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A version of this question on stackexchange got a few comments from one person and no answers.

Let $e_k$ be the $k$th-degree elementary symmetric polynomial in variables $x_1,\ldots,x_n$ (so in particular $e_k=0$ if $k>n$). I don't know much about these polynomials.

Consider the rational function $$ \frac{e_1+e_3+e_5+\cdots}{e_0+e_2+e_4+\cdots}.\tag{1} $$

The simplest case is with just two variables: $$x_1\circ x_2=\dfrac{x_1+x_2}{1+x_1 x_2}.\tag{2}$$

  • Is it known, in the sense of being in the refereed literature or at least widespread folklore, that the binary operation defined by $(2)$ is associative?
  • Is it known that $x_1\circ \cdots\circ x_n$ equals the expression in $(1)$ (the proof is trivial, but is the fact mentioned in the literature)?
  • Is it known that $\pm1$ are absorbing elements for this operation, so that if just on of the $x$s in $(1)$ equals $1$ then the whole expression in $(1)$, depending on $x_1,\ldots,x_n$, equals $1$, and similarly $-1$? (If $x_1=1$ and $x_2=-1$, then you get $0/0$ and if I'm not mistaken, the singularity is not removable.)
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  • $\begingroup$ A theorem says if you have associativity plus moderately civilized behavior the operation is isomorphic to multiplication on $[0,\infty]$ (also with two absorbing elements, that don't "absorb" if both arguments are present in a product). In this case that happens if the only values the $x$s can take are non-negative. Then, if $f:[0,\infty]\to[-1,1]$ and $x=f(u)=(1-u^2)/(1+u^2)$, then $f(u_1 u_2)=f(u_1)\circ f(u_2)$. But the mapping $f$ on the whole line $[-\infty,\infty]$ is a two-to-one function into [-1,1]$, so that's not an isomorphism. $\endgroup$
    – Michael Hardy
    Commented Feb 26, 2013 at 17:42

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The fact that the binary operation defined by (2) is associative can be found in the Wikipedia article on formal groups (section 2) http://en.wikipedia.org/wiki/Formal_group. Presumably one might find more in the literature of formal groups.

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  • $\begingroup$ The article mentions that it comes from the addition formula for the hyperbolic tangent function. So now I'm wondering why I hadn't remembered that after fiddling with this operation for a few hours, and figuring out in particular that it comes from a "multiplication formula" for $f(u) = (1-u^2)/(1+u^2)$. $\endgroup$
    – Michael Hardy
    Commented Feb 27, 2013 at 1:20
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Not a complete answer, but note that \begin{align} (*) \;\;\;e_k =e_k(x_1,\ldots,x_n)=\sum x_{i_1}x_{i_2}\cdots x_{i_k}, \end{align} where the sum is over $1 \leq i_1 < i_2 < \ldots < i_k \leq n$. Moreover, for $1 \leq k< n$, $$e_k|_{x_j=1}=e_k(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_n)+e_{k-1}(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_n),$$ and $$e_n|_{x_j=1}=e_{n-1}(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_n).$$ The $e_k$ term above comes from those summands in $(*)$ for which j does not occur among the $i_1,\ldots,i_k$, and the $e_{k-1}$ term comes from the remaining terms. So setting one of the $x_j=1$ in (1) above is clearly 1. (I'd have to think about plugging in $-1$ a little more).

I don't know off-hand where the best place is to look these particular things up, but Macdonald's "Symmetric Functions and Hall Polynomials" (the 2nd edition) would be a good start.

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    $\begingroup$ Doesn't setting $x_j=-1$ just change the coefficient of $e_{k-1}$ to $-1$ (same for $e_{n-1}$. In that case, setting one $x_j=-1$ gives you -1 in (1). $\endgroup$
    – David Hill
    Commented Feb 26, 2013 at 18:00

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