I am a bit stuck about figuring out that the result of a differential map $f_{\ast p}$ defined on a tangent space is a derivation in the "other" tangent space.

Suppose $f \colon N \to M$ is a map between a $n$-dimensional manifold $N$ and a $m$-dimensional manifold $M$. Define $f_{\ast p}$ as the map

$v_p \in T_p(N) \mapsto f_{\ast p}(v_p) \in T_{f(P)}(M)$ such that $f_{\ast p}(v_p)(g) = v_p(g \circ f)$

where $g$ is a differentiable function from $M$ to $\mathbb{R}$. Let $(U,u)$ be a chart around $P \in N$ and let $(V,v)$ a chart around $f(p) \in M$. I am a bit skeptical because, following the definition, it seems to have defined just another derivation on $T_p(N)$ (well, actually it is), i. e.

$\frac{\partial}{\partial x_i}(g\circ f\circ u^{-1})(u(P))$.

My question is: is $f_{\ast p}(v_p)$ a derivation on $T_p(M)$ too because of the relation

$g\circ f\circ u^{-1} = (g \circ v^{-1})(v\circ f\circ u^{-1})$ ?

If that is the case, then it seems to work because we have a derivation that sends a point in $\mathbb{R}^m$ (that is $v(P)$) to a point of $\mathbb{R}$ but I'm not so sure. Thanks.

  • $\begingroup$ To be clear: $f_{*p}(v_p)$ is a derivation of $C^\infty(M)$ at $p$. How would it act on $C^\infty(N)$? $\endgroup$ – Dylan Moreland Sep 24 '11 at 3:36
  • $\begingroup$ What do you mean with "derivation of $C^{\infty}(M)$ at $p$" ? $\endgroup$ – user14174 Sep 24 '11 at 5:23
  • $\begingroup$ @Lmn6 Try explicitly writing down your definition of a derivation. That will probably help you see what Dylan means. $\endgroup$ – yasmar Sep 24 '11 at 10:11
  • $\begingroup$ @yasmar: well, in the original message you can find the definition and the local form too. Am I missing something? $\endgroup$ – user14174 Sep 24 '11 at 12:35
  • $\begingroup$ You have defined $f_{*p}(v_p)$ by the way it acts on an element $g \in C^\infty(M)$, i.e., it is a derivation of $C^\infty(M)$. Dylan then asks you how you propose to define its action on an element $h \in C^\infty(N)$. $\endgroup$ – yasmar Sep 24 '11 at 14:01

it seems to have defined just another derivation on $T_p(N)$

No, we have defined a derivation on $T_{f(p)}M$, which is a device that takes elements of $C^\infty(M)$ and outputs a number (subject to certain rules). It works in two steps:

  1. Given a function on $ M$, pull it back to $N$ via composition with $f$.
  2. Apply the existing derivative to this composition; the result is a number.

Without looking at the internal structure of this device, all we see is: functions on $M$ go in, numbers come out. So, it's a derivation on $M$.


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