Tangent vectors as linear functionals on F(M)

[Rasalhague]

Let M be an n-dimensional manifold, with tangent spaces T_pM for each point p in M. Let F(M) be the vector space of smooth functions M --> R, over R, with the usual definitions of addition and scaling. Tangent vectors in TM can be defined as linear functionals on F(M) (Fecko: Differential Geometry..., 2.2). What is the relationship between the n-dimensional tangent spaces T_pM and the (presumably infinite dimensional) dual space, F(M)^*, of F(M)? What can be said about the zero vector of F(M)^*? Since the T_pM are subsets of F(M)^*, and are each vector spaces in their own right, over the same field (and with the same addition and scaling functions?), they should each contain the zero vector; and yet the tangent spaces are said to be disjoint.

 

[Fredrik]

A tangent vector at p can be defined as a member v of F(M)* that satisfies v(fg)=v(f)g(p)+f(p)v(g) for all f,g in F(M). Clearly F(M)* contains a lot more than just the members of the union of all the tangent spaces. The 0 vector of F(M)* takes every f in F(M) to 0. So the 0 vector of F(M)* is also the 0 vector of each T_pM. The best answer I can think of right now to the question about the "relationship" is that the T_pM can be thought of as n-dimensional subspaces of the infinite-dimensional vector space F(M)*, but you seem to have realized that already.

If you define the tangent spaces this way, they are clearly not disjoint, but there are other definitions. One is to define an equivalence relation on the set of smooth curves that go through p at parameter value 0. We say that two such curves B and C are equivalent if (x\circ B)'(0)=(x\circ C)'(0), where x is some coordinate system whose domain includes p. (This definition is independent of the choice of coordinate system). We can then define T_pM as the set of equivalence classes. The vector space structure is defined by mapping the curves to ℝⁿ and using the standard vector space structure there:

b<b>+c[C]= [x^{-1}\circ (a(x\circ C)+b(x\circ B))]</b>

( and [C] are the equivalence classes that contain the curves B and C respectively. This definition is also independent of the choice of coordinate system). These tangent spaces are clearly disjoint, since they consist of equivalence classes of curves that go through p at parameter value 0.

Let's use the notation D_pM instead of T_pM for the set of all v in F(M)* such that v(fg)=v(f)g(p)+f(p)v(g). It's not too hard to show that the equality

\phi[C](f)=(f\circ C)&#039;(0)

defines an isomorphism \phi:T_pM\rightarrow D_pM.

So the tangent spaces are disjoint if you want them to be, i.e. if you choose to consider T_pM the "true" tangent space at p, and D_pM just a space that's isomorphic to the "true" tangent space at p (and also easier to work with).

 

[Rasalhague]

Thank you for your detailed and clear answer, Fredrik. That's very interesting. So, when these various definitions of tangent vectors are said to be "equivalent", they're only equivalent in some respects? Or more precisely, each tangent space, defined in one way, is isomorphic to the corresponding tangent space, defined in one of the other ways; but the different definitions lead to different concepts of the tangent bundle? Or is the disjointness of fibres (tangent spaces) a required part of what it means to be a tangent bundle, so that the definition of tangent vectors as derivations can't be used to define the tangent bundle?

 

[Fredrik]

Yes, they're equivalent in the sense that they're isomorphic. It really doesn't matter which n-dimensional vector space we think of as the tangent space at p, since all n-dimensional vector spaces (over the same field) are isomorphic.

For the tangent bundle to satisfy the definition of a fiber bundle, there must exist a projection map \pi:TM\rightarrow M that takes each point in the tangent bundle to the point in the manifold that's associated with it. If we define the tangent bundle as the union of all the T_pM, we can define the projection map by \pi[C]=C(0). In other words, if v is in T_pM, then \pi(v)=p. This clearly doesn't work if we define the tangent bundle as the union of all the D_pM, because the zero vector will mess things up. But there's a simple workaround: We define the tangent bundle as the set of all pairs (p,v) such that v is in D_pM, and define the projection map by \pi(p,v)=p.

By the way, I've seen that some authors (http://www.worldscibooks.com/etextbook/3812/3812_chap1_2.pdf ) define the tangent vectors as functionals acting on equivalence classes of smooth functions instead of on the smooth functions themselves. I haven't really understood the point of that, but I can't say that I've tried.

