Abstract. We prove that the deep holes of , , , the sixteen-dimensional
Barnes–Wall lattice, and in every even dimension are precisely the minimizers of
every Gaussian lattice sum. We determine the exact minima and equality points,
which also gives the minimizers for every convergent positive mixture of Gaussian
potentials. The analytic core is a sharp hyperbolic inequality for geometric means;
reduction to two-point measures proves the bound, and a power inequality extends
it to the entire family. Projection and diagonal gluing transfer sharp bounds to and
Barnes–Wall, while a positive cubic theta identity gives the theorem. A uniform
rounding argument identifies the geometric equality sets of and .
1. Introduction
For a Euclidean lattice and , consider the periodic Gaussian sum
Its minimizers are called cold spots. When is large, the nearest lattice points dominate
the sum, so the geometry of the deepest holes governs the limiting problem. A cold spot
that remains a minimizer for every must also control the contributions of all the more
distant shells. This is the universal cold-spot problem.
Bachoc, Moustrou, Vallentin, and Zimmermann [1] proved that the deep holes of several
highly symmetric lattices are stable cold spots; they minimize the Gaussian sum once is
sufficiently large. They asked whether these minima persist for every positive parameter.
We answer that question for four exceptional lattices and obtain an infinite family as
well.
Theorem 1.1.For each of the lattices
the minimizers of , for every , are exactly the deep holes of .
Here a deep hole is a point at maximal distance from the lattice. The root lattices are
normalized to have minimal squared norm two, is the dual of that normalization of , and
has minimal squared norm four. We use
This union is a lattice precisely when is even, and . It is integral when and even when
; no integrality assumption is needed for the theorem.
The proofs give exact polarization values and explicit equality loci. For reference, the
squared covering radii and the numbers of deep-hole classes modulo lattice translations are
For , the deep holes of have a particularly simple description; precisely half of
their coordinates are integers and half are half-integers. The additional equality
families at will explain both the exceptional count and the transfer to the other
lattices.
Corollary 1.2.Let be one of the lattices in Theorem 1.1, and let
for a nonzero positive measure , with the lattice sums convergent. Then the
minimizers of are exactly the deep holes of .
Proof.Tonelli’s theorem integrates the pointwise Gaussian inequality. It is strict off
the common minimum set for every positive parameter, so integration introduces no
additional equality points. □
The main analytic step is a geometric-mean inequality on a hyperbolic arc. Classical
quadratic theta identities put the four one-dimensional factors of the sum on this arc, and
the inequality compares an arbitrary collection of coordinates with its diagonal values.
Once this eight-dimensional inequality is available for arbitrary probability measures, a
power inequality extends it to every even dimension above eight. We give this reduction
and its equality analysis in Sections 2 and 3, before proving the analytic inequality in
Sections 4and 5.
The remaining lattices are connected by two operations that preserve sharp
Gaussian bounds. On a suitable affine slice of , the two fiber theta functions agree;
projection therefore gives the complete minimum set of . For Barnes–Wall, a
diagonal gluing of two copies of turns the theta sum into an inner product of
coset profiles, and equality is a finite character calculation. The argument uses a
different identity, arising from the three-point Fourier transform over the Eisenstein
integers. Its positive terms give a short minimum iteration and determine equality
without a separate theorem on triangular cold spots. These constructions occupy
Sections 6–8.
2. Quadratic theta identities
Fix and define
Write , , and . The function is even and one-periodic; is even and satisfies . Poisson
summation gives
Each variable factor is strictly increasing in and positive. Thus the minima of occur
exactly at , and its maxima exactly at . The analogous product for shows for , with ; in
particular .
2.1. Duplication and the hyperbolic arc. For a fixed , abbreviate
Set , , , and . Splitting the double sums for by parity of the sum of their integer
indices, and using half-sum and half-difference coordinates, gives
These relations will reduce the four positive theta factors to a single parameter.
