# Does every manifold admit a *flat* Riemannian metric?

It is well-known that every smooth manifold $$M$$ admits a Riemannian metric, and the construction is straightforward. Let $$\left\{U_\alpha, \phi_\alpha\right\}$$ be a smooth atlas on $$M$$ admitting a partition of unity $$\left\{\psi_\alpha\right\}$$. On each coordinate chart $$U_\alpha$$, we can define a Riemannian metric $$g_\alpha$$ locally by setting $$g_\alpha = \phi_\alpha^*\overline g$$, where $$\overline g$$ is the Euclidean metric on $$\mathbb R^n$$. Using the partition of unity, we can construct a global Riemannian metric by setting $$g = \sum_\alpha \psi_\alpha g_\alpha$$.

My first question is: Is this metric flat?

I believe the answer is no. A metric is flat iff for each point there is a coordinate chart that is locally isometric to Euclidean space. However, it is not obvious to me that the above constructed metric is isometric to Euclidean in the overlaps of different coordinate charts. A counter-example to this could be $$M = \mathbb R^2$$, and your atlas could be $$U_1 = \mathbb R^2$$, $$\phi_1 = \mathrm{Id}$$, and $$U_2 = \left\{ (x,y) : x > 0\right\}$$, $$\phi_2(x,y) = (\log x, y)$$, with partition of unity $$\psi_1 \equiv \psi_2 \equiv \frac 1 2$$. Since $$\phi_2^* \overline g = x^{-2} dx^2 + dy^2$$, the metric you would get from this patching in the right half-plane would be $$g = \frac 1 2 \left(1+x^{-2}\right)dx^2 + dy^2$$, which is clearly not flat.

My second question is: Given an atlas $$\{U_\alpha, \phi_\alpha\}$$ of a smooth manifold $$M$$, does there exist a Riemannian metric on $$M$$ for which every chart $$\phi_\alpha : U_\alpha \to \mathbb R^n$$ is a local isometry?

This question seems harder to answer, and it seems unlikely, but an obvious counter-example to this eludes me. (But a paper I'm reading seems to propose such a construction, but it skips many details so I'm skeptical that I'm interpreting it correctly.)

My third question is: Weaker than the previous question, does every manifold admit a flat metric?

No, most manifolds do not admit flat metrics. For instance, in the case of surfaces, the Gauss-Bonnet theorem says that for a closed Riemannian surface $$M$$, the integral of the curvature over the whole manifold is equal to $$2\pi\chi(M)$$. So, for any closed surface whose Euler characteristic is nonzero (i.e., any closed surface besides a torus or Klein bottle), any metric must have nonzero average curvature and in particular cannot be flat.
• Is this the only obstruction to having a flat metric? That is, does every connected closed manifold $M$ such that $\chi(M) = 0$ admit a flat metric? Commented May 24, 2020 at 19:07
• @pyon: No. For instance, if a connected closed manifold has a flat metric, its universal cover is $\mathbb{R}^n$, so its higher homotopy groups must be trivial. So for instance, $S^3$ has no flat metric even though $\chi(S^3)=0$. Commented May 24, 2020 at 19:45
• @pyon: In some sense, yes. To be more precise, every complete connected flat manifold is a quotient of $\mathbb{R}^n$ by a properly discontinuous group of isometries. Commented May 25, 2020 at 14:31