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Let $M$ be a Rieamnnian manifold with metric $g: X(M) \times X(M) \to C^{\infty}(X)$, where $X(M)$ are the vector fields of $X$.

As is well known, we can induce a bilinear pairing $$ \langle \cdot , \cdot \rangle_g: \Omega^1(M) \times \Omega^1(M) \to C^{\infty}(M) $$ by setting $$ \langle \omega, \omega' \rangle_g = g(\omega^{\sharp}, (\omega')^{\sharp}), $$ where, as usual, $\sharp$ is defined by $g(\omega^\sharp, X) = \omega(X)$, for $X \in X(M)$.

On the other hand, as is also well known, there exists a unique element $\omega_g \in \Omega(M) \otimes \Omega(M)$ such that, for $(X,Y) \in X(M) \times X(M)$, $$ \omega_g (X,Y) = g(X,Y), $$ where $\omega_g$ is applied to $(X,Y)$ in the obvious way.

Thus, we have a pairing on $T^\ast(M)$ and an element of $\Omega(M) \times \Omega(M)$ both coming from $g$. I would like to know if there exists a simple relationship between these two objects. (By simple, I suppose I mean something global and algebraic, free from messy local expressions.)

Moreover, what does metric compatibility for a connection look like for either of these?

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What is the "action" of $g$ on $T(M) \times T(M)$? If you just mean applying $g$ to two vector fields, in what sense is this an action? – Spiro Karigiannis Feb 8 '11 at 17:53
@Spiro: I've edited according to your comment. Is this better? – John McCarthy Feb 8 '11 at 18:33
Yes, I understand now. Both Deane and Dick are correct, though, this is linear algebra. Everything here is defined pointwise and is tensorial. When you understand what is happening at a point, you understand what's happening on the whole manifold. These two objects are exactly the same thing, as long as you view the space of symmetric bilinear forms on $V$ as a subspace of $V^* \otimes V^*$. – Spiro Karigiannis Feb 8 '11 at 19:35
Yes, I am interested in K\"ahler structures as it happens, thanks for the reference. – John McCarthy Feb 8 '11 at 21:10
I'm an old-fashioned differential geometer, so I don't understand all of this. Any chance you could explain how you define a Riemannian or Kahler metric as "something global and algebraic, free from messy local expressions"? – Deane Yang Feb 8 '11 at 21:33

It is worth noting that this is really a question in linear algebra. An inner product on a vector space $V$ defines an element in $ \omega \in V^*\otimes V^ * $. The same inner product also induces an inner product on $V^* $. How is the inner product on $V^* $ related to $\omega$? It is, I suppose, a reasonable question, but you should be able to figure it all out using a basis of $V$, the dual basis of $V^*$, and the matrices representing the inner products as well as the tensor $\omega$.

[ADDED] As for the relationship between $\omega_g$ and the inner product on the cotangent bundle, isn't it the same as the relationship you give between $g^*$ and the inner product on the tangent bundle? In other words $$ \omega_g(\theta^\sharp, \phi^\sharp) = g^*(\theta, \phi) $$

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Yes of course we can do this for a vector space. The trouble is that that the vector space structure of $T(X)$ is infinte dimensional and doesn't admit any managable basis. It's only its module structure that is managable. One can of course look at everything locally, but this is messy. What I asked for in the question was a global solution to the problem. – John McCarthy Feb 8 '11 at 19:14
It is not really an infinite dimensional question, and it is a "pointwise" and not a local question about the manifold since everything is happening fiber by fiber. Thus, as Deane says, it IS a finite dimensional vector space question. – Dick Palais Feb 8 '11 at 19:21
Fair enough for the distinction between pointwise and local - so to rephrase - Is there a global way to do it that is neither pointwise nor local, in other words, can we do this using the global module structure, not the pointwise vector space structure. – John McCarthy Feb 8 '11 at 19:47
@John: I cannot understand why you would WANT to use a "global" method other than the obvious one that applies the pointwise identification fiber by fiber. – Dick Palais Feb 8 '11 at 20:19
But if you are thinking about global objects, then your inner product takes values in $C^\infty(M)$ and not $\mathbb{R}$, right? If that's the case, I think you need to rewrite your question completely, because then I have no idea what you are asking. – Deane Yang Feb 8 '11 at 20:55

What comes to my mind is a co-algebra structure, however, I'm not sure if it is of big importance. The coproduct $\Delta: T^\ast (M)\to T^\ast (M)\otimes T^\ast (M)$ is a homeomorhpism which preserves the inner product.

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[This is off-topic of course, but I can't resist] The (shuffle) coalgebra structure is indeed of big importance, but nobody seems to notice. Consider the "total" covariant derivative iterated to the k-th total covariant derivative. Apply this to a tensor product. In classical tensor calulus this gives a huge mess. But using the comultiplication you get a nice and short generalization of the general Leibniz product rule (except the binomial coefficients are now hidden in the comultiplication). Simplest nontrivial application: Curvature is a tensor. – Martin Gisser Feb 8 '12 at 13:13

In some cases, the notion of a Riemann manifold is overly general. If should happen that the problem(s) one has in mind admit a Kählerian complex structure, then the vanishing of the Nijenhuis tensor—which expresses an integrability condition—can provide the "non-messy" global structure that is requested. Associated to the vanishing Nijenhuis tensor is a closed symplectic form that in dynamical models usefully specifies Hamiltonian vector fields.

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Thanks for the answer. Could you be a little more specific on why it provides the required global structure please? – John McCarthy Feb 8 '11 at 21:55
Like several other posters, the requested answer that is "global and algebraic, free from messy local expressions" is a bit hazy. It was just a guess on my part, that no matter what nice structures you have in mind, the Kahlerian compatible triple (that is, the compatible metric, symplectic, and complex structures) and/or the Hamiltonian flows thereby associated to smooth functions, will be helpful in constructing them. Further reading in (e.g.) Moroianu's "Lectures on Kahler Geometry" and Flaherty's (older) "Hermitian and Kahlerian Geometry in Relativity" might be helpful too. – John Sidles Feb 8 '11 at 22:13
Ok, thanks a lot, I'll have a look at those references. – John McCarthy Feb 8 '11 at 22:17
The original question has nothing to do with complex or Kahler structures. This answer seems non-responsive. – Dan Lee Mar 2 '11 at 18:31

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