Why no flat metric on a sphere?

What is a simple proof that there is no flat metric on a sphere, but there is a flat metric on a torus?

Ideally such a proof would clearly distinguish between the sphere and the torus. Naively one might think neither can be endowed with a flat metric. I'm hoping for some insight that reveals this naive view to be incorrect.

• Is Gauss-Bonnet "simple"? – Neal Jun 8 '20 at 13:37
• In dimension 2, the Gauss-Bonnet formula states that if $S$ is a compact surface with Gauss curvature $k$, then $\int_S k = 2\pi\chi(S)$, where $\chi(S)$ is the Euler characteristic of the surface $S$. Thus, a surface that can have a flat metric is of Euler characteristic $0$. – Didier Jun 8 '20 at 13:44
• Regarding the torus, the standard way to exhibit a flat metric is to think of it as a quotient of $\mathbb{R}^2$ by isometries. However, if one prefers a "less abstract" way of getting a flat metric on a torus, one can think of the induced metric on $S^1 \times S^1$ from the standard embedding in $\mathbb{R}^4$. The intuition tells us that by trying to think of a torus as embedded in $\mathbb{R}^3$, we won't be able to make the major radius be equal to the minor radius but if we are ready to go up one dimension, the situation becomes more symmetric and we can embed it in such a way that both – levap Jun 8 '20 at 13:49
• radii are equal and then verify directly that this is a flat metric. – levap Jun 8 '20 at 13:49
• Here's an idea of an approach I have not carried out: Show the total curvature of a compact surface is constant under deformations of the metric, without appealing to any topology. – Phillip Andreae Jun 8 '20 at 14:15

As the other answer also states, a simple proof that a flat metric on the torus is possible can be seen by pushing forward the flat metric on $$\mathbb{R}^2$$ by the cover $$\pi: \mathbb{R}^2 \to S^1 \times S^1$$, which can be done since the metric in $$\mathbb{R}^2$$ is invariant by translations.

An alternative to Gauss-Bonnet to see that the sphere admits no flat metric is by using that a flat metric on a simply connected manifold has trivial holonomy, which allows us to define a global non-vanishing section of the tangent bundle by taking the parallel transports of some chosen non-zero tangent vector, a contradiction with the hairy-ball theorem.

• I like this reliance on holonomy! Thanks. – Joseph O'Rourke Jun 8 '20 at 17:37

The torus is the quotient of $$\mathbb{R}^2$$ by $$f(x,y)=(x+1,y); g(x,y)=(x,y+1)$$ and $$f$$, $$g$$ preserves the standard Euclidean metric of $$\mathbb{R}^2$$ and induce a metric on $$\mathbb{T}^2$$.

A flat metric on $$S^2$$ induces an atlas on $$S^2$$ whose coordinate change are affine transformations. To see this, remark that we can find an atlas $$(f_i:U_i\rightarrow\mathbb{R}^2)$$ such the restriction of the metric to $$U_i$$ is flat or equivalently is the pullback of the Euclidean metric of $$\mathbb{R}^2$$ by $$f_i$$. The coordinate change of this atlas preserve the Euclidean metric of $$\mathbb{R}^2$$ thus are affine transformations. This implies that the tangent bundle of $$S^2$$ is flat and this is not true since the Euler characteristic of $$S^2$$ is not zero

Another way to show that This implies is to remark that given an affine atlas on $$S^2$$ one can extends a chart defines by this atlas to a local diffeomorphism (The developing map) $$D:S^2\rightarrow \mathbb{R}^2$$ since $$S^2$$ is simply connected. The standard way to do that if to fix $$x_0\in S^2$$, with an affine chart $$(U_0,f_0)$$ which contains $$x_0$$, for every element $$x\in S^2$$ consider a path $$c:I\rightarrow S^2$$ such that $$c(0)=x_0, c(1)=x$$ and a subdivision $$[t_1,t_2],...[t_{n-1},t_n=1]$$ such that $$c([t_j,t_{j+1}])\subset$$ a chart $$(f_j:U_j\rightarrow \mathbb{R}^2$$ and to set $$D(x)=g_1...g_nf_n(x)$$ where $$g_j=f_{j-1}\circ f^{-1}_j$$, this is well defined since $$S^2$$ is simple connecetd. this is impossible since $$S^2$$ is compact.

Certainly one can cite Gauss-Bonnet. Let $$K$$ denote the Gaussian curvature of a metric. As the sphere's Euler characteristic is $$2$$, any metric must have $$2 = \frac{1}{2\pi}\int_{\mathbb{S}^2} K\ dV$$ so no metric can have Gaussian curvature $$K$$ identically zero.

Meanwhile one can construct a flat metric on the torus by exhibiting it as a quotient of $$\mathbb{R}^2$$ by a lattice of isometries.

I do not recall any proof that does not appeal to Gauss-Bonnet but would love to read some in other answers. If the audience is not familiar with Gauss-Bonnet, perhaps it would be illustrative to walk through the proof of Gauss-Bonnet, inducing the audience to think carefully about triangulations and how one can turn the integral of a smooth local quantity into a global counting problem.

Edit 2: I take back what I said earlier about not recalling an alternate proof. Any space that has a flat metric has a universal cover with a flat metric. By the classification of simply connected constant-curvature space forms, any simply connected flat space is isometric, hence homeomorphic, to Euclidean space. As the sphere is simply connected and not homeomorphic to Euclidean space it cannot admit a flat metric. (thank you Aloizio Macedo for pointing out my original convolutions)

Lee Mosher points out below that one can in fact rely on the Cartan-Hadamard theorem instead of the classification of space forms.

I don't think the classification of space forms relies on Gauss-Bonnet but I'm not certain of this. I am more confident that Cartan-Hadamard does not rely on Gauss-Bonnet.

• Thanks for this nice perspective. It would be useful to see a path that does not directly invoke G.-B. But maybe such a path would necessarily include G.-B. in a disguised form. – Joseph O'Rourke Jun 8 '20 at 14:01
• @JosephO'Rourke See my edit, I think we can actually appeal to the classification of constant-curvature simply connected spaces. – Neal Jun 8 '20 at 14:12
• One can also replace the classification of simply connected constant curvature space forms with a simpler differential geometry result, namely the Cartan-Hadamard Theorem: a simply connected Riemannian $n$-manifold whose sectional curvatures are all $\le 0$ is diffeomorphic to $\mathbb R^n$. – Lee Mosher Jun 8 '20 at 14:14
• @Neal By appealing to the classification of constant-curvature simply connected spaces, why talk about the universal cover (which is $S^2$ itself) and not apply it directly to $S^2$? – Aloizio Macedo Jun 8 '20 at 14:15
• Thank you for the suggestions! – Neal Jun 8 '20 at 14:30