Is the Hubble parameter a frequency and if it is what is it a frequency for? Logic and a bit of math lead to the conclusion that the Hubble parameter is a frequency since its unit is
$$ \frac {\mathrm{km}}{\mathrm{Mpc} \cdot s} $$
where km and  Mpc are both units of length, so if you transform them accordingly you can cancel them out to
$$ \text s^{-1} = \mathrm{Hz}. $$
But what is this frequency for?
 A: The large-scale structure of the universe is one on which lengths scale in proportion to a time-dependent function called the scale factor, denotes $a(t)$. If over some period $a$ doubles, that just means the large-scale structure of the universe doubles its length scales. The Hubble parameter $H=\frac{\dot{a}}{a}$ is in general time-dependent. In fact, the only way for it to be constant is for $a$ to vary exponentially over time (viz. $a\propto e^{Ht}$, in which cases $1/H$ is the time $a$ takes to multiply by $e$). In most of the universe's history, we have had $a\propto t^p$ with $p$ approximately constant for long time periods and $t$ the time since the Big Bang. Then $H=\frac{p}{t}$ is inversely proportional to the age of the universe.
A: The way Hubble constant is presented is quite misleading.The reason for that was "to fit"  Hubble data misinterpreted via Doppler shift with R dot over R term of Friedman solution of Einstein equations.  The simplest way to see what Hubble constant is, one should to calculate it in  METERS . Take MKS units, velocity m/s, distance m . Then you will see that H has dimension 1/m.
H squared has dimension 1/m squared which is the (negative) Gaussian curvature of Lobachevskian Universe understood as Poincare ball model of Lobachevskian geometry. As a reference see " Expansion of Universe -Mistake of Edwin Hubble ..." Acta Physica Polonica B ,vol 39,nr.8, 2008.
