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Let $M$ be a Riemannian manifold, and let $p \in M$.

I know that the tangent space at $p$, $T_pM$, is isomorphic as a vector space to the cotangent space at $p$, $(T_pM)^*$. Thus, in some sense, as vector spaces, they are indistinguishable.

Since $M$ is a Riemannian manifold, $T_pM$ has an inner product, the Riemannian metric, and thus is not just a vector space, but also a Hilbert space. Thus elements $v \in T_pM$ have notions of angle and length, and can be thought of as "geometric vectors", i.e. arrows pointing from some origin.

Do elements $w \in (T_pM)^*$ also have notions of length and angle, allowing them to be thought of as geometric vectors? Or is $(T_pM)^*$ just a metrizable topological vector space, allowing one to think of $w \in (T_pM)^*$ as "points", objects for which there is a meaningful notion of distance and a "Cartesian grid", as well as notions of addition and scalar multiplication, but not of length or angle.

Overarching question: do the tangent space at $p$, $T_pM$, and the cotangent space at $p$, $(T_pM)^*$, have any different structures defined on them which are not defined on their counterpart? And for those structures which are defined on both sets, for which are they isomorphic?

  1. Is the cotangent space also a Hilbert space? If so, are the tangent and cotangent spaces isomorphic as Hilbert spaces?
  2. Is the cotangent space also a Banach space? If so, are they isomorphic as Banach spaces?
  3. Are they both metric spaces? If so, are they isomorphic as metric spaces?
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    $\begingroup$ Is $M$ here finite dimensional? $\endgroup$ – John Hughes Oct 20 '16 at 17:29
  • $\begingroup$ @JohnHughes I guess it has to be right or else the question doesn't make sense -- I'll change it $\endgroup$ – Chill2Macht Oct 20 '16 at 17:30
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    $\begingroup$ Well...the question makes sense, but the answer, in the infinite-dimensional-case would be "Yes, they're distinguishable". :) $\endgroup$ – John Hughes Oct 20 '16 at 17:33
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    $\begingroup$ $T^*M$ has a natural symplectic structure. $\endgroup$ – Moishe Kohan Oct 20 '16 at 17:33
  • $\begingroup$ @JohnHughes Oh I guess I spoke too soon then -- I will change the question back to its original version then. Sorry about that -- I would like to hear about this distinction in the infinite-dimensional case. $\endgroup$ – Chill2Macht Oct 20 '16 at 18:09
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The Riemannian metric provides a natural identification of the cotangent space with the tangent space. Therefore the cotangent space is also naturally equipped with a Euclidean (or if you prefer Hilbert-space) metric. The simplest way of thinking of the identification is to send a tangent vector $v$ to the covector given by the inner product with $r$, namely $\langle v,\cdot\rangle$.

So they are naturally isomorphic as Hilbert spaces, Banach spaces, and metric spaces.

Expressed in terms of indices, the relation is simply this. If $v_i$ represents a vector and $\alpha^j$ a covector then the relation $v_i=g_{ij}\alpha^j$ (summation over a repeated index as usual) gives a way of passing between a vector and its corresponding covector. Here $g_{ij}$ is the Riemannian metric.

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  • $\begingroup$ Is this identification the metric tensor? en.wikipedia.org/wiki/Riemannian_manifold#Riemannian_metrics Wikipedia doesn't say anything else about cotangent spaces on Riemannian manifolds, and the article on cotangent spaces only refers to differentiable manifolds, where both the tangent and cotangent spaces are a priori only vector spaces (in a way that is compatible with the manifold's structure). $\endgroup$ – Chill2Macht Oct 20 '16 at 18:06
  • $\begingroup$ Using the elementary definition without tensors, and just as an inner product on the tangent space, can one define it naturally on the cotangent space by using adjoints and identifying the tangent space with its double dual (which works because it is finite-dimensional)? $\endgroup$ – Chill2Macht Oct 20 '16 at 18:08
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    $\begingroup$ I am not sure what you are asking exactly, but I will add a formula hopefully to make things clearer. $\endgroup$ – Mikhail Katz Oct 21 '16 at 9:54
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Let $M$ be a Riemannian manifold, and let $p \in M$.

