On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube
It is proved the existence of a regular hypersimplex inscribed into the (4n-1)-dimensional cube under condition that some system of 4n-2 algebraic equations with 4n-2 unknown quantities y0, y'0, y1; y'1,..., y2n-2, y'2n-2 has at least one solution with a real value of y0 or any other...
Збережено в:
| Опубліковано в: : | Журнал математической физики, анализа, геометрии |
|---|---|
| Дата: | 2006 |
| Автор: | |
| Формат: | Стаття |
| Мова: | English |
| Опубліковано: |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2006
|
| Онлайн доступ: | https://nasplib.isofts.kiev.ua/handle/123456789/106581 |
| Теги: |
Додати тег
Немає тегів, Будьте першим, хто поставить тег для цього запису!
|
| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Цитувати: | On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube / A.I. Medianik // Журнал математической физики, анализа, геометрии. — 2006. — Т. 2, № 1. — С. 62-72. — Бібліогр.: 4 назв. — англ. |
Репозитарії
Digital Library of Periodicals of National Academy of Sciences of Ukraine| id |
nasplib_isofts_kiev_ua-123456789-106581 |
|---|---|
| record_format |
dspace |
| spelling |
Medianik, A.I. 2016-09-30T19:22:29Z 2016-09-30T19:22:29Z 2006 On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube / A.I. Medianik // Журнал математической физики, анализа, геометрии. — 2006. — Т. 2, № 1. — С. 62-72. — Бібліогр.: 4 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106581 It is proved the existence of a regular hypersimplex inscribed into the (4n-1)-dimensional cube under condition that some system of 4n-2 algebraic equations with 4n-2 unknown quantities y0, y'0, y1; y'1,..., y2n-2, y'2n-2 has at least one solution with a real value of y0 or any other yi ≠ 0, i ≥ 1. en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України Журнал математической физики, анализа, геометрии On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube Article published earlier |
| institution |
Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| collection |
DSpace DC |
| title |
On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube |
| spellingShingle |
On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube Medianik, A.I. |
| title_short |
On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube |
| title_full |
On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube |
| title_fullStr |
On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube |
| title_full_unstemmed |
On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube |
| title_sort |
on existence of a regular hypersimplex inscribed into the (4n - 1)-dimensional cube |
| author |
Medianik, A.I. |
| author_facet |
Medianik, A.I. |
| publishDate |
2006 |
| language |
English |
| container_title |
Журнал математической физики, анализа, геометрии |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| format |
Article |
| description |
It is proved the existence of a regular hypersimplex inscribed into the (4n-1)-dimensional cube under condition that some system of 4n-2 algebraic equations with 4n-2 unknown quantities y0, y'0, y1; y'1,..., y2n-2, y'2n-2 has at least one solution with a real value of y0 or any other yi ≠ 0, i ≥ 1.
|
| issn |
1812-9471 |
| url |
https://nasplib.isofts.kiev.ua/handle/123456789/106581 |
| citation_txt |
On Existence of a Regular Hypersimplex Inscribed into the (4n - 1)-Dimensional Cube / A.I. Medianik // Журнал математической физики, анализа, геометрии. — 2006. — Т. 2, № 1. — С. 62-72. — Бібліогр.: 4 назв. — англ. |
| work_keys_str_mv |
AT medianikai onexistenceofaregularhypersimplexinscribedintothe4n1dimensionalcube |
| first_indexed |
2025-11-26T11:41:47Z |
| last_indexed |
2025-11-26T11:41:47Z |
| _version_ |
1850619779265593344 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2006, vol. 2, No. 1, pp. 62�72
On Existence of a Regular Hypersimplex Inscribed
into the (4n� 1)-Dimensional Cube
A.I. Medianik
B. Verkin Institute for Low Temperature Physics and Engineering
National Academy of Sciences of Ukraine
47 Lenin Ave., Kharkov, 61103, Ukraine
E-mail:medianik@ilt.kharkov.ua
Received March 24, 2005
It is proved the existence of a regular hypersimplex inscribed into the
(4n� 1)-dimensional cube under condition that some system of 4n� 2 alge-
braic equations with 4n�2 unknown quantities y0; y
0
0
; y1; y
0
1
; : : : ; y2n�2; y
0
2n�2
has at least one solution with a real value of y0 or any other yi 6= 0, i � 1.
