From: Pertti Lounesto
Newsgroups: sci.math
Subject: Re: Hmmmmmm!
Date: Thu, 07 Jan 1999 00:10:04 +0200
Keywords: Clifford algebras and quadratic spaces
Robin Chapman asked for proof or refutation of the following:
> Let Q be a quadratic form on a finite dimensional vector space V over F_2, the
> field of two elements. That is Q: V -> F_2 is a function such that the
> induced map B : V x V -> F_2 defined by B(x, y) = Q(x+y) - Q(x) - Q(y)
> is bilinear. Suppose also that B is non-singular, that is, B(x, V) = 0
> implies x = 0.
>
> The Clifford algebra C of Q is the unital F_2-algebra with generators the
> elements of V and relations x^2 = Q(x) (for x in V). The even Clifford algebra
> C_0 of Q is the subalgebra of C generated by the elements xy for x, y in V.
>
> Then the centre of C_0 is non-trivial (i.e., it contains more than
> 2 elements) if and only if the number of x in V with Q(x) = 1 exceeds the
> number of x in V with Q(x) = 0.
>
> Then the centre of C_0 is a field if and only if the number of x in V with
> Q(x) = 1 exceeds the number of x in V with Q(x) = 0.
The assertion in Robin Chapman's problem is true; this
assertion can be deduced from the following information.
The words "quadratic space" mean a vector space which has finite
dimension over a field, and is provided with a non degenerate
quadratic form. Quadratic subspaces are subspaces on which this
quadratic form is still non degenerate; completely singular (or
isotropic) subspaces are subspaces on which this quadratic form
vanishes.
(A) Information about the Clifford algebra Cl of a quadratic
space V over a field K of characteristic 2.
As in Chapman's post, Q is the non degenerate quadratic form
on V, and B the associated symmetric bilinear form, which is here
also a symplectic form.
The center Z(Cl_0) of the even subalgebra has dimension 2, it
is spanned by 1 and some element z which has the following properties:
z^2 -z belongs to K, and vz = (z+1)v for all v in V.
This center is not a field if K contains an element k such that
z^2-z = k^2-k; indeed Z(Cl_0) is then the direct sun of the
ideals K(z-k) and K(z-k+1) (observe that (z-k)(z-k+1)=0 ).
But Z(Cl_0) is a field if such a k does not exist (indeed the
polynomial x^2-x-(z^2 z) of degree 2 in x is irreducible in
K[x], and we obtain a field if we add to K a root of this
polynomial).
This element z can be constructed in the following way: V is an
orthogonal direct sum of quadratic planes P_1, P_2, ... , P_m (and
consequently the dimension of V is an even integer 2m); each plane
P_j is spanned by two vectors a_j and b_j such that B(a_j, b_j) =1;
each product a_jb_j satisfies these properties:
(a_jb_j)^2 - a_jb_j = Q(a_j)Q(b_j),
v a_jb_j = (a_jb_j +1) v if v belongs to P_j,
v a_jb_j = a_jb_j v if v belongs to a P_i with i not= j;
consequently we may choose z equal to the sum of all products a_jb_j;
the only other admissible choice would be z+1; moreover z^2 -z
is equal to the sum of all products Q(a_j)Q(b_j).
The quadratic space V is called neutral if it is the direct sum
of two completely singular subspaces; this condition is equivalent
to the existence of at least one completely singular subspace of
dimension m (the half of dim V). An orthogonal direct sum of two
isomorphic quadratic subspaces is always neutral; it is sufficient
to prove this assertion for an orthogonal direct sum of two isomorphic
quadratic planes P and P'; let (a,b) be a basis of P such that
B(a,b)=1 , and (a',b') the basis of P' which is the image of (a,b)
by an isomorphism from P onto P'; the vectors a+a' and b+b' span
a completely singular plane; if you want a supplementary completely
singular plane, take the plane spanned by Q(b)(a+a')+b and
Q(a)(b+b')+a'.
