# The Unapologetic Mathematician

## 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 $v$.

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

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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4. […] Armstrong: Derivatives in Coordinates, The Derivative, Coordinate Vectors Span Tangent Spaces, Tangent Vectors at a […]

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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13. […] this pullback of we must work out how to push forward vectors from . That is, we must work out the derivative of […]

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14. […] the derivative, we see […]

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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