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While working on a project, I came across a paper that includes this sum on page 15 as the definition of a support function $r$ for a surface with tetrahedral symmetry:

$$ r(ξ, \bar{ξ}) = \frac{1}{\#𝒢} \sum_{g ∈ 𝒢} r_0\bigl(g(ξ), g(\bar{ξ})\bigr) $$

where

  • $𝒢$ is "the tetrahedral group", "a discrete subgroup of isometries $𝒢 ⊂ O(3)$";
  • $r_0(R) = -\frac{a + b R^2 + (3 - b) R^4 + (1 - a) R^6}{(1 + R^2)^3} + c$ is a rational support function of a 3D surface with rotational symmetry;
  • $ξ$ is a complex number, "the local complex coordinate on the unit 2-sphere in $𝔼^3$ obtained by stereographic projection from the south pole"; and
  • $a$, $b$, and $c$ are real number parameters that affect the shape and scale of the generated surface.

I am trying to figure out how to evaluate this sum to get an algebraic expression for $r(ξ, \bar{ξ})$. However, I'm currently stuck on the fact that $r_0$ is defined as a single-variable function, yet appears to be called with two input variables.

Am I misreading the notation here? Or is there a convention for how to handle this with which I am unfamiliar?

The sum in this question is found on page 15 of the paper, in Proposition 4.6, and $r_0$ is stated as "being equal to the support function in Example 4.2" from page 13.

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    $\begingroup$ you asked this before, why delete and ask again, losing all the comments you received? mathoverflow.net/q/463816/11260 --- if you wish to improve the questions, you do that by editing, not by deleting and reposting. $\endgroup$
    – Carlo Beenakker
    Commented Feb 15 at 19:31
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    $\begingroup$ Voted to close because this is a duplicate of an easily un-deletable question. $\endgroup$
    – Steven Landsburg
    Commented Feb 15 at 19:48
  • $\begingroup$ @CarloBeenakker The old question had been edited so much that the comments weren't at all connected to the latest content, and the comments hadn't been helpful in solving my problem anyway. Someone commented that due to all the changes and the lack of relevancy with old comments I should start over with a new question, so I did. $\endgroup$
    – Lawton
    Commented Feb 15 at 19:56
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    $\begingroup$ isn't the answer simply that $r_0(R)$ means a function of the two variables $x_1$ and $x_2$ that depends only on the combination $R=\sqrt{x_1^2+x_2^2}$, so you can write it equivalently as a function of a single variable; the equation you write down in terms of $r_0(g(\xi),g(\bar{\xi}))$ applies to any function $r_0(x_1,x_2)$, while the specific example 4.2 is for a radially symmetric function. $\endgroup$
    – Carlo Beenakker
    Commented Feb 15 at 21:07
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    $\begingroup$ just look at section 4.1: it refers to a rotationally symmetric function around the $x_3$ axis in terms of the radial parameter $R$, which must imply $R^2=x_1^2+x_2^2$, because that is what radial symmetry means $\endgroup$
    – Carlo Beenakker
    Commented Feb 15 at 22:10

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I am copying my comment in the answer box, to have more space to address the questions of the OP.

My answer is simply that $r_0(R)$ means a function of the two variables $x_1$ and $x_2$ that depends only on the combination $R=\sqrt{x_1^2+x_2^2}$, so you can write it equivalently as a function of a single variable; the equation on page 15 of the cited paper in terms of $r_0(g(\xi),g(\bar{\xi}))$ applies to any function $r_0(x_1,x_2)$, while the specific example 4.2 is for a radially symmetric function.

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