13
$\begingroup$

Let $D(\mathbb R) $ be the set of all differentiable functions $f: \mathbb R \to \mathbb R$. Then obviously $D(\mathbb R)$ forms a semigroup under usual function composition. Can we characterize (up to semigroup isomorphism) all finite subsemigroups of $D(\mathbb R)$ which do not contain any constant function ?

$\endgroup$
1
  • 3
    $\begingroup$ Why do you think not all finite semigroups are there? $\endgroup$
    – user6976
    Aug 5, 2018 at 7:05

2 Answers 2

8
$\begingroup$

Yes, we can! ;-)

In fact there are only two finite subsemigroups of $D(\mathbb{R})$ which do not contain constant functions.


Every finite subsemigroup of $D(\mathbb{R})$ necessarily contains an idempotent $f$, i.e. a function $f$ such that $f \circ f = f$. Let us examine the properties of $f$.

Let $A$ be the range of $f$. By continuity of $f$, $A$ is an interval. Since $f(f(x)) = f(x)$ for every $x \in \mathbb{R}$, we have $f(x) = x$ for $x \in A$.

If $f$ is non-constant, then $A$ has non-empty interior. We claim that in this case $A = \mathbb{R}$.

Suppose, contrary to this claim, that $A$ is bounded from above, and denote the right endpoint of $A$ by $b$. Then $f(b) = \lim_{x \to b^-} f(x) = \lim_{x \to b^-} x = b$, and so $b \in A$. Thus, $f$ attains a local maximum at $b$. Since $f$ is differentiable at $b$, we have $f'(b) = 0$. On the other hand, $f'_-(b) = \lim_{x \to b^-} f'(x) = 1$, a contradiction. We conclude that $A$ is not bounded from above. Similarly, $A$ is unbounded from below.

Thus, either $f$ is constant or $f$ is the identity function. It follows that any finite subsemigroup of $D(\mathbb{R})$ contains either a constant function or the identity function.


Suppose that $X$ is a finite subsemigroup of $D(\mathbb{R})$ with no constant function and $g \in X$. Then the subsemigroup of $X$ generated by $g$ contains an idempotent, and hence — the identity function. In other words, $g^{\circ n}$ is the identity function for some $n$.

Todd Trimble already pointed out in his answer that necessarily $n = 1$ or $n = 2$, and if $n = 2$, then $g$ is decreasing. Here is a shorter variant of his argument that does not require differentiability:

  1. $g$ is invertible and continuous, and hence strictly monotone; $g \circ g$ is thus strictly increasing;

  2. if $g(g(x)) > x$ for some $x$, then $g^{\circ 2n}(x) > x$, a contradiction; similarly, if $g(g(x)) < x$ for some $x$, then $g^{\circ 2n}(x) < x$; therefore, $g(g(x)) = x$ for all $x$;

  3. if $g$ is increasing, then in a similar way $g(x) = x$ for all $x$.

Finally, if $g, h \in X$ and none of them is the identity function, then both are decreasing, and so $g \circ h$ is an increasing function in $X$. Therefore, $g \circ h$ is the identity function, and consequently $g = h^-1 = h$.


We have thus proved that that any discrete subsemigroup of $D(\mathbb{R})$ with no constant function contains the identity function and at most one strictly decreasing function $g$ such that $g = g^{-1}$.

$\endgroup$
0
5
$\begingroup$

This is of course nowhere near a full answer, but in response to Mark Sapir's question, I think the only invertible elements in the monoid of finite order are of order $1$ or $2$ (involutions).

A diffeomorphism, by which I mean an invertible element in the monoid of (not necessarily continuously) differentiable functions on $\mathbb{R}$, clearly cannot have zero derivative anywhere, by the chain rule. Nor can the derivative ever change sign, since a derivative function satisfies the intermediate value property: if it is positive at one point and negative at another, then it is $0$ somewhere in between (even in the non-continuously differentiable case). So, by the mean value theorem, a diffeomorphism is either strictly monotone increasing or strictly monotone decreasing.

In the increasing case, the only element $f$ of finite order is the identity. For if $x < f(x)$ for any $x$, then by strict monotonicity of $f$ we have $x < f(x) < f f(x) < f f f (x) < \ldots$, and a similar argument applies if $f(x) < x$.

In the case where $f$ is decreasing, suppose $f$ has order $n$. Then $f^2 = f \circ f$ is increasing and has order $n/2$ if $n$ is even, and order $n$ if $n$ is odd. By the preceding paragraph, it follows that $n/2 = 1$ or $n = 1$ respectively (and of course the latter case doesn't happen since the identity function is not monotone decreasing).

$\endgroup$
1
  • 1
    $\begingroup$ @KeithKearnes Ah ha, good point. I guess I was thinking of submonoids. $\endgroup$
    – Todd Trimble
    Aug 5, 2018 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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