A quadratic plane is either neutral or anisotropic (this means
that 0 is the only singular (or isotropic) vector); and every neutral
quadratic space is an orthogonal direct sum of neutral planes; moreover
if V is neutral, the center of Cl_0 is isomorphic to K^2; these three
assertions are true for fields of any characteristic.
(B) Additional information when K has two elements.
All anisotropic quadratic planes over the field K=F_2 are
isomorphic; consequently a direct sum of two anisotropic planes is
always neutral. This implies that for each integer m>0 there are
only two isomorphy classes of quadratic spaces of dimension 2m:
either neutral spaces of dimension 2m, or direct orthogonal sums
of an anisotropic plane and a neutral space of dimension 2m-2.
If V is neutral, it is already known that Z(Cl_0) is not a field
(indeed z^2-z = 0); if V is not neutral, it is easy to check that
z^2-z = 1, and therefore Z(Cl_0) is a field (obtained by adjunction
to F_2 of a non trivial cubic root of 1; indeed z^3 =1).
Let d_m be the proportion of singular vectors in a neutral
quadratic space V of dimension 2m, that is, the number of v in V
such that Q(v)=0, divided by the number 4^m of all elements of V .
It is easy to prove that the proportion of singular vectors in a
non neutral quadratic space of dimension 2m is the complementary
proportion 1 - d_m. Thus the assertion of your colleague is
equivalent to this one: d_m is always greater than 1/2.
The sequence of numbers d_m begins in this way:
d_0 = 1/1, d_1 = 3/4, d_2 = 10/16, d_3 = 36/64, ... ...
It is easy to prove the following induction formula:
d_{m+1} = 1/4 + d_m/2;
whence d_m = 1/2 + 1/2^{m+1}.
This finishes the proof.
Pertti Lounesto
==============================================================================
From: Robin Chapman
Newsgroups: sci.math
Subject: Re: Hmmmmmm!
Date: Thu, 07 Jan 1999 10:26:57 GMT
In article <3693DF3B.2019@hit.fi>,
Pertti.Lounesto@hit.fi wrote:
> Robin Chapman asked for proof or refutation of the following:
>
> > Let Q be a quadratic form on a finite dimensional vector space V over F_2, the
> > field of two elements. That is Q: V -> F_2 is a function such that the
> > induced map B : V x V -> F_2 defined by B(x, y) = Q(x+y) - Q(x) - Q(y)
> > is bilinear. Suppose also that B is non-singular, that is, B(x, V) = 0
> > implies x = 0.
> >
> > The Clifford algebra C of Q is the unital F_2-algebra with generators the
> > elements of V and relations x^2 = Q(x) (for x in V). The even Clifford algebra
> > C_0 of Q is the subalgebra of C generated by the elements xy for x, y in V.
> >
> > Then the centre of C_0 is non-trivial (i.e., it contains more than
> > 2 elements) if and only if the number of x in V with Q(x) = 1 exceeds the
> > number of x in V with Q(x) = 0.
> >
> > Then the centre of C_0 is a field if and only if the number of x in V with
> > Q(x) = 1 exceeds the number of x in V with Q(x) = 0.
Well done! This is of course the Clifford algbra characterization of the
Arf invariant. Only one quibble:
> The center Z(Cl_0) of the even subalgebra has dimension 2, it
> is spanned by 1 and some element z which has the following properties:
> z^2 -z belongs to K, and vz = (z+1)v for all v in V.
I have to admit it's far from obvious to mee that the centre of C_0 is
2-dimensional. It took me a bit of effort to prove this, but of course
I make no claim of expertise in Clifford algebras. So is there a simple
way of seeing this?
Anyway, for the followup: how about a proof or refutation of the following
application of Clifford algebras to Gauss's theory of composition of
binary quadratic forms which I found in the literature several years ago.
[I hope you are not only interested in Clifford algebras over *fields*].
