Given a simplex $S$ with a vertex $v$ by the solid angle at this vertex I mean the value $\hbox{vol}(B \cap S)/\hbox{vol}(B)$ where $B$ is a small enough ball centered at $v$ (for example, in the plane, 1/4 corresponds to $90^°$). Let $\rho_{min}(S)$ denotes the minimum of the solid angles of $S$.

My question is what values can $\rho_{min}(S)$ attain when it ranges over simplices in dimension $d$?

Comments: Since it is easy to let $\rho_{min}(S)$ to be close to $0$, the question above is equivalent with asking what is the least upper bound on $\rho_{min}(S)$.

In the plane, the question is trivial (the least upper bound is 1/6), thus more interesting cases occur when $d \geq 3$. In higher dimensions, it is relatively easy to show that $\rho_{min}(S) \leq \frac1{2(d+1)}$. But I suspect that this value should be much smaller.

A natural candidate for the least upper bound is the size of the solid angle in the regular simplex. Is this value the least upper bound?

Actually, I would be also happy with a weaker upper bound (than the one from the regular simplex) such as $\rho_{min}(S) \leq \frac{1}{2^d}$ (this is the solid angle in the $d$-cube).

  • $\begingroup$ Interesting question. Here is an initial quick thought: the cone associated with the solid angle at the vertex is the polar dual of the normal cone at the vertex (the cone of normal vectors at the vertex). For the normal cone, the answer to the equivalent question is easy, because the total solid angle of the normal cones must add to 1. There are bounds for the product of the volumes of a figure and its polar dual. Putting those two together might give you a result. $\endgroup$ – Yoav Kallus Jul 10 '13 at 19:13
  • $\begingroup$ Yes, from this follows that there is a normal cone at some vertex with angle at least $\frac{1}{d+1}$. I was not able to derive that its dual has small volume though. Could you be more specific with the bounds? $\endgroup$ – Martin Tancer Jul 10 '13 at 21:26
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    $\begingroup$ I'm actually not so sure now that these bounds are known. I was thinking of bounds on the Mahler product $|K||K^0|$, where $K$ is an origin-symmetric convex body, and I was thinking that there should be an equivalent form for cones. But now I'm not so sure, because I don't know if the product $|C||C^0|$ (where $|\cdot|$ is the solid angle) is a linear invariant for cones (is it?). In any case, if $a$ is the upper bound for this product, then the sum of inverses of solid angles is $\ge1/a$, and so the smallest solid angle is $\le a(n+1)$. Not sure if this would give you anything better than ... $\endgroup$ – Yoav Kallus Jul 10 '13 at 22:38
  • $\begingroup$ ... what you already have. $\endgroup$ – Yoav Kallus Jul 10 '13 at 22:41
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    $\begingroup$ maybe someone can finish the following argument: Let $AB$ be the diameter of the simplex. WLOG at least half of the vertices, $V_1, \dots, V_k$, are closer to $B$ than to $A$. Then the angle between $AB$ and $AV_i$, $i=1,\dots,k$, is at most $\pi/3$. I have a feeling that this should imply an upper bound $1/c^n$ on the size of the solid angle at $A$, for some $c>1$. $\endgroup$ – Jan Kyncl Jul 11 '13 at 6:52

This answer is based on a comment by J. Kynčl. It gives a bound roughly $\rho_{min}(S) \leq \left(\frac{1}{1.074}\right)^d$ which is already exponentially decreasing, but perhaps still far from the optimum. Thus, a better bound would still be of interest.

Let $AB$ one of the longest edges of $S$. Without loss of generality, at least half of the remaining vertices of $S$ are not farther from $B$ than from $A$. Let $V_1, \dots, V_k$ be such vertices ($k \geq (d-1)/2$) and $U_1, \dots, U_{\ell}$ be the remaining vertices (closer to $A$ than to $B$). As J. Kynčl points out, the angles $V_iAB$ are at most $60^°$. We also need that angles $U_iAB$ are at most $90^°$ since $BU_i$ is at most as long as $AB$.

