Gaussian curvature

Definition 1

The Gaussian curvature is the product of the principal curvatures:

\begin{matrix} (1) & K = κ_{1} \cdot κ_{2} \end{matrix}

In this sense, it is the determinant of the shape operator.
Picture from @needham2021visual page 136 (he calls it $K_{e x t}$ , and it is usually known as extrinsic curvature. If we are not in a flat ambient manifold, it doesn't have to be the same as the intrinsic curvature of the other definitions).
Pasted image 20220415084807.png

Definition 2

It is the local area magnification of the Gauss map
Pasted image 20220415085919.png
(@needham2021visual page 132)

Sometimes it is called extrinsic curvature.

Definition 3

At a point $p$ , $K (p)$ can be defined as the local angular excess per unit area (@needham2021visual page 18):

K (p) = lim_{Δ_{p} \to p} \frac{E (Δ_{p})}{A (Δ_{p})}

Here $E (Δ)$ is the sum of the angles of the triangle $Δ$ minus $π$ . This is rather informal, but I think it can be formalized as the angle defect of parallel transport about a geodesic triangle. It is related to Harriot theorem. Also related to the local Gauss-Bonnet theorem.

Equivalences of definitions (incomplete)

See this question and this answer in mathoverflow.

Proving that definition 1 and definition 2 are equivalents is easy. A visual aid is in @needham2021visual page 134
To visualize the equivalence of definition 2 and 3:

There is a physical "proof" of this fact which I learned from Mark Levy; it is in his book "THE MATHEMATICAL MECHANIC: Using Physical Reasoning to Solve Problems".
Imagine that you keep the axis of a bicycle wheel and move it in such a way that the bicycle wheel lies in the tangent plane to the surface. In the initial position the wheel stays still; you go along a loop in the surface and stop. After that your wheel rotates by some angle α with respect to the initial position. If your loop was triangular this angle is its defect; this is definition 3.
The parallel motion does not rotates the wheel, so the same result will be the same if you only rotate the axis without moving the center of the wheel. This tells you that α depends only on the spherical image of the loop and from here it is easy to see that it is proportional to the algebraic area of the domain bounded by the spherical image of the loop. I.e., the area of the sphere that Gauss map traces out on a region; this is definition 2.

Equivalence between 1 and 3 is done, visually, in @needham2021visual page 256.

Facts

To compute it we can use the first fundamental form and the second fundamental form. Coming from notation here, we have

\begin{aligned} K & = \det (F_{1}^{- 1} F_{2}) \\ = \det {(F_{1})}^{- 1} \det (F_{2}) \\ = \frac{L N - M^{2}}{E G - F^{2}} . \end{aligned}

It is not strange that we use the second fundamental form if we take into account Definition 1 and Definition 2 (extrinsic definitions).

But surprisingly enough, it does not depend on the second fundamental form, is an intrinsic property! This can be observed in definition 3. This was stated by Gauss in his Theorema Egregium. This can also be concluded by (or it is used in) the local Gauss-Bonnet theorem.
Therefore, it can be computed only with the first fundamental form. In an orthogonal coordinate system (i.e., $F = 0$ ) we have:

K = - \frac{1}{2 \sqrt{E G}} (\frac{\partial}{\partial u} \frac{G_{u}}{\sqrt{E G}} + \frac{\partial}{\partial v} \frac{E_{v}}{\sqrt{E G}})

Following notation introduced here it can be computed with

K = - \frac{1}{A B} (\partial_{v} [\frac{\partial_{v} A}{B}] + \partial_{u} [\frac{\partial_{u} B}{A}])

This is called the metric curvature formula. It is shown in @needham2021visual page 268, by using the holonomy#Local chart and the fact that the curvature is ultimately equal to the holonomy per unit area. But it also appears in @needham2021visual page 452 by using Cartan's method of moving frames.

Gaussian curvature has a simpler expression in isothermal coordinates.
Non orthogonal coordinates. According to Wikipedia page, we have the Brioschi formula

K = \frac{| \begin{array}{ccc} - \frac{1}{2} E_{v v} + F_{u v} - \frac{1}{2} G_{u u} & \frac{1}{2} E_{u} & F_{u} - \frac{1}{2} E_{v} \\ F_{v} - \frac{1}{2} G_{u} & E & F \\ \frac{1}{2} G_{v} & F & G \end{array} | - | \begin{array}{ccc} 0 & \frac{1}{2} E_{v} & \frac{1}{2} G_{u} \\ \frac{1}{2} E_{v} & E & F \\ \frac{1}{2} G_{u} & F & G \end{array} |}{{(E G - F^{2})}^{2}}

It is related to geodesics by means of Jacobi equation.
The total curvature in a region $Ω$ coincides with the holonomy of the loop $\partial Ω$ . See @needham2021visual page 246

Visualizations:

