The Derivative
It turns out that the tangent bundle construction is actually a functor. Given a smooth map 
between smooth manifolds, we will get a smooth map 
. Yes, we’d usually write 
for a functor’s action on a map, but the 
notation is pretty classical.
So if we’re given a tangent vector 
we want to get a tangent vector 
. And since we already have 
sending points of 
to points of 
, it only makes sense to ask that 
. That is, in terms of the tangent bundle projection functions, we can write 
. In other words, the projection 
will be a natural transformation from the tangent bundle functor to the identity functor.
Anyway, this means that for each 
we’ll get a map 
. 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 is the analogue of the same concept on manifolds, and so we will call it the derivative of .
So here’s our definition: if 
is a differentiable map in some open set 
and if 
, then we define our map by
=v(\phi\circ f)](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cleft%5Bf_%7B%2Ap%7D%28v%29%5Cright%5D%28%5Cphi%29%3Dv%28%5Cphi%5Ccirc+f%29&bg=e6e6e6&fg=333333&s=0&c=20201002)
where 
is any smooth function on a neighborhood of 
. That is, 
is a linear functional on 
; if 
represents a germ at 
we can compose it with to represent a germ at 
, and then we can apply 
itself to this germ. It should be immediately clear that this construction is linear in .
April 6, 2011 -
Posted by John Armstrong |
Differential Topology, Topology
The Unapologetic Mathematician 
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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).
Yes, Carl, that’s it. Geometrically we just watch how the map
takes a curve passing through 
and maps it to a curve passing through 
, and we observe that the tangent vector of the target curve at 
is independent of everything but the tangent vector of the original curve at 
.
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?
Well, some of this seems a little confused; I’m really not sure what you’re trying to do here.
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.
Yes, the curve does not live in the tangent space. A curve is a function from a parameter interval to a manifold.
I see; thanks again, I’ll go back to the books to clear things up.