We deal with binary quadratic forms over the integers Z. A binary quadratic
form (A, Q) is a free module A of rank 2 over Z and map Q: A -> Z where
1) Q(ax) = a^2 Q(x) for all a in Z and x in A,
2) B: A x A -> Z defined by B(x, y) = Q(x + y) - Q(x) - Q(y) is bilinear.
Such a form (A, Q) is non-degenerate if B(x, A) = 0 implies x = 0 and
primitive if the ideal of Z generated by Q(A) is all of Z.
As always, we define Clifford algebras C(A, Q) and C_0(A, Q) in the obvious
way: C(A, Q) is a unital Z-algebra with generators A and relations x^2 = Q(x)
for x in A, while C_0(A, Q) is the unital subalgebra of C(A, Q) generated by
the elements xy for x, y in A.
Let (A_1, Q_1), (A_2, Q_2) and (A, Q) be binary quadratic forms. A composition
map is a bilinear map m : A_1 x A_2 -> A satisfying
Q(m(x_1, x_2)) = Q_1(x_1) Q_2(x_2)
for all x_j in A_j.
Theorem:
Let (A_1, Q_1), (A_2, Q_2) and (A, Q) be non-degenerate primitive binary
quadratic forms and m: A_1 x A_2 -> A be a composition map.
Then there are uniquely determined algebra homomorphisms
f_j: C_0(A_j, Q_j) -> C_0(A, Q) [j = 1, 2] such that
m(c_1 x_1, c_2 x_2) = f_1(c_1) f_2(c_2) m(x_1, x_2)
for all c_j in C_0(A_j, Q_j) and x_j in A_j.
Robin Chapman + "They did not have proper
SCHOOL OF MATHEMATICal Sciences - palms at home in Exeter."
University of Exeter, EX4 4QE, UK +
rjc@maths.exeter.ac.uk - Peter Carey,
http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda, chapter 20
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
==============================================================================
From: Pertti Lounesto
Newsgroups: sci.math
Subject: Re: Hmmmmmm!
Date: Tue, 19 Jan 1999 13:17:41 +0200
Robin Chapman wrote:
> Pertti Lounesto wrote:
>
> > The center Z(Cl_0) of the even subalgebra has dimension 2, it is
> > spanned by 1 and some element z which has the following properties:
> > z^2 -z belongs to K, and vz = (z+1)v for all v in V.
>
> I have to admit it's far from obvious to mee that the centre of C_0 is
> 2-dimensional. It took me a bit of effort to prove this, but of course
> I make no claim of expertise in Clifford algebras. So is there a simple
> way of seeing this?
About the center of the even Clifford subalgebra.
If E is a quadratic space of even dimension over a field K,
it is well known that the center of the subalbebra of even elements
in its Clifford algebra has dimension 2 ; but the proof of this fact
is not quite easy, and depends on whether the characteristic of K
is equal or not equal to 2 ; here I shall give a proof when the
characteristic is 2 .
When E is a plane, the even Clifford subalgebra has
dimension 2 and is commutative; thus all is evident. When the
dimension of E is > 2, we decompose E in an orthogonal direct sum
of planes, and proceed by induction on the number of planes ;
the Clifford algebra is the twisted tensor product of the Clifford
algebras of the planes, and here a twisted tensor product is an
ordinary tensor product. The induction uses the following lemma .
LEMMA . Let A and B be two graded algebras, and C = A\ot B their
tensor product: A = A_0+A_1 , B = B_0+B_1 (direct sums),
C = C_0+C_1 with C_0 = (A_0\ot B_0)+(A_1\ot B_1)
and C_1 = (A_0\ot B_1)+(A_1\ot B_0) ;
we suppose that A_1 (or B_1) contains invertible elements, and that
the centers of A_0 and B_0 have dimension 2, and are spanned by 1
and another element u, respectively v, such that
xu = (u+1)x for all x in A_1 ,
resp. yv = (v+1)y for all y in B_1 ;
then the center of C has dimension 2 and is spanned by 1 (that is
1\ot 1) and w = u\ot 1 + 1\ot v ; moreover it is stated that
zw = (w+1)z for all z in C_1 .