Key words: multidimensional cube, regular simplex, Hadamard's matrix,
circulant matrix, antipodal n-gons, generating polinomial, idempotancy,
necessary and su�cient conditions.
Mathematics Subject Classi�cation 2000: 05B20, 52B.
1. Introduction
We obtained in [1] necessary and su�cient conditions for the existence of
Hadamard's matrix of order 4n of the half-circulant type, which had introduced
in [2], and announced the existence theorem for a regular hypersimplex inscribed
into the (4n � 1)-dimensional cube. The theorem will be proved in this paper.
But �rst we recall some results from [1] and their consequences.
Convex n-gons P and P 0, inscribed into the regular (2n�1)-gon, are said to be
antipodal, if the total number of their diagonals and sides of the same length equals
n for all, without exception, admissible lengths. If the long radius of the regular
(2n � 1)-gon is 1 and its centre coincides with the origin of coordinates of the
complex plane, it is no loss of generality to assume that its vertices are monomial
z
k, k = 0; 1; 2; : : : ; 2n � 2, where z = e
2�i
2n�1 . Then n-gon P is represented by
the generating polynomial p(z) =
P2n�2
k=0 xkz
k, where xk = 1 if the vertex of
the regular (2n � 1)-gon with number k belongs to P , and xk = 0 in otherwise.
c
A.I. Medianik, 2006
On Existence of a Regular Hypersimplex...
Accordigly, n-gon P
0 is represented by a polynomial p
0(z) =
P2n�2
k=0 x
0
kz
k. It
follows from Lemma 1 in [1] that
jpj2 = n+ 2
n�1X
k=1
dk cos
2�k
2n� 1
;
where dk is the number of equal diagonals and sides of P , for which the vision angle
(from the regular (2n�1)-gon centre) equals 'k = 2�k
2n�1 , k = 1; 2; : : : ; n�1. There
is similar equality (with replacement dk by d
0
k) for the generating polynomial p0(z).
Since for antipodal n-gons P and P
0 by de�nition dk + d
0
k = n, 1 � k � n � 1,
their generating polynomials satisfy relation jpj2 + jp0j2 = n by Theorem 3 from
[1].
Hadamard's matrix H of order 4n (every its entry equals 1 or �1 and its rows
are pairwise orthogonal) is said to be half-circulant if it has the following form:
H =
0
BBBB@
1 � � � 1 � � �
... A
... B
1 � � � �1 � � �
... B
... �A
1
CCCCA ; (1)
where A and B are square matrices of order 2n� 1, being by circulants [2]. More
precisely, A is an usual circulant [3, p. 272], which we will call the right circulant.
And B is the left circulant, obtained (in contrast to A) by the cyclic permutation
of entries of each subsequent row, starting from the �rst, on the left (but not to
the right as in usual circulant). Hence, all the entries of the principal diagonal of
the right circulant equal the �rst entry of the �rst row, while all the entries of the
secondary diagonal of the left circulant equal to the last entry of the same row.
In [1, p. 58] the following necessary and su�cient condition for the existence
of half-circulant Hadamard matrix of order 4n was obtained.
Theorem 1. The half-circulant Hadamard matrix of order 4n exsists if and
only if there exist antipodal n-gons P and P
0
inscribed into the regular (2n� 1)-
gon.
Recall that the right circulant A = (aim), 1 � i;m � 2n� 1 is determined by
coe�cients xk, k = 0; 1; 2; : : : ; 2n� 2 of the polynomial p(z), which represents P ,
while the left circulant B = (bim), 1 � i;m � 2n� 1 is determined by coe�cients
x
0
k of the p0(z), which represents P 0, namely:
aim = 1� 2xjm�ij; bim = 1� 2x0jm+i�2j: (2)
Here the sign of modulus means the least nonnegative residue modulo 2n� 1.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 63
A.I. Medianik
The rows of the matrix �H, obtained from Hadamard matrix H by removing
its �rst column (consisting from +1 only), are, obviously, coordinates of vertices
of a regular hypersimplex in E4n�1, inscribed into a hypercube with edge 2, whose
centre coincides with the origin. This implies
Corollary. In order that one can inscribed a regular hypersimplex into a
cube of dimension 4n � 1 it is su�cient that there exist antipodal convex n-gons
inscribed in the regular (2n� 1)-gon.
