# Dual Riemannian metric and the Dual Metric Form

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