PROOF in four steps.
First step : the elements of A_0\ot B that commute with
all elements of A_0\ot K , are the elements of (K+Ku)\ot B .
Indeed take any x in A_0 and some y_n in a basis (y_1,y_2,...,
y_q) of B ; let us calculate the commutator (or bracket) of x\ot y_n
and any x'\ot 1 with x' in A_0 ; we find that
(x\ot y_n)(x'\ot 1) - (x'\ot 1)(x\ot y_n) = (xx'-x'x)\ot y_n ;
it is clear that a linear combination of elements like x\ot y_n
commutes with all x'\ot 1 if and only if it belongs to the tensor
product of the center of A (as first factor) and B (as second factor).
Second step : the elements of A_0\ot B_0 that commute with
all elements of (A_0\ot K)+(K\ot B_0) , are the elements of the
subalgebra spanned by 1 , u\ot 1, 1\ot v and u\ot v .
Indeed, they must be in the intersection of (K+Ku)\ot B_0 and
A_0\ot (K+Kv) .
Third step : w belongs to the center of C_0 and
zw = (w+1)z for all z in C_1 .
This results from straightforward calculations.
Fourth step : the center of C_0 is K+Kw .
Indeed it is clear that every element in the subalgebra spanned
by 1, u\ot 1, 1\ot v and u\ot v is the sum of an element of K+Kw and
some k\ot v + k'u\ot v , with k and k' in K; it lies in the center
of C_0 if and only if k\ot v + k'\ot v lies in it; let us choose an
invertible x in A_1 , so that x and xu are linearly independant in
A_1 ; for any non zero y in B_1, the equalities xu=(u+1)x and
yv = (v+1)y lead to
(x\ot y)(k\ot v + k'u\ot v) - (k\ot v + k'u\ot v)(x\ot y)
= kx\ot y + k'(xu\ot y + x\ot vy) ;
since x\ot y belongs to C_0, the element k\ot v + k'u\ot v lies in
the center of C_0 if and only if first k'=0, secondly k=0.
End of the proof.
Pertti Lounesto
==============================================================================
From: Pertti Lounesto
Newsgroups: sci.math
Subject: Re: Hmmmmmm!
Date: Fri, 05 Mar 1999 14:53:48 +0200
Robin Chapman wrote:
>Anyway, for the followup: how about a proof or refutation of the following
>application of Clifford algebras to Gauss's theory of composition of
>binary quadratic forms which I found in the literature several years ago.
>[I hope you are not only interested in Clifford algebras over *fields*].
>
>We deal with binary quadratic forms over the integers Z. A binary quadratic
>form (A, Q) is a free module A of rank 2 over Z and map Q: A -> Z where
>1) Q(ax) = a^2 Q(x) for all a in Z and x in A,
>2) B: A x A -> Z defined by B(x, y) = Q(x + y) - Q(x) - Q(y) is bilinear.
>
>Such a form (A, Q) is non-degenerate if B(x, A) = 0 implies x = 0 and
>primitive if the ideal of Z generated by Q(A) is all of Z.
>
>As always, we define Clifford algebras C(A, Q) and C_0(A, Q) in the obvious
>way: C(A, Q) is a unital Z-algebra with generators A and relations x^2 = Q(x)
>for x in A, while C_0(A, Q) is the unital subalgebra of C(A, Q) generated by
>the elements xy for x, y in A.
>
>Let (A_1, Q_1), (A_2, Q_2) and (A, Q) be binary quadratic forms. A composition
>map is a bilinear map m : A_1 x A_2 -> A satisfying
>Q(m(x_1, x_2)) = Q_1(x_1) Q_2(x_2)
>for all x_j in A_j.
