In (2D) Cartesian coordinates, the Euclidean metric...

$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

...is flat space. If the diagonal elements are exchanged for other real numbers greater or less than zero, would this still count as flat space even though it is no longer Euclidean? I think the answer is yes but just want to make sure.

Also, when a physicist says that the local tangent space on a manifold is flat, do they imply that the metric is locally Euclidean, or can the diagonal elements be any non-zero real number in that local tangent space?

  • 1
    $\begingroup$ I would do what QMechanic says and drop your metric into a computer code and compute the Riemann curvature tensor. If it vanishes, you're dealing with a flat geometry. $\endgroup$
    – kηives
    Dec 30, 2012 at 23:42
  • $\begingroup$ I would be more carefull with wording. A matric is not a space! $\endgroup$
    – MBN
    Jan 12, 2016 at 17:19

3 Answers 3


1) OP is asking about the use of the word flat metric. It means a pseudo-Riemannian metric (of arbitrary signature) whose corresponding Levi-Civita Riemann curvature tensor vanishes.

2) However, the word Euclidean space may potentially cause confusion among mathematicians and physicists. For a mathematician an Euclidean space is always an affine space, while a physicists often use it as just another word for Riemannian manifold, which is not necessarily affine.

In short, the word Euclidean refers for a physicist to a positive signature (typically as opposed to Minkowski signature), while a mathematician use the word Euclidean to refer to an affine structure.

A mathematician calls a pseudo-Riemannian manifold with Minkowski signature for a Lorentzian manifold.


If you use different units on different coordinates and if you use a different unit on the distance metric then the diagonal elements of the metric matrix can differ from $1$. For example if you want to measure distances in meters, but you use inches for the $x$ dimension and centimeters for the $y$ dimension, then the 2D metric would be: $$\begin{bmatrix} \mathrm{meters^2/inch^2} & 0 \\ 0 & \mathrm{meters^2/centimeter^2} \end{bmatrix}$$

So in general the diagonal values can differ from $1$, but it is much more sensible to use the same units for $x, y$ and distance and in that case the diagonal elements will all be $1$.

Now in the Minkowski 4D space-time $(t,x,y,z)$ of our universe, the metric is usually written like this:

$$\begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

And in this case the $-1$ is for the time direction and it is significant. That cannot be changed to a $+1$ and is what gives significance to the speed of light - in particular light always travels a "proper" distance of 0 in the full 4D space-time. Note that a total distance of 0 is only possible if at least one of the diagonal elements has the opposite sign of the other diagonal elements. In fact, the $-1$ is often written as $-c^2$ which is just another indications that the unit of measurement for time differs from the units for the spatial coordinates. Again using the "same" units can change the diagonal element to $-1$ instead.


When we say "flat" space, it can mean more than just Euclidean space. In relativity, a flat spacetime is Minkowskian--there has to be some difference in signs between that of the time coordinate and those of the spatial coordinates.

Whether the local tangent space is Euclidean or Minkowskian depends largely on context.


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