Lattice-based coding schemes and Diophantine approximations · 2020. 12. 16. · Integer-Forcing (IF) receiver’s goal is to decode V := AX. Let aT k be the k-th row of the matrix

Lattice-based coding schemes and Diophantineapproximations

Evgeniy Zorin

University of York

23 January 2018

E.Zorin (York) Lattices and Diophantine approximations 23 January 2018 1 / 36

Single-user receiver

y1...ym

= Hx1...

xn

+z1...

zn

,where H ∈ Mm×n(R)

Messages xk ∈ Rd , k = 1, . . . , n, are subject to the power constraint interms of the Signal-to-Noice Ratio (SNR):

‖xk‖2 ≤ n · SNR, ∀k = 1, . . . , n.

The mutual information in this case is equal to

C = log det(Im + SNR ·Ht ·H

).

y1...ym

= Hx1...

xn

+z1...

zn

‖xk‖2 ≤ n · SNR, ∀k = 1, . . . , n.

).

y1...ym

= Hx1...

xn

+z1...

zn

‖xk‖2 ≤ n · SNR, ∀k = 1, . . . , n.

).

y1...ym

= Hx1...

xn

+z1...

zn

1 Zero-forcing,

2 Minimum mean-squared error (MMSE).

y1...ym

= Hx1...

xn

+z1...

zn

1 Zero-forcing,

2 Minimum mean-squared error (MMSE).

B. Nazer and M. Gastpar, “Compute-and-forward: Harnessing interferencethrough structured codes”, IEEE Transactions on Information Theory(2011)The decoder can directly recover a linear combination of interfering datastreams.

Each data stream is drawn from the same lattice codebook. The codebookstructure ensures that any integer combination of codewords is itself acodeword, and thus decodable at high rates

Key step: selection of an integer matrix A to approximate the channelmatrix H.

B. Nazer and M. Gastpar, “Compute-and-forward: Harnessing interferencethrough structured codes”, IEEE Transactions on Information Theory(2011)The decoder can directly recover a linear combination of interfering datastreams.Each data stream is drawn from the same lattice codebook. The codebookstructure ensures that any integer combination of codewords is itself acodeword, and thus decodable at high rates

Integer-Forcing linear receiver

J. Zhan, B. Nazer, U. Erez, and M. Gastpar, “Integer-forcing linearreceivers”, IEEE Transactions on Information Theory (2014)

Integer-Forcing (IF) receiver’s goal is to decode V := AX.

Let aTk be the k-th row of the matrix A.

The receiver first performs MMSE estimate of the vectors vk = aTk X from

the output Y

To make the Integer-Forcing scheme to be successful, decoding over allsub-channels should be correct.

the output Y

We define the effective SNR at the kth sub-channel as

SNReff ,k :=(aTk

(I + SNRHTH

)ak)−1

,

andSNReff := min

k=1,...,nSNReff ,k .

Theorem (Zhan, Nazer, Ordentlich and Gastbar)

Integer-forcing can achieve any rate

RIF <1

2n log (SNReff ) .

Remark

The optimal choice of the matrix A is given by

Aopt := argminA∈Zn×ndetA 6=0

maxk=1,...,n

aTk

(I + SNRHTH

)ak .

SNReff ,k :=(aTk

(I + SNRHTH

)ak)−1

,

andSNReff := min

RIF <1

2n log (SNReff ) .

Remark

maxk=1,...,n

aTk

(I + SNRHTH

)ak .

SNReff ,k :=(aTk

(I + SNRHTH

)ak)−1

,

andSNReff := min

RIF <1

2n log (SNReff ) .

Remark

maxk=1,...,n

aTk

(I + SNRHTH

)ak .

SNReff ,k :=(aTk

(I + SNRHTH

)ak)−1

,

andSNReff := min

RIF <1

2n log (SNReff ) .

Remark

maxk=1,...,n

aTk

(I + SNRHTH

)ak .

Can we estimate SNReff ?

O. Ordentlich and U. Erez, “Precoded Integer-Forcing Universally Achievesthe MIMO Capacity to Within a Constant Gap”, IEEE Transactions onInformation Theory (2015)

for any positive definite m ×m matrix Q, defineMn(Q) := min

a∈Zn\{0}aT ·Q · a

Theorem (Ordentlich and Erez)

1

n2Mn(In + SNR ·HT ·H) < SNReff ≤ Mn(In + SNR ·HT ·H).

