Why do the $c$, $b$ & $t$ quarks have strong hypercharge?

Question

From the $SU(3)$ flavour symmetry we have quantum numbers for the up, down and strange quark, associated with the Eigenvalues of the Gell-Mann matrices $\lambda_3$ and $\lambda_8$. Namely $\lambda_3$ gives $I_3$ and $\lambda_8$ gives $Y$.

As I understand it there is no $SU(6)$ symmetry covering every quark flavour due to the large mass differences between the quarks. How then do these heavier quarks have a hyper charge if they are not part of the triplet and so cannot act on these operators?

My initial thought was that these heavier quarks transformed under some U(1) and you could write each quark in a sextuplet where the heavier quarks act on the identity. I.e. that the operator associated with hypercharge would be $$ \hat Y \propto \begin{bmatrix}\lambda_8&\\&I_{3x3}\end{bmatrix} $$ But as the heavier quarks have no isospin this would require $$ \hat I_3 \propto \begin{bmatrix}\lambda_3&\\&\textbf{0}\end{bmatrix} $$ which seems to invalidate this idea as there doesn't seem to be a clear reason from this argument why isospin should be zero if hyper charge is non zero. Something like there can only be one quantum number associated with a $U(1)$ symmetry perhaps? Any corrections to my assumptions or explanation would be appreciated.

Answer 1

Strong hypercharge is a "practically obsolete" global symmetry of the strong interactions, which conserve it: they don't change flavor.

It is basically a zero-point shift in the association of $T_3$ with charge, via the Gell-Mann−−Nishijima formula, $$ Y=2(Q-T_3). $$ So, beyond the first generation, it is but a plain reader of twice the charges of the quarks; for the s,c,b,t, $$ \hat Y= \frac{2}{3} \operatorname{diag} (-1,2,-1,2) , $$ which some formal dandies choose to redefine through the mnemonic $$ Y=B +S+C+B'+T $$ confusing the innocent who forget strangeness S and bottomness B' enter with a negative sign for the quark! B is just the baryon number.

In consequence, If you wished to array your 6 quarks in a 6x6 matrix extending flavor SU(3) to a larger classification (not degeneracy!) scheme, you'd have to write (instead of your wrong expression), $$ \hat Y =\frac{1}{3} \operatorname{diag} (1,1,-2,4,-2,4)~, $$ the proper extension of $\lambda_8$, which, however, is not traceless.

It indeed commutes with isospin, an accidental property of the first generation only (traceable to mere lightness thereof, not degeneracy!), $$ \hat I_3 = \frac{1}{2} \operatorname{diag} ( \tau_3,0,0,0,0) ~. $$