I am following these notes https://webspace.science.uu.nl/~ban00101/lecnotes/lie2002prereq.pdf (Tangent vectors are defined using equivalent clases of curves) and on page 4 it says that the map $T_xf $ is linear which should follow from lemma 1.1 and corollary 1.2 which are:
I am having trouble to see why it is so. According to the diagram:
$T_xf = T_{f(x)}\circ D(\psi \circ f \circ \chi^{-1})(\chi(x))\circ T_x\chi$
From Corollary 1.2 , $T_x \chi$ and $T_{f(x)} \psi$ are linear. They act on tangent vectors like this:
if $\gamma'(0) \in T_x X$, then $T_x \chi (\gamma'(0) ) = \frac{d}{dt}(\chi \circ \gamma)(0)$
I don't think we need lemma 1.1. (do we?) Moreover it looks to me that $D(\psi \circ f \circ \chi^{-1})(\chi(x))$ is a constant function because it is a function (the derivative of the transition map) between euclidean spaces evaluated at a fixed point , $(\chi(x))$ and therefore it is not linear. What am I doing wrong? Maybe this is not a constant function, but then how does it acts on its arguments?
It looks to me that the statement in yellow is wrong too, shouldn't that be a multiplication rather that a composition? (maybe this is part of the problem? But that seems to be only part of the proof of corollary 1.2, I interpreted it as multiplication so that for any two charts $\chi$ and $\psi$ differ by a multiplicative constant factor)
Can someone shed some light?
Edit:
Edit 2
In page 3 of the notes it is defined the following notation : $T_x\chi c'(0) = \frac{d}{dt}[\chi\circ c](0)$. Using this equation $(2)$ can be rewritten as:
$T_x\psi c'(0)=D(\tau)(\chi(x))T_x\chi c'(0)$.
And just as for instance when I have function $f$ and $g$, $f(x)=g(x) \forall x$ in their domains I can write $f=g$, In the above example I can leave out the argument $c'(0)$ and write:
$T_x\psi =D(\tau)(\chi(x))T_x\chi $
which should be the expression in yellow were it not by that $\circ$ that is mysteriously popping out of nowhere, which I think I am proving it should be a multiplication
Furthermore the equation in (2) must be a multiplication between $D(\tau)(\chi(x))$ and $\frac{d}{dt}\chi \circ c(t)|_{t=0}$ because a multiplication is what I get from applying the chain rule and equation (1) in the notes. Meaning that my proof of this...
is as follows:
$\frac{d}{dt}\psi \circ c(t)|_0=\frac{d}{dt}\psi\circ\chi^{-1}\circ \chi \circ c(t)|_0 =\frac{d}{dt}\tau \circ \chi \circ c(t)|_0 = D(\tau)(\chi(c(0)))\frac{d}{dt}\chi \circ c (t)|_{t=0} = D(\tau)(\chi(x))\frac{d}{dt}\chi \circ c (t)|_{t=0}$
The author considers charts $\chi : U \to U'$ with $x \in U$ and $\psi : V \to V'$ with $f(x) \in V$ such that $f(U) \subset V$. This yields a differentiable map $\psi \circ f \circ \chi^{-1} : U' \to V'$ between open subsets of Eucidean spaces. At each point $q \in U'$ it has a derivative denoted by $D(\psi \circ f \circ \chi^{-1} ) \mid_q$ or $D(\psi \circ f \circ \chi^{-1})(q)$. This is a linear map $\mathbb R^n \to \mathbb R^m$. It is in general not constant.
If you accept that diagram $(3)$ commutes, then you get $$T_xf = (T_{f(x)}\psi)^{-1} \circ D(\psi \circ f \circ \chi^{-1})(\chi(x))\circ T_x\chi .$$ Here it is essential that $T_{f(x)}\psi$ is a linear isomorphism as proved in Lemmas 1.1 (bijectivity) and 1.2 (linearity).
Let us now recall some fact from multivariable calculus.