Problem

Assume that the channel matrix H has coefficients distributed with respectto a given law (e.g. Gaussian distribution independently for each entry).Let κ ∈ (0, 1).Find the best possible value of s ≥ 0 such that the event SNReff ≥ s isrealised with probability greater than κ.

O. Ordentlich and U. Erez, “Precoded Integer-Forcing Universally Achievesthe MIMO Capacity to Within a Constant Gap”, IEEE Transactions onInformation Theory (2015)for any positive definite m ×m matrix Q, define

Mn(Q) := mina∈Zn\{0}

aT ·Q · a

1

Problem

aT ·Q · a

1

Problem

aT ·Q · a

1

Problem

Refined version of the problem

Define

Hm,n (C0,SNR) :={H ∈ Rn×m : log det

(Im + SNR ·HT ·H

)= C0

}.

Problem

Assume that the channel matrix H is chosen randomly from the setHm,n (C0,SNR), according to any given probability distribution on this set.Let κ ∈ (0, 1).Find the best possible value of s ≥ 0 such that the event SNReff ≥ s isrealised with probability greater than κ; equivalently, determine thecumulative distribution function of the quantity SNReff seen as a randomvariable.

Refined version of the problem

Define

Hm,n (C0,SNR) :={H ∈ Rn×m : log det

(Im + SNR ·HT ·H

)= C0

}.

Problem

Assume that the channel matrix H is chosen randomly from the setHm,n (C0,SNR), according to any given probability distribution on this set.Let κ ∈ (0, 1).Find the best possible value of s ≥ 0 such that the event SNReff ≥ s isrealised with probability greater than κ; equivalently, determine thecumulative distribution function of the quantity SNReff seen as a randomvariable.

Mathematical interpretation

So, we would like to find cumulative distribution function of

Mm(Im + SNRHTH)

when

H ∈ Hm,n (C0,SNR) :={H ∈ Rn×m : log det

(Im + SNR ·HT ·H

)= C0

}.

Let S++d denotes the set of positive definite (symmetric) matrices indimension d ≥ 2. Define

Σ++d :={

Σ ∈ S++d : det(Σ) = 1}

Problem (Mathematical version of the previous problem)

For a given probability measure µ on the set Σ++d , estimate the probabilityµ (Md (Σ) ≤ δ) as a function of δ > 0.

Mm(Im + SNRHTH)

when

(Im + SNR ·HT ·H

)= C0

}.

Let S++d denotes the set of positive definite (symmetric) matrices indimension d ≥ 2.

Define

Σ++d :={

Σ ∈ S++d : det(Σ) = 1}

Mm(Im + SNRHTH)

when

(Im + SNR ·HT ·H

)= C0

}.

Σ++d :={

Σ ∈ S++d : det(Σ) = 1}

Mm(Im + SNRHTH)

when

(Im + SNR ·HT ·H

)= C0

}.

Σ++d :={

Σ ∈ S++d : det(Σ) = 1}

Mathematical Problem 2

Σ++d :={

Σ ∈ S++d : det(Σ) = 1}

Problem (Problem 2)

For a given probability measure µ on the set S++d , estimate the probabilityµ (Md (S) ≤ δ) as a function of δ > 0.

Mathematical Problem 2

Σ++d :={

Σ ∈ S++d : det(Σ) = 1}

Problem (Problem 2)

For a given probability measure µ on the set S++d , estimate the probabilityµ (Md (S) ≤ δ) as a function of δ > 0.

Mathematical simplification

The quantityMd (Q) := min

a∈Zd\{0}aT ·Q · a

does not change if we replace the matrix Q by ATQA, for anyA ∈ SLd (Z).

Because of this, in our problem we can restrain the considerations to theset of reduced symmetric quadratic forms Σ++d ,red .

Then, there is a bijection between the locally symmetric set

Xd := SLd (Z)\SLd (R)/SOd (R)

and Σ++d ,red , given by

φ : g ∈ Xd 7→ g T · g ∈ Σ++d ,red .

