Can sine be made into a homomorphism? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:27:23Z http://mathoverflow.net/feeds/question/123041 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123041/can-sine-be-made-into-a-homomorphism Can sine be made into a homomorphism? Jon Cohen 2013-02-26T22:58:34Z 2013-02-26T23:53:27Z <p>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?</p> <p>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). </p> <p>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). </p> <p>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. </p> http://mathoverflow.net/questions/123041/can-sine-be-made-into-a-homomorphism/123046#123046 Answer by Eric Wofsey for Can sine be made into a homomorphism? Eric Wofsey 2013-02-26T23:53:27Z 2013-02-26T23:53:27Z <p>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.</p> <p>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 <code>$\bigcap f^n(X)=\{0\}$</code>. 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.</p> <p>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 $f:X\to X$.</p> <p>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.</p>