Theorem 1 gives geometrical necessary and su�cient conditions for the exis-
tence of half-circulant Hadamard matrices. To �nd adequate analitical condi-
tions a homogeneous polynomial of third degree w = w(y) = w
3
0 +
Pn�1
m=1(w
3
m +
w
3
2n�1�m) was introduced in [1]. Here y is the vector with coordinates y0, y1; : : :,
y2n�2 and
w0 =
1
p
2n� 1
(y0 +
p
2
n�1X
j=1
yj);
wm =
1
p
2n� 1
[y0 +
p
2
n�1X
j=1
(yj cos
2�mj
2n� 1
+ y2n�1�j sin
2�mj
2n� 1
)]; (3)
w2n�1�m =
1
p
2n� 1
[y0 +
p
2
n�1X
j=1
(yj cos
2�mj
2n� 1
� y2n�1�j sin
2�mj
2n� 1
)];
and also the matrix of system (3) is orthogonal T = (tik), det T = 1. Thus, our
system can be written in the form: wi =
P2n�2
k=0 tikyk, i = 0; 1; 2; : : : ; 2n� 2.
Note that in [1, Th. 6] it was established what the surface of third order
w = n is a non-reducible smooth hypersurface in the projective space P
2n�1.
Another characteristic property of the polynomial w(y) is as follows: the system
of equations
y =
1
3
rw; (4)
where rw is the vector with coordinates @w
@yi
, i = 0; 1; 2; : : : ; 2n�2, is idempotent.
This means that the solution of (4) with respect to w
2
i = 1
3
@w
@wi
has the form:
w
2
i =
P2n�2
k=0 tikyk = wi, i.e., every wi is equal to its square, and consequently it
can assume only values 0 or 1. This follows from the next property: the matrix
of the arising linear system coincides with the trasposed matrix T
0 of system
(3), which is inverse of T , as det T = 1 (see [1, p. 62]). Moreover, if in the left
hand side of (4) the vector y with coordinate y0; y1; : : : ; y2n�2 is replaced by a
vector with coordinates y0 + Æ; y1; : : : ; y2n�2, then it follows from the idempotent
property that the solution of system (4) with respect to w
2
i has next form: w
2
i =P2n�2
k=0 tikyk + ti0Æ = wi +
Æp
2n�1 for all i = 0; 1; 2; : : : ; 2n � 2, because in (3)
64 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
On Existence of a Regular Hypersimplex...
all coe�cients ti0 at y0 are equal to 1p
2n�1 . In particular, if yi = 0 for all i, then
w
2
i = Æp
2n�1 for i = 0; 1; 2; : : : ; 2n� 2.
Assume that in (3) wi = xi, where xi is 0 or 1, i.e., the values which every
wi can take for any solution y0; y1; : : : ; y2n�2 of system (4). Then we may obtain
the following solution of (3), viewed as a linear system with respect to unknown
quantities y0; y1; : : : ; y2n�2:
y0 =
1
p
2n� 1
2n�2X
i=0
xi;
yj =
r
2
2n� 1
[x0 +
n�1X
m=1
(xm + x2n�1�m) cos
2�jm
2n� 1
]; (5)
y2n�1�j =
r
2
2n� 1
n�1X
m=1
(xm � x2n�1�m) sin
2�jm
2n� 1
:
Since xi = 0 or 1 for all i = 0; 1; 2; : : : ; 2n�2, it follows that system (4) has 22n�1
solutions. Besides, all its solutions are real (see Lem. 2 in [1]). Among them
there are C
n
2n�1 solutions such that
P2n�2
i=0 xi = n. Obviously, these solutions
induce convex n-gons, inscribed into the regular (2n � 1)-gon, with generating
polynomials
P2n�2
k=0 xkz
k.
Now, let y = fy0; y1; : : : ; y2n�2g and �y = f�y0; �y1; : : : ; �y2n�2g be two solutions
of system (4), moreover, the �rst one has form (5) and the second one is given by
(5) in which xi, i = 0; 1; 2; : : : ; 2n� 2 are replaced by �xi, where �xi = 1� xi (this
is possible, since it was proved earlier that every xi is 0 or 1, as well as 1� xi).