G := SLd (R),Γ := SLd (Z),

H := SOd (R)

So,Xd := Γ\G/H.

The set Xd can be equipped with a natural G –invariant probabilitymeasure µXd arising from the G –invariant probability measure µΓ\G on thespace of lattices Γ\G .With the help of the bijection φ, the measure µXd can be pushed forwardto a probability measure φ∗µXd on the space Σ

++d ,red .

Consider

pXd (δ) = (φ∗µXd )({

Σ ∈ Σ++d ,red : Md (Σ) ≤ δ})

= µXd ({g ∈ Xd : Md (φ(g)) ≤ δ})

G := SLd (R),Γ := SLd (Z),

H := SOd (R)So,

Xd := Γ\G/H.

++d ,red .

Consider

Σ ∈ Σ++d ,red : Md (Σ) ≤ δ})

= µXd ({g ∈ Xd : Md (φ(g)) ≤ δ})

G := SLd (R),Γ := SLd (Z),

H := SOd (R)So,

Xd := Γ\G/H.

The set Xd can be equipped with a natural G –invariant probabilitymeasure µXd arising from the G –invariant probability measure µΓ\G on thespace of lattices Γ\G .

With the help of the bijection φ, the measure µXd can be pushed forwardto a probability measure φ∗µXd on the space Σ

++d ,red .

Consider

Σ ∈ Σ++d ,red : Md (Σ) ≤ δ})

= µXd ({g ∈ Xd : Md (φ(g)) ≤ δ})

G := SLd (R),Γ := SLd (Z),

H := SOd (R)So,

Xd := Γ\G/H.

++d ,red .

Consider

Σ ∈ Σ++d ,red : Md (Σ) ≤ δ})

= µXd ({g ∈ Xd : Md (φ(g)) ≤ δ})

G := SLd (R),Γ := SLd (Z),

H := SOd (R)So,

Xd := Γ\G/H.

++d ,red .

Consider

Σ ∈ Σ++d ,red : Md (Σ) ≤ δ})

= µXd ({g ∈ Xd : Md (φ(g)) ≤ δ})

D.Y. Kleinbock and G.A. Margulis, “Logarithm laws for flows onhomogeneous spaces”, Inventiones Mathematicae (1998).

Vd =πd/2

Γ(

d2 + 1

) and Ad = 2πd/2Γ(

d2

)Theorem (Kleinbock & Margulis, 1998)

The following inequalities hold for any δ > 0 :

Vd2ζ(d)

δd/2 − cdV 2d4δd ≤ pXd (δ) ≤

Vd2ζ(d)

δd/2· (1)

Here, ζ denotes the Riemann zeta function and cd a strictly positiveconstant which, when d ≥ 3, can be taken to be

cd =1

ζ(d) · ζ(d − 1)·

Vd =πd/2

Γ(

d2 + 1

d2

)

Theorem (Kleinbock & Margulis, 1998)

Vd2ζ(d)

δd/2· (1)

cd =1

ζ(d) · ζ(d − 1)·

Vd =πd/2

Γ(

d2 + 1

d2

)Theorem (Kleinbock & Margulis, 1998)

Vd2ζ(d)

δd/2· (1)

cd =1

ζ(d) · ζ(d − 1)·

Final Estimate

Adiceam and Z., “On the Minimum of a Positive Definite Quadratic Formover Non–Zero Lattice points. Theory and Applications”, Journal deMathmatiques Pures et Appliques (2018)

Let f : S++d → R+ be a density function supported on S++d . The

corresponding measure is denoted by νf .

mf (δ) := νf({

Q ∈ S++d : Md (Q) ≤ δ})

Theorem (Adiceam and Z.)

Let δ ∈ (0, 1). Then,

0 ≤ 1−∫ ∞√δ

gf ≤ mf (δ) ≤ 1−∫

Id (δ)Gf ≤ 1, (2)

where

Id (δ) :=(√

δ, +∞)d.

Final Estimate

mf (δ) := νf({

Q ∈ S++d : Md (Q) ≤ δ})

Let δ ∈ (0, 1). Then,

0 ≤ 1−∫ ∞√δ

gf ≤ mf (δ) ≤ 1−∫

Id (δ)Gf ≤ 1, (2)

where

Id (δ) :=(√

δ, +∞)d.

