What unit of measurement is GV? I am reading this paper; at page 12, it saids:

Lockwood et al. (1986) found that the recovery time was independent of rigidity in the range ~ 2 to ~ 5 GV

what is GV? I thought that rigidity was measured in T m (Tesla meter).
 A: "GV" is gigavolts.
The Handbook of Geophysics (slightly older version here) explains it:

The magnetic rigidity P of a particle is a measure of its resistance to a magnetic force that deflects the particle from a straight-line trajectory. The rigidity, with units of momentum per unit charge, is defined as
$P = \frac{pc}{q}$
where q is the charge of the particle. If pc is electron volts, then q is the number of electronic charge units and P is in volts. Convenient units are MV ($10^6$ V) and GV ($10^9$ V.

Note that a factor of $c$ was added to the over-the-line 'momentum' to give it energy units. 
A: Magnetic rigidity is defined in slightly different ways depending on the field.
When discussing an apparatus, you'll often see this definition of rigidity: $R \equiv B \,\rho = \frac{p}{|q|}$, where $B$ is the magnetic field and $\rho = \frac{p}{|q|B}$ is the gyroradius of a charged particle with relativistic momentum $p$ in that field. The units here are tesla-meters. Note that $1~{\rm T\,m} \approx 0.3\, {\rm GV}/c$, where $c$ is the speed of light.
When one cares more about the momentum of the particles, such as with cosmic rays in the paper you cite, you'll see a slightly different definition, $R \equiv \frac{pc}{Ze} $, where $Z$ is the atomic number of the ion, so that the units come out as just volts. I think it just comes down to not wanting to write "GV/$c$" every time.
When dealing with highly relativistic particles, $R = \frac{E}{Ze}$, and it's natural to think of momentum and energy as very similar—though not equivalent!—quantities. Translating to rigidity is a simple matter of dividing by the charge.