2.2. Product and quartic identities. Changing the two summation indices to their
sum and difference gives, for every ,
For either parity of the sum, the remaining difference sum is ; this accounts for the
constant. Similarly, combining the integer and half-integer diagonal cosets gives
Separation into even and odd indices gives . Together with duplication, these formulas
imply
All rearrangements are justified by absolute, locally uniform convergence of the
Gaussian series.
3. The family and the exceptional dimension eight
The four theta factors in Section 2 satisfy two quadratic relations. Their common
parametrization leads to the following inequality, whose proof occupies Sections 4 and 5.
We state it for probability measures because this makes the passage from dimension eight
to higher dimensions immediate.
For a nonnegative function on a probability space, write , with .
Theorem 3.1 (Hyperbolic geometric means).For every probability measure on , where
,
Equality holds exactly when is a point mass or a balanced reflected pair .
Proposition 3.2.For every probability measure on , with the notation of Section 2,
Its equality measures are precisely
Proof.Define by and , which are consistent because . Since , define by
For the range assertion, the first identity initially gives the hyperbolic cosine; the
second would give a negative value for if . In particular,
Applying Theorem 3.1 to the pushforward measure gives
The expression inside the geometric mean is the diagonal theta value
by (4) and (5). Equality in this last step restricts to ; their distinct hyperbolic
parameters and the equality statement of Theorem 3.1 then give exactly (8). □
3.1. The Gaussian minimum in every even dimension. For , parity decomposition
gives
Fold each coordinate as , where , , and . Then and are the original nonalternating
factors, while are the absolute values of their alternating counterparts. Therefore
Equality requires each nonzero signed alternating product to have negative
sign.
Theorem 3.3.For every even and every ,
When , the complete minimum set is
When , the minimum set consists of the following three families
(i)
;
(ii)
points with four integer and four half-integer coordinates;
(iii)
, where has odd coordinate sum and has an even number of negative signs.
All these minimum sets are independent of .
Proof.Set and
Proposition 3.2, applied to , gives . Moreover and . For and ,
For , equality forces , , and equality in (7). Of its four equality measures, only
the balanced endpoint measure has both alternating geometric means zero. Hence
exactly half the folded coordinates are zero and half are . Conversely, every such
point has both alternating products zero and both nonalternating products , so it
attains equality regardless of integer-part parity.
For , all four measures in (8) are possible. The all-zero and all-half folded tuples
require odd integer-part sum, and together form . For four zero and four half
coordinates, both alternating products vanish, so no sign restriction remains. For
the all-quarter tuple, the first alternating product has sign and the second has sign
. Both are negative exactly under the conditions in (iii). These families attain the
bound, proving completeness. □
3.2. A uniform covering argument. The equality loci above can be identified
geometrically by a single rounding estimate. In dimension eight, both endpoints of the
estimate are sharp, accounting for the extra families of deep holes.
Proposition 3.4.For every even , the squared covering radius of is . Its deep
holes are exactly the points in Theorem 3.3.
Proof.Let be the distance of from the nearest integer, and set
Rounding to the integer lattice costs . If the rounded point has odd coordinate
sum, changing the choice in a coordinate with maximal repairs parity at additional
cost . The analogous repair for the half-integer coset costs at most . Hence
The secant inequality now gives
The first affine function is nondecreasing in and the second nonincreasing, so their
minimum is at most their common value at . That value is
The coefficient of is nonpositive, and the remaining expression is convex in . Its
endpoint maximum is .
If , equality forces , hence , , and . Both affine slopes in (12) are then nonzero, so
equality requires . Equality in the secant estimate requires every to be or , giving exactly
of each. The parity repairs cost nothing at tied rounding coordinates, so every point of
(11) has squared distance .
For , expression (13) becomes , whose maximum one occurs precisely at or . The
second case gives four zero and four half folded coordinates as before. In the first, all . For
, both affine slopes are nonzero and equality requires , so . For this quarter-grid
case, both nearest-coset distances equal one exactly when both parity repairs are
required; writing , this means odd and an even number of negative signs in
.