Final Estimate

mf (δ) := νf({

Q ∈ S++d : Md (Q) ≤ δ})

Let δ ∈ (0, 1). Then,

0 ≤ 1−∫ ∞√δ

gf ≤ mf (δ) ≤ 1−∫

Id (δ)Gf ≤ 1, (2)

where

Id (δ) :=(√

δ, +∞)d.

Final Estimate

mf (δ) := νf({

Q ∈ S++d : Md (Q) ≤ δ})

Let δ ∈ (0, 1). Then,

0 ≤ 1−∫ ∞√δ

gf ≤ mf (δ) ≤ 1−∫

Id (δ)Gf ≤ 1, (2)

where

Id (δ) :=(√

δ, +∞)d.

Let T ++d be the group of upper triangular matrices with strictly positivediagonal entries.

the following map is bijective

ϕchol : L ∈ T ++d 7→ LT L ∈ S++d (3)

(this map corresponds to Cholesky decomposition)Let λk be the Lebesgue measure on Rk .

p :=d(d − 1)

2.

Given any β = (β1, . . . , βd ) ∈ (R>0)d ,

Gf (β) := 2d ·

d∏i=1

βd−i+1i ·∫Rp

(f ◦ ϕchol ) (β,u) · dλp(u).

and, given any β1 > 0

gf (β1) :=

∫(R>0)d−1

Gf (β1, β̃) · dλd−1(β̃). (4)

Let T ++d be the group of upper triangular matrices with strictly positivediagonal entries.the following map is bijective

ϕchol : L ∈ T ++d 7→ LT L ∈ S++d (3)

p :=d(d − 1)

2.

Given any β = (β1, . . . , βd ) ∈ (R>0)d ,

Gf (β) := 2d ·

d∏i=1

βd−i+1i ·∫Rp

gf (β1) :=

∫(R>0)d−1

Gf (β1, β̃) · dλd−1(β̃). (4)

ϕchol : L ∈ T ++d 7→ LT L ∈ S++d (3)

(this map corresponds to Cholesky decomposition)

Let λk be the Lebesgue measure on Rk .

p :=d(d − 1)

2.

Given any β = (β1, . . . , βd ) ∈ (R>0)d ,

Gf (β) := 2d ·

d∏i=1

βd−i+1i ·∫Rp

gf (β1) :=

∫(R>0)d−1

Gf (β1, β̃) · dλd−1(β̃). (4)

ϕchol : L ∈ T ++d 7→ LT L ∈ S++d (3)

p :=d(d − 1)

2.

Given any β = (β1, . . . , βd ) ∈ (R>0)d ,

Gf (β) := 2d ·

d∏i=1

βd−i+1i ·∫Rp

gf (β1) :=

∫(R>0)d−1

Gf (β1, β̃) · dλd−1(β̃). (4)

ϕchol : L ∈ T ++d 7→ LT L ∈ S++d (3)

p :=d(d − 1)

2.

Given any β = (β1, . . . , βd ) ∈ (R>0)d ,

Gf (β) := 2d ·

d∏i=1

βd−i+1i ·∫Rp

gf (β1) :=

∫(R>0)d−1

Gf (β1, β̃) · dλd−1(β̃). (4)

ϕchol : L ∈ T ++d 7→ LT L ∈ S++d (3)

p :=d(d − 1)

2.

Given any β = (β1, . . . , βd ) ∈ (R>0)d ,

Gf (β) := 2d ·

d∏i=1

βd−i+1i ·∫Rp

gf (β1) :=

∫(R>0)d−1

Gf (β1, β̃) · dλd−1(β̃). (4)

Final EstimateAdiceam and Z., “On the Minimum of a Positive Definite Quadratic Formover Non–Zero Lattice points. Theory and Applications”, Journal deMathmatiques Pures et Appliques (2018)

Let f̃ : Σ++d → R+ be a density function supported on Σ++d . The

corresponding measure is denoted by ν̃f .

m̃f (δ) := ν̃f({

Q ∈ Σ++d : Md (Q) ≤ δ})

