3
$\begingroup$

The question is just about terminology. "Commuting diagrams" generalize both associativity and commutativity laws. They don't seem specifically related to commutativity, so why do we say the diagram "commutes"?

Clarification:

Both "commutativity" and "associativity" are interchange laws. The first says path ab can be interchanged with path ba. The second says that the path (gf)h and the path g(fh) can be interchanged. It seems arbitrary that the name of only one of these interchange laws (i.e. "commutativity") was chosen to describe the general case.

$\endgroup$
1
  • $\begingroup$ commutes means either order yields the same result $\endgroup$ – J. W. Tanner Nov 4 '20 at 17:23
5
$\begingroup$

The etymology of the word "commute" comes from "to interchange two things", from Latin. When we talk about, say, a group commuting, we talk about interchanging the orders of the elements ($ab = ba$). When we talk about a graph commuting, we talk about interchanging the paths we travel through the graph ($f = g \circ h$).

$\endgroup$
3
  • 1
    $\begingroup$ In this sense, "associativity" also interchanging law: it says that the path (gf)h and the path g(fh) can be interchanged. It seems arbitrary that the name of only one of these interchange laws (i.e. "commutativity") was chosen to describe the general case. Anyway perhaps this is flogging a dead horse. $\endgroup$ – JRC Nov 5 '20 at 11:28
  • 2
    $\begingroup$ @JRC: I think the point of the answer was that while both commutativity and associativity in mathematics correspond to some interchanging laws, the word "commute" has linguistically broader meaning and the naming "commutative diagrams" may refer to the more general meaning of the word as opposed to the particular commutativity in algebraic structures. $\endgroup$ – user87690 Nov 5 '20 at 13:21
  • $\begingroup$ @user87690 Thanks. I'm accepting this answer, due to your additional reasoning. $\endgroup$ – JRC Nov 5 '20 at 14:36
8
$\begingroup$

Actually, it is quite related to commutativity. $\require{AMScd}$ \begin{CD} A@>{f}>> A\\ @V{g}VV @VV{g}V\\ A @>{f}>> A \end{CD} Let $f,g$ be endomorphisms of some algebraic structure $A$. Then $f\circ g=g\circ f$ if and only if the diagram above commutes. So, in some cases commutativity of these operators is equivalent to commutativity of the diagram.

Anyway, this is my interpretation.

Edit: To address the comment, I will only mention that associativity is also a sort of commutativity. For instance, we could define associativity to be the commutativity of the following diagram: \begin{CD} A\times A\times A@>{\mu\times \operatorname{Id}}>> A\times A\\ @V{\operatorname{Id}\times \mu}VV @VV{\mu}V\\ A\times A @>{\mu}>> A \end{CD} where $\mu:A\times A\to A$ is given by $\mu(x,y)=x\cdot y$, and $\cdot$ is a binary operation on $A$. Then the commutativity of the diagram says that for all $x,y,z\in A$ $$x\cdot(y\cdot z)=(\mu\circ (\operatorname{Id}\times \mu))(x,y,z)=(\mu\circ (\mu\times \operatorname{Id}))(x,y,z)=(x\cdot y)\cdot z.$$ So, it is reasonably to consider associativity as a "special case" of a commutativity condition.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks. But diagrams also generalize associativity, but we don't say that the diagram "associates". $\endgroup$ – JRC Nov 5 '20 at 9:33
  • $\begingroup$ That's sort of true, but you can think of associativity as a certain commutativity of diagrams. $\endgroup$ – Alekos Robotis Nov 7 '20 at 19:20
  • $\begingroup$ Thanks so much for the elaboration. As expressed by @user87690 "while both commutativity and associativity in mathematics correspond to some interchanging laws, the word "commute" has linguistically broader meaning and the naming "commutative diagrams" may refer to the more general meaning of the word as opposed to the particular commutativity in algebraic structures". Your edit shows that associativity can be expressed as diagramatic commutativity, but this was acknowledged in para 1 of the question. My question concerned the original (non-categorical) commutativity. Sorry if I was unclear. $\endgroup$ – JRC Nov 9 '20 at 10:06
4
$\begingroup$

If $\mathcal{M} = (M, \cdot)$ is a monoid, regarded as a category with a single object $\star$ (so the morphisms of the category are the elements of the monoid and the composition is given by the monoid operation), then two elements $a,b \in M$ commute (in the sense that $a \cdot b = b \cdot a$) if and only if the following diagram commutes:

$$ \begin{array}{ccc} \star & \xrightarrow{a} & \star \\ {\scriptsize b} \downarrow ~ && ~ \downarrow {\scriptsize b} \\ \star & \xrightarrow[a]{} & \star \end{array} $$

So in this sense, commutativity for diagrams generalises the usual notion of commutativity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.