Differential of transition map.

Question

I'm a beginner in the calculation of differentials in smooth manifolds. I was thoughtful about the case of the transition map differential. For example, in the complex projective line $\mathbb C P_1$, the transition map $\phi_1 \circ \phi_2^{-1}$ is given by $\frac{1}{z}$. To calculate the differential of this map at an arbitrary point $x$ just perform the conventional derivative calculation on $\mathbb C$? That is, $d_x (\phi_1 \circ \phi_2^{-1}) = \frac{-1}{x^2}$, it seems to me that this makes sense, since $\phi_1 \circ \phi_2^ {-1}$ is a map of $\mathbb C$ to $\mathbb C$, and tangent spaces can be identified with $\mathbb C$. Is my reasoning correct?

Answer 1

Well, strictly speaking, the differentials (after identifying $T_x\Bbb{C}$ and $T_{(\phi_1\circ \phi_2^{-1})(x)}\Bbb{C}=T_{1/x}\Bbb{C}$ with $\Bbb{C}$ canonically) are linear maps $d(\phi_1\circ \phi_2^{-1})_x:\Bbb{C}\to\Bbb{C}$, so they can't be equal to the number $(\phi_1\circ\phi_2^{-1})'(x)=-\frac{1}{x^2}$. But they're very similarly related. $d(\phi_1\circ \phi_2^{-1})_x$ is the linear transformation $h\mapsto (\phi_1\circ\phi_2^{-1})'(x) \cdot h=-\frac{h}{x^2}$. So, the linear transformation is simply scalar multiplication by the usual complex derivative.

In other words, one can make an extra identification: $\text{Hom}(\Bbb{C},\Bbb{C})\cong \Bbb{C}$, where the isomorphism is $T\mapsto T(1)$. More generally, for any vector space $V$ over a field $\Bbb{F}$, we have $\text{Hom}(\Bbb{F},V)\cong V$ simply by $T\mapsto T(1)$.