Let δ ∈ (0, 1). Then,

0 ≤ 1−∫ ∞√δ

g̃f̃≤ m̃

f̃(δ) ≤ 1−

∫∆d−1(δ)

G̃f̃≤ 1, (5)

∆d−1(δ) :=

{β′ ∈ (R>0)d−1 :

(∀i = 1, . . . , d − 1, βi >

√δ)∧

(d−1∏i=1

βi <1√δ

)}.

m̃f (δ) := ν̃f({

Q ∈ Σ++d : Md (Q) ≤ δ})

Let δ ∈ (0, 1). Then,

0 ≤ 1−∫ ∞√δ

g̃f̃≤ m̃

f̃(δ) ≤ 1−

∫∆d−1(δ)

G̃f̃≤ 1, (5)

∆d−1(δ) :=

{β′ ∈ (R>0)d−1 :

(∀i = 1, . . . , d − 1, βi >

√δ)∧

(d−1∏i=1

βi <1√δ

)}.

m̃f (δ) := ν̃f({

Q ∈ Σ++d : Md (Q) ≤ δ})

Let δ ∈ (0, 1). Then,

0 ≤ 1−∫ ∞√δ

g̃f̃≤ m̃

f̃(δ) ≤ 1−

∫∆d−1(δ)

G̃f̃≤ 1, (5)

∆d−1(δ) :=

{β′ ∈ (R>0)d−1 :

(∀i = 1, . . . , d − 1, βi >

√δ)∧

(d−1∏i=1

βi <1√δ

)}.

m̃f (δ) := ν̃f({

Q ∈ Σ++d : Md (Q) ≤ δ})

Let δ ∈ (0, 1). Then,

0 ≤ 1−∫ ∞√δ

g̃f̃≤ m̃

f̃(δ) ≤ 1−

∫∆d−1(δ)

G̃f̃≤ 1, (5)

∆d−1(δ) :=

{β′ ∈ (R>0)d−1 :

(∀i = 1, . . . , d − 1, βi >

√δ)∧

(d−1∏i=1

βi <1√δ

)}.

Interference Alignment

and

Diophantine approximations

MISO channel

y = h1x1 + h2x2.

y =

0 if x1 = x2 = 0

h1 if x1 = 0 and x2 = 1

h2 if x1 = 1 and x2 = 0

h1 + h2 if x1 = x2 = 1 ,

y = h1x1 + h2x2 + z .

Picture become more complicated if we transmit x1 and x2 from biggerinterval, e.g. x1, x2 ∈ [1, . . . , 16].

MISO channel

y = h1x1 + h2x2.

y =

0 if x1 = x2 = 0

h1 if x1 = 0 and x2 = 1

h2 if x1 = 1 and x2 = 0

h1 + h2 if x1 = x2 = 1 ,

y = h1x1 + h2x2 + z .

MISO channel

y = h1x1 + h2x2.

y =

0 if x1 = x2 = 0

h1 if x1 = 0 and x2 = 1

h2 if x1 = 1 and x2 = 0

h1 + h2 if x1 = x2 = 1 ,

y = h1x1 + h2x2 + z .

MISO channel

y = h1x1 + h2x2.

y =

0 if x1 = x2 = 0

h1 if x1 = 0 and x2 = 1

h2 if x1 = 1 and x2 = 0

h1 + h2 if x1 = x2 = 1 ,

y = h1x1 + h2x2 + z .

MISO channel

y = h1x1 + h2x2.

y =

0 if x1 = x2 = 0

h1 if x1 = 0 and x2 = 1

h2 if x1 = 1 and x2 = 0

h1 + h2 if x1 = x2 = 1 ,

y = h1x1 + h2x2 + z .

Precoding:

x1 = h22u1 + h12v1

x2 = h21u2 + h11v2

Precoding:

x1 = h22u1 + h12v1

x2 = h21u2 + h11v2

y1 = (h11h22)u1 + (h21h12)u2 + (h11h12)(v1 + v2)

y2 = (h21h12)v1 + (h11h22)v2 + (h21h22)(u1 + u2) .

y1 = (h11h22)u1 + (h21h12)u2 + (h11h12)(v1 + v2)

y2 = (h21h12)v1 + (h11h22)v2 + (h21h22)(u1 + u2) .

