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For $n\geq 2$, let $\Delta^n$ be a regular $n$-dimensional simplex in $\mathbb{R}^n$ centered at the origin ${\bf 0}$ and inscribed in the unit sphere $S^{n-1}$.

Let ${\bf v}_0,{\bf v}_1,\ldots,{\bf v}_n\in S^{n-1}$ be the vertices of $\Delta^n$. It is well known that the angle $\alpha_n$ subtended by any two vertices of $\Delta^n$ through its center (i.e. ${\bf 0}$) is given by

$$\alpha_n=\textrm{arc}\cos\Big(-\frac{1}{n}\Big).$$

For $j=0,\ldots,n$, let $\Delta_j^n$ denote the convex hull in $\mathbb{R}^{n}$ of the $n+1$ points ${\bf 0},{\bf v}_0,{\bf v}_1,\dots,{\bf v}_{j-1},\widehat{{\bf v}_j},{\bf v}_{j+1},\ldots,{\bf v}_n$ (where the hat means omission). Thus $\bigcup_{j=0}^n\Delta_j^n=\Delta^n$.

Take two nonzero vectors ${\bf x},{\bf y}\in\mathbb{R}^n$ such that ${\bf x},{\bf y}\in\Delta_j^n$ for some $j=0,\ldots,n$.

Question 1: Is it true that

$$\frac{{\bf x}}{|{\bf x}|}\cdot \frac{{\bf y}}{|{\bf y}|}\geq-\frac{1}{n}?$$

Question 2: If so, does equality occur in the last inequality only if the vectors ${\bf x},{\bf y}$ are multiples of some ${\bf v}_i,{\bf v}_k$ (necessarily for $i\neq j\neq k\neq i$)?

Thank you.

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1 Answer 1

Yes to both questions. Let's fix $j$ and let $S=\{x/|x|:x\in \Delta_j^n\}$. The minimum in question 1 is equal to $\min\{x\cdot y:x,y\in S\}$. For a fixed $y$ the map $x\mapsto x\cdot y$ is the orthogonal projection onto the line through $y$. Every hyperplane of the form $\{x:x\cdot y=a\}$ meets either both $S$ and $\Delta_j^n$, or neither. (This can be justified by decomposing $x=ay+n$ with $y\cdot n=0$ and then scaling $n$ appropriately.) Therefore, we can consider $\min\{x\cdot y:x,y\in \Delta_j^n\}$ instead. But this is the minimum of a linear function on a polyhedron, so it must be attained at some vertex $v_k$. Since for every $y$ the minimum is realized by $x$ at a vertex, it follows that the minimum over both $x$ and $y$ is attained by a pair of vertices. This answers Q1.

Regarding Q2, suppose that the minimum is also attained by some $x',y'$ where $y'$ is not a vertex. We can still assume that $x$ is a vertex $v_k$, by the above. Since $y\mapsto x\cdot y$ also attains the minimum at some vertex $v_\ell$, it follows that $\Delta_j^n$ contains a line segment that is orthogonal to $v_k$ and contains $v_\ell$. This is ruled out by inspection of $\Delta_j^n$.

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