Cyclic algebras: a tool for Space-Time Codingusers.cms.caltech.edu/~schulman/Courses/0506cs286/F.Oggier.CMI.1.… · Cyclic algebras: a tool for Space-Time Coding CMI seminar,

Cyclic algebras:a tool for Space-Time Coding

CMI seminar, Caltech

January 20th, 2005

Frederique Oggier

joint work withJean-Claude Belfiore and Ghaya Rekaya, ENST, Paris, France

Emanuele Viterbo, Politecnico di Torino, Torino, Italy

1

Cyclic algebras: a tool for Space-Time Coding • January 20th, 2005

The problem we are interested in

x x3 4

x1

x2

22h

12h

21h

11h

X H Y=HX+Z

y1

y2

y y3 4

I Codes for multiple antennas, with M transmit and M receive antennas.

I Also called Space-Time Codes.

2


The 2 × 2 MIMO channel

x x3 4

x1

x2

22h

12h

21h

11h

X H Y=HX+Z

y1

y2

y y3 4

I X : 2 × 2 matrix codeword from a space-time code

C =

{

X =

(

x1 x2

x3 x4

)

|x1, x2, x3, x4 ∈ C

}

the xi are functions of the information symbols taken from a constellation S (e.g. PSK,QAM).

I H : 2 × 2 channel matrix is a complex Gaussian matrix with independent, zero mean, entries.

I Z: 2 × 2 complex Gaussian noise matrix.3


The code design

The goal is the design of the codebook C:

C =

{

X =

(

x1 x2

x3 x4

)

|x1, x2, x3, x4 ∈ C

}

the xi are functions of the information symbols taken from a constellation S (e.g. PSK, QAM).

I The pairwise probability of error of sending X and decoding X 6= X is upper bounded by

P (X → X) ≤ const

| det(X − X)|2M.

I We assume the receiver knows the channel (this is called the coherent case).

4


A simplified problem

Find a family C of M × M matrices such that

det(Xi − Xj) 6= 0, Xi 6= Xj ∈ C.

5


A simplified problem

I Find a family C of M × M matrices such that

det(Xi − Xj) 6= 0, Xi 6= Xj ∈ C.

Such a family C is said fully-diverse.

I Furthermore| det(Xi − Xj)|2 ≥ const, Xi 6= Xj ∈ C.

6


The idea behind division algebras

I The difficulty in building C such that

det(Xi − Xj) 6= 0, Xi 6= Xj ∈ C,

comes from the non-linearity of the determinant.

I An algebra of matrices is linear, so that

det(Xi − Xj) = det(Xk),

Xk a matrix in the algebra.

7


The idea behind division algebras

I The problem is now to build a family C of matrices such that

det(X) 6= 0, 0 6= X ∈ C.

or equivalently, such that each 0 6= X ∈ C is invertible.

I By definition, a field is a set such that every (nonzero) element in it is invertible.

I Take C inside an algebra of matrices which is also a field.

I A division algebra is a non-commutative field.

8


The leitmotiv

Let C be a subset of an algebra of matrices which is a division algebra, then

det(Xi − Xj) 6= 0, Xi 6= Xj ∈ C.

9


Outline

I Division algebras do exist: the Hamiltonian Quaternions

I Introducing number fields

I Introducing cyclic algebras

I Encoding and Rate

I Full diversity

10


The Hamiltonian Quaternions: definition

I Let {1, i, j, k} be a basis for a vector space of dimension 4 over R.

I We have the rule that i2 = −1, j2 = −1, and ij = −ji.

I The Hamiltonian Quaternions is the set H defined by

H = {x + yi + zj + wk | a, b, c, d ∈ R}.

11


Hamiltonian Quaternions are a division algebra

I Define the conjugate of a quaternion q = x + yi + wk:

q = x − yi − zj − wk.

I Compute thatqq = x2 + y2 + z2 + w2, x, y, z, w ∈ R.

