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PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 131, Number 10, Pages 3039–3042
S 0002-9939(03)06880-1
Article electronically published on May 5, 2003
THE MINIMUM NUMBER OF ACUTE DIHEDRAL ANGLES
OF A SIMPLEX
GANGSONG LENG
(Communicated by Wolfgang Ziller)
Abstract. For any n-dimensional simplex Ω ⊂ Rn , we confirm a conjecture
of Klamkin and Pook (1988) that there are always at least n acute dihedral
angles in Ω.
Let Ω be an n-simplex in Rn with vertices A1 , A2 , · · · , An+1 (i.e., Ω =
hA1 , A2 , · · · , An+1 i), let Ωi = hA1 , · · · , Ai−1 , Ai+1 , · · · , An+1 i denote its facet
which lies in a hyperplane πi , and let ei be the unit outer normal vector of πi (i =
1, 2, · · · , n + 1). Denote by θij the supplement of the angle between ei and ej ; we
call θij the dihedral angle between Ωi and Ωj . dihedral angles. In [1], Klamkin
Obviously, for any n-simplex, there exist n+1
2
shows that there are always at least three acute dihedral angles in any tetrahedron.
Further, he posed the question: can one generalize this result to an n-simplex?
Pook [1] conjectures that n is the required minimum number of acute dihedral
angles in any n-simplex.
In this paper, we confirm this supposition.
Theorem 1. There exist at least n acute dihedral angles in any n-simplex, and
there exists an n-simplex which has only n acute dihedral angles.
It is clear that Theorem 1 can be replaced equivalently by the following statement.
Theorem 10 . There exist at most 12 n(n−1) obtuse dihedral angles in any n-simplex,
and there exists an n-simplex which has only 12 n(n − 1) obtuse dihedral angles.
To prove Theorem 10 , we need the following three lemmas.
Lemma 1. Let {x1 , x2 , · · · , xn } be a given linearly independent set of vectors from
Rn . Then there exists an n-simplex Ω which takes kxx11 k kxx22 k , · · · , kxxnn k as n unit
outer normal vectors of facets.
Proof. Since x1 , x2 , · · · , xn are linearly independent, there exist n linearly independent vectors v1 , v2 , · · · , vn such that
hxi , vj i = δij kxi k2 (i, j = 1, 2, · · · , n),
Received by the editors May 31, 2000 and, in revised form, August 8, 2001.
2000 Mathematics Subject Classification. Primary 52A20.
Key words and phrases. Simplex, dihedral angle, dual basis.
This work was supported by the National Natural Sciences Foundation of China (10271071).
c
2003
American Mathematical Society
3039
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3040
GANGSONG LENG
where δij is the Kronecker delta symbol and h, i denotes the ordinary inner product
of Rn .
−−−−−→
Without loss of generality, we may assume that vi = An+1 Ai (i = 1, 2, · · · , n).
We now consider the n-simplex Ω = hA1 , A2 , · · · , An+1 i. Since xi ⊥ vj (i 6= j), it
follows that
xi ⊥ πi (i = 1, 2, · · · , n),
where πi is the hyperplane spanned by the facet Ωi . This completes the proof of
Lemma 1.
Lemma 2 (Yang and Zhang [3]). Let Ω and Ω0 be two n-simplices with dihedral
0
(1 ≤ i < j ≤ n + 1), respectively. If
angles θij and θij
0
(1 ≤ i < j ≤ n + 1),
θij ≤ θij
then Ω and Ω0 are similar, namely,
0
(1 ≤ i < j ≤ n + 1).
θij = θij
Let {u1 , · · · , un } and {s1 , · · · , sn } be two bases of Rn . They are called dual if
hui , sj i = δij (1 ≤ i ≤ j ≤ n).
It is well known that there exists a unique dual basis for a given basis of Rn .
Lemma 3. Let U = {u1 , · · · , un } be a basis of Rn and {s1 , · · · , sn } be the dual
basis of U. If hui , uj i < 0 (i 6= j), then hsi , sj i > 0 for all i, j.
Proof. We first consider n = 2. Suppose that hu1 , u2 i < 0 and {s1 , s2 } is the dual
basis. Then s1 = hs1 , s1 iu1 + hs1 , s2 iu2 . Hence
0 = hs1 , u2 i = hs1 , s1 ihu1 , u2 i + hs1 , s2 i ku2 k2 .
Thus hs1 , s2 i > 0.
Turning to the case of general n, let
H1 = {x ∈ Rn |hu1 , xi = 0}.
