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Why is it that symmetries of a system are usually realized as coming from groups (and categorified versions of it) and not other algebraic structures? Why are the terms "symmetry monoid" or "symmetry ring" not frequent (or existent) in the literature? Wouldn't, for instance, the former allow for a description of irreversible transformations? Is a ring structure for symmetries "too much" to require?

Qmechanic
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Alonso Perez-Lona
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3 Answers3

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Symmetries form groups because (a) they have an identity ("do nothing"), and (b) each operation is invertible. Non-invertible transformations would form a monoid, but such transformations do not describe a "symmetry" because some property of the system is destroyed by the transformation (you can't get back to the original state -- if you could you would have an inverse).

Rings and algebras do have applications in physics, but because they involve multiple operations they go beyond the simple requirements of a symmetry.

Eric Smith
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When we talk about "symmetries" of some object - we mean that the object "stays the same" under certain transformations. So, let us identify a set of such transformations $G$ and objects $X$. An element of the transformation set $g\in G$ should be able to act on an object $x \in X$. I'll denote this action as function application $g(x) \in X$.

From here, we naturally are able to able to compose these transformations: $$ g_1,g_2 \in G\quad \Rightarrow \quad g_1g_2 \in G : (g_1g_2)(x) = g_1(g_2(x))$$ This composition is associative and has a unique unit $e \in G : \forall x\in X. e(x) = x$
(Notice the uniqueness of $e$ - important in the following.)

We can get the inverse axiom if we assume that $G$ is finite. In that case for every $g \in G$ there are two powers $a > b > 0$ such that $g^a = g^b$ as a result we've got $g^{a-b} = e$ so we've got the inverse $g^{a-b-1}$. This line of argument should also work for compact manifold structures to get the Lie groups like $SU(2)$.

In my experience, when physicists talk about "symmetry groups" they assume finiteness/compactness without stressing (or, sometimes, being aware of) it.

If we don't assume finiteness/compactness of $G$ then, generally, we don't have the inverse axiom. The most obvious example is fixed-increment translations: $$T(f(x)) = f(x + 1)$$ Constant functions $f$ are symmetric under $T$, but $T$ and $e$ don't generate a group.

Kostya
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  • I'm not sure your last example makes sense. There is a group in question - the group $\mathbb{Z}$ of the integers under addition. – jacob1729 Apr 02 '21 at 13:57
  • In particular, whilst physicists do often care about finite groups (in a crystallography setting) or compact groups (in a field theory setting) - this is more for reasons that depend on the specific case at hand, than an avoidance of non-compact groups. The Lorentz and Poincare groups are non-compact and physicists are perfectly happy with them. – jacob1729 Apr 02 '21 at 13:59
  • @jacob1729 check out the difference between a monoid of natural numbers $\mathbb{N}$ and the group of integers $\mathbb{Z}$. I intentionally stated that only using $e$ and $T$ you won't be able to generate a group. – Kostya Apr 02 '21 at 14:21
  • @jacob1729 My point was that quite often physicists state that "symmetry" implies invertibility. And that this is formally correct only when the transformation set is finite/compact. – Kostya Apr 02 '21 at 14:31
  • Ah, this wasn't clear that you were only considering right translations (in group theory language you would definitely say that $\mathbb{Z}^+$ is generated by $1$). I'm not really clear what the point of this example is though since the left translations are also symmetries. Intuitively, if doing $X$ is a symmetry of a set $S$ then it should at least preserve the cardinality $|S|$ and is therefore a bijection and so has an inverse. – jacob1729 Apr 02 '21 at 16:04
  • @jacob1729 $\mathbb{Z}^+$ is not a group, though, not even a monoid under addition. So, yeah left translations are symmetries but they don't form a group. I agree that if you assume invertibility then the whole discussion is trivial. I understood the OP question as "why we are assuming invertibility" - tried to address it. – Kostya Apr 02 '21 at 16:23
  • Apologies for notation, I meant $(\mathbb{Z},+)$ which I agree could be confusing. Nonetheless, the word 'symmetry' is a natural language term, not a mathematical term. And I think its weird to consider symmetries that are not bijections - your example does not furnish one and so I think its artificial. – jacob1729 Apr 02 '21 at 18:14
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Physics uses mathematics as a tool to have theories that describe existing data and can predict new ones. It depends on observations and measurements.

The SU groups came into the study of data slowly. First SU(2), to describe the two baryon states of proton and neutron, that are almost of the same mass. It was useful in nuclear physics studies.

SU(3) first appeared as useful in describing particle data in the eightfold way,

bardec

The $Ω^-$ was a prediction , its discovery it validated the quark model .

The SU(3) color was found crucial for fitting strong interaction data. And so on.

If in the future experiments and data more complicated mathematical concepts will be needed to get a fit and predict new situations, they will be used.

The choice of the tools is indicated by the data, and not the other way around, at least until/if we have a theory of everything.

anna v
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