The endpoints must be treated separately, because one affine slope vanishes. An
integer point has distance one from exactly when its coordinate sum is odd; the
all-half case has the same condition on its integer part. These are precisely . Thus
every equality family agrees with the analytic classification, completing the proof.
□
3.3. Translation classes and symmetry. For , each half-coordinate subset of size has
two integer-parity classes modulo . Translation by complements the subset and pairs these
classes, giving classes modulo . Coordinate permutations, even sign changes, and
translations act transitively; an odd integer-part parity can be reversed by changing the
signs of one half-integer and one integer coordinate.
For , family (i) gives one class. In family (ii), the choices of half-coordinate
support each have two integer parities, and translation by identifies complementary
supports, leaving classes. Family (iii) has allowed sign vectors, paired with
their negatives by the same half-translation, leaving classes. Thus the total is
.
These classes form one affine Weyl orbit. Reflection in the root sends to . Reflection of
this point in a half root with first sign positive, three further positive signs, and four
negative signs gives a vector with four half coordinates and four zeros. Permutations, even
sign changes, and lattice translations then give all the equality families described above.
This proves Theorem 1.1 for and , subject only to the hyperbolic inequality proved
next.
4. Localization and an aligned two-point inequality
Theorem 3.1 is a minimization statement over probability measures. Its first variation has
a restricted shape; after division by a positive factor, the second derivative is strictly
convex. This forces any minimizing measure to have at most two support points. The
resulting two-point inequality has two regimes, one controlled by divided differences and
the other by variation of the interval length.
4.1. Reduction to two atoms. For a probability measure on , put
where , and define
Thus the desired inequality is .
Lemma 4.1.Every global minimizer of is supported on at most two points.
Proof.The probability measures on form a weakly compact space. The quantities
and the last integral are continuous. The function , extended by at zero, is a
decreasing limit of continuous truncations, so its integral, and hence , is upper
semicontinuous; the same holds for . Thus is lower semicontinuous. Moreover,
so . Consequently is finite and lower semicontinuous and attains a minimum.
Let be any minimizer. On define
omitting a term when its coefficient or is zero. Variation toward a point mass
gives
This formula remains valid when ; its logarithmic integral is then and remains
so along the variation for every , leaving the exponential identically zero. The same
applies to .
Extend by endpoint limits. At zero the limit is if and finite otherwise, with
the analogous statement at . The variation inequality extends to these limits, so
everywhere. Since this lower bound is the -average, is supported on the global
minimum set of .
Write . Differentiation gives
Each displayed quotient is strictly convex. For , writing and gives
For , with ,
Reflection treats the remaining quotients. Since , the right side of (14) is strictly
convex even when or vanishes. Its negative set is an interval, possibly empty or
touching an endpoint. Therefore is strictly convex on the two outer pieces and
strictly concave on the intervening piece. Each outer piece contributes at most one
global minimum, and the middle contributes none except possibly its endpoints.
Isolated zeros cannot produce a flat interval, so there are at most two minima.
□
It remains to prove the two-point assertion with its equality cases. For and , write
and normalize it by
Lemma 4.2.The ratio is at least one. Equality holds exactly when is a point
mass, or when and .
The point-mass case follows from . Reflection sends to without changing , so we may
assume and . We first treat ; Section 5 treats the complementary range and completes the
proof.
4.2. Two arithmetic deficits. The numerator of separates into two terms after the
geometric mean in its denominator is replaced by the corresponding arithmetic
mean. The weights appear in opposite orders in these terms. We compare their
deficits from the arithmetic bound, keeping the distance between the two support
points fixed; one deficit decreases under translation, while the sum of the two is
positive.
Fix and . For put , and define
The two facts we need are
The derivative assertion extends to strict decrease on by continuity.
For the first, let and set
Then and
Expansion gives
because is strictly decreasing; its logarithmic derivative is , as is increasing. At , the
term tends to zero while remains positive, so strictness persists.
