The Unapologetic Mathematician

Mathematics for the interested outsider

The Derivative

It turns out that the tangent bundle construction is actually a functor. Given a smooth map f:M\to N between smooth manifolds, we will get a smooth map f_*:\mathcal{T}M\to\mathcal{T}N. Yes, we’d usually write \mathcal{T}f for a functor’s action on a map, but the f_* notation is pretty classical.

So if we’re given a tangent vector v\in\mathcal{T}_pM we want to get a tangent vector f_*(v)\in\mathcal{T}_qN. And since we already have f sending points of M to points of N, it only makes sense to ask that q=f(p). That is, in terms of the tangent bundle projection functions, we can write f(\pi(v))=\pi(f_*(v)). In other words, the projection \pi:\mathcal{T}M\to M will be a natural transformation from the tangent bundle functor to the identity functor.

Anyway, this means that for each p\in M we’ll get a map f_{*p}:\mathcal{T}_pM\to\mathcal{T}_{f(p)}N. Since these are both vector spaces, it only stands to reason that we’d have a linear map. We haven’t yet established the connection between our “tangent vectors” and the geometric notion, but we do have a notion from multivariable calculus of a linear map that takes tangent vectors to tangent vectors: the Jacobian, which we saw as a certain extension of the notion of the derivative. We will find that our map f_* is the analogue of the same concept on manifolds, and so we will call it the derivative of f.

So here’s our definition: if f:U\to N is a differentiable map in some open set U\subseteq M and if p\in U, then we define our map f_{*p}:\mathcal{T}_pM\to\mathcal{T}_{f(p)}N by

\displaystyle\left[f_{*p}(v)\right](\phi)=v(\phi\circ f)

where \phi\in\mathcal{O}(V) is any smooth function on a neighborhood of f(p)\in N. That is, f_{*p}(v) is a linear functional on \mathcal{O}_{f(p)}; if \phi represents a germ at f(p) we can compose it with f to represent a germ at p, and then we can apply v itself to this germ. It should be immediately clear that this construction is linear in v.

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April 6, 2011 - Posted by | Differential Topology, Topology

23 Comments »

  1. [...] take the derivative and see what it looks like in terms of coordinates. Say we have a smooth manifold and a smooth map [...]

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  2. [...] of the Derivative We’ve said that the tangent bundle construction is a functor with the derivative as the action on morphisms. But we haven’t actually verified that it obeys the conditions of [...]

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  3. [...] we have a canonical tangent vector in , we can hit it with the derivative and see what happens. We get a tangent [...]

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  5. [...] to more general manifolds. We know that the proper generalization of the Jacobian is the derivative of a smooth map , where is an open region of an -manifold and is another -manifold. If the [...]

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  6. [...] map of manifolds is called an “immersion” if the derivative is injective at every point . Immediately we can tell that this can only happen if [...]

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  7. [...] be a smooth map between manifolds. We say that a point is a “regular point” if the derivative has rank ; otherwise, we say that is a “critical point”. A point is called a [...]

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  8. [...] key observation is that the inclusion induces an inclusion of each tangent space by using the derivative . The directions in this subspace are those “tangent to” the submanifold , and so these [...]

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  9. [...] of increasing . That is, includes the interval into “at the point “, and thus its derivative carries along its tangent bundle. At each point of an (oriented) interval there’s a [...]

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  10. [...] be a smooth map between manifolds, with derivative , and let and be smooth vector fields. We can compose them as and , and it makes sense to ask if [...]

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  11. [...] from back to itself, and in particular it has the identity as a fixed point: . Thus the derivative sends the tangent space at back to itself: . But we know that this tangent space is canonically [...]

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  12. [...] because smooth maps push points forward. It turns out that vectors push forward as well, by the derivative. And so we can define the pullback of a -form [...]

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  15. [...] forms are entirely made from contravariant vector fields, so we can pull back by using the derivative to push forward vectors and then [...]

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  16. [...] The orientation on a hypersurface consists of tangent vectors which are all in the image of the derivative of the local parameterization map, which is a singular [...]

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  17. Hi, a quick followup please. And , again, great site; keep it up. Sorry for the ASCII; ignore it if it is too hard,sorry don’t know how to do otherwise.

    We’re given a map F:M–>N between manifolds.

    So,please , let me see if I understand f_*p well, using the curve def. of tangent vector v based at p.

    Consider a chart (U,Phi) containing p in M:, and a curve:

    C(t):(-1,1)–>M ; C(0)=p we define a tangent vector v in TpM in terms of the derivative (living in R^n) given by: d/dt( (Phi)oC)(0), so that
    .
    C(t) represents a tangent vector v_p in TpM (specifically, v_p representis all curves C_i with the same number value d/dt(PhioC_i)(0) ) .

    Then F: M–>N sends this vector/curve-class v_p in M to the curve FoC in N. Now, this curve is itself

    a tangent vector based at F(p) in N; we consider a chart (W, Csi) containing F(p). Then F_* (v_p) is the class given by

    d/dt(Csi oFo C)(0).

    Is that It?

    Thanks again for the nice site. Let me know if you write a book ( I guess it would be an E-book).

    Comment by carl | May 30, 2013 | Reply

  18. Yes, Carl, that’s it. Geometrically we just watch how the map F takes a curve passing through p\in M and maps it to a curve passing through F(p)\in N, and we observe that the tangent vector of the target curve at F(p) is independent of everything but the tangent vector of the original curve at p.

    Comment by John Armstrong | May 30, 2013 | Reply

  19. Thanks, John:

    But, in practice, say we have the usual setup: C:(-1,1)–>M c(0)=p and (U,Phi) a chart containing p. Then we consider the curve

    C’:=Phi oC landing in R^n. Do we actually consider C’ as a curve in R^n, or do we transplant/pullback C’ from R^n into T_pM by the

    chart maps?

    Comment by carl | May 30, 2013 | Reply

  20. Well, some of this seems a little confused; I’m really not sure what you’re trying to do here.

    Comment by John Armstrong | May 30, 2013 | Reply

  21. O.K, I thought I was using the tangent plane as a sub for R^n ; the curve in R^n could live instead in
    the tangent space, which is a copy of R^n. But maybe I am confused.

    Comment by carl | May 30, 2013 | Reply

  22. Yes, the curve does not live in the tangent space. A curve is a function from a parameter interval to a manifold.

    Comment by John Armstrong | May 30, 2013 | Reply

  23. I see; thanks again, I’ll go back to the books to clear things up.

    Comment by carl | May 30, 2013 | Reply


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