If $M$ is a smooth manifold and $TM$ is the tangent bundle clearly $T_pM\cong T_qM$ (as vector spaces) for every $p,q\in M$. Nobody ensures that the previous vector spaces isomorphism is natural (or canonical). In $\mathbb R^n$ we have that $T_p\mathbb R^n$ and $T_q\mathbb R^n$ are naturally isomorphic to $\mathbb R^n$ so we can differentiate a vector field along a direction, in the usual way so taking the directional derivatives of each component.

If the isomorphism between tangent spaces in different point isn't natural, why can't we differentiate a vector field in the usual way? The problem is comparing vectors belonging in different (isomorphic) vector spaces; but we can send the two vectors, with an isomorphism, in a common vector space and then subtract them. Where is the importance of a natural isomorphism?

  • 2
    $\begingroup$ If you change what you mean by "with an isomorphism in a common vector space" then your result pulled back to the original tangent spaces will change. That is, what you are proposing will depend on how you make your two tangent spaces isomorphic to a common vector space. There are lots of isomorphisms between two vector spaces of the same dimension. $\endgroup$ – KCd Nov 19 '12 at 22:35
  • $\begingroup$ It is true that there are lots of isomorphisms between vector spaces, but there are also lots of natural isomorphisms. For example canonical automorphisms of $V$ are the elements of $Z(GL(V))$, so homothetic transformations. $\endgroup$ – Dubious Nov 20 '12 at 19:24
  • $\begingroup$ So even if I choose a natural isomorphism, the directional derivative will not depends on the basis imposed on tangent spaces, but it depends on the choice of the natural isomorphism. $\endgroup$ – Dubious Nov 20 '12 at 19:28
  • 1
    $\begingroup$ What Kofi says is not quite right (e.g., a finite-dimensional vector space and its double dual are naturally isomorphic), but I suppose what Kofi means is that two vector spaces of the same dimension are generally not isomorphic in a canonical way. Galoisfan, do you think tangent spaces at different points of a general (abstract) manifold are naturally isomorphic in some way? What does the term "natural isomorphism" mean to you? $\endgroup$ – KCd Nov 20 '12 at 21:13
  • 1
    $\begingroup$ There is no such thing as a canonical isomorphism in general between two vector spaces and it is hopeless to prove there are "lots" of them between two fixed vector spaces. As an analogy, there is no such thing as a natural inner product on a (finite-dim.) real vector space. Let $V$ be the vector space of real polynomials $a+bx+cx^2$ that vanish at 45. There are many inner products on $V$, but none is "natural" or "canonical", right? $\endgroup$ – KCd Nov 21 '12 at 16:01

If a manifold has curvature, then parallel transport of vectors depends on the path. In other words, any map between tangent spaces at different points is dependent on an arbitrary choice which is particular for each pair of points and each particular manifold. It is the same problem as when relating a finite-dimensional vector space $V$ with its dual $V^*$. In order to do so, we need to fix a basis for $V$.

In regards to the importance of naturality, informally a transformation is called natural if it is "independent of its source" in a sense. For example, the isomorphism between $V$ and $V^{**}$, the double algebraic dual, is natural. We send $v\in V$ to $v^{**}\in V^{**}$ such that $v^{**}(f)=f(v)$ for all $f\in V^*$. Note that this is independent not only of the basis of $V$, but also does not reference anything particular about $V$, such as a basis, and so this isomorphism can be carried out consistently over all finite-dimensional vector spaces over a common field. Quoting Wikipedia, a transformation is not natural if it cannot be extended consistently over the entire category in question.

  • $\begingroup$ many books say, without explaining the meaning of the adjective "natural", that the parallel transport give a way to naturally identify tangent spaces on points which can be connected by a path. $\endgroup$ – Dubious Nov 25 '12 at 11:54
  • $\begingroup$ I will explain better my doubt: If on a smooth manifold we haven't a natural identification beetwen tangent spaces, than we can't define a derivative of a vector field as "a limit of a quotient". Once we define a parallel transport, then a covariant derivative of a vector field $V$ along a curve $\gamma$ is $$lim_{t\to t_0}\frac{P^{-1}_{t_0,t}V(t)- V(t_0)}{t-t_0}$$ So i guess that $P_{t_0,t}$ is a natural isomorphism because allows to write the covariant derivative as a limit of a quotient. $\endgroup$ – Dubious Nov 25 '12 at 12:11
  • $\begingroup$ just quoting the book "Manifold and Differential Geometry - Jeffrey M. Lee" on page 502, talking about the Koszul connection: "...The definition of this connection takes advantage of the natural identification of tangent spaces..." $\endgroup$ – Dubious Nov 25 '12 at 13:11
  • $\begingroup$ I haven't read Lee's book, and I am not too familiar this this construction, but this isomorphism seems to depend on the choice of metric for the manifold and so cannot be natural. Maybe he means that for a manifold with this extra structure, the tangent spaces have natural isomorphisms between them (I cannot see why this would be true), but then we are no longer talking about general differentiable manifolds, and there is in general several connections on a given manifold. $\endgroup$ – Espen Nielsen Nov 25 '12 at 14:12

Why don't you let $M = S^2 \subset \mathbb{R}^3$ ? The equation for the Tangent plane is $$T_{(a,b,c)}S^2 = \{ (x-a)a+(y-b)b+(c-z)c = 0 \}$$ This is a collection of planes parameterized by points $(a,b,c): a^2 + b^2 + c^2 = 1$.

We can define an initial vector $\vec{x}=(x,y,z) \in T_{(a,b,c)}S^2$. Then we can ask $$\vec{x}(t + \delta t) = A(t) \vec{x}(t) \in T_{(a,b,c) + \vec{v}\delta t}S^2$$ For any time-paramterized matrix $A$, we should get a connection on the sphere that necessarily Levi-Civita connection, since this flow is not the geodesic flow.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.