I know that the tangent space at $p$, $T_pM$, is isomorphic as a vector space to the cotangent space at $p$, $(T_pM)^*$. Thus, in some sense, as vector spaces, they are indistinguishable.

That's true to an extent, but the isomorphism is still not "wholeheartedly natural". If $\alpha = g(v, \cdot)$, and if the metric $g$ is replaced by a constant multiple $\lambda g$ (i.e., the unit of length with respect to $g$ changes by a factor of $\lambda$), then $$ \alpha = \lambda g(v/\lambda, \cdot),\qquad \lambda\alpha = \lambda g(v, \cdot). $$ These relationships are obvious, profound, and (possibly) under-appreciated: If your scale of length (metric) gets longer, your tape measure (cotangent vector) gets longer while every object you measure (tangent vector) gets shorter.

Overarching question: do the tangent space at $p$, $T_pM$, and the cotangent space at $p$, $(T_pM)^*$, have any different structures defined on them which are not defined on their counterpart? And for those structures which are defined on both sets, for which are they isomorphic?

Arguably, this question (while a good one) does not address the whole story. One must also ask how the tangent and cotangent spaces (or better, the bundles) behave with respect to mappings.


Tangentially (no pun), tangent and cotangent bundles in the holomorphic category are, in general, genuinely distinct (e.g., they may have different Chern classes; one may have non-trivial holomorphic sections while the other doesn't; etc.). An Hermitian metric, being conjugate-linear in the second argument, does not define an isomorphism of holomorphic vector bundles unless its components are constant.

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Based on Moishe Cohen's comment, and some reading on Wikipedia, I have come to the conclusion, that if symplectic forms do in fact induce volume forms, then one way to distinguish the tangent and cotangent bundle of a differentiable manifold is via the fact that the cotangent bundle has notions of orientation and volume, i.e. signed volume, and thus following from that a notion of integration, and then from that a notion of differentiation.

This in particular explains why phase space formulations of classical mechanics usually concern themselves primarily with the cotangent bundle, because it has symplectic structure and thus provides notions of volume to work with in phase space, leading e.g. to Liouville's theorem.

I have mused on this, also incorrectly and hand-wavingly, here and here.

The reason why the tautological one-form, hence symplectic structure, although natural to the cotangent bundle, is not natural to the tangent bundle, seems to be possibly related somehow to this statement made in the Wikipedia article on cotangent bundles:

Note that the tautological one-form is not a pullback of a one-form on the base M.

EDIT: The key point here seems to be page 7 of this document, which says:

The spaces $V$ and $V^*$ are closely related and are in fact isomorphic, though it’s important to observe that there is no canonical isomorphism between them. Isomorphisms between $V$ and $V^*$ do arise naturally from various types of extra structure we might add to $V$.

Thus unless we have a Riemannian manifold or a fixed global frame (compatible choice of basis for every vector space of the manifold) there is no canonical way to relate the tangent and cotangent space -- thus there is no way to canonically use the natural symplectic form on the cotangent bundle to induce a symplectic form on the tangent bundle. Thus, for manifolds without additional, extra structure, differentiation and integration are only naturally defined for the cotangent bundle, thus distinguishing it from the tangent bundle.

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    $\begingroup$ Symplectic geometry is a fascinating subject but your original question started with a Riemannian metric which immediately leads to the identification of the two spaces you are interested in. If you have a question about symplectic geometry, you should ask it separately, or else clarify the intention of this question. $\endgroup$ – Mikhail Katz Oct 21 '16 at 10:11
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    $\begingroup$ William, in Riemannian geometry the two spaces are indeed canonically isomorphic. Without a metric this would not be true, but here we do have a metric! The document you quote is not relevant. $\endgroup$ – Mikhail Katz Nov 6 '16 at 16:09
  • $\begingroup$ @MikhailKatz Yes you are right, there does not seem to be a way to distinguish the two in Riemannian geometry. This answer is only relevant to manifolds without a Riemannian metric, but my question was originally asking about Riemannian manifolds, which I had been reading about the time, whereas my class so far is only about arbitrary smooth manifolds. So this provides one way to conceptually distinguish the two. Anyway I will accept your answer because it is correct and answers the question as I had asked it -- I did not realize at the time how special Riemannian manifolds are. $\endgroup$ – Chill2Macht Nov 6 '16 at 16:31

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