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What ideas/formulas are required to solve this? Exercise: If a and b are two geodesics from point p to q, how do you prove that M is not simply connected? M is a regular surface in R3 and has negative Gaussian curvature K. Points p & q are two distinct points on surface M.

I know surface is simply connected if any simple closed curve can be shrunk to a point continuously in the set. So a sphere is simply connected, and a donut is not simply connected. And I know that since the curvature is negative, and since K=det(Sp)=w*z, so eigenvalues w and z have opposite signs. I also have the formula K=(eg-f^2)/(EG-F^2) based on the 1st fundamental forms and derivatives for solving for E,F,G,e,f,g. Not sure what if anything that is used to solve. And the only geodesic equations I can think of off hand are

d/dt(E u' + F v')=1/2[E_u u'^2 + 2F_u u'v' +G_u v'^2 ], d/dt(Fu'+Gv')=1/2[E_v u'^2 +2F_u u'v' +G_v v'^2 ], and geodesic curvature c=, and Gauss-Bonnet integrals but not sure how that can be applied. If they can be applied, please let me know. Maybe the fact that the normal of any point on the geodesic arc also is on the same normal vector to the surface M at that point helps? What ideas and equations are required to solve the exercise?

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You are calling $a$ and $b$ geodesics, but you are also calling $a$ and $b$ principal curvatures. It would be good to use different notations for these things. – treble Nov 1 '12 at 0:17
thanks @treble, i fixed it to say w and z for principal curvatures. maybe it is not even necessary for me to mention them at all since Will seems to have solved the exercise without them – user47735 Nov 1 '12 at 16:55

EDIT, Thursday evening: a good illustration for the bit about negative curvature is the inner circle in the "hole" in a torus of revolution, which is a "closed geodesic" and cannot be contracted continuously to a point. The closed geodesic occurs in the (inner) part of the torus where the Gauss curvature is negative, while the outer portion has positive curvature. Meanwhile, no compact surface in $\mathbb R^3$ can have negative Gauss curvature everywhere, for which Neal gave a simple answer in comments. It is, of course, possible to have noncompact surfaces with negative curvature, such as the catenoid of revolution. However, for any such surface of revolution with negative curvature, the absolute value of the curvature goes to $0$ as we get far from the central axis. So Mariano asked whether there are any infinite surfaces in $\mathbb R^3$ with Gauss curvature that is bounded away from $0,$ and the answer is no, not for a $C^2$ surface. Link to the question MARIANO QUESTION.

ORIGINAL: This is Gauss-Bonnet for polygons on an oriented surface. The familiar case should be this: the area of a geodesic triangle on the unit sphere is $\alpha + \beta + \gamma - \pi,$ where $\alpha, \beta, \gamma$ are the angles at the vertices of the triangle. Next, we define the external angles $\theta_1 = \pi - \alpha, \; \theta_2 = \pi - \beta, \; \theta_3 = \pi - \gamma, $ according to Figure 4-25 of do Carmo:


enter image description here


The unit sphere has curvature $K=1,$ so the integral of $K$ over a polygon is just its area. Furthermore, the boundary arcs are geodesics, with geodesic curvature $k_g = 0.$ So the statement about this area is equivalent to $$ \int \int_T \; K d \sigma \; + \theta_1 + \theta_2 + \theta_3 = 2 \pi. $$

Compare this with Gauss-Bonnet, formula (1) here:


enter image description here


You may or may not be told this: in the simply connected hyperbolic plane of constant curvature $-1,$ the area of a triangle is $ \pi - (\alpha + \beta + \gamma),$ where $\alpha, \beta, \gamma$ are the angles at the vertices of the triangle. However, because $K = -1,$ this is once again the same as $$ \int \int_T \; K d \sigma \; + \theta_1 + \theta_2 + \theta_3 = 2 \pi. $$

For your exercise, you are asked about an (oriented) "diangle" $D,$ with angles $0 \leq \alpha, \beta \leq \pi.$ If an angle is equal to $0,$ that is what the author calls a "cusp," see Figure 4-26, the right half. We take the external angles as before, and get $$ \int \int_D \; K d \sigma \; + \theta_1 + \theta_2 = 2 \pi, $$ $$ \int \int_D \; K d \sigma \; + (\pi - \alpha) + (\pi - \beta) = 2 \pi, $$ $$ \int \int_D \; K d \sigma \; = \alpha + \beta. $$ What does this mean? We have $K < 0.$ The right hand side of the equation is nonnegative. If there is any interior to the "diangle," the integral is strictly negative. The only legal possibility is that the two geodesics are identical, both vertices are cusps, the two geodesic arcs are exactly the same, once out, once back.

Otherwise, and here is where the topology comes in, the two geodesic arcs do not bound a piece of surface as I was assuming, and the loop cannot be deformed homotopically to a single point, thus the surface is not simply connected.

Well, I hope that works for you. There is quite a bit of detail that could be added about orientation. Furthermore, I do not think a closed surface in $\mathbb R^3$ can have negative Gauss curvature. I know that a closed surface in $\mathbb R^3$ cannot have constant negative Gauss curvature.

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A closed surface in $\mathbb{R}^3$ can have negative Gauss curvature at some points (e.g. on the line $s = 0$ of the embedding $(\cos{s}\cos{t},\cos{s}\sin{t},\sin{s})$ of $T^2\hookrightarrow\mathbb{R}^3$). A closed surface must have at least one point of positive Gauss curvature (enclose the surface in a sphere of minimal radius, which must be tangent to the surface at some point; the surface has positive curvature at that point). – Neal Nov 1 '12 at 2:56
@Neal, works for me. Right, the version I recall is the farthest point from the origin. – Will Jagy Nov 1 '12 at 2:58
But I'd guess, for some reason, that a closed surface (closed in that it is a closed subset of $R^3$) can be of negative curcature. I dunno! – Mariano Suárez-Alvarez Nov 1 '12 at 3:03
@MarianoSuárez-Alvarez, I do not see how, but my dissertation was in minimal surfaces, which is a special case and may be too restrictive. I'm thinking Ian Agol would know. If there is an example, it is awfully damn crinkly. – Will Jagy Nov 1 '12 at 3:23
I asked on MO for Ian to answer :-P – Mariano Suárez-Alvarez Nov 1 '12 at 3:35

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