For the second fact put , , , , and
Define
Differentiating the deficit, with , gives
As , its negativity is equivalent to
For , let denote the second divided difference. With ,
The Hermite–Genocchi formula is
where , , and has total mass . Cancelling positive factors reduces (18) to
The pushforward density under is
Write . First take exponent . The contribution of is positive; on , pair with .
After omitting the common positive denominator, the paired integrands sum to
Thus . For every , subtracting times this integral from (19) leaves the nonnegative
integrand . This proves the second fact in (17).
4.3. The aligned range. Suppose , , and . Use the preceding deficits with
The hypothesis is . The difference between and
is , which is positive by (17). Weighted arithmetic–geometric mean therefore gives
.
When , the arithmetic bound is an identity, and the last arithmetic–geometric mean is
an equality precisely when . Reflection gives the corresponding statement for and . It
remains to treat , , and .
Instead of differentiating simultaneously in the two support points, hold their positions
relative to the midpoint fixed and vary the interval length. A Cauchy representation
makes the sign of this derivative explicit. For half of the weights, the limiting
configuration already gives the required lower bound; the other weights reduce to an
endpoint configuration, where a stationary minimum has a decreasing inward
direction.
5.1. A Cauchy representation. Fix and set
Thus , , and the last condition in (20) is . The open admissible range is , equivalently
.
Let , with its branch analytic off and positive for . Since the exponents sum to four, it
continues to real positive values for as well. If is its polynomial part at infinity, the upper
boundary value on the cut has imaginary part
The Cauchy formula applied to therefore gives
The endpoint powers are integrable. At integer , is a polynomial and the integral
coefficient vanishes.
Substituting the exponential formulas for the hyperbolic functions in (16) gives
If , then , independent of . Thus all variation is carried by the resolvent term. For its
image under is
with derivative
All denominators are positive in the admissible range. Using in (21)–(22) yields
5.2. The sign of the kernel integral. We claim when and . If , every term in (23) is
negative. Otherwise and . Pair with by reciprocation. Since
the paired integrand for is
Put and . Then , since
and
The paired contribution is therefore negative, and the leftover interval also contributes
negatively. This proves the claim.
Equation (24) now gives a complete monotonicity statement at fixed centered support;
decreases strictly with for , increases strictly for , and is independent of for
.
5.3. The limit at large scale. Let be the random variable taking values with
probabilities . Expansion at in (22) gives
Indeed, and the coefficient of in is , while and .
The function is strictly convex on , and is strictly concave there. For the latter,
the derivative of is negative because . Since , Jensen’s inequality twice gives
For distinct atoms, simultaneous equality requires opposite values with equal
probabilities. Hence the limit in (25) is strictly greater than one throughout (20).
Monotonicity proves for .
5.4. The endpoint configuration. For , monotonicity reduces the remaining case to
the smallest admissible . Since , that endpoint has . Fix the weight and set . The variables
parametrize , . Put
Direct substitution gives the boundary ratio
Here . With and , the useful factorizations are
For example, and , which give the formulas immediately.
Lemma 5.1.A stationary point of in has a strictly negative inward derivative in
for the original ratio , with fixed.
Proof.Write
At a stationary point, logarithmic differentiation of (26) gives
The needed derivatives are
The equation gives
Substitution into the equation yields
Eliminating between these equations gives
Now differentiate the original inward in . Because , the derivative of the term
containing vanishes at zero, and the other terms give
□
Lemma 5.2.For fixed , if the infimum of is below one, it is attained at an interior
point of .
Proof.All boundary limits except the corner follow directly from (26). As , the ratio
tends uniformly to one. As with bounded away from one, it also tends to one. If
and , it tends to , where . If with bounded away from one, it tends to infinity.
At the remaining corner write , . Then
If , expansion gives
by concavity of the power .