Lemma 1. If y = fy0; y1; : : : ; y2n�2g is a solution of idempotent system (4),
then �y = f
p
2n� 1� y0;�y1; : : : ;�y2n�2g is also its solution.
Indeed, summing termwise equations (5) for solution y and similar equations
for solution �y, we have in view of the conditions xi + �xi = 1 for all admissible i
and the identity 1
2
+
Pn�1
m=1 cos
2�jm
2n�1 � 0 (see [1, p. 55]):
y0 + �y0 =
1
p
2n� 1
2n�2X
i=0
(xi + �xi) =
p
2n� 1;
yj + �yj =
r
2
2n� 1
(1 + 2
n�1X
m=1
cos
2�jm
2n� 1
) = 0;
y2n�1�j + �y2n�1�j =
r
2
2n� 1
n�1X
m=1
0 � sin
2�jm
2n� 1
= 0;
whence the Lemma 1 follows.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 65
A.I. Medianik
Solutions of system (4), corresponding by (5) to sets fxig and f�xig, for which
xi + �xi = 1 at any admissible i, are said to be conjugate. They possess some
property, which follows immediately from the relations obtained above.
Lemma 2. If a solution y = fy0; y1; : : : ; y2n�2g of idempotent system (4) sat-
is�es the equality y0+
P2n�2
i=1 �iyi = 0, where each �i is any real number, then the
conjugate solution �y = f�y0; �y1; : : : ; �y2n�2g satis�es the equality �y0+
P2n�2
i=1 �i�yi =p
2n� 1. Conversely, if y0 +
P2n�2
i=1 �iyi =
p
2n� 1, then �y0 +
P2n�2
i=1 �i�yi = 0.
It should be observed, that in [1, equalities (14)] formulas for all derivatives
involved in (4), were given. We need now one of them, with index i = 0, namely:
@w
@y0
=
3
p
2n� 1
2n�2X
i=0
y
2
i ; (6)
which one can also obtain with the aid of (3).
Recall at last that by Lemma 3 from [1], convex n-gons P and P
0, inscribed
in the regular (2n� 1)-gon and represented by polynomials p(z) =
P2n�2
k=0 xkz
k
and p
0(z) =
P2n�2
k=0 x
0
kz
k, are antipodal ones, if and only if the coordinates of
the corresponding vectors y and y
0, represented by the equality (5) and similar
equality with primes, satisfy for all j = 1; 2; : : : ; n� 1 antipodal conditions
y
2
j + y
2
2n�1�j + y
0 2
j + y
0 2
2n�1�j =
2n
2n� 1
: (7)
Hence we have obtained analytical necessary and su�cient conditions for the
existence of Hadamard's matrices (see [1, Th. 5]).
Theorem 2. A half-circulant Hadamard matrix of order 4n exists if and only
if system (4) has two solutions y = fy0; y1; : : : ; y2n�2g and y0 = fy00; y
0
1; : : : ; y
0
2n�2g
such that y0 = y
0
0 = np
2n�1 and so that the rest coordinates of vectors y and y
0
should satisfy antipodal conditions (7).
The solutions, given in the theorem, correspond, obviously, to points of the
cubic surface w = n.
From Theorem 2 it follows that the question about the existence of half-
circulant Hadamard matrices of order 4n, and consequently the existence of a
regular hypersimplex inscribed into the (4n � 1)-dimensional cube, reduces to
the question about the solvability in the real number of a system of quadratic
equations, in which the number of equations (which equals 5n � 3) exceed the
number of unknown quantities (4n� 2). Thus, the existence of even one solution
is not guaranteed. Therefore we modify it so that these numbers should coincide,
that will provide the existence of complex solutions and, eventually, the existence
of real solutions of the modi�ed system under some additional conditions.
66 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
On Existence of a Regular Hypersimplex...
