I am new to differential geometry and I am trying to understand Gaussian curvature. The definitions found at Wikipedia and Wolfram sites are too mathematical. Is there any intuitive way to understand Gaussian curvature?
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4 Answers
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For a intuitive understanding, imagine a flat sheet of paper (or just grab one in your hand). It has zero Gaussian curvature. If you take that sheet and bend it or roll it up into a tube or twist it into a cone, its Gaussian curvature stays zero.
Indeed, since paper isn't particularly elastic, pretty much anything you can do to the sheet that still lets you flatten it back into a flat sheet without wrinkles or tears will preserve its Gaussian curvature.
Now take that sheet and wrap it over a sphere. You'll notice that you have to wrinkle the sheet, especially around the edges, to make it conform to the sphere's surface. That's because a sphere has positive Gaussian curvature, and so the circumference of a circle drawn on a sphere is less than $\pi$ times its diameter. The wrinkles on the paper are where you have to fold it to get rid of that excess circumference.
Similarly, if you tried to wrap the sheet of paper over a saddle-shaped surface, you'd find that you would have to tear it (or crumple it in the middle) to make it lie on the surface. That's because, on a surface with negative Gaussian curvature, the circumference of a circle is longer than $\pi$ times its diameter, and so, to make a flat sheet lie along such a surface, you either have to tear it to increase the circumference, or wrinkle it in the middle to reduce the radius.
Indeed, in nature, plants can produce curved or wrinkled leaves simply by altering the rate at which the edges of the leaf grow as compared to the center, which alters the Gaussian curvature of the resulting surface, as in this picture of ornamental kale:
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For more nice illustrations, see for example these two articles.
answered Oct 6, 2011 at 1:25
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3$\begingroup$ I quite like the paper demonstration. :) $\endgroup$ Oct 6, 2011 at 2:19
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2$\begingroup$ Other nice illustrations can be found in the "Oddest book title of the year"-winning book Crocheting Adventures with Hyperbolic Planes. $\endgroup$ Oct 6, 2011 at 7:59
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1$\begingroup$ Wonderful explanation! I like this style very much! 精彩的评论！ $\endgroup$– TomDec 8, 2016 at 15:52
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I know you're looking for an intuitive explanation, but I've always believed that intuition ought to come from concrete mathematical facts, if possible. Otherwise, you have no way of knowing whether or not the intuition someone feeds you actually matches the formal mathematics. (On that note, +1 for Joseph O'Rourke's answer.)
The Gaussian curvature, $K$, is given by $$K = \kappa_1 \kappa_2,$$ where $\kappa_1$ and $\kappa_2$ are the principal curvatures. Just from this definition, we know a few things:
For $K$ to be a large positive number, then $\kappa_1$ and $\kappa_2$ should both be large and have the same sign (i.e. both positive or both negative).
For $K$ to be zero, either $\kappa_1 = 0$ or $\kappa_2 = 0$.
For $K$ to be a large negative number, then $\kappa_1$ and $\kappa_2$ should both be large but have opposite signs.
Now recall that $\kappa_1(p)$ and $\kappa_2(p)$ are the maximum and minimum normal curvatures of all curves passing through $p.$ So:
$K(p) > 0$ means that the curves through $p$ of extremal normal curvature "curve the same way" (such as the red curve and the green curve). So, points in the purple region have $K > 0$. In some sense, the surface is shaped like an elliptic paraboloid there (like a bowl).
$K(p) = 0$ means that one of the curves through $p$ of extremal curvature has zero normal curvature (such as the yellow curve). So, points along the yellow curve have $K = 0$. In some sense, the surface is shaped like a parabolic cylinder there (like a bent piece of paper).
$K(p) < 0$ means that the curves through $p$ of extremal curvature "curve in opposite ways" (such as the blue curve and green curve). For example, points in the gray region have $K < 0$. In some sense, the surface is shaped like a hyperbolic paraboloid there (a saddle).
