Euclidean ball and sphere

Let

  •   B^n\ :=\ \,\{x\in R^n : |x| \le 1\}
  •   S^{n-1} := \{x\in R^n : |x| = 1\}
  •   \Delta^n := \{(x\ x) : x \in B^n\}

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I’d like to compute explicitly the linearly determined function  F : B^n\times B^n \setminus \Delta^n \rightarrow S^{n-1} such that  F(a\ b) = b  whenever  b\in S^{n-1}. I want to show its well known continuity.

Let  a\ b\ \in\ B^n,  and  a \ne b.  Let

c_t\ :=\ (1-t)\cdot a + t\cdot b

I’d like to find  t := t_a \le 0  such that  |c_t|=1,  i.e.  c_t^2=1.  This last equation is equivalent to the following ones:

  •   (1-t)^2\cdot a^2 + 2\cdot(1-t)\cdot t\cdot a\cdot b + t^2\cdot b^2\ \ =\ \ 1
  •   (a - b)^2\cdot t^2 + 2\cdot a\cdot (b-a)\cdot t + a^2-1\ \ =\ \ 0
  •   t^2 + 2\cdot\frac{a\cdot (b-a)}{(a-b)^2}\cdot t + \frac{a^2-1}{(a-b)^2}\ \ =\ \ 0

Define   \delta\ :=\ \left(\frac{a\cdot (b-a)}{(a-b)^2}\right)^2 + \frac{1 - a^2}{(a-b)^2}.  Both summands are non-negative (since  |a| \le 1).  Thus  \delta  is a non-negative real number. The two solutions  t_a\ t_b  of the above quadratic equation are given by

  •   t_a := -\frac{a\cdot (b-a)}{(a-b)^2} - \delta^{\frac 12}
  •   t_b := -\frac{a\cdot (b-a)}{(a-b)^2} + \delta^{\frac 12}

Thus  t_a \le 0  (of course), and  t_a = 0  ⇔  |a| = 1   —   indeed, when   |a| = 1 \ge |b|   then

a\cdot(b-a) = a\cdot b - a^2 = a\cdot b - 1 \le |a|\cdot|b| - 1 \le 0

and it’s clear that  t_a = 0. The inverse implication is simple.

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