To this end let us introduce by analogy with w = w(y) another polynomial
w
0 = w
0
0
3
+
Pn�1
m=1(w
0
m
3
+ w
0
2n�1�m
3
), where w0
i =
P2n�2
k=0 tiky
0
k, and consider the
following system of 4n� 2 equations:
8>>>>>>>><
>>>>>>>>:
@w
@yi
= 3yi; i = 1; 2; : : : ; 2n� 2;
(y0 +
p
2
n�1P
j=1
yj)
2(y0 +
p
2
n�1P
j=1
yj �
p
2n� 1)2 + (y00 �
np
2n�1)
4
+
n�1P
j=1
(y2j + y
2
2n�1�j + y
0 2
j + y
0 2
2n�1�j �
2n
2n�1)
2 = 0;
@w0
@y0
i
= 3y0i; i = 0; 1; : : : ; 2n� 2:
(8)
This system consists from two subsystems similar to (4), namely, one is in
terms of the vector y = fy0; y1; : : : ; y2n�2g (�rst 2n � 2 equations) and another
is in terms of the vector y0 = fy00; y
0
1; : : : ; y
0
2n�2g (last 2n� 1 equations), and also
common for its the middle equation, which takes into account antipodal conditions
(7). In [1, p. 66] the following existence theorem was announced.
Theorem 3. The system of 4n� 2 nonhomogeneous algebraic equations with
4n� 2 unknown quantities y0; y
0
0; : : : ; y2n�2; y
0
2n�2 (8) has for any integer n > 1
24n�1 solutions, with regard their multiplicity. If y0 takes a real value at least for
one of this solutions, then a half-circulant Hadamard matrix of order 4n exists
and into the (4n � 1)-dimensional cube one can inscribe a regular simplex of the
same dimension.
2. Proof of the existence theorem
Let us transform system (8) to the homogeneous one by two sets of variables
with the aid of additional coordinates y2n�1 and y
0
2n�1 (with regard that w and
w
0, by de�nition, are homogeneous polynomials of third degree).
8>>>>>>>>>><
>>>>>>>>>>:
@w
@yi
= 3yiy2n�1; i = 1; 2; : : : ; 2n� 2;
(y0 +
p
2
n�1P
j=1
yj)
2 � (y0 +
p
2
n�1P
j=1
yj �
p
2n� 1y2n�1)2y0 42n�1
+(y00 �
np
2n�1y
0
2n�1)
4
y
4
2n�1 +
n�1P
j=1
[(y2j + y
2
2n�1�j)y
0 2
2n�1
+ (y0 2j + y
0 2
2n�1�j)y
2
2n�1 �
2n
2n�1 y
2
2n�1y
0 2
2n�1]
2 = 0;
@w0
@y0
i
= 3y0iy
0
2n�1; i = 0; 1; : : : ; 2n� 2:
(9)
Actually, every equation of system (9) is homogeneous both with respect to vari-
ables y0; y1; : : : ; y2n�1 and also variables y
0
0; y
0
1; : : : ; y
0
2n�1, and besides the �rst
2n� 2 equtions are homogeneous of second degree with respect to the �rst set of
variables and zero degree with respect to the second ones, but for the last 2n� 1
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 67
A.I. Medianik
equations opposite property is valid. Therefore one can consider system (9) on the
product of two projective spaces P 2n�1�P
0 2n�1(y0; : : : ; y2n�1; y00; : : : ; y
0
2n�1). By
well-known Bezout's theorem (see [4, p. 268�269]) this system of 4n�2 equations
with 4n unknown quantities has 24n�1 solutions with regard their multiplicity (if
the number of its solutions is �nite, what will follow from the present proof), that
equals the intersection index
X
ki1 � � � ki2n�1
k
0
j1
� � � k0j2n�1
= 22n � 22n�1 = 24n�1;
where km, k
0
m are the homogeneous degrees of the equation with number m in
(9) with respect to coordinates in P
2n�1 or P
0 2n�1 respectively and the sum
is over all permutations (i1 : : : i2n�1 j1 : : : j2n�1) of integers 1; 2; : : : 4n � 2 with
i1 < i2 < � � � < i2n�1; j1 < j2 < � � � < j2n�1, where a nonzero contribution gives
only one summand, with i2n�1 = 2n � 1. Underline that only nonzero solutions
are taken into account, i.e., the solutions in which are not vanish all coordinates
into P
2n�1 or those into P
0 2n�1, and all proportional one another solutions are
taken just the same solution [4, p. 272].
First, let us prove that for any nontrivial solution of homogeneous system (9)
both y2n�1 and y
0
2n�1 do not vanish, i.e., every its solution induces the corre-
sponding one of nonhomogeneous system (8).
Actually, it follows from the smoothness of the cubic surface w0 = n in P 0 2n�1
(see Introduction) that y
0
2n�1 6= 0 (in the opposite case, all @w0
@y0
i
are vanishing
according to (9), and consequently there are singular points on this hypersurface).
