# Flat manifold and metric

Is "X is a flat Riemannian manifold" equivalent to "for any metric g on X, there is a change of coordinates that can transform g to a tensor with only constants on its diagonal"?

• Hm. Flatness depends heavily on the choice of a metric, if I am not mistaken. A manifold which is flat with respect to a metric might not be flat with respect to another. So your question does not make sense. – Giuseppe Negro Jan 3 '13 at 3:39
• No, it means $X$ is already equipped with a metric with zero curvature. That is, "flat" is a condition on Riemannian manifolds, not a condition on manifolds. – Qiaochu Yuan Jan 3 '13 at 3:39
• I found that in a seemingly reputable astronomy book. "If it is possible to find a frame in which all the components of g are constant, the space is 'flat'". – Frank Jan 3 '13 at 3:47
• Maybe let's talk only about Riemannian manifolds. I'll update the question. – Frank Jan 3 '13 at 3:49
• Maybe what you meant was: Is “$(M,g)$ is a flat Riemannian manifold for some metric tensor $g$” equivalent to “For all metric tensors $g$ on $M$, there exists a local change of coordinates that puts $g$ into constant diagonal form”. – Haskell Curry Jan 3 '13 at 4:14

From the comments: "I found that in a seemingly reputable astronomy book. "If it is possible to find a frame in which all the components of g are constant, the space is 'flat'"." --Frank

This question makes sense, but it is phrased in a new-user-hostile way. We have a space $(M,g)$ on our hands. Now consider an open neighborhood $U$ of some point and start picking local frames $(\xi_1, \dots, \xi_n)$ of the tangent space $T_M$ on $U$. The statement in the book says that if you can find a frame such that $g(\xi_j,\xi_k)$ are constant functions on $U$ for all $j,k$, then $g$ is flat on $U$.

The condition says that the $\xi_j$ are parallel with respect to $g$, or that $\nabla \xi_j = 0$, where $\nabla$ is the Levi-Civita connection of $g$. This implies that $$R(\xi_j,\xi_k) \xi_l = \nabla_{\xi_j} \nabla_{\xi_k} \xi_l - \nabla_{\xi_k} \nabla_{\xi_j} \xi_l - \nabla_{[\xi_j,\xi_k]} \xi_l = 0$$ for all $j, k, l$. Here $R$ is the curvature tensor of $g$. Since $R$ is a tensor on the space of tangent fields and $(\xi_1, \dots, \xi_n)$ is a frame on $U$, this implies that $R = 0$ on $U$, so $g$ is flat.

• I didn't realize that it was phrased in a "new-user-hostile" way - what do you mean by this? I just phrased the question in the shortest possible way I could think of. This being mathematics, I thought we didn't need much fluff around the question :-) – Frank Jan 3 '13 at 16:27
• Gunnar - Also, can you point me to get a good reference to learn differential geometry - I need some serious education on that topic. – Frank Jan 3 '13 at 16:33
• @Frank: I didn't mean that your phrasing was user-hostile, but that the book's was; I meant no great offense, just that this is the sort of hand-waving remark people make when they know a subject that beginners stumble over. For references I liked Lee's "Riemannian manifold", Gallot, Sylvelstre & Lafontaine's "Riemannian geometry", and you should definitely have a look at the first two or three volumes of Spivak's "A comprehensive introduction to differential geometry" (if nothing else, read volume 2 a few times). – Gunnar Þór Magnússon Jan 4 '13 at 7:11
• @Frank A side remark from the perspective of a frequent answerer on this site: "the shortest possible way" is usually not the best possible way. The time spent unpacking and clarifying tersely worded questions (usually in a series of back-and-forth comments) greatly exceeds the time spent reading a long post. – user53153 Jan 7 '13 at 3:49
• @GunnarÞórMagnússon I'm a little confused because given a riemannian manifold $(M,g)$ we can always find, for instance, a local orthonormal frame $\zeta_1 , \cdots , \zeta_n$. So this would satisfy the $g(\zeta_i, \zeta_j)= \delta_{ij}$ being constant. However, the metric $g$ is flat if and only this local frame comes from coordinates. Am I missing something? – Ashley Mar 6 '17 at 22:06

No. In particular, the second statement is always false. For if $X$ is a smooth manifold, then in local coordinates (diffeomorphic to $\mathbb R^n$, say) on $X$ choose a non-flat metric and then extend it arbitrarily to $X$. Then there does not exist a coordinate change (on the chosen local coordinates) that makes $g$ identically constant.

The correct statement is "$(X,g)$ is a flat Riemannian manifold if and only if about each point $x \in X$ there exists a neighborhood $U \ni x$ and local coordinates on $U$ in which $g$ is the standard Euclidean metric." (The correct statement is always more of a mouthful, isn't it?)

The point is that saying $X$ is a Riemannian manifold implies that $X$ carries with it a fixed Riemannian metric $g$. You do not get to choose $g$ as in your initial statement.

From the standpoint of definitions, saying that $(X,g)$ is flat means precisely that the Riemann curvature tensor is identically zero, and then from this one proves the statement in my second paragraph (see Petersen's or Do Carmo's Riemannian Geometry books, for instance).