 

[Rasalhague]

To help me keep track of things, I'd like a notation that distinguishes between tangent vectors, (p,v), and images, v, of tangent vector fields. Would the following (I hope slight) modification of what you said work?

Let "equivalence classes of (p,v)" : F(M) --> R, for fixed p, be linear and obey the Leibniz rule, and let v : F(M) --> F(M) such that, for all p, (v(f))(p) = (p,v)(f).

Now as elements of F(M)^*, (p₁,0) = (p₂,0) = 0, so that we can't define the tangent bundle as the set of linear and Leibniz elements of F(M)^*, because then the projection would be undefined at the zero vector. (Obviously there can only be one zero vector.)

But we can define the tangent bundle as the set of pairs (p,v), all of which are elements of F(M)^*, though not all distinct elements of F(M)^*. So T_pM = {p} x {all v}, rather than the subset of F(M)^* which can be represented by elements of T_pM = {p} x {all v}.

 

[Fredrik]

To help me keep track of things, I'd like a notation that distinguishes between tangent vectors, (p,v), and images, v, of tangent vector fields.

I like to use lower case u,v for tangent vectors, and upper case X,Y,Z for vector fields. Not sure why. I write the value of X at p as X_p. I would say that in the pair (p,v), v is a member of T_pM, so that it's v that should be called a tangent vector, but obviously, you can define a vector space structure on {(p,v)|v in T_pM} that makes it isomorphic to T_pM, so it certainly makes sense to think of the pairs (p,v) as tangent vectors too.

Let "equivalence classes of (p,v)" : F(M) --> R, for fixed p, be linear and obey the Leibniz rule, and let v : F(M) --> F(M) such that, for all p, (v(f))(p) = (p,v)(f).

I don't understand what you're saying here, especially the first part. Are you defining (p,v)(f) by saying that it's equal to v(f)?

Now as elements of F(M)^*, (p₁,0) = (p₂,0) = 0, so that we can't define the tangent bundle as the set of linear and Leibniz elements of F(M)^*, because then the projection would be undefined at the zero vector. (Obviously there can only be one zero vector.)

Sounds right.

But we can define the tangent bundle as the set of pairs (p,v), all of which are elements of F(M)^*, though not all distinct elements of F(M)^*. So T_pM = {p} x {all v}, rather than the subset of F(M)^* which can be represented by elements of T_pM = {p} x {all v}.

You lost me again. I'm not saying that you're wrong, only that I don't quite understand what you're saying.

If we use my definition with the D_pM above, the fiber over p is \pi^{-1}(p)=\{p\}\times D_pM. Maybe that's what you meant.

 

[Rasalhague]

I don't understand what you're saying here, especially the first part. Are you defining (p,v)(f) by saying that it's equal to v(f)?

No, because I defined v as a map from F(M) to F(M). By F(M), I mean the set of smooth maps f:M-->R. So v(f) belongs to F(M), whereas I defined (p,v) as a map from F(M) to R, so (p,v)(f) belongs to R. (Not necessarily a distinct map for distinct values of p and v.) Rather I meant to define (p,v)(f) as equal to the value of v(f) at p, that is, v(f)(p).

You lost me again. I'm not saying that you're wrong, only that I don't quite understand what you're saying.

If we use my definition with the D_pM above, the fiber over p is \pi^{-1}(p)=\{p\}\times D_pM. Maybe that's what you meant.

I think so, yes, except that I'm trying to make this consistent with Spivak's notation and terminology in Differential Geometry, Vol. I, beginning of Ch. 3. There, he seems to treat (p,v) itself as an element of the tangent space T_pM, and of the tangent bundle TM. I'm trying to say something like, "Let tangent vectors (when defined in terms of derivations) be to linear and Leibniz elements of F(M)^* as fractions are to the set of rational numbers." So, (p,0) and (q,0) would be distinct elements of TM, just as 2/4 and 1/2 are distinct fractions.

(Hmm, I see now Spivak has some more general definitions later in that chapter, and, in fact, redefines some of the terms he used at the beginning, so I should probably read the rest before trying to work this out.)