In an $SO(10)$ GUT, how many gauge fields acquire masses that are above the electro-weak scale?

The $$SO(10)$$ group is spontaneously broken down into the Standard Model gauge group at around $$10^{16}$$ GeV and the electroweak scale is ~246 GeV. I think there should be 45 gauge fields predicted by an $$SO(10)$$ GUT (from $$\frac{1}{2}N(N-1)$$ generators, with $$N=10$$). How many of them would experience SSB and therefore acquire mass at scales above the electroweak scale?

• Did you count? What is your guess? You know the number of them for the SM, right? Jan 8, 2022 at 23:22
• @CosmasZachos Well I was hoping it wouldn't require a guess. To put my question more formally, how many of 45 generators are broken and thus result in massive gauge bosons?
– user319271
Jan 8, 2022 at 23:24
• How is this not a simple subtraction (45 - # of generators of the SM gauge group)? Jan 8, 2022 at 23:26
• @ACuriousMind Would that not just tell you how many gauge bosons in total are predicted above the electroweak scale but below the GUT scale - not the number of massive gauge bosons in that range?
– user319271
Jan 8, 2022 at 23:33
• But what does "GUT scale" mean if not that at that point all gauge symmetries except the usual SM gauge symmetries are broken, and hence massive? Jan 8, 2022 at 23:38

Above the SM SSB scale of 1/4 TeV, there are 8+3+1 = 12 massless gauge bosons; so, in your model, all other massive elementary vector bosons correspond to SSBroken generators, which makes them 45-12= 33 in all, all the way to "the" GUT scale, past which there are no more massive vectors. The GUT scale may involve several steps of partial braking to SU(5)$$\times$$ U(1), etc., but the takeaway point is that a few orders of magnitude below all of these scales, you must have 33 massive vectors.