# Reading 4 – A Quick and Dirty Introduction to Differential Geometry

Your reading assignment for Thursday is to read Chapter 2 of our course notes on Discrete Differential Geometry, “A Quick and Dirty Introduction to Differential Geometry.” This chapter provides a different perspective from the topological picture we’ve been studying in class: rather than slowly building everything “from the ground up” (sets, topology, smooth structure…) it skips immediately to the juicy stuff: curved surfaces with real geometry sitting in space. We’ll fill in everything “in between” in the next few lectures.

To test your own understanding, try computing (in explicit coordinates) the metric induced by the immersion
$f(u,v) := \frac{1}{u^2+v^2+1}(2u,2v,u^2+v^2-1),\qquad\qquad$
which maps the plane $$\mathbb{R}^2$$ to the unit sphere in $$\mathbb{R}^3$$. Do you notice anything special about this metric?

Submission: As usual, please send an email to kmcrane@cs.cmu.edu and nsharp@cs.cmu.edu no later than 10:00 AM on Thursday, February 4th including the string DDGSpring2016 in your subject line.  Your email for readings should always include:

1. a short (2-3 sentence) summary of what you read, and
2. at least one question about something you found confusing / interesting / incomplete / not addressed.

## 1 thought on “Reading 4 – A Quick and Dirty Introduction to Differential Geometry”

1. Keenan says:

Here’s a solution to the problem above: for an immersion $$f: \mathbb{R}^2 \to \mathbb{R}^3$$, for any two vectors $$X$$ and $$Y$$ the induced metric is given by
$g(X,Y) := \langle df(X), df(Y) \rangle,$
where $$\langle \cdot, \cdot \rangle$$ is the Euclidean inner product. We can think of $$df(X)$$ as either the pushforward of a tangent vector $$X$$ (i.e., how does the vector $$X$$ get stretched out as we go from $$\mathbb{R}^2$$ to $$\mathbb{R}^3$$), or as the good old-fashioned directional derivative with respect to the parameter $$X$$ (i.e., “how much does the position $$f$$ change if we move a little bit in the direction $$X$$?”). If we want to work out an explicit expression for the metric, we can work out these derivatives in a pair of orthogonal coordinate directions $$u$$ and $$v$$; all other derivatives are then easily obtained via linear combinations.

The derivative of the leading coefficient with respect to $$u$$ is given by
$\tfrac{\partial}{\partial u}(u^2+v^2+1)^{-1} = -2u(u^2+v^2+1)^{-2},$
and likewise
$\tfrac{\partial}{\partial v}(u^2+v^2+1)^{-1} = -2v(u^2+v^2+1)^{-2}.$
The partial derivatives of the vector term with respect to $$u$$ and $$v$$ are
$$(2,0,2u)$$ and $$(0,2,2u)$$, respectively.
Applying the product rule and simplifying yields
$df(u) = \frac{(2(1-u^2+v^2),-4uv,4u)}{(1+u^2+v^2)^2}$
and
$df(v) = \frac{-4uv,2(1+u^2-v^2),4u)}{(1+u^2+v^2)^2}.$
The entries of the metric are then
$\begin{array}{rcl} \langle df(u), df(u) \rangle &=& \frac{4}{(1+u^2+v^2)^2}, \\ \langle df(u), df(v) \rangle &=& 0 \\ \langle df(v), df(u) \rangle &=& 0 \\ \langle df(v), df(v) \rangle &=& \frac{4}{(1+u^2+v^2)^2}, \end{array}$
which we can also write in matrix form as
$I = \left[ \begin{array}{cc} \frac{4}{(1+u^2+v^2)^2} & 0 \\ 0 & \frac{4}{(1+u^2+v^2)^2} \end{array} \right].$
Or, if we let $$r := \sqrt{u^2+v^2}$$ be the distance to the origin, then the metric can be expressed as simply
$g(u,v) = \frac{4}{(1+r^2)^2}\langle u, v \rangle.$
Here it becomes clear that the immersion $$f$$ is \emph{conformal}, i.e., the only difference between the inner product on $$\mathbb{R}^2$$ and on the surface is a uniform scale factor at each point, which means that angles are preserved. This metric is the one induced by stereographic projection.