TITLE: Is there any easy way to understand the definition of Gaussian Curvature?
QUESTION [81 upvotes]: 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?

REPLY [36 votes]: 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.