How do you tell if a metric is curved? I was reading up on the Kerr metric (from Sean Carroll's book) and something that he said confused me. 
To start with, the Kerr metric is pretty messy, but importantly, it contains two constants - $M$ and $a$. $M$ is identified as some mass, and $a$ is identified as angular momentum per unit mass. He says that this metric reduces to flat space in the limit $M \rightarrow 0$, and is given by
$$ds^2 = -dt^2 + \frac{r^2 + a^2 \cos^2\theta}{r^2 + a^2}dr^2 + \left(r^2 + a^2 \cos^2\theta \right)d\theta^2 + \left(r^2 + a^2 \right)\sin^2\theta d\phi^2 $$
and $r$, $\theta$ and $\phi$ are regular spherical polar co-ordinates. 
But I don't understand why this space is obviously flat. The Schwarzschild metric also contains terms involving $dt^2$, $dr^2$, $d\theta^2$ and $d\phi^2$ but is curved. I always thought that a metric with off diagonal elements implied a curved space, but clearly I was very wrong.
Question: How do you tell if a metric is curved or not, from it's components?
 A: You tell if a space (or spacetime) is curved or not by calculating its curvature tensor. Or more unambiguously one of the curvature scalars (e.g. Ricci, or Kretschmann) since these don't depend on the coordinate system, but all of the information in the scalars is also contained in the Riemann tensor.
It is not necessarily obvious whether a given metric is curved or flat. You can take a perfectly flat spacetime and express it in some bizarre coordinate system, in which the metric has nonconstant off-diagonal terms. It's a simple exercise to take flat space and use the tensor transformation laws for the metric, with some arbitrary weird coordinate transformation that you just made up. You'll see what I mean.
A: The flat space time refers here to the spacetime of Minkowski written with the spherical coordinates (I think one of your sign is wrong in your equation) 
$$
ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + r^2 sin^2 \theta d\phi^2.
$$
where the metric is diagonal and has constant coefficients $g_{\mu \nu} = ( -1,1,1,1)$. 
I would say that the conditions for a flat space-time, regarding only its metric, refer to its diagonal shape and constant (at least constant) 
A: In the limit where $M \to 0$, the Kerr metric reduces to the spherical coordinates form of the Minkowskian metric. In that sense, we recognize it and say it is 'obvious' that is is flat. (The Schwarzschild metric is also flat in the limit $M \to 0$.)
But to show that any given metric is curved or not we have to calculate a curvature invariant. For example usually we calculate the Ricci curvature  $R= R^i{}_i = R^{ki}{}_{ki}$ where the first $R$ is the Ricci curvature, the second $R$ the Ricci curvature tensor and the third $R$ the Riemann tensor. If it's $0$ the space is curved, otherwise it's not. Carroll has it in his book.
A: Carroll points out that, after the (a=fixed,M->0) limit "we recognize the spatial part of this as flat space in ellipsoidal coordinates", so in order to had realized that it was a flat space by inspection you should have been aware of the aspect of a flat metric in ellipsoidal coordinates, touche.
