I've read that in a negatively curved space (hyperbolic space), the measured circumference of a circle is greater then the expected circumference. But I can't just imagine that virtually. Can anyone help?
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I think if you draw a (small, somewhat awkwardly distorted) circle on a saddle-type surface (the point $X$ in the image below), you will find that it has the desired property.
You can imagine to yourself that going from $A$ to $B$ in a circular path passing through point $C$ is a much longer path than the half-circumference of a cirlce with diameter $XC$. This is because of how the region between the lines $XB$ and $XC$ are "flayed out"
In a hyperbolic space, every point behaves locally like the saddle.
Not an answer, but very often, the behavior in hyperbolic space differs from Euclidean in the opposite way to the behavior on the sphere. Like: the sum of the angles of a triangle is greater than $\pi$ on the sphere, less in hyperbolic space. It’s the same with the circumference of a circle: on the unit sphere, if the your radius-segment (on the surface) is of length $\ell$, the distance (through the solid ball) between the ends of the diameter is $2\sin\ell$. (If you have trouble seeing this, slice the ball through this diameter, so that you’re talking about the ends of an arc of length $2\ell$.) The circle on the surface of the ball has genuine radius $\sin\ell$, so that the circumference is $2\pi\sin\ell$, always less than $2\pi\ell$.
Now, you expect that the same formula will work in the hyperbolic plane, except that you replace the trig function with a hyperbolic function. And indeed, the circumference of the circle now is $2\pi\sinh\ell$, always greater than $2\pi\ell$.