15
$\begingroup$

When functions $f$ and $g$ have the property that $f(g(x)) = g(f(x))$ for all $x$ in the domains I call this property 'commutativity'. (usually both functions map from $\mathbb{R}$ to $\mathbb{R}$, so the problem of domain/range doesn't matter)

However, commutativity is actually when: $a*b=b*a$

I use it knowing that it probably isn't the right term... but I've never found out what I should call it.

$\endgroup$
  • 2
    $\begingroup$ It definitely is the right term. Physicists talk about commutators $[f,g] = (f\circ g) - (g\circ f)$ all the time, which basically measures how far two operators are from commuting. $\endgroup$ – leftaroundabout Aug 18 '16 at 9:59
36
$\begingroup$

The term is OK, and in fact refers to the same property: The set $\Bbb R^{\Bbb R}$ of all maps $\Bbb R\to\Bbb R$ is endowed with a binary operation $\circ$, the composition of functions: If $f$ and $g$ are maps $\Bbb R\to\Bbb R$, then so is $f\circ g$ (which is defined by $(f\circ g)(x):=f(g(x))$). This binary operation has many interesting properties, such as

  • there is a neutral element, namely the identity function $\operatorname{id}\colon \Bbb R\to\Bbb R, x\mapsto x$. We have $\operatorname{id}\circ f=f\circ\operatorname{id}=f$ for all $f$
  • it is associative: We have $f\circ(g\circ h)=(f\circ g)\circ h$. In fact, in many cases we use this associativity to prove the associativity of any other given operation.
  • Sometimes, we have $f\circ g=g\circ f$. We say that these elements commute, just as we would with any other algebraic structure.
$\endgroup$
13
$\begingroup$

The terminology "$f$ and $g$ commute" is perfectly fine and commonly used. For example, in linear algebra, it is a useful fact that two diagonalizable operators can be simultaneously diagonalized if and only if the operators commute. Another example is given by the Lie bracket of two vector fields, a very important object in geometry that measures the extent to which the flows of the vector fields commute locally.

$\endgroup$
11
$\begingroup$

Composition is a binary operation on functions that is not necessarily commutative. However, if $f \circ g = g \circ f$, then $f$ and $g$ are said to “commute” with respect to $\circ$.

$\endgroup$
8
$\begingroup$

Indeed, commute is the correct terminology: See the Wikipedia entry for function composition. To be more precise (or verbose), you might call it "commute with respect to composition" but when we are speaking of functions $f$ and $g$, then it is clear what is meant by the phrase "$f$ and $g$ commute" as the operation that is implied is composition, not multiplication--functions operate on each other as compositions.

$\endgroup$
  • $\begingroup$ This is not correct; you don't say that f and g are commutative, because that should be an INHERENT property, whereas here we're talking about something that true of f RELATIVE to a particular g. For that reason, we say that f and g commute (with each other), as the other answers explain. $\endgroup$ – Tom Church Aug 17 '16 at 16:46
  • $\begingroup$ @TomChurch Yes, now that you mention it...I've corrected the post. Thanks. $\endgroup$ – heropup Aug 17 '16 at 16:49
  • 2
    $\begingroup$ I don't think it's because of inherent or relative though. We use "commutativity" to mean a property of the operator ($\circ$ in this case), or of some structure (such as a group) involving that operator. We say "they commute" to mean a property of some elements. If $f$ and $g$ (together with the identity) formed a subgroup then we'd say the subgroup is commutative or abelian, but not that the elements of the subgroup "are commutative". Function composition isn't inherently commutative, but it's commutative on that set of functions. $\endgroup$ – Steve Jessop Aug 18 '16 at 8:05

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.