Suppose . If is bounded along a subsequence, then , , , and . Dropping from the
numerator gives
If instead , then and . Arithmetic–geometric mean in the numerator gives
Every sequence at the corner has a subsequence of one of these types. Thus the
lower limit on the whole boundary is at least one, which proves the assertion by
compactness. □
5.5. Completion of the hyperbolic inequality. Fix and suppose some configuration
has . The aligned range is already excluded. In the unaligned range, decreasing at fixed
centered support reduces to a no-larger endpoint value. Hence any infimum
below one over all configurations of this weight equals the boundary infimum. By
Lemma 5.2, it is attained at an interior boundary point, where it is stationary in .
Lemma 5.1 then gives an inward perturbation with a strictly smaller value, a
contradiction.
This proves . In the unaligned range the inequality is strict; for this follows from the
strictly larger limiting value, and for a finite interior value is strictly larger than its
boundary value. A nondegenerate boundary equality in the latter range would itself be a
global minimum and contradict Lemma 5.1. Together with the aligned equality analysis
and reflection, this proves Lemma 4.2.
Finally, Lemma 4.1 places every minimizing measure for on at most two points. The
two-point inequality therefore proves Theorem 3.1; conversely, every equality measure is a
global minimizer and hence has the two-point equality form. This also covers endpoint
atoms and vanishing geometric means.
6. Projection and the dual lattice
An orthogonal projection need not preserve a Gaussian minimum, because different fiber
cosets generally contribute different weights. The following criterion identifies exactly the
situation needed here; on one affine slice, all fiber weights coincide.
Proposition 6.1 (Projection transfer).Let be a full lattice in an orthogonal
decomposition , both subspaces rational for . Put , , and , and write the finite glue
decomposition as
with distinct projected cosets. Suppose satisfies
If has a universal minimum set meeting , then has universal minimum set
and its minimum is . If is the deep-hole set of , then is the deep-hole set of .
Proof.Orthogonal factorization gives the exact identity
The asserted minimum and equality set follow because the slice meets .
Equality of the fiber theta functions for all parameters implies equality of their
nearest squared distances, say , by taking the leading exponential as . Therefore
If the slice contains a deep hole of , this gives and identifies all projected deep
holes. □
Take , , and . Then , is , and . Unimodularity gives ; integrality gives the inclusion in
the dual, while
The two fiber cosets are and . At their Gaussian factors agree by reflection, so
The deep hole lies on this slice, since . Proposition 6.1 therefore yields
Theorem 6.2.For every ,
The equality points are exactly the deep holes of , with squared radius .
Proof.The minimum follows from (30). The nearest squared fiber distance is , so the
radius formula is . □
6.1. The equality classes. Let denote projection onto , and put . Since ,
Modulo , all the deep holes are
where complementary subsets give the same class.
To see completeness, impose on the equality families the fiber congruence . Integer
holes project to . A four-half-coordinate hole with odd integer-part sum projects to . A
quarter-coordinate hole satisfies the congruence precisely when the number of negative
quarter signs is divisible by four; subtracting then turns it into one of the preceding two
types without changing its projection.
The classes in (31) are distinct except for complements. Indeed , so . For , the
coordinate differences of include both integers and half-integers, excluding membership
even in ; the same observation gives . There are therefore classes.
They form one affine-isometry orbit. The vector is an root of squared norm two, and .
Reflection in sends to , which gives the required class modulo .
7. Diagonal gluing and the Barnes–Wall lattice
The second transfer uses an inner product of theta profiles. It applies to a pair of lattices
of the same rank, rather than to an orthogonal projection.
Proposition 7.1 (Diagonal gluing).Let have finite index , and define
If the Gaussian minima of at a fixed parameter are , respectively, then
Equality holds exactly when both minimize the -sum and, in each coset , at least
one of minimizes the -sum.
Proof.Put . Then
Summing gives the bound and its equality conditions. □
If both lattices have universal minima and the equality conditions can be met
simultaneously, their diagonal gluing inherits universal minima. For Barnes–Wall, the
relevant coset profiles take only two values, and their compatibility becomes a character
calculation.
7.1. The lattices and the glue. Let be the binary code of affine functions on . Thus
has parameters and has parameters . Set
Equivalently,
This is the Barnes–Wall lattice scaled to minimal squared norm two and covolume ; the
usual integral normalization is [2, Section 7].