Therefore one can assume y
0
2n�1 = 1. Then last 2n � 1 equations of system
(9) coincide with corresponding equations of system (8), which form a system
of type (4). By virtue of its idempotency, we have w
0 2
i =
P2n�2
k=0 tiky
0
k = w
0
i,
i = 0; 1; 2; : : : ; 2n � 2. Thus, this subsystem of system (8) has, as the system
of type (4), exactly 22n�1 real solutions, and this equals the degree product of
corresponding equations of (9).
The remaining 2n � 1 equations of system (9) one can consider as a system
in P
2n�1, if in its last equation y
0
0; y
0
1; : : : ; y
0
2n�2; y
0
2n�1 will be replaced by any
solution of the lower subsystem of system (9) with y
0
2n�1 = 1.
Let us assume that y2n�1 = 0 for a nontrivial solution of the considered system
in P
2n�1. Then we obtain8><
>:
@w
@yi
= 0; i = 1; 2; : : : ; 2n� 2;
(y0 +
p
2
n�1P
j=1
yj)
4 +
n�1P
j=1
(y2j + y
2
2n�1�j)
2 = 0:
(10)
Introduce in this system one more equation (number one): @w
@y0
= 3t2
p
2n� 1,
where t is an unknown parameter, whose value will determine later.
All these equations, except the last, form a system of type (4) with the vector
68 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
On Existence of a Regular Hypersimplex...
y = ft2
p
2n� 1; 0; : : : ; 0g. We have in view of the idempotent property of system
(4): w2
i =
P2n�2
k=0 tik �0+ti0 �t2
p
2n� 1 = t
2, i.e., for any admissible value of index
i wi = �t. Let us assume that xi = �wi, where �wi is one of admissible values of
wi. Then we obtain from (5) yi = cit, i = 0; 1; : : : ; 2n�2, where all coe�cients ci
are real numbers and besides c0 6= 0 (since any sum of 2n� 1 summands, which
equals 1 or �1, obviously, cannot vanish). Substituting these values yi into the
last equation of system (10), we have
t
4[(c0 +
p
2
n�1X
j=1
cj)
4 +
n�1X
j=1
(c2j + c
2
2n�1�j)
2] = 0;
hence t = 0, and so y0 = y1 = : : : = y2n�2 = 0. Since by assumption y2n�1 = 0
too, then the given solution is trivial. Obtained contradiction proves that y2n�1 6=
0, and consequently one can assume that y2n�1 = 1. With regard that y02n�1 = 1
too, every solution of homogeneous system (9) induces the corresponding solution
of nonhomogeneous system (8), i.e., the last one has 24n�1 solutions. This is the
�rst to be proved.
Let us assume z = y0 +
p
2
Pn�1
j=1 yj and prove that all solutions of the upper
subsystem of system (8) satisfy the following relation:
@w
@y0
= 3(y0 � z +
z
2
p
2n� 1
): (11)
In fact, summing termwise the �rst n�1 equations of system (8) and di�eretiating
the polynomial w, we obtain
3
n�1X
j=1
yj = 3w2
0
p
2(n� 1)
p
2n� 1
+
n�1X
m=1
[3(w2
m + w
2
2n�1�m)
p
2
p
2n� 1
n�1X
j=1
cos
2�mj
2n� 1
]
=
3
p
2
p
2n� 1
[(n�
1
2
)w2
0 �
1
2
(w2
0 +
n�1X
m=1
(w2
m + w
2
2n�1�m))]
=
1
p
2
[
3
p
2n� 1
(y0 +
p
2
n�1X
j=1
yj)
2 �
@w
@y0
]:
After transfering @w
@y0
to the left hand side of obtained equality, we have
@w
@y0
+ 3
p
2
n�1X
k=1
yj =
3
p
2n� 1
(y0 +
p
2
n�1X
j=1
yj)
2 =
3z2
p
2n� 1
;
and relation (11) follows.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 69
A.I. Medianik
Adding equality (11) with a parameter z to system (8), we obtain the subsys-
tem of 2n� 1 equations of type (4) with the vector y = fy0 � z + z2p
2n�1 ; y1; : : : ;
y2n�2g. By the idempotent property for all i = 0; 1; 2; : : : ; 2n� 2 we have
w
2
i =
2n�2X
k=0
tikyk +
1
p
2n� 1
(�z +
z
2
p
2n� 1
) = wi �
z
p
2n� 1
+
z
2
2n� 1
:
This quadratic equation has following roots
2wi = 1� (1�
2z
p
2n� 1
):
Consequently, we obtain for every admissible i: wi =
zp
2n�1 or wi = 1 � zp
2n�1 .
