How to decompose E6?

Hi I'm trying to understand decomposition of $E_6\to SO(10)\otimes U(1)_\psi\to SU(5)\otimes U(1)_\psi\otimes U(1)_\chi\to SU(3)_C\otimes SU(2)_L\otimes U(1)_Y\otimes U(1)^\prime$. I believe I mostly understood the 27 dim vector representation part which decomposes as $27_{Q_\psi}=16_1+10_{-2}+1_4$ where 16 becomes matter fields, 10 EW Higgs fields $10_{-2} = 5_{-2}(\hat{H}_u,\hat{D})\oplus \overline{5}_{-2}(\hat{H}^c_d,\hat{D}^c)$ (and exotic quarks) and 1 is the U1 singlet. However, I can not say that I completely understood adjoint representation decomposition where $78_{Q_\chi}=45_0+16_{-3}+\overline{16}_{3}+1_0$, 1 and 16s will clearly become u1 singlet and fermion fields respectively but what about 45? As far as I understood it decomposes as $45=24_0+10_?+\overline{10}_?+1_0$ where 1 will become B' i believe and 24 is SM generators aka gauge fields B, W, Y, Isospin but what about 10's? are those again EWhiggs fields?

Note that $Q_\chi$ and $Q_\psi$ are assumed to be normalized.

Thanks

• Could you add more info on what $\psi$, $\chi$, and $Q$ stand for? – Kosm Sep 14 '17 at 22:59
• Sorry my bad; $Q_{\psi,\eta}$ stands for the charge coming from $U(1)_{\psi,\eta}$ respectively, just like hypercharge where they mix at the lower energies and become $Q^\prime=Q_\psi\sin\theta_{E_6}+Q_\eta\cos\theta_{E_6}$ thus 78 adjoint and $27+\overline{27}$ vector representation mix and make a complete MSSM+U(1). (as far as I understand) – jackaraz Sep 14 '17 at 23:15
• what does it mean, "U(1) singlet"? Zero U(1) charge? – Kosm Sep 14 '17 at 23:27
• It's not in Slansky's appendix? – Cosmas Zachos Sep 14 '17 at 23:43
• @CosmasZachos I found all these information from there and some other papers and thesis. However, I couldn't find physical correspondence of 10's in 45. I would very much appreciate if you can show me where does it say it. Thanks – jackaraz Sep 15 '17 at 0:22