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Say that a set $X \subset \mathbf{S}^{d-1}$ is ortho-closed if for any set $\{x_1,\dots,x_{d-1}\} \subset X$ there exists $x \in X$ such that $\langle x,x_i \rangle = 0$ for $i=1,\dots,d-1$. (Apologies for the terrible name - other suggestions are welcome.) Thus, as a silly example, the set of standard basis vectors is ortho-closed.

My rather vague question is, what (non-trivial) finite ortho-closed sets are there? Much more concretely (and actually what I'm most interested in), with $d=3$, is there a finite ortho-closed set containing the normalizations of the seven non-zero 0-1 vectors in $\mathbb{Z}^3$? (Or indeed is there a finite ortho-closed set containing an arbitrary fixed set?)

It seems to me that the answer ought to be `no', and that one ought to be able to construct an infinite sequence of vectors that would have to be in any such set. However I can't yet see how to do this, and it occurred to me that this is a fairly natural property which might already be known. Thus, any references would also be greatly appreciated.

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  • $\begingroup$ Probably a silly remark, but you might as well take unit vectors (replace each $x$ with $x/\|x\|$). Then you're looking at sets of vectors on the unit ($d-1$)-sphere, which might help with a density/measure/compactness argument. $\endgroup$ – Joe Silverman Oct 23 '15 at 13:45
  • $\begingroup$ Thanks Joe, that's a much clearer formulation - I've edited the question. $\endgroup$ – paul Oct 23 '15 at 13:55
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We identify opposite vectors (or we may factor $S^{2}$ by $\{\pm 1\}$).

We claim that all finite ortho-closed sets on $S^2$ are coplanar after removing one vector. (I think the same argument works in higher dimension as well.)

Firstly, one useful property: If $a,b,c\in X$, where $c$ is not orthogonal to $a$, then the (normalized) projection of $c$ onto the plane spanned by $a$ and $b$ also lies in $X$. Indeed, $X$ contains the vector $x\perp a,b$; next, it contains the vector $y\perp x,c$; finally, it contains $z\perp x,y$, and $z$ is the required projection.

Now, starting from any two vectors $a,x\in X$, we find in $X$ the vectors $b\perp a,x$ and $c\perp a,b$, thus $X$ contains an orthogonal triple $a,b,c$. If all vectors from $X$ distinct from $a,b,c$ lie in exactly one of the planes $\langle a,b\rangle$, $\langle a,c\rangle$, and $\langle b,c\rangle$, then our claim is true. Otherwise, using the property, we find two such planes containing some other vectors from $X$. We ay assume that these vectors are $u_0\in\langle a,b\rangle$ and $v_0\in\langle a,c\rangle$. Notice that $a$ forms acute angles with $u_0$ and $v_0$.

Now, let $a_0$ be the projection of $a$ onto $\langle u_0,v_0\rangle$, and let $u_1$ and $v_1$ be the projections of $a_0$ onto $\langle a,b\rangle$ and $\langle a,c\rangle$, respectively. Then $\angle (a,u_1)<\angle (a,u_0)$ and $\angle(a,v_1)<\angle(a,v_0)$. Repeating the process we find an infinite number offvectors in $X$.

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You can get a nice combinatorial proof of Ilya's result if you use the Friendship Theorem.

As Ilya did, we can identify opposite vectors. Your condition says that for any two distinct vectors $v_1$ and $v_2$ of $X$, there exists a vector $v \in X$ which is orthogonal to both $v_1$ and $v_2$. However, since you are in $\mathbb{R}^3$, the vector $v$ must be unique.

Now let us consider the vectors in our set $X$ to be vertices of a graph $G$, such that two vectors are adjacent if they are orthogonal. Then, by the above, this graph has the property that every pair of vertices has a unique common neighbor. The Friendship Theorem says that any such graph is a windmill graph, meaning that $G$ consists of a vertex $v$ adjacent to all other vertices of $G$, and the remaining vertices form a perfect matching.

Translating the orthogonalities encoded in the edges of $G$ back to the vector setting gives you exactly Ilya's result.

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  • $\begingroup$ Wow! Occasionally, I did not know this theorem, thanks! $\endgroup$ – Ilya Bogdanov Oct 23 '15 at 20:42
  • $\begingroup$ Gosh, that is very pretty! $\endgroup$ – paul Oct 26 '15 at 9:51
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The $0$-$1$ vectors for $d=3$ are indeed not contained in a (finite) ortho-closed set. I'll work with non-normalized vectors to represent directions. We must put $(1,-1,1)\perp (1,1,0), (0,1,1)$ into our set, and then the direction orthogonal to $(1,1,0), (1,-1,1)$ is $(-1,1,2)$. Now in general, if we were forced to put any vector into the set, then we also obtain its permutations, by symmetry.

Thus the forced presence of $v_n=(-1,1,n)$ gives us $(1,n+1,-1)$ (because this is the direction orthogonal to $v_n$ and $(1,0,1)$), but then also $v_{n+1}$. All $v_n$ must be in our set.

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  • $\begingroup$ Thanks Christian, that's exactly the sort of construction that I was looking for. $\endgroup$ – paul Oct 23 '15 at 16:51

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