Hence, if z is a complex number, then the imaginary parts of wi and wm at m 6= i
may di�er only by sign. Besides, assuming in (5) xi = wi, we see, that for all
solutions of our system
P2n�2
i=0 wi =
p
2n� 1 y0 holds. Since by conditions of our
theorem y0 is a real number, it follows that all wi are real too (because the sum
of odd number of complex summands whose imiginary parts have equal absolute
values cannot be a real number). It follows then, that z is a real number, if
y0 is real, and consequently y1; y2; : : : ; y2n�2 are real too. And by the proved
earlier unknown quantities y00; y
0
1; : : : ; y
0
2n�2 may take so only real values. Thus
all summands of the middle eqution of system (8) should vanish, as nonnegative
quantities.
From the vanishing of the �rst summand in this equation it follows that z =
y0+
p
2
Pn�1
j=1 yj = 0 or z =
p
2n� 1. Then from the relation (11) in any case we
have @w
@y0
= 3y0 and our solution satis�es the �rst equation of system (4), which
is obtained at i = 0. Therefore coordinates of vector y and, by the proved above,
also coordinates of vector y0 satisfy all the equtions of system (4). Moreover, y
and y
0 satisfy antipodal conditions (7), what follows from the vanishing of the
last summand of the middle equation of system (8). Next, from the vanishing of
its second summand, it follows that y00 = np
2n�1 , i.e., the vectors y è y
0 satisfy
all the conditions of Theorem 2, except the last one for y0. Let us show that this
condition for y0 may be assumed also satis�ed.
Indeed, since @w
@y0
= 3y0, we obtain the following quadratic equation for y0
from equality (6):
y
2
0 �
p
2n� 1y0 +
2n�2X
i=1
y
2
i = 0: (12)
Similar equality is true for y00; y
0
1; : : : ; y
0
2n�2 (it goes in the lower subsystem of (8)
with i = 0). Since y
0
0 = np
2n�1 , then
P2n�2
i=1 y
0 2
i =
p
2n� 1y00 � y
0 2
0 =
n(n�1)
2n�1 .
70 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
On Existence of a Regular Hypersimplex...
Summing termwise antipodal conditions (7), which are valid for our real-valued
solution, we have
2n�2X
i=1
y
2
i =
2n(n� 1)
2n� 1
�
2n�2X
i=1
y
0 2
i =
n(n� 1)
2n� 1
:
Substituting this in equation (12), we determine admissible values for y0: y0 =
np
2n�1 or y0 =
n�1p
2n�1 . But in case y0 =
n�1p
2n�1 , there is for the conjugate solution
�y = f
p
2n� 1 � y0;�y1; : : : ;�y2n�2g by Lemma 1: �y0 =
p
2n� 1 � n�1p
2n�1 =
np
2n�1 . So that if y0 6= np
2n�1 , one can replace a given solution by the conjugate
one. And then it will satisfy the middle equation of system (8) together with
y
0 = fy00; y
0
1; : : : ; y
0
2n�2g; since its �rst summand vanish too by Lemma 2 (at
�i =
p
2n� 1 for all i = 1; 2; : : : ; n � 1), and besides the rest summands do not
change at all their forms. Consequently, without loss of generility we may assume
that for our real-valued solution y0 =
np
2n�1 .
Thus vectors y = fy0; y1; : : : ; y2n�1g and y
0 = fy00; y
0
1; : : : ; y
0
2n�2g, inducing
together the given real solution of system (8), satisfy all conditions of Theorem 2.
Therefore we obtain the existence of a half-circulant Hadamard matrix of order
4n hence, and that is, of a regular hypersimplex inscribed into the (4n � 1)-
dimensional cube as well. This concludes the proof.
