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?
4 Answers
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 saddleshaped 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:
For more nice illustrations, see for example these two articles.
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3I quite like the paper demonstration. :) – J. M. ain't a mathematician Oct 06 '11 at 02:19

7This answer is awesome. – Matt Montag Oct 06 '11 at 03:31

2Other nice illustrations can be found in the ["Oddest book title of the year"winning](http://www.telegraph.co.uk/culture/books/bookprizes/7520047/CrochetingAdventureswithHyperbolicPlaneswinsoddestbooktitleaward.html) book [Crocheting Adventures with Hyperbolic Planes](http://www.amazon.com/CrochetingAdventuresHyperbolicPlanesTaimina/dp/1568814526). – Hans Lundmark Oct 06 '11 at 07:59

1Nice explanation! – johngreen May 12 '16 at 06:22

1Wonderful explanation! I like this style very much! 精彩的评论！ – Tom Dec 08 '16 at 15:52
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 2ndorder Taylor expansion) we can make rigorous the notion of being "locally like" an elliptic paraboloid, a cylinder, or a hyperbolic paraboloid.
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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. – t.b. Oct 06 '11 at 07:00

@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? – Jesse Madnick Oct 06 '11 at 07:03

2I 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. – t.b. Oct 06 '11 at 07:04

Oh, interesting! I've never heard that definition before. I usually think about these things in terms of the Darboux frame, and hence in terms of geodesic and normal curvatures. – Jesse Madnick Oct 06 '11 at 07:05

I see. What I was alluding to is quite well explained on the [Wikipedia page on principal curvature](http://en.wikipedia.org/wiki/Principal_curvature). – t.b. Oct 06 '11 at 07:10

@t.b.: I don't think Wikipedia is correct in this case. I think that is only true when the curve in question is a geodesic. – Jesse Madnick Oct 06 '11 at 07:14

2It 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. – t.b. Oct 06 '11 at 07:17

Sure, but the yellow curve has zero normal curvature, so its osculating circle should have... infinite radius? Like I said, that can't be right. I believe your formula is only true for geodesics. – Jesse Madnick Oct 06 '11 at 07:19

1Yes, 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 :) – t.b. Oct 06 '11 at 07:29
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.
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1I 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? – Oct 26 '16 at 13:45

1My description above parallels the description in my own book, [*Geometric Folding Algorithms: Linkages, Origami, Polyhedra*](http://www.gfalop.org/). But that doesn't go much deeper than above. I'm not quite sure which are the best sources; sorry. – Joseph O'Rourke Oct 26 '16 at 17:09

Please interpret my comment as a followup 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](http://www.ams.org/publicoutreach/featurecolumn/fcarcsphericon2)  think inflated balloon. However, the [BertrandDiguePuiseaux thrm](https://en.wikipedia.org/wiki/Bertrand–Diguet–Puiseux_theorem) 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. – Antoni Parellada Apr 22 '20 at 17:48

@AntoniParellada: Maybe this will help? [Area of a circle on sphere](https://math.stackexchange.com/a/1832201/237). – Joseph O'Rourke Apr 22 '20 at 19:14

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](https://math.stackexchange.com/a/1300712/152225). – Antoni Parellada Apr 22 '20 at 21:02
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.
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