Problem

Estimate outage probability of this scheme.

y1 = (h11h22)u1 + (h21h12)u2 + (h11h12)(v1 + v2)

y2 = (h21h12)v1 + (h11h22)v2 + (h21h22)(u1 + u2) .

Problem

Estimate outage probability of this scheme.

1 ...

2 A. S. Motahari, S. Oveis-Gharan, M.A. Maddah-Ali, A. K. Khandani.Real interference alignment: exploiting the potential of single antennasystems, IEEE Trans. Inform. Theory 60 (2014), no. 8, 4799–4810.

3 A. S. Motahari, S. Oveis–Gharan, M.A. Maddah–Ali, A. K. Khandani.“Real Interference Alignment”, 2010.

4 ...

Definition

Given a real positive function ψ : R+ → R+ with ψ(r)→ 0 as r →∞ (i.e.a so called approximating function), let

B(ψ) := {x ∈ R : |qx − p| > ψ(q) for a.b.f.m. (q, p) ∈ N× Z} ,

where ’for a.b.f.m.’ reads ‘for all but finitely many’.

Theorem (Khintchine, 1924)

Let ψ be an approximating function. Then

|B(ψ)| =

FULL if

∞∑q=1

ψ(q)

Definition

Given a real positive function ψ : R+ → R+ with ψ(r)→ 0 as r →∞ (i.e.a so called approximating function), let

B(ψ) := {x ∈ R : |qx − p| > ψ(q) for a.b.f.m. (q, p) ∈ N× Z} ,

where ’for a.b.f.m.’ reads ‘for all but finitely many’.

|B(ψ)| =

FULL if

∞∑q=1

ψ(q)

|B(ψ)| =

FULL if

∞∑q=1

ψ(q)

Example

For almost all x ∈ R we have

‖qx‖ > 1q log+(q)2

(6)

for a.b.f.m. q ∈ N.

However if we are interested in just one value q = 1, the inequality (6)fails for all x ∈ R.Moreover, the inequality

‖qx‖ > 10−100

q log+(q)2(7)

fails for all integers q ∈ [−1024; 1024] on a set of x of strictly positiveLebesgue measure. Indeed, it fails on, say,

x ∈ [2− 10−100

1024 log(1024)2; 2 +

10−100

1024 log(1024)2]

Example

‖qx‖ > 1q log+(q)2

(6)

for a.b.f.m. q ∈ N.However if we are interested in just one value q = 1, the inequality (6)fails for all x ∈ R.

Moreover, the inequality

‖qx‖ > 10−100

q log+(q)2(7)

x ∈ [2− 10−100

1024 log(1024)2; 2 +

10−100

1024 log(1024)2]

Example

‖qx‖ > 1q log+(q)2

(6)

for a.b.f.m. q ∈ N.However if we are interested in just one value q = 1, the inequality (6)fails for all x ∈ R.Moreover, the inequality

‖qx‖ > 10−100

q log+(q)2(7)

x ∈ [2− 10−100

1024 log(1024)2; 2 +

10−100

1024 log(1024)2]

Mutidimensional case

Definition

Let m ∈ N and ψ : (0,∞)→ (0,∞), define

Bm(ψ) := {x ∈ Rm : ‖a.x‖ > ψ(|a|) for a.b.f.m. a ∈ Zm, a 6= 0} ,

Theorem

Let m ∈ N, ψ be an approximating function and

Sm(ψ) :=∞∑

q=1

qm−1ψ(q) . (8)

Then for any m > 1

|B(ψ)|m =

FULL if Sm(ψ)

Mutidimensional case

Definition

Let m ∈ N and ψ : (0,∞)→ (0,∞), define

Bm(ψ) := {x ∈ Rm : ‖a.x‖ > ψ(|a|) for a.b.f.m. a ∈ Zm, a 6= 0} ,

Theorem

Let m ∈ N, ψ be an approximating function and

Sm(ψ) :=∞∑

q=1

qm−1ψ(q) . (8)

Then for any m > 1

|B(ψ)|m =

FULL if Sm(ψ)

Corollary

Suppose that ψ is an approximating function and Sm(ψ) 0 such that

‖a.x‖ > κ(x)ψ(|a|) for all a ∈ Zm, a 6= 0 . (9)

Question

Is it possible to replace κ(x) > 0 in the above corollary with an absoluteconstant κ independent of x?

