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Currently I'm studying differential geometry, and more specifically the Gauss map. I'm using the (in)famous Do Carmo Differential Geometry and Surfaces book. I'm having a hard time understanding the use of the Gauss map. I understand what it's doing, but not why. I also do not understand why this map has useful properties (as the second fundamental form can be derived from it). To give you an "insight" to my mind, this is how I see the map (correct me if I'm wrong)

My view of the Gauss map (not a formal definition)

The Gauss map maps the normal vector at each point of a curve to the unit sphere. Therefore, it provides a mapping from every point of the curves to the unit sphere.

So, why do we need this map? Why is this useful and why do we need this map to compute important properties such as the second fundamental form? For me, every curve would just be put into an unit circle. If I need to clarify my thinking, please tell me!

Maurice
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1 Answers1

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First, some results:

The second fundamental form is sometimes defined using the Gauss map, so that's an obvious reason that it's useful. It's used together with the first fundamental form in calculating things like Gaussian curvature. In particular, there is the following theorem:

Theorem: The Gaussian curvature $K$ of an oriented surface $M\subset\mathbb{R}^3$ is the Jacobian of its Gauss map.

This also allows you to calculate the total Gaussian curvature of an oriented surface in terms of its image under the Gauss map. So, the Gauss map is inherently tied into the calculation of an extremely important invariant. You might be interested in Barrett O'Neil's book on elementary differential geometry, especially pages 307-309. In this section, he proves the above result and provides some nice visualization.

Those are some results, so now let's think about the intuition (why it's related to these results). The pushforward of the Gauss map is the negative of the shape operator, which provides a mathematical description of the shape of a surface. In case you haven't seen it before, the shape operator is the self adjoint map $S_p:T_p(M)\rightarrow T_p(M)$ for a given $p\in M$ defined as follows: if $p\in M$, then for any $v\in T_p(M),$ $$S_p(v)=-\nabla_v U,$$ where $U$ is the unit normal vector field on a neighborhood of $p$ in $M$, and $\nabla$ is the covariant derivative. This describes the shape of the surface in the sense that we are measuring how the unit normal changes in the $v$ direction, which describes how the tangent planes are varying in the $v$ direction, hence describing the curving of the surface in $\mathbb{R}^3$ locally. This might make the Gauss map's connection to curvature more believable.

cmk
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  • I will read that, thanks! However, since it has theorems related to the map I see it is useful, but I cannot see why the map itself is useful. Also, I do not see why these kind of theorems necessarily depend on the Gauss map, isn't it as easy/difficult to formulate these theorems without such a map? – Maurice Jun 04 '19 at 20:19
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    Evidently, it's not as easy! The Gauss map describes a fairly fundamental sort of idea, so it isn't surprising that it's so useful. Here's another interpretation: the push forward of the Gauss map is the negative of the shape operator, which as its name indicates, described the "shape" of a surface in $\mathbb{R}^3$. – cmk Jun 04 '19 at 20:24
  • @R4000 I've added some stuff related to my post about how the Gauss map relates to curvature that you might find more useful! – cmk Jun 04 '19 at 20:38
  • Thanks for taking your time. I do however still think it is an unintuitive mapping, but it clarifies the relation to other theorems and applications :) – Maurice Jun 04 '19 at 20:48
  • Never mind, rereading it, I, also intuitively, understand it. Thanks for the elaborate answer – Maurice Jun 05 '19 at 12:21
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    @R4000 Awesome, I'm glad that it helped! – cmk Jun 05 '19 at 12:33