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.
On
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.
On
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.
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.