Angular excess

I think that from the local Gauss-Bonnet theorem can be deduced, in a constant curvature surface:

K = \frac{E (Δ)}{A (Δ)}

where $Δ$ is a triangle made of geodesics, $E (Δ)$ is the angular excess $\sum α_{i} - π$ , and $A (Δ)$ is the area of this triangle. Imagine a triangle of unit area painted on a sphere, it will have an angular excess given, visually, by the non-straightness of the edges. If we inflate the sphere we obtain a surface with less curvature, and a triangle with the same area will be then, relatively, smaller, and the edge will look more straight (less angle excess)
Pasted image 20211214190134.png

If the surface is not of constant curvature for a point $p$ :

K_{p} = lim_{Δ_{p} \to p} \frac{E (Δ_{p})}{A (Δ_{p})}

where $Δ_{p} \to p$ means that we have a succession of triangles shrinking down toward $p$ . See [Needham 2021].

Pasted image 20211214190231.png

This can be generalized to $n$ -gon with

E (n - g \circ n) \equiv [angle sum] - (n - 2) π,

Area excess

From Wikipedia.
Gaussian curvature is the limiting difference between the area of a geodesic disk and a disk in the plane

K = lim_{r \to 0^{+}} 12 \frac{π r^{2} - A (r)}{π r^{4}}

The particular case of a sphere: @needham2021visual page 20. Proof in general: @needham2021visual page 278.

Length excess

Gaussian curvature is the limiting difference between the circumference of a geodesic circle and a circle in the plane

K = lim_{r \to 0^{+}} 3 \frac{2 π r - C (r)}{π r^{3}}

The particular case of a sphere: @needham2021visual page 20, again. Proof in general: @needham2021visual page 278.

Relation to the curvature of a connection

What follows is a very informal annotation.

Consider a surface $S$ and the tangent space $T S$ . The Riemannian metric of $S$ , i.e., the first fundamental form, let us define the Levi-Civita connection which, at the end of the day, is a distribution on $T S$ , that is, a subbundle of $T (T S)$ . How is this? Well, given a vector $v \in T_{p} S$ , parallel transport let us define two "infinitesimal curves" in $T S$ leaving $v$ , being one of them the parallel transport along the first coordinate direction $x$ and the other the parallel transport along the second coordinate direction $y$ . These two curves are transversal to $V$ , the vertical subbundle of $T (T S)$ , because we are leaving $p$ , and let us take their tangent vectors to define a subvector space of $T_{v} T S$ : the horizontal space of the distribution-connection, in such a way that $T (T S) = V \oplus H$ .
Pasted image 20211229094631.png
The curvature of this connection (in the sense of performing the bracket of the vector fields of the horizontal distribution and then project to the vertical distribution) should be the same as the Gaussian curvature of the surface or, at least, be directly related. For example, observe that if the surface has non-zero Gaussian curvature, then parallel transports do not, in general, commute. This corresponds to the fact that $H$ does not posses integral manifolds, because if we leave $v$ travelling along curves tangent to $H$ (parallel transports) we may arrive to a different $v^{'} \in T_{p} S$ . In other words, flat surfaces (in the sense of Gaussian curvature) let us create copies of it in every $v$ , so we can foliate $T S$ with these copies.

More formally

In a Riemannian surface $(S, g)$ consider the Levi-Civita connection $\nabla$ corresponding to the metric $g$ . Suppose we have an orthonormal frame ${e_{1}, e_{2}}$ with dual coframe ${ω^{1}, ω^{2}}$ . We know that $\nabla$ is described here by the connection forms

Θ = (\begin{matrix} 0 & - T_{12}^{1} ω^{1} - T_{12}^{2} ω^{2} \\ T_{12}^{1} ω^{1} + T_{12}^{2} ω^{2} & 0 \end{matrix}),

where $d ω^{1} = T_{12}^{1} ω^{1} \land ω^{2}$ and $d ω^{2} = T_{12}^{2} ω^{1} \land ω^{2}$ (see here).

The curvature of the connection is a matrix of 2-forms $Ω = d Θ$ . Indeed, by mere computations we arrive to

Ω = - (e_{1} (T_{12}^{2}) - e_{2} (T_{12}^{1}) + (T_{12}^{1})^{2} + (T_{12}^{2})^{2}) (\begin{matrix} 0 & ω^{1} \land ω^{2} \\ - ω^{1} \land ω^{2} & 0 \end{matrix}) .

Gauss' Equation says that

d Θ_{2}^{1} = K ω^{1} \land ω^{2}

Then $d Θ_{2}^{1} (e_{1}, e_{2}) = - (e_{1} (T_{12}^{2}) - e_{2} (T_{12}^{1}) + (T_{12}^{1})^{2} + (T_{12}^{2})^{2})$ is indeed the Gaussian curvature of the surface.