We have and . To identify geometrically, pair the eight coordinates by their last binary
coordinate and apply the orthogonal map with block
on each pair. Then . For , the integral vector has equal parities in each pair, and its
four pair parities have even sum; these are the words of , hence the words of
independent of the last coordinate. For , each pair has opposite parities and the
first-entry parities again have even sum, giving the remaining words. The inverse
transformation verifies the converse. Thus is isometric to and has squared covering radius
.
Every class of has a unique expression
The code is self-dual; it has dimension four, and intersections of supports of
affine functions have even size. This also shows that addition in (32) is binary, so
.
The Plotkin description gives
Indeed, the half-coordinate words in the two blocks are independently or its
complement, and their integer-part parities are both . This is exactly the even total integer
parity required by , and proves the converse as well.
7.2. The sharp bound. Retain and , and write , . The theorem and the similarity
above give
The last equality follows from (3) and . At , the quartic identities give , so
Proposition 7.1 yields
This is attained at , . For the representatives (32), direct substitution in (9) gives
There are eight cosets of each parity, so their inner product is exactly the
bound.
The equality criterion is already independent of ; both must be deep holes of , and
every quotient coset must make at least one of an -deep hole. We now identify this set
with all the deep holes of .
7.3. Eligible holes and fifteen characters. Call an -deep hole eligible if one
of its translates by is an -deep hole. The eligible classes modulo are exactly
where the roots are and the vectors with an even number of negative signs.
Here is a direct verification from Theorem 3.3. Integer and all-half -holes lie in , so are
not -deep. An -hole with four half coordinates has support word of weight four; it lies in
if , and otherwise has squared distance from . In the latter case its binary distance from
is two; self-duality supplies a codeword meeting its support oddly, and complementation
makes the intersection size three. Every quarter-coordinate -hole has squared
distance from , since its nearest integer point attains the lower bound from the
half-integer coordinate grid. These are all the eligible holes. A half root divided by two
is a quarter vector, and addition of gives the required odd parity. A divided
integer root has two half coordinates; addition of a half-codeword gives the eligible
four-half pattern. The distance-two argument also proves the converse, establishing
(35).
For a root , define
The exponent is integral for and changes by an even integer under , so this
is a character of . These are exactly the fifteen nontrivial characters, and two
root halves represent the same class modulo precisely when their characters
agree.
For an integer root , the character on (32) is . The seven nonzero linear functionals on
all occur. Equality of these functionals is, by self-duality, equality of the corresponding
even binary words modulo , which is equality of the root-half classes. For a half root with
even negative-sign word ,
The even binary code modulo has dimension three, giving eight further characters, all
nontrivial on the factor. Again their equality is equivalent to equality of the
root-half classes. The two lists are disjoint and exhaust the nonzero characters of
.
7.4. Theta and distance profiles. For every root and ,
For a half root, the theta formula is the quarter-coordinate sign calculation. For an
integer root, subtraction of a half-codeword gives two, four, or six half coordinates,
so the alternating products vanish. Four half coordinates give ; two or six give
using and . The four-half case is exactly . Rounding these same coordinate patterns
gives the distance formula. Each value occurs in eight cosets, and translating by replaces by .
7.5. All Barnes–Wall deep holes. For arbitrary , choose a coset nearest to in . The
two covering radii give
The complementary profiles of the point attaining (34) attain this distance as
well.
If equality holds, the same argument in each block forces both to be -deep. Choosing a
nearest coset for the other block then shows each is eligible. Their classes are therefore
among (35). Distinct characters have intersecting affine near-coset hyperplanes, which
would give total squared distance one in some coset. Equal characters give depth exactly
when the two near-coset hyperplanes are complementary. Thus all deep holes are
This is also exactly the equality set of Proposition 7.1, since at least one profile in each
coset must equal . Every deep hole, and only a deep hole, is therefore a Gaussian
minimizer for every positive parameter.