In fact, using Dupin's Indicatrix (which is really just a 2nd-order Taylor expansion) we can make rigorous the notion of being "locally like" an elliptic paraboloid, a cylinder, or a hyperbolic paraboloid.
answered Oct 6, 2011 at 6:31
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$\begingroup$ Maybe it would be worth mentioning what the principal curvatures are ($1/r$ and such), I think this would help people who are not conversant with the technical terms understand what you are talking about. $\endgroup$– t.b.Oct 6, 2011 at 7:00
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$\begingroup$ @t.b.: What do you mean by "$1/r$ and such"? Do you mean the actual computations of the principal curvatures of the red/yellow/green/blue curves? $\endgroup$ Oct 6, 2011 at 7:03
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2$\begingroup$ I find "principal curvature = reciprocal of the radius osculating circle" the most intuitive way of thinking about the principal curvatures. That's what I meant by $1/r$ and such. No it's not about the computation, but what does it mean for the principal curvature to be large? The radius of the osculating circle has to be small. $\endgroup$– t.b.Oct 6, 2011 at 7:04
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2$\begingroup$ It is correct (I haven't learned it from there). There are two osculating circles (provided you're not at an umbilical point): one for each principal curvature. They intersect orthogonally, one is tangent to the direction where the surface curves the most and one tangent to the direction in which it curves the least. In your picture the circles indicated are the osculating circles, the curvatures are the inverses of their radii and the sign of the curvature is determined whether the center is in direction of the normal or in the opposite direction. $\endgroup$– t.b.Oct 6, 2011 at 7:17
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1$\begingroup$ Yes, sorry I messed up in my previous comment. You go on the geodesics in direction of the principal curvatures and draw the circles with radii corresponding to their curvature. It's a long time since I last thought about this stuff :) $\endgroup$– t.b.Oct 6, 2011 at 7:29
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One way to view the Gaussian curvature $K$ is as an area deficit, a comparison between the area $\pi r^2$ of a flat disk of radius $r$, to the area of a geodesic disk on the surface with intrinsic radius $r$. Let $A(r)$ be this latter area centered on a point of the surface. Then $$K = \lim_{r \rightarrow 0} \frac {12 (\pi r^2 - A(r) )}{\pi r^4}$$ You can see in the numerator the term $\pi r^2 - A(r)$ is exactly the area deficit. So at a point where $A(r)$ is smaller than the flat area, $K$ is positive.
This formulation was discovered by Diquet. There is a similar formulation based on a circumference deficit, due to Bertrand and Puiseux.
answered Oct 6, 2011 at 0:21
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1$\begingroup$ I love this answer. This was the approach I was looking for. My teacher encouraged me to understand the Gaussian curvature and geodesic curvature from this "area setting". Do you know any source where I can learn more about this? $\endgroup$– user378947Oct 26, 2016 at 13:45
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1$\begingroup$ My description above parallels the description in my own book, Geometric Folding Algorithms: Linkages, Origami, Polyhedra. But that doesn't go much deeper than above. I'm not quite sure which are the best sources; sorry. $\endgroup$ Oct 26, 2016 at 17:09
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$\begingroup$ Please interpret my comment as a follow-up quick question / clarification: The fine point to consider is that the stretching from the parametric Euclidean plane to the surface would naturally increase the area as in here - think inflated balloon. However, the Bertrand-Digue-Puiseaux thrm calls for "The geodesic circle of radius r centered at p is the set of all points whose geodesic distance from p is equal to r," which decreases area by "retracting" the radii. $\endgroup$ Apr 22, 2020 at 17:48
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$\begingroup$ @AntoniParellada: Maybe this will help? Area of a circle on sphere. $\endgroup$ Apr 22, 2020 at 19:14
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$\begingroup$ Thank you. Yes, by "retracting" the radii I was trying to convey the visual of a constant radius, $r,$ on the flat surface, bent along a geodesic on a curved surface in $\mathbb R^3.$ In the case when the surface is a sphere of radius $R,$ the area demarcated by $r$ as measured along the surface is going to be $2\pi R^2\left(1 - \cos(r/R)\right)$ as opposed to $\pi r^2,$ as nicely presented also here. $\endgroup$ Apr 22, 2020 at 21:02
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The Gaussian curvature is the ratio of the solid angle subtended by the normal projection of a small patch divided by the area of that patch.
The fact that this ratio is based totally on the definition of distance within the surface (independent of the embedding of the surface; that is, bending and twisting, etc.) is Gauss' Theorema Egregium.
