Are all baryons from decuplet resonances?? Because all resonances decay by strong interaction; but omega minus baryon can't decay via strong, because any other bound state with strangeness -3 exists. Thanks
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No, only some baryons form a decuplet under SU(3) flavor symmetry, specifically those 10 spin 3/2 baryons formed from up, down,and strange quarks,depicted in the following diagram (figure credit Wikipedia baryon article , figure listed as public domain):
On the diagram $Q$ is electric charge, $I_3$ is isospin, and $S$ is the strangeness quantum number.
If you think about just adding the charm quark and think about baryons containing a charm quark, you see that not all baryons can fall into a decuplet, because the flavor symmetry is SU(4) once we introduce charm. And so on for the bottom baryons.
Or on a more familiar note, neither the neutron or proton, the baryons that are most familiar to non-physicists, are part of a decuplet. They are part of an octet of spin 1/2 baryons.
The reason the $\Omega^-$ can't decay under the strong interaction is that the strangeness quantum number is conserved by strong decays . So since the decay of the particle involves a change in strangeness quantum number, it must decay by weak interaction.
By flavor symmetry, I meant symmetry under exchange of different types (flavors) of quarks. There are six known flavors or types of quarks, and all have been experimentally verified to exist now: up, down,strange, charm, top, bottom. SU(3) and SU(4) are terms from group theory , which is a mathematical description of symmetry, and is used to express the symmetry of how things transform when exchanged. That's a very qualitative description to say the least.
