general relativity - Non-integer $k$ value in Friedman-Robertson-Walker model?

I understand that $k$ describes positive, negative, or no curvature. However, why can't there be, for example, +0.5 (semi-positive) curvature, etc?

Answer

The continuum of curvatures does exist, but we find it more convenient to put it elsewhere. The crucial part of the metric that encodes the curvature is a factor $1-K r^2$, where $r$ is the radial coordinate (using any point as the origin), and $K$ is any real number, positive or negative. By dimensional analysis, there is some length $L$ such that either $K=1/L^2$ or $K=-1/L^2$, so we can rewrite our formula as $1 - k (r/L)^2$, where $k$ is just the sign of $K$, or $0$ if $K=0$. The case $k=0$ corresponds to $L$ equal to infinity.

This new variable $k$ can only have the values $\pm 1$ or $0$, but that's okay because $L$ still can be any length, so we have the whole range of curvatures. $L$ is known as the radius of curvature of the universe, and a larger $L$ implies a smaller curvature. $k$ determines whether this curvature is positive or negative.

Now, and this is a bit of a technical point, we can make that $L$ go away if we measure our coordinate $r$ in units of $L$. In our formula, we can set $x = r/L$ to get just $1-kx^2$. The continuum of curvatures is still there but it is hidden inside of $x$, because the physical interpretation of $x$ depends on $L$: $x$ is how many times $L$ fits in your distance. So if for example $L = 1\ \text{light-year}$, $x=2$ is a distance of $2$ light-years, but if $L = 3$ light-years then $x=2$ is actually a distance of $6$ light-years. Mathematically, the price we pay is that $L$ now shows up elsewhere in the formulas, in the part we use to calculate lengths.

To sum up, the curvature can indeed take any value: the closer it is to zero, the closer space is to being flat. The fact that $k$ can only be $\pm 1$ or zero is just a matter of convenience: we use $k$ to label the three qualitatively different scenarios of positive/negative/zero curvature. $L$ just sets the size scale for the universe.