Then the vectors u2 , · · · , un are on the opposite side of H1 from u1 . For j ≥ 2, let
uj = yj + zj , yj ⊥ H1 , zj ∈ H1 ; then
yj =
hu1 , uj i
hu1 , uj i
u 1 , zj = u j −
u1 .
ku1 k2
ku1 k2
If i 6= j and i, j ≥ 2, then
hu1 , ui i
hu1 , uj i
u1 , uj −
u1 i
ku1 k2
ku1 k2
hu1 , ui ihu1 , uj i hu1 , ui ihu1 , uj i
+
= hui , uj i − 2
ku1 k2
ku1 k2
hu1 , ui ihu1 , uj i
= hui , uj i −
< 0.
ku1 k2
Note that sk ∈ H1 for k ≥ 2, and
hu1 , ui i
u1 , sk i = huj , sk i = δjk
hzj , sk i = huj −
ku1 k2
for k, j ≥ 2.
Since {z2 , · · · , zn } and {s2 , · · · , sn } are biorthognal sets of n − 1 vectors in the
(n − 1)-dimensional subspace H1 , they must be dual bases in H1 , and hzi , zj i < 0
hzi , zj i = hui −
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ACUTE DIHEDRAL ANGLES OF A SIMPLEX
3041
when i 6= j. If dim H1 > 2, then we can continue projecting to lower-dimensional
subspaces until {u1 , · · · , un } is mapped to a basis G = hgn−1 , gn i in a two-dimensional subspace with hgn−1 , gn i < 0 and with {sn−1 , sn } the dual basis of G. But
then in dimension 2, hsn−1 , sn i > 0, completing the proof.
Proof of Theorem 10 . We will first construct an n-simplex that has exactly 12 n(n−1)
obtuse dihedral angles. Denote by Ω an n-simplex all of whose dihedral angles are
acute (obviously, such an Ω exists, e.g., regular simplex).
Let u1 , u2 , · · · , un be unit outer normal vectors of n facets of Ω. Then
hui , uj i < 0, i 6= j,
and U = {u1 , · · · , un } is a basis of Rn . Further let {s1 , · · · , sn } be the dual basis of
U. By Lemma 1, there exists an n-simplex Ω which takes e1 , · · · , en as unit outer
normal vectors of n facets, where ei = kssii k . By Lemma 3, we have
hei , ej i =
(1.1)
1
hsi , sj i > 0, 1 ≤ i, j ≤ n.
ksi k ksj k
Let en+1 be the unit outer normal vector of the (n + 1)-th facet of Ω, and fi the
(n − 1)-dimensional volume of the facet Ωi of Ω. By Minowski’s realization theorem
(see [2, p. 390]), we have
n+1
X
fi ei = 0.
i=1
Therefore, for j ∈ {1, 2, · · · , n},
hen+1 , ej i = h−
n
X
i=1
(1.2)
=−
n
X
i=1
ei
fi
, ej i
fn+1
hei , ej i
fi
fn+1
< 0.
By (1.1) and (1.2), we know that Ω has 12 n(n − 1) obtuse dihedral angles and n
acute dihedral angles.
Assume that there exists an n-simplex whose obtuse dihedral angles are more
than 12 n(n − 1). Without loss of generality, we may assume that
α1 , · · · , α 12 n(n−1) , α 12 n(n−1)+1 , · · · , α 12 n(n−1)+j
are its obtuse dihedral angles, while α 12 n(n−1)+j+1 , · · · , α 12 n(n+1) are acute. Thus,
we can construct a rectangular n-simplex with acute dihedral angles α 12 n(n−1)+j+1 ,
· · · , α 12 n(n+1) . In fact, let {e1 , · · · , en } be a standard base of Rn , and let
s
en+1 = −
1−λ
e1 −· · ·−
j
s
1−λ
ej −cos α 12 n(n−1)+j+1 ej+1 −· · ·−cos α 12 n(n+1) en ,
j
where
λ = cos2 α 12 n(n−1)+j+1 + · · · + cos2 α 12 n(n+1) .
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3042
GANGSONG LENG
Then the rectangular simplex that takes e1 , · · · , en , en+1 as n+1 unit normal vectors
of facets has n acute dihedral angles
s
s
1−λ
1−λ
, · · · , arccos
, α 12 n(n−1)+j+1 , · · · , α 12 n(n+1) .
arccos
j
j
By Lemma 2, this is contradictory. This completes the proof.
Acknowledgements
The author would like to thank the referees for many valuable suggestions and
comments.
References
[1] M. S. Klamkin, L. P. Pook, Acute dihedral angles, Problems 1281, Math. Mag. 61, 5 (1988),
320.
[2] R. Schneider, Convex Bodies: the Brunn–Minkowski Theory, Cambridge University Press,
Cambridge, 1993. MR 94d:52007
[3] L. Yang and J. Z. Zhang, A necessary and sufficient condition for embedding a simplex with
prescribed dihedral angles in E n , Acta Math. Sinica 26 (1983), 250–256. MR 84g:51023
Department of Mathematics, Shanghai University, Shanghai, 200436, People’s Republic of China
E-mail address: [email protected]
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