Note that for construction of Hadamard's matrix of order 4n, whose exis-
tence follows from Theorem 3, it is necessary to assume in (2) xjm�ij = wjm�ij,
x
0
jm+i�2j = w
0
jm+i�2j, where ws and w
0
s are obtained from (3) by substitution
y0 =
np
2n�1 ; y1; : : : ; y2n�2 and y
0
0 =
np
2n�1 ; y
0
1; : : : ; y
0
2n�2 in its. The coordinates of
all vertices of a regular hypersimplex inscribed into the (4n�1)-dimensional cube
answer to rows of the matrix �H, which is obtained by removing from Hadamard's
matrix H its �rst column.
R e m a r k 1. It follows from our proof of the existence theorem that
its assertion holds if the �rst summand in the middle equation of system (8)
is replaced by the following one: (y0 +
p
2
Pn�1
j=1 yj)
4 or (y0 +
p
2
Pn�1
j=1 yj �p
2n� 1)4, since antipodal conditions (7) are correct for vectors �y = f
p
2n� 1�
y0;�y1; : : : ;�y2n�2g and y
0 as soon as they are true for vectors y = fy0; y1; : : : ;
y2n�2g and y
0, because at i � 1 for the conjugate solution �y2i = (�yi)2 = y
2
i . This
is directly related with the existence theorem announced in [1].
As we have seen, unknown quantity y0 plays a special role in Theorem 3. But
this is not quite the right.
In fact, as it was determined in the proof of Theorem 3, for any admissible
value i = 0; 1; 2; : : : ; 2n � 2 wi =
zp
2n�1 or wi = 1 � zp
2n�1 , where z = y0 +
p
2
Pn�1
j=1 yj. So that wi� 1
2
= �( zp
2n�1�
1
2
) = ��, where � = zp
2n�1�
1
2
. At the
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 71
A.I. Medianik
same time from equalities (5) it follows, with regard of xk = wk(y0; y1; : : : ; y2n�2)
and the identity 1
2
+
Pn�1
m=1 cos
2�jm
2n�1 � 0, that
8>>>>>><
>>>>>>:
y0 =
1p
2n�1
2n�2P
i=0
[(wi � 1
2
) + 1
2
] =
p
2n�1
2
+�c0;
yj =
q
2
2n�1 [w0 � 1
2
+
n�1P
m=1
[(wm � 1
2
) + (w2n�1�m � 1
2
)] cos 2�jm
2n�1 ] = �cj ;
y2n�1�j =
q
2
2n�1
n�1P
m=1
[(wm � 1
2
)� (w2n�1�m � 1
2
)] sin 2�jm
2n�1 = �c2n�1�j;
(13)
where c0; cj ; c2n�1�j are some real numbers, depending on the signs of di�erences
wk � 1
2
for k = 0; 1; 2; : : : ; 2n� 2, which are determined by the solution of system
(8) itself. Consequently, if at least one of yi 6= 0 at i � 1 is a real number, then
all of the rest unknown quantities y0; : : : ; yi�1; yi+1; : : : ; y2n�2 are real too, since
� = zp
2n�1 �
1
2
, that is z, will be then real numbers.
Theorem 4. The assertion of Theorem 3 holds if the condition for y0 to be
real is replaced by that for any other yi 6= 0; 1 � i � 2n� 2.
R e m a r k 2. Strictly speaking, homogeneous system (9), using in the
proof of the existence theorem, has exactly 24n�1 solutions, if it is in general case.
Otherwise, the number of its solutions is in�nite. But it follows from relations (13)
that this number is �nite, since the number � must satisfy an equation of fourth
degree, which arises from the middle equation of system (8) after the substitution
all yi from (13). So that system (9) is in general case and has by Bezout's theorem
exactly 24n�1 solutions.
References
[1] A.I. Medianik, Antipodal n-gons inscribed into the regular (2n � 1)-gon and half-
circulant Hadamard matrices of order 4n. � Mat. �z., analiz, geom. 11 (2003),
45�66. (Russian)
[2] A.I. Medianik, Regular simplex inscribed into a cube and Hadamard's matrix of
half-circulant type. � Mat. �z., analiz, geom. 4 (1997), 458�471. (Russian)
[3] R. Bellman, Introduction to matrix analysis. Nauka, Moscow (1969). (Russian)
[4] I.R. Shafarevich, Foundations of algebraic geometry. Nauka, Moscow (1972). (Rus-
sian)
72 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
|