Modulo , subtract the common diagonal translation and reduce modulo . There are
fifteen root-half classes and eight quotient classes with , giving deep-hole classes. For ,
replace by in (34) and multiply hole coordinates by ; the squared covering radius
becomes three. This proves the Barnes–Wall part of Theorem 1.1.
8. A cubic theta identity and
The argument rests on a positive identity for three translates of the triangular theta
function. Positivity allows the product to be compared with itself at a dilated argument,
and a minimum iteration then gives the sharp bound. The underlying three-point Fourier
transform is the Eisenstein counterpart of the half-sum and half-difference transformation
used earlier; cubic theta identities of this kind have a classical literature, including
[3].
Let , , and put
The triangular lattice has minimal squared norm one, deep-hole classes , and squared
covering radius , as its regular hexagonal Voronoi cell shows. Also . Fix and write
Proposition 8.1.For every ,
Consequently,
with equality exactly when .
8.1. The Fourier identity.For define
and set , , , and
We first prove
Use the frequency lattice , where and . Reduction modulo is
The product on the left of (40) is invariant under , so only total frequencies occur. Fix
. For , put
This is a bijection onto with . Its inverse is
The one congruence makes all three expressions integral because . Orthogonality of the
three-point Fourier transform gives
Hence the translation phase is .
Define the two residue sums
For , the three residue pairs contribute . For either nonzero residue of , their
contribution is . Since and , grouping the total frequencies gives
Here multiplication by on frequencies has matrix and . The index-three Fourier filter
is
The matrix is multiplication by in these coordinates; evenness removes the minus sign.
Both sides of (40) have degree three in theta values, giving (38). Absolute convergence
justifies all regroupings.
8.2. Minimum iteration and equality. Put . Arithmetic–geometric mean in (38)
gives
The positive continuous periodic function has a positive minimum . At a minimizing
point, , so ; equality is attained at zero.
If , the same chain forces and
Multiplication gives . Every Fourier coefficient of is positive, so its maximum occurs
exactly at lattice points. Thus , equivalently . Conversely, these three classes all attain .
This proves Proposition 8.1 without using a separate minimum theorem for
.
8.3. Polarization and the deep holes of . Let
Then is the standard root lattice. For completeness, a simple-root basis is
Its Gram graph is the diagram with arms of lengths two, two, and one. The full arms
generate two copies of , and supplies the remaining direction in the third copy, so the
roots generate .
The three-coset decomposition and Proposition 8.1 give
A permutation of attains equality. Conversely, scalar equality places each coordinate in
one of the phases modulo . If their multiplicities are , the three summands are . Since ,
equality in arithmetic–geometric mean requires . Hence the complete minimum set is
and the minimum is . There are two classes modulo , interchanged by negation.
The same positive identity identifies all the deep holes. For a fixed ,
one isolates the finitely many nearest points and bounds the Gaussian tail. Applying
this limit to (38), with , gives
Equality requires both and to be triangular deep holes. Their two phase conditions
intersect only at , so equality is exactly .
Therefore
The points in (42) attain . Equality forces each coordinate into the three phase classes
and then forces the phase counts to be equal, so these are all deep holes. Scaling by gives
squared radius for ; replace by in its minimum formula and scale the hole coordinates
likewise. This proves the remaining case of Theorem 1.1.
References
[1]
C. Bachoc, P. Moustrou, F. Vallentin, and M. C. Zimmermann, Polarization of lattices:
stable cold spots and spherical designs, Forum Math. Sigma 14 (2026), e25. DOI
10.1017/fms.2026.10169.
[2]
A. E. Brouwer, Lattices, lecture notes, 2002, Section 7. Author’s notes.
[3]
R. Chapman, Cubic identities for theta series in three variables, Ramanujan J. 8 (2004), 459–465.arXiv math/0009187.
[4]
F. W. J. Olver et al., eds., NIST Digital Library of Mathematical Functions, Chapter 20,
Section 20.5. dlmf.nist.gov/20.5.