Is the surface of an infinitely large sphere a plane?

Question

A pictorial analogy is the difference of the curvature between two points on the surface of a soccer ball and the earth. The curvature on the surface of the soccer ball is immediately noticeable, whereas this isn't the case when you're looking around in a wide, flat region with no hills and mountains on the earth's surface.

In a bit more mathematical terms let $r$ be the radius sphere and the curvature be defined as $\frac 1 r$. With $\lim_{r \to \infty} \frac 1 r$ the sphere becomes infinitely large such that the curvature converges to 0.

Does this mean that under this circumstances a sphere becomes a plane, or is my intuition just wrong?

Answer 1

Consider a sphere $S_R$ of radius $R>0$ in ${\mathbb R}^3$ touching the $(x,y)$-plane from above at $(0,0,0)$. The equation of this sphere is $$x^2+y^2+(z-R)^2=R^2\ .$$ Solving for $z$ as a function of $x$ and $y$ and taking the solution giving $z=0$ at $(x,y)=(0,0)$ we obtain $$z={x^2+y^2\over R+\sqrt{R^2-x^2-y^2}}={x^2+y^2\over 2R}+{\rm higher \ order\ terms}\ .$$ It follows that $z=0$ is a fine approximation to the spherical surface near $(x,y)=(0,0)$ when $R\gg1$.

Answer 2

You are right; a sphere is a differentiable manifold. Rather than take the radius to $\infty$, take the radius of your neighborhood to $0$. There is a diffeomorphism from a small neighborhood of a point on the sphere with the unit disc in $\Bbb{R}^2$, and in the limit that diffeomorphism approaches an isometry.