set of notes that defines a Riemannian metric as a map on the cotangent bundle: $$g: T^*M \times T^*M \rightarrow \mathbb{R}$$ so that at a particular point $p \in M$, $$g_p: T^*_pM \times T^*_pM \rightarrow \mathbb{R}.$$

At first I thought this might be a typo, but on the Wikipedia page for metric tensors it mentions an isomorphism between the cotangent space and tangent space that allows us to define a metric on the cotangent space. I previously had the impression that this isomorphism gives the inverse metric tensor (i.e. "raising and lowering" indices), but now after reading these notes I am wondering if I am wrong and it instead gives an equivalent way of defining a Riemannian metric.

Which one is correct? Should a Riemannian metric be defined on the tangent space or the cotangent space?

CBBAM
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    Everyone defines a Riemannian metric to be a choice of inner product on each tangent space. It is then a simple matter of linear algebra to see that this is equivalent to specifying an inner product on each cotangent space. See here for the details. So, can a Riemannian metric be defined on the cotangent spaces? Sure, but no one starts out that way. – peek-a-boo Sep 19 '23 at 00:31
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    Also, nitpicky comment, $g$ is not a map $T^M\times T^M$. If anything, it should be a map from out of the fiber product $T^M\times_MT^M$, or what amounts to the same thing, the direct sum of these vector bundles $T^M\oplus T^M$. – peek-a-boo Sep 19 '23 at 00:36
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    Keep in mind that this is really a question in linear algebra. An inner product on a finite-dimensional vector space $V$ induces a linear isomorphism $V \rightarrow V^$, where $v \in V$ maps to $\ell \in V^$, where $$ \ell(w) = \langle v, w\rangle.$$ Using this isomorphism, you get an inner product on $V^$. Since $(V^)^* = V$, an inner product on $V^*$ defines an inner product on $V$. Your question is about the case when $V = T_pM$. – Deane Sep 19 '23 at 01:20
  • @peek-a-boo Thank you for your comment. I saw the construction in the thread you linked, where you set $$h(\phi, \psi) := g \left( g^{\sharp}(\phi), g^{\sharp}(\psi)\right)$$ and this defines an inner product on $V^* \times V^*$. I was able to understand this construction if we already have an inner product $g$ on $V \times V$. How would we do this if we didn't have such an inner product to start with only had the inner product $h$, for example? – CBBAM Sep 19 '23 at 02:25
  • @Deane Thank you. I understand how, given a vector $v$, the Riesz representation theorem gives us an isomorphism via $v \mapsto \langle v, \cdot \rangle$, but I think I need to read more to understand how one can construct an inner product on $V^$ from this, and conversely how to obtain an inner product on $V$ from one on $V^$. – CBBAM Sep 19 '23 at 02:30
  • You have an inner product $h$ on $V^$. Applying the same logic gives you an inner product $\gamma$ on $V^{}$. But now recall the famous double duality isomorphism $\iota:V\to V^{}$. So, with this, you can transport the inner product on $V^{}$ to an inner product $g$ on $V$ (so, $g$ is the pullback of $\gamma$ under $\iota$ or more explicitly, for all $v,w\in V$, $g(v,w):=\gamma(\iota(v),\iota(w))$). Btw, a remark about terminology, you write “$h$ defines an inner product on $V^\times V^$” and “an inner product $g$ on $V$”, but really they are inner products on $V^,V$ respectively. – peek-a-boo Sep 19 '23 at 02:35
  • @peek-a-boo I get it now, thank you very much! I think this is what Deane was getting at. – CBBAM Sep 19 '23 at 02:37
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    There is an exposition here on the relationship with mechanics. The kinetic energy formula $K=\frac{1}{2}mv^2 =\frac{1}{2m}p^2$ is exactly this relationship, if $p=mv$, where $v$, the velocity lives on the tangent bundle and the momentum $p$ lives on the cotangent bundle. The mass metric $m$ is considered a map between the tangent bundle and the cotangent bundle. The pairing on the cotangent bundle is the cometric. – tensor_and_manifold Sep 23 '23 at 22:04
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    @tensor_and_manifold Thank you, this is exactly what I was looking for as I came across this point in the context of classical mechanics. – CBBAM Sep 24 '23 at 17:10

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