Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the usual sine function $\mathbb{R}\rightarrow \mathbb{R}$. Is there some (single) group structure we can put on $\mathbb{R}$ with respect to which sine becomes a homomorphism?

I suspect the answer is either no for a trivial reason, or yes by a simple set-theoretic argument (probably providing a great many such group structures of no interest).

This latter seems plausible if I can "replace" the reals (and the sine function) by some arbitrary equal-sized set (and sufficiently similar function). Indeed, if I only asked that we have a pair of group structures $*_1$ and $*_2$ so that sine is a homomorphism from one to the other, then such an argument does go through (if I'm not mistaken).

So I'm putting this question forward mostly in case there's a pleasant surprise. Depending on the answer, one could of course ask for further restrictions (abelian, continuous (probably impossible), torsion, torsion-free etc), but for now I'll leave as is.

share|improve this question
4  
No. $\sin{t} = \sin{\pi - t}$ for all real numbers $t$, so the kernel of your homomorphism would contain the difference $\pi - 2t$ for all real numbers $t$. Since the kernel is all of $\mathbb{R}$, the homomorphism must be trivial. You can try to remedy this by taking an appropriate linear combination of sine and cosine, but the only examples of this which should work are those which are scalar multiplies of $\cos{t} + i \cdot \sin{t} = e^{i \cdot t}$. –  DavidLHarden Feb 26 '13 at 23:13
10  
There's no reason that the difference between $t$ and $\pi-t$ should be $\pi-2t$ with respect to the group structure on the domain. –  Eric Wofsey Feb 27 '13 at 0:00
    
A much more general and perhaps interesting (though perhaps intractible) question: Can you combinatorially characterize the (set-theoretic) dynamical systems of group-homomorphisms? There are a few obvious necessary conditions (a fixed point to be the identity, all nonempty preimages have the same cardinality), but it's far from obvious that they are sufficient. –  Eric Wofsey Feb 27 '13 at 0:05

1 Answer 1

up vote 6 down vote accepted

If you require the group structure to be continuous, this is impossible. Indeed, in that case, the image $[-1,1]$ would have to be a group as well. But $[-1,1]$ is not homogeneous, so it cannot be a topological group.

Without requiring continuity, it is possible. Let me first give a construction where you restrict from $\mathbb{R}$ to $X=[-\pi/2,\pi/2]$; I will write $f:X\to X$ for the sine function. In that case, the (set-theoretic) dynamical system given by $f$ has a particularly simple structure: $f$ is injective with a unique fixed point 0, $X_n=f^n(X)\setminus f^{n+1}(X)$ has cardinality $2^{\aleph_0}$ for all $n$, and $\bigcap f^n(X)=\{0\}$. Such a dynamical system is completely determined by giving the sequence of sets $X_n$ for $n\geq 0$ with bijections $X_n\to X_{n+1}$. Let $Y$ be the free $\mathbb{Q}$-vector space on $X$; then it is easy to see that the induced homomorphism $\tilde{f}:Y\to Y$ will have the same properties. Choosing compatible bijections $X_n\to Y_n$, we get a group structure on $X$ for which $f$ is a homomorphism.

Extending this to all of $\mathbb{R}$ is now easy: identify $\mathbb{R}$ with $X\times \mathbb{Z}$ in such a way that the projection $X\times\mathbb{Z}\to X$ sends each $x\in \mathbb{R}$ to the unique $y\in [-\pi/2,\pi/2]$ such that $\sin x=\sin y$. Then the sine map $\mathbb{R}\to \mathbb{R}$ can be identified with the projection to $X$ followed by $f:X\to X$.

This construction is quite flexible if you want the group structure to have various properties. For instance, you could replace $Y$ with the free group on $X$, or the free abelian group on $X$, or the free $\mathbb{F}_p$-vector space on $X$, or many other constructions; you could also replace $X\times\mathbb{Z}$ with any semidirect product of $X$ with a countable group.

share|improve this answer
    
This is great: the question had been bothering me for a while and the dynamical sysyems perspective is particularly elegant. +1 –  Vidit Nanda Feb 27 '13 at 1:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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