>
>Theorem:
>
>Let (A_1, Q_1), (A_2, Q_2) and (A, Q) be non-degenerate primitive binary
>quadratic forms and m: A_1 x A_2 -> A be a composition map.
>Then there are uniquely determined algebra homomorphisms
>f_j: C_0(A_j, Q_j) -> C_0(A, Q) [j = 1, 2] such that
>
>m(c_1 x_1, c_2 x_2) = f_1(c_1) f_2(c_2) m(x_1, x_2)
>
>for all c_j in C_0(A_j, Q_j) and x_j in A_j.
Here Z is a commutative ring with unit element 1 such that
1+1 does not vanish, and without divisors of zero; thus Z is a
subring of its field of fractions Z', in which 1/2 exists; of
course Z may be the ring of integers. We consider three
free modules A_1, A_2 and A of rank 2 over Z, provided with
non degenerate quadratic forms Q_1, Q_2 and Q ; they are non
degenerate in the weak sense (for instance Q determines an injective
map from A into its dual A*); moreover the ideal of Z generated
by Q_1(A_1) is Z, and also the ideal generated by Q_2(A_2).
It is assumed that there is a bilinear map m from A_1\times A_2
into A such that (for all x_1 and x_2)
Q(m(x_1,x_2)) = Q_1(x_1) Q_2(x_2) ;
these hypotheses imply the existence of two algebra morphisms
f_j : Cl_0(A_j,Q_j) \arrow Cl_0(A,Q) (j=1,2)
involving the even Clifford subalgebras, and such that
m(c_1x_1, c_2x_2) = f_1(c_1)f_2(c_2) m(x_1,x_2)
for all c_1 in Cl_0(A_1,Q_1), c_2 in Cl_0(A_2,Q_2), x_1 in A_1
and x_2 in A_2 .
I give a proof in three steps. All indices "prime" mean that
an extension of scalars from Z to Z' has been done; A_1 is embedded
in the vector space A'_1 over the field Z', which is provided with
a non degenerate quadratic form Q'_1 extending Q_1, the Clifford
algebra Cl(A_1,Q_1) is embedded in the Clifford algebra Cl'(A'_1,Q'_1),
and so forth ... The bilinear maps associated to the quadratic maps
are denoted B_1, B_2 and B ; for instance B(x,y)=Q(x+y)-Q(x)-Q(y) .
FIRST STEP .
Let k be a non zero element of Z and g a linear map from A_1 into A
such that Q(g(x_1)) = k Q_1(x_1) for all x_1 in A_1 ; there exists
a unique algebra morphism f from Cl_0(A_1,Q_1) into Cl'_0(A',Q') such
that the following equality holds in Cl'(A',Q') for all x_1 in A_1 and
all c_1 in Cl_0(A_1,Q_1) : g(c_1x_1) = f(c_1)g(x_1) ;
moreover kf(c_1) belongs to Cl_0(A,Q) for all c_1 in Cl_0(A_1,Q_1).
PROOF . Let (a_1,b_1) be a basis of A_1 ; Cl_0(A_1,Q_1) is
a free module with basis (1, a_1b_1) ; let f be the linear map
from Cl_0(A_1,Q_1) into Cl'_0(A',Q') which maps 1 to 1, and a_1b_1
to g(a_1)g(b_1)/k ; f is an algebra morphism because there is a
polynomial X^2+uX+v in Z[X] which gives 0 when X is replaced by
a_1b_1 or by g(a_1)g(b_1)/k (exactly u = B_1(a_1,b_1) and v =
Q_1(a_1)Q_1(b_1)) . It remains to check the equalities
k g((a_1b_1)a_1) = (g(a_1)g(b_1)) g(a_1) ,
k g((a_1b_1)b_1) = (g(a_1)g(b_1)) g(b_1) ;
this is done by straightforward calculations.
The unicity of f results from the following fact: if c is an element
of Cl'_0(A',Q') such that cg(a_1)=cg(a_2)=0, then c