# Polynomial maps between noncommutative groups

Below I will give some definitions. My question is: do these appear in the literature, and if so, under what name?

Let $G$ and $H$ be groups that may not be commutative. For $y\in G$, define $R_y:G\to G$ by $R_y(x)=xy$. Let $f$ be a function from $G$ to $H$. Define $\delta_n(f):G^n\to H$ by \begin{align*} \delta_0(f) &= 1 \\\\ \delta_1(f)(x) &= f(1) f(x)^{-1} \\\\ \delta_2(f)(x,y) &= f(1) f(x)^{-1} f(xy) f(y)^{-1} \\\\ \delta_3(f)(x,y,z) &= f(1) f(x)^{-1} f(xy) f(y)^{-1} f(yz) f(xyz)^{-1} f(xz) f(z)^{-1} \end{align*} and in general $$\delta_{n+1}(f)(x_1,\dotsc,x_{n+1}) = \delta_n(f)(x_1,\dotsc,x_n) \delta_n(f\circ R_{x_{n+1}})(x_1,\dotsc,x_n)^{-1}.$$ This has one term for each subset $J\subseteq\{1,\dotsc,n\}$, with exponent $(-1)^{|J|}$. The order of the terms corresponds to the Binary Reflected Gray Code (see Wikipedia, for example). One could imagine using other orders such as lexicographic, but the BRGC order seems to do the right thing for the examples that I am considering.

I'll say that $f$ is polynomial of degree at most $n$ if $\delta_{n+1}(f)$ is the constant function with value $1$. Clearly $f$ is polynomial of degree at most $0$ iff it is constant, and it is polynomial of degree at most $1$ iff it is a constant times a homomorphism. The commutative case is fairly well-known, and is consistent with the usual meaning of 'polynomial' for maps $\mathbb{Z}^p\to\mathbb{Z}^q$. I know of a 1971 paper by Andreas Dress, but it would not surprise me if there were earlier references. However, I have never seen the noncommutative case.

Even in the commutative case, it takes some work to prove that any composite of polynomial maps is polynomial. I do not know whether that holds in the noncommutative case.

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I see that $x\mapsto x^2$ does not have degree two! – Tom Goodwillie Dec 14 '12 at 14:03
I also see that when $G=\lbrace 1,g\rbrace$ has order two then $f:G\to H$ has degree n if and only if $((f(1)f(g)^{-1})^{2^n}=1$. – Tom Goodwillie Dec 14 '12 at 14:06
Is there a similar definition you can make for homogeneous degree $n$ maps? You could then ask if a degree $n$ map is a product of homogeneous maps of each degree $\leq n$. – Eric Wofsey Dec 14 '12 at 20:56
@Eric: Even in the abelian case it does not go that way. A degree two map $\mathbb Z\to \mathbb Z$ has the form $f(n)=a+bn+cn^2$ where $a$, $2b$, and $b+c$ are integers. What is the homogeneous quadratic part of $f(n)=\frac{n+n^2}{2}$? – Tom Goodwillie Dec 15 '12 at 3:49

Polynomial mappings of groups were investigated by Leibman in connection with density Ramsey theory.

Definition. Let $G$ be a group and $f:\mathbb{Z}\to G$ any map. The discrete derivative of $f$ is the map $D_a f(b) = f(b+a) f(b)^{-1}$. The map is called polynomial of degree $d$ if all its $d+1$-th discrete derivatives vanish identically (i.e. equal $1_G$).

One could replace $\mathbb{Z}$ by a more general group here; I stick to the integers for simplicity.

Theorem Assume that $G$ is nilpotent. Then the polynomial mappings form a group under pointwise operations.

Different proofs can be found in Leibman's article, an appendix in an article of Green, Tao and Ziegler and here. The appendix contains probably the most systematic approach based on Host-Kra cube groups, in particular it is shown that the composition of polynomial maps between nilpotent groups is also polynomial.

Note that the theorem does not say that the polynomial mappings of a given degree form a group, to recover a result of this kind one has to refine the notion of degree, see the references above.

The theorem fails already for the dihedral group $D_3$ that is the smallest non-nilpotent group. Indeed, let δ be a rotation and σ a reflection in $D_3$ so that δ³ = σ² = (σδ)² = 1. The sequences (…, σ, 1, σ, 1,…) and (…, σδ, 1, σδ, 1,…) are polynomial (of degree 1) but their pointwise product (…, δ, 1, δ, 1,…) is not polynomial (of any degree).

I am not sure whether this presents an obstruction to polynomiality of composition of polynomial maps between arbitrary groups.

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Thanks, that's great. I am mostly interested in the nilpotent case. – Neil Strickland Dec 16 '12 at 8:42

See a paper of Anokhin. He considered only the case where $H$ is abelian (and $G$ is arbitrary) and his definition of degree is slightly different. For instance, he says that the degree of $f$ is at most one if $$f(x)-f(xy)+f(xyz)-f(xz)=0$$ for all $x,y,z\in G$ (see Lemma 2).

In that paper, it was proved, e.g., that

all functions from a nontrivial group $G$ to a nontrivial abelian group $H$ are polynomial iff, for some prime $p$, $G$ is a finite $p$-group and $H$ is an abelian $p$-group (Theorem 2).

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Thanks, I'll see what I can extract from that, though I don't read Russian. – Neil Strickland Dec 15 '12 at 10:03
They have a link to the English translation: dx.doi.org/10.1023%2FA%3A1025001013073 (full text is freely available). – Anton Klyachko Dec 15 '12 at 18:58