Given a map $F : U' \to V'$ between open subsets $U' \subset \mathbb R^n$ and $V' \subset \mathbb R^m$, the derivative of $F$ at a point $p \in U'$ (if it exists) is linear map $DF(p) : \mathbb R^n \to \mathbb R^m$. One also writes $DF \mid_p$ for this map. We can represent the linear map $DF(p)$ by the Jacobian matrix $JF(p) = JF \mid_p = \begin{pmatrix} \dfrac{\partial F_i}{\partial x_j}(p) \end{pmatrix}$ of $F$ at $p$. If we understand elements of Euclidean spaces as column vectors, we get $DF \mid_p(v) = JF \mid_p \cdot v$, where $\cdot$ denotes matrix multiplication.
If $n =1$, then $DF \mid_p$ is a linear map $\mathbb R \to \mathbb R^m$. It is uniquely determined by the value $DF \mid_p(1) \in \mathbb R^m$ which is nothing else than the "ordinary derivative" $F'(p) = \dfrac{dF}{dt}(p)$ with components $F'_i(p) = \dfrac{dF_i}{dt}(p)$.
Given another map $G : V' \to W'$, the chain rule says that $D(G \circ F)(p) = DG(F(p)) \circ DF(p)$ or $D(G \circ F) \mid_p = DG \mid_{F(p)} \circ DF \mid_p$. The symbol $\circ$ denotes composition of functions. In terms of Jacobians this says $J(G \circ F) \mid_p = JG \mid_{F(p)} \cdot JF \mid_p$. Moreover, if $n =1$, the chain rule is equivalent to $$\dfrac{d(G \circ F)}{dt}(p) = D(G \circ F) \mid_p(1) = (DG \mid_{F(p)} \circ DF \mid_p)(1) = DG \mid_{F(p)}(DF \mid_p(1)) \\= DG \mid_{F(p)}(\dfrac{dF}{dt}(p)) .$$
Let us come to the yellow statement.
For each chart $\chi : U \to U'$ with $x \in U$ the author defines $$T_x\chi :T_xM \to \mathbb R^n, T_x\chi(c'(0)) = \frac{d (\chi \circ c)}{dt}(0) = \frac{d (\chi \circ c)}{dt} \mid_{t=0}.$$ Here $c'(0)$ denotes the equivalence class $[c] \in T_x M = \mathcal C_x/\sim$ of the curve $c$, it is not a derivative in the usual sense. Such only exist for curves with range $\mathbb R^n$. For example, $\chi \circ c$ is such a curve and we could write $(\chi \circ c)'(0)$ instead of $\frac{d (\chi \circ c)}{dt}(0)$. I guess the autors wants to avoid this because of his convention that such expressions denote equivalence classes of curves.
If $\psi : V \to V'$ is another chart with $x \in V$, then we have the transition function $\tau = \psi \circ \chi^{-1}$ which is differentiable. Then $\tau \circ (\chi \circ c) = \psi \circ c$ and the chain rule gives $$D(\psi \circ c) \mid_{t=0} = D\tau \mid_{\chi(c(0))} \circ D(\chi \circ c)\mid _{t=0} = D\tau \mid_{\chi(x)} \circ D(\chi \circ c)\mid _{t=0}$$ or $$\frac{d(\psi \circ c)}{dt}\mid_{t=0} = D\tau \mid_{\chi(x)}\left(\frac{d(\chi \circ c)}{dt}\mid _{t=0}\right) .$$ This is the formally correct version of $(2)$. The author writes it a bit sloppy as $\frac{d(\psi \circ c)}{dt}\mid_{t=0} = D\tau \mid_{\chi(x)}\frac{d(\chi \circ c)}{dt}\mid _{t=0}$. Anyway, inserting the definitions of $T_x\chi, T_x\psi$ we get $$T_x\psi(c'(0)) = D\tau \mid_{\chi(x)}(T_x\chi(c'(0))) = (D\tau \mid_{\chi(x)} \circ T_x\chi)(c'(0)) .$$ Without the argument $c'(0)$ this reads as $$T_x\psi = D\tau \mid_{\chi(x)} \circ T_x\chi = D\tau(\chi(x)) \circ T_x\chi .$$