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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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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. – Michael Hardy Feb 26 '13 at 17:42

The fact that the binary operation defined by (2) is associative can be found in the Wikipedia article on formal groups (section 2) Presumably one might find more in the literature of formal groups.

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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 $f(u) = (1-u^2)/(1+u^2)$. – Michael Hardy Feb 27 '13 at 1:20

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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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). – David Hill Feb 26 '13 at 18:00

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