Although this question arose out of a physics textbook, I believe it is better addressed on math stackexchange.
In Hartle's "Gravity", there was an exposition on differential geometry given. Initially, the text considered the following line element for the sphere:
$$ dS^2 = a^2 (d\theta^2 + \sin^2 \theta\ d\phi^2)$$
Later, the text considered the more general line element
$$ dS^2 = a^2 (d\theta^2 + f^2(\theta)\ d\phi^2)$$
It was computed that:
- the pole-to-pole distance is $a \int_0^\pi d\theta=\pi a$, which is independent of $f(\theta)$
- the circumference of of a circle at constant $\theta$ is $C(\theta)=\int_0^{2\pi} af(\theta)\ d\phi = 2\pi a f(\theta)$
The book then considered the example where $f(\theta)=\sin\theta (1-\frac{3}{4}\sin^2\theta)$. By thinking about how $C(\theta)$ varies (which is something like a smoothened M), the text then claimed that we can represent this geometry by a surface in $\mathbb{R}^3$ that looks like a peanut. See diagram below:
However, when I thought further about it, I reached a contradiction. If I fix $\phi$ and travel along the peanut surface from $\theta=0$ to $\theta=\pi$, I don't think that the line element satisfies $dS=a\ d\theta$ on the peanut surface. For one, it is clear that the peanut surface is not at distance $a$ away from the origin at all $\theta$ angles.
- Am I right to say that this drawing is a simplification to visualise $C(\theta)$ but in reality this geometry is not simply described by a surface like this in Euclidean $\mathbb{R}^3$?
- Bonus: What functions of $f(\theta)$ will lead to a geometry that can be represented by a surface in Euclidean $\mathbb{R}^3$? If I think about the lengths of paths along fixed $\phi$, it seems that only $f(\theta)=\sin \theta$ (corresponding to $S^2$) works.