I The inverse of the quaternion q is given by

q−1 =q

qq.

12


Hamiltonian Quaternions: a matrix formulation

I Any quaternion q = x + yi + zj + wk can be written as

(x + yi) + j(z − wi) = α + jβ, α, β ∈ C.

I Now compute the multiplication by q:

(α + jβ)(γ + jδ) = αγ + jαδ + jβγ + j2βδ

= (αγ − βδ) + j(αδ + βγ)

I Write this equality in the basis {1, j}:(

α −β

β α

)(

γ

δ

)

=

(

αγ − βδ

αδ + βγ

)

13


Hamiltonian Quaternions and Cyclic Algebras

A handwaving parallel:(

α −β

β α

)

↔(

x0 γσ(x1)

x1 σ(x0)

)

C ↔ a number fieldx ↔ σ(x)

−1 ↔ γ

14


Number fields: the idea

I The set Q is easily checked to be a field.

I Take i, such that i2 = −1. One can build a new field “adding” i to Q, the same way i is addedto R to create C.

I To get a field, we add all the multiples and powers of i. We obtain Q(i).

I Note we can start with the field Q(i) and add√

5, we get a new field, denoted by Q(i,√

5).

I We say that Q(i,√

5) is an extension of Q(i), which is itself an extension of Q.

15


Number field: the definition

I If L/K is a field extension, then L has a natural structure as a vector space over K

I An element x ∈ Q(i,√

5) can be written as w = x+y√

5 , where {1,√

5} are the basis “vectors”and x, y ∈ Q(i) are the scalars.

I Also w = (a + ib) +√

5(c + id), a, b, c, d ∈ Q. Thus Q(i,√

5) is a vector space of dimension 2over Q(i), or of dimension 4 over Q

I A finite field extension of Q is called a number field.

16




α −β

β α

)

↔(

a0 +√

5b0 γσ(a1 +√

5b1)

a1 +√

5b1 σ(a0 +√

5b0)

)

, a0, a1, b0, b1 ∈ Q(i)

C ↔ Q(i,√

5)

x ↔ σ(x)

−1 ↔ γ

17


Number field and polynomial

I A way to describe i is to say it is the solution of the equation X 2 + 1 = 0. Building Q(i), wethus add to Q the solution of a polynomial equation.

I Such a polynomial is called the minimal polynomial

I The polynomial X2+1 is the minimal polynomial of i over Q. Similarly, X 2−5 is the minimalpolynomial of

√5 over Q(i).

18


Defining automorphisms

I We define automorphisms of a number field L using the roots of the minimal polynomial.

I For Q(√

5), X2 − 5 = (X +√

5)(X −√

5), there are thus two automorphisms

σ1 : Q(√

5) → Q(√

5)

a + b√

5 7→ a + b√

5

σ2 : Q(√

5) → Q(√

5)

a + b√

5 7→ a − b√

5

I True for a, b ∈ Q(i) or Q.

I Note that σ2(σ2(a + b√

5)) = a + b√

5.

19




α −β

β α

)

↔(

a0 +√

5b0 γσ(a1 +√

5b1)

a1 +√

5b1 σ(a0 +√

5b0)

)

, a0, a1, b0, b1 ∈ Q(i)

C ↔ Q(i,√

5)

x ↔ σ(a0 +√

5b0) = a0 −√

5b0

−1 ↔ γ

20


Cyclic algebras

I Let L/K be a Galois extension of degree n such that its Galois group G = Gal(L/K) is cyclic,with generator σ. Denote by K∗ (resp. L∗) the non-zero elements of K (resp. L), and choosean element γ ∈ K∗. We construct a non-commutative algebra, denoted A = (L/K, σ, γ), asfollows:

A = L ⊕ eL ⊕ . . . ⊕ en−1L

such that e satisfiesen = γ and λe = eσ(λ) for λ ∈ L.