Corollary

Question

Is it possible to replace κ(x) > 0 in the above corollary with an absoluteconstant κ independent of x?

Corollary

Question

Is it possible to replace κ(x) > 0 in the above corollary with an absoluteconstant κ > 0 independent of x?

Definition

Bm(Ψ, κ) := {x ∈ Rm : ‖a.x‖ > κΨ(a) for all a ∈ Zm, a 6= 0}

Theorem

Let m ∈ N and µ be a probability measure on Rm absolutely continuouswith respect to the Lebesgue measure on Rm. Let Ψ : Zm → R+ be anyfunction, MΨ := maxa∈Zm\{0}Ψ(a) and

Σ(Ψ) :=∑a∈Zm

Ψ(q).

Then for any δ ∈ (0, 1) there is an (explicit) constant κ > 0 depending onµ, Σ(Ψ) and δ only such that

µ (Bm(ψ, κ)) ≥ 1− δ.

κ :=1

2min

{1

MΨ,

(δ

2Σ(Ψ)Σf

)1/m}.

Definition

Theorem

Let m ∈ N and µ be a probability measure on Rm absolutely continuouswith respect to the Lebesgue measure on Rm.

Let Ψ : Zm → R+ be anyfunction, MΨ := maxa∈Zm\{0}Ψ(a) and

Σ(Ψ) :=∑a∈Zm

Ψ(q).

µ (Bm(ψ, κ)) ≥ 1− δ.

κ :=1

2min

{1

MΨ,

(δ

2Σ(Ψ)Σf

)1/m}.

Definition

Theorem

Σ(Ψ) :=∑a∈Zm

Ψ(q).

µ (Bm(ψ, κ)) ≥ 1− δ.

κ :=1

2min

{1

MΨ,

(δ

2Σ(Ψ)Σf

)1/m}.

Definition

Theorem

Σ(Ψ) :=∑a∈Zm

Ψ(q).

µ (Bm(ψ, κ)) ≥ 1− δ.

κ :=1

2min

{1

MΨ,

(δ

2Σ(Ψ)Σf

)1/m}.

Definition

Theorem

Σ(Ψ) :=∑a∈Zm

Ψ(q).

µ (Bm(ψ, κ)) ≥ 1− δ.

κ :=1

2min

{1

MΨ,

(δ

2Σ(Ψ)Σf

)1/m}.

Example

Assume that x follows the standard normal distribution, N (0, 1), and thatψ(n) = 1

n·log(n)2 , n ≥ 2, with ψ(1) = 1.

One can readily verify that

S(ψ) < 3.11.Let δ takes the values 0.5, 0.25, 0.1 and 0.01. Then we have fromprevious calculations:

δ 0.5 0.25 0.1 0.01

N 2 2 3 4

κ 0.00224 0.00112 0.00011 3.03455 · 10−6

In particular, see that for 99% of the values of the random variable x thathas normal distribution N (0, 1) the inequality

‖nx‖ > 3 · 10−6

n · log(n)2

holds for all n ∈ N.

Example

n·log(n)2 , n ≥ 2, with ψ(1) = 1. One can readily verify thatS(ψ) < 3.11.

Let δ takes the values 0.5, 0.25, 0.1 and 0.01. Then we have fromprevious calculations:

δ 0.5 0.25 0.1 0.01

N 2 2 3 4

κ 0.00224 0.00112 0.00011 3.03455 · 10−6

‖nx‖ > 3 · 10−6

n · log(n)2

Example

n·log(n)2 , n ≥ 2, with ψ(1) = 1. One can readily verify thatS(ψ) < 3.11.Let δ takes the values 0.5, 0.25, 0.1 and 0.01. Then we have fromprevious calculations:

δ 0.5 0.25 0.1 0.01

N 2 2 3 4

κ 0.00224 0.00112 0.00011 3.03455 · 10−6

‖nx‖ > 3 · 10−6

n · log(n)2

Example

n·log(n)2 , n ≥ 2, with ψ(1) = 1. One can readily verify thatS(ψ) < 3.11.Let δ takes the values 0.5, 0.25, 0.1 and 0.01. Then we have fromprevious calculations:

