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Let $M$ and $N$ be smooth manifolds with or without boundary and let $F : M \to N$ is a smooth map. Show that $dF_p : T_pM \to T_{F(p)}N$ is linear.

Now what I'd try to normally do to dhow that some map $f : V \to W$ between vector spaces is linear is to show $f(v + w) = f(v) + f(w)$ and $f(cv) = cf(v)$ for $v, w \in V$ and $c \in \mathbb{F}$ the base field for the vector spaces.

So pick $v, w \in T_pM$ and $c \in \mathbb{R}$. $v$ and $w$ are derivations at the point $p \in M.$ But I'm not sure what $v + w$ would even look like, in the sense that I don't know how to manipulate it. I'm sorry if this is somethign really basic but I don't know what $dF_p(v + w) = ???$. The only fact I can conclude at the moment about that is that $dF_p(v+w) \in T_{F(p)}N$. So $dF_p(v+w) \in T_{F(p)}N$ is itself a linear map from $C^{\infty}(N) \to \mathbb{R}$. And sure $v, w : C^{\infty}(M) \to \mathbb{R}$ and both are linear maps. And sure the differential satisfies the following rule :

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And the operator $dF_p(v) : C^{\infty}(N) \to \mathbb{R}$ is linear, but I can't use any of those facts to show $dF_p : T_pM \to T_{F(p)}N$ is linear can I?

How can I show that $dF_p : T_pM \to T_{F(p)}N$ is linear.?

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Just do it: $$\begin{align} dF_p(v_1+\lambda v_2)(f) &= (v_1+\lambda v_2)(f \circ F) \\ &= v_1(f\circ F) + \lambda v_2(f\circ F) \\ &= dF_p(v_1)(f) + \lambda dF_p(v_2)(f) \\ &= (dF_p(v_1)+\lambda dF_p(v_2))(f).\end{align}$$

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  • $\begingroup$ But that's $dF_p(\alpha)$ 'acting' on $f$ for some $\alpha \in T_pM$. Why is is that to prove that $dF_p$ is linear we have to pick a smooth function $f \in C^{\infty}(N)$ , and show that $dF_p$ is linear in the sense of taking inputs of tangent vectors and 'acting' on other smooth functions on $N$? What I mean by this is why can't I just pick $v_1, v_2 \in T_pM$ and $\lambda \in \mathbb{R}$ and show $dF_p(v_1 + \lambda v_2) = dF_p(v_1) + \lambda dF_p(v_2)$ somehow? Like I would do for a normal linear map between vector spaces. $\endgroup$ – Perturbative Dec 6 '17 at 10:02
  • $\begingroup$ You want to prove that $dF_p(v_1+\lambda v_2) = dF_p(v_1)+\lambda dF_p(v_2) $ right? So it suffices to check that they act the same over any test function $f$. $\endgroup$ – Ivo Terek Dec 6 '17 at 10:04
  • $\begingroup$ Okay I get what you're saying now, so what you're saying is analgous to showing that two functions $f, g$ between sets $A, B$ are the same if and only if $f(x) = g(x)$ for $x \in A$ (i.e. their function evaluations for a given input are the same). In this case $dF_p(v_1 + \lambda v_2)$ and $dF_p(v_1) + \lambda dF_p(v_2)$ are both functions mapping $C^{\infty}(N) \to \mathbb{R}$, so to show that they are equivalent you have to pick a $f \in C^{\infty}(N)$ and show that $dF_p(v_1 + \lambda v_2)(f) = (dF_p(v_1) + \lambda dF_p(v_2))(f)$. $\endgroup$ – Perturbative Dec 6 '17 at 10:11
  • $\begingroup$ Yes, that's pretty much it $\endgroup$ – Ivo Terek Dec 6 '17 at 10:12

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