Tristan Needham Visual Differential Geometry,pg-32

In the beginning, I understood the metric as the factors by which the length of displacement on surface and on the plane relate. But, in the book the following formula is given and suggests to me that separation vectors on the surface are related to separation vectors in the plane:

$$ d \hat{s} = \lambda(z, \gamma)dz$$

With, $dz = e^{i \gamma} ds$

My doubt is that the separation vector on the surface is existing in the tangent plane at the base point $\hat{z}$ and is three dimensional, so how could it possibly be related to the complex separation vector $dz$ existing in the flat plane?

## Best Answer

$\newcommand{\Cpx}{\mathbf{C}}$The formula $$ d\hat{s} = \lambda(z, \gamma)\, dz $$ must be interpreted not as equating a spatial displacement with a scalar multiple of the "flat" displacement $dz$, but as an equality between an infinitesimal displacement $d\hat{s} = \hat{q} - \hat{z}$ in the tangent plane to the hemisphere at $\hat{z}$ and the image of the "flat" displacement $dz = q - z$ under the differential of (the inverse of) central projection from the plane to the hemisphere.

More precisely, if $F:\Cpx \to H$ denotes the inverse projection from the plane to the open hemisphere, then $\hat{z} = F(z)$ and (speaking infinitesimally) $$ \hat{z} + d\hat{s} = \hat{q} = F(q) = F(z + dz) = F(z) + (DF(z))(dz); $$ consequently, $d\hat{s} = (DF(z))(dz)$.

(The identification of $(DF(z))(dz)$ with a scalar multiple of $dz$ presumably represents the differential with respect to some basis.)