δ 0.5 0.25 0.1 0.01

N 2 2 3 4

κ 0.00224 0.00112 0.00011 3.03455 · 10−6

‖nx‖ > 3 · 10−6

n · log(n)2

holds for all n ∈ N.E.Zorin (York) Lattices and Diophantine approximations 23 January 2018 30 / 36

MANIFOLDS

Let M be a manifold parametrized by f : U → Rm, where U ⊂ Rd is anopen set (might be all Rd ). Let l ∈ N and assume that at every pointx ∈ U among all partial derivatives of the order up to l , ∂k f

∂xa11 ...∂x

add

(x),

a1, . . . , ad ∈ N ∪ {0}, k = a1 + · · ·+ ad , 1,≤ k ≤ l , there are d linearlyindependent vectors. Then we say that manifold M is non-degenerate.

Example

Consider the simplest case, when M is a planar curve, defined by afunction f : I → R, where I ⊂ R is an interval. Then the condition thatthe curve M is non-degenerate means that at every point x at least one ofderivatives of f of an order up to l is non-zero.

Let Ψ : Zm \ {0} → (0,∞) be a function satisfying

Σ(Ψ) :=∑

q∈Zm\{0}

Ψ(q) 0.

Bm(Ψ, κ,M) := {x ∈ U : ‖a.f(x)‖ > κΨ(a) for all a ∈ Zm, a 6= 0} .

Theorem (Adiceam, Beresnevich, Leveseley, Velani and Z.)

Let l ∈ N and let M be a compact d–dimensional C l+1 submanifold of Rnand assume it is l–non–degenerate at every point. Let µ be a probabilitymeasure supported on M absolutely continuous with respect to | . |M. LetΨ : Zn → R+ be a monotonically decreasing function in each variablesatisfying (10). Then there exist positive constants κ0,C0,C1 dependingon Ψ and M only such that for any 0 < δ < 1, the inequality

µ(Bn(Ψ, κ) ∩M) ≥ 1− δholds with

κ := min{κ0, C0Σ

−1Ψ δ, C1δ

d(n+1)(2l−1)}.

Let Ψ : Zm \ {0} → (0,∞) be a function satisfying

Σ(Ψ) :=∑

q∈Zm\{0}

Ψ(q) 0.

Bm(Ψ, κ,M) := {x ∈ U : ‖a.f(x)‖ > κΨ(a) for all a ∈ Zm, a 6= 0} .Theorem (Adiceam, Beresnevich, Leveseley, Velani and Z.)

Let l ∈ N and let M be a compact d–dimensional C l+1 submanifold of Rnand assume it is l–non–degenerate at every point. Let µ be a probabilitymeasure supported on M absolutely continuous with respect to | . |M. LetΨ : Zn → R+ be a monotonically decreasing function in each variablesatisfying (10). Then there exist positive constants κ0,C0,C1 dependingon Ψ and M only such that for any 0 < δ < 1, the inequality

µ(Bn(Ψ, κ) ∩M) ≥ 1− δholds with

κ := min{κ0, C0Σ

−1Ψ δ, C1δ

d(n+1)(2l−1)}.

One more example

So, admitting a small error δ > 0 we can always find a constant κ suchthat for m-tuple of integers a ∈ Zm \ {0} we have

‖a.f(x)‖ > κΨ(a).

For example for any δ > 0, we can provide a constant κ such that withprobability > 1− δ we have

‖a.f(x)‖ > κ∏i max(1, ai log(ai )

2)

Remark

If we transmit only such m-tuples a = (a1, . . . , am) that |ai | ≤ Q,i = 1, . . . ,m, we can optimize this result.

One more example

Define

Ψ(a1, . . . , am) :=

{1

Qm , if |ai | ≤ Q for all i = 1, . . . ,m.0 otherwise.

We have SΨ = 1. Then we have that for any δ > 0 there exists anexplicite constant κ such that for any a = (a1, . . . , am) such that |ai | ≤ Q,i = 1, . . . ,m we have

‖a.f(x)‖ > κQm

.

Thank you!