Such an algebra is called a cyclic algebra.

I The algebra A is defined as a direct sum of copies of L, thus an element x in the algebra iswritten

x = x0 + ex1 + . . . + en−1xn−1,

with xi ∈ L.

I Since the algebra is noncommutative, the rule λe = eσ(λ) explains how to do the computa-tion if the element e is multiplied by the left.

21


Cyclic algebras: matrix formulation

We illustrate the computation on an example. For n = 2, we have

xy = (x0 + ex1)(y0 + ey1)

= x0y0 + x0ey1 + ex1y0 + ex1ey1

= x0y0 + eσ(x0)y1 + ex1y0 + γσ(x1)y1,

since e2 = γ. In matrix form, this yields

xy =

(

x0 γσ(x1)

x1 σ(x0)

)(

y0

y1

)

.

22


Cyclic algebras: matrix formulation

I

A = L ⊕ eL

such that e satisfiese2 = γ and λe = eσ(λ) for λ ∈ L.

I

x = x0 + ex1 ∈ A ↔(

x0 γσ(x1)

x1 σ(x0)

)

I

e ∈ A ↔(

0 γ

1 0

)

23


Cyclic Algebras: encoding and rate

I Information symbols are from QAM constellations, or HEX constellations.

I Since QAM symbols are in Z[i] ⊆ Q(i).

I Consider our example, where L = Q(i,√

5). Then

C =

{

(

a0 +√

5b0 γ(a1 −√

5b1)

a1 +√

5b1 a0 −√

5b0

)T

| a0, a1, b0, b1 ∈ QQAM

}

.

I Codes made from cyclic algebras are said full rate: n2 information symbols encoded for n2

symbols transmitted.

24


Introducing the notion of norm

I We have defined automorphisms of a number field L using the roots of the minimal polyno-mial.

I For Q(√

5), X2 − 5 = (X +√

5)(X −√

5), there are thus two automorphisms

σ1 : Q(√

5) → Q(√

5)

a + b√

5 7→ a + b√

5

σ2 : Q(√

5) → Q(√

5)

a + b√

5 7→ a − b√

5

I True for a, b ∈ Q(i) or Q.

I The norm of x is defined by

NL/K(x) =

n∏

i=1

σi(x)

25


Full diversity

I Remember that codes coming from cyclic algebras satisfy

det(Xi − Xj) = det(X), Xi 6= Xj, X ∈ C.

I We want det(X) 6= 0 for all X 6= 0.

I If n = 2, we have

det

(

x0 γσ(x1)

x1 σ(x0)

)

= x0σ(x0) − γx1σ(x1) = NL/K(x0) − γNL/K(x1).

Thusdet(X) = 0 ⇐⇒ γ = NL/K

(

x0

x1

)

,

26


Full diversity

Theorem. Let L/K be a cyclic extension of degree n with Galois group Gal(L/K) =< σ >.If γ, γ2, . . . , γn−1 ∈ K∗ are not a norm, then the cyclic algebra A = (L/K, σ, γ) is a divisionalgebra.

27


Summary of the construction

Suppose you want a code for M antennas.

I Take a number field of degree M , say Q(i,√

5) for M = 2. (with cyclic Galois group).

I Build the code

C =

{

(

a0 +√

5b0 γ(a1 −√

5b1)

a1 +√

5b1 a0 −√

5b0

)T

| a0, a1, b0, b1 ∈ QQAM

}

.

I Choose γ which is not a norm, to get a division algebra.

28


And next?

I In this talk, I explained how to build a cyclic division algebra.

I There are more properties one can get for these codes.

1. a lower bound on the diversity2. the same average transmit enery per antenna3. shaping gain

29



Documents

Cyclic algebras: a tool for Space-Time Codingusers.cms.caltech.edu/~schulman/Courses/0506cs286/F.Oggier.CMI.1.… · Cyclic algebras: a tool for Space-Time Coding CMI seminar,