Let $h$ be the hyperplane perpendicular to $AB$ passing through $A$ and $h^+$ be the halfspace containing $B$ with the boundary hyperplane $h$. Let $C$ be the cone with apex $A$ determined by $S$ (that is, our task is to determine which fraction of $C$ belongs to a ball $B(A,\varepsilon)$ with center $A$ and small radius $\varepsilon$. From the discussion above it follows that $C$ is fully contained in $h^+$. Furthemore, let $\kappa$ be the affine $(k+1)$-space determined by $A$, $B$, $V_1$, $\dots$, $V_k$. We also need another $(k+1)$-dimensional cone $C_{60}$ which is formed by all points $X$ in $\kappa$ such that the angle $XAB$ is at most $60^°$. From the discussion above it follows that $V_i \in C_{60}$, and consequently $C \cap \kappa \subseteq C_{60}$. It is not too difficult to compute that $\hbox{vol}_{k+1}(B(A,\varepsilon) \cap C_{60})/\hbox{vol}_{k+1}(B(A,\varepsilon)) \leq \left(\frac{\sqrt 3}2\right)^{k+1}$, where $\hbox{vol}_{k+1}$ is the $(k+1)$-dimensional volume in $\kappa$. (See the left picture below.)

enter image description here

Now, let us consider a small enough ball $B(A, \varepsilon)$ and let us estimate $\hbox{vol}(B(A, \varepsilon) \cap C)/ \hbox{vol}(B(A, \varepsilon))$. For this let us consider a $(k+1)$-space $\kappa'$ parallel with $\kappa$. The task is to show that

$$(*) \hskip{2cm} \frac{\hbox{vol}_{k+1}(B(A, \varepsilon) \cap C \cap \kappa')}{\hbox{vol}_{k+1}(B(A, \varepsilon) \cap \kappa')} \leq \left(\frac{\sqrt 3}2\right)^{k+1}.$$ As soon as we show $(*)$ we get the same bound on $\hbox{vol}(B(A, \varepsilon) \cap C)/ \hbox{vol}(B(A, \varepsilon))$ by the Fubini theorem.

In order to show $(*)$, let us first reailize that $C \cap \kappa'$ is either empty or it equals to $(C \cap \kappa) + Y$, where $Y$ is the intersection point of $\kappa'$ and the $\ell$-dimensional cone determined by $A$ and $U_1, \dots, U_\ell$ (here, for simplicity, we assume that $A$ is the origin). Thus, in particular, $Y \in h^+$ and $C \cap \kappa' \subseteq Y + C_{60}$. (See the right picture above.)

The final step is thus to show that $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Y + C_{60})) \leq \hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Z + C_{60}))$ where $Z$ is the center of $B(A, \varepsilon) \cap \kappa'$. This inequality is best shown by the picture below. (First shift $Y$ to $Y'$ on $h$. Then bound $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Y' + C_{60}))$ by decomposing $B(A, \varepsilon) \cap (Y' + C_{60})$ into two parts as on the middle and right picture.)

enter image description here

Finally, $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Z + C_{60}))/\hbox{vol}_{k+1}(B(A, \varepsilon)) \leq \left(\frac{\sqrt 3}2\right)^{k+1}$ as in the case of $C_{60}$. This gives the final bound $$\rho_{min}(S) \leq \left(\frac{\sqrt 3}2\right)^{(d+1)/2} \leq \left(\frac{1}{1.074}\right)^d.$$

  • $\begingroup$ nice! ... a small correction: in the 6th paragraph, "or it equals to $C \cap \kappa + Y$" should perhaps be "or it equals to $C \cap (\kappa + Y)$ $\endgroup$ – Jan Kyncl Jul 11 '13 at 18:29
  • $\begingroup$ No, I indeed mean $(C \cap \kappa) + Y$. I should say that $A$ coincides with the origin however; otherwise I would have to subtract $A$. Now it is corrected. Thanks for pointing out and thanks for your idea! $\endgroup$ – Martin Tancer Jul 11 '13 at 19:09
  • $\begingroup$ oh, now I see... both expressions are correct, but $(C \cap \kappa) + Y$ is stronger and actually useful. $\endgroup$ – Jan Kyncl Jul 11 '13 at 23:27

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