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Pick fixed $a=(a_1,a_2,\dots,a_d)\in\{\pm1\}^d$.

Consider map $F_a:\underbrace{\Bbb R^n\times\dots\times \Bbb R^n}_d\rightarrow\Bbb R^n$ given by $F(x_1,\dots,x_d)=\sum_{i=1}^da_ix_i$.

Denote $S_n\subset\Bbb R^n$ as unit sphere at origin, $B_r\subset\Bbb R^n$ to be closed ball (not sphere) of radius $r$ at origin.

Fix $K\in(0,1)$.

Algebraically or geometrically, what does the space $$F_a^{-1}\big(B_K\big)\bigcap\underbrace{S_n\times\dots\times S_n}_d=F_a^{-1}\big(B_K\big)\bigcap S_n^d$$ look like?

What is its measure in $S_n^d$?

Is there a $K\in(0,1)$ such that $\bigcup_{a\in\{\pm1\}^d} \Big(F_a^{-1}\big(B_K\big)\bigcap S_n^d\Big)\subsetneq S_n^d$?

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  • $\begingroup$ Also, technically speaking $F_a^{-1}$ of any set is automatically contained in $S^d_n$, so all intersections are redundant. I can see where this comes from, but it may make the question a bit more readable. $\endgroup$ Jul 19, 2015 at 11:16
  • $\begingroup$ The question certainly does not appear to have anything to do with algebraic geometry. If you have reason to believe it does, feel free to explain how and add the tag back. $\endgroup$ Jul 19, 2015 at 12:15
  • $\begingroup$ I thought if you want to describe the space nicely, we have to use polynomial relations? That was the reasoning. $\endgroup$
    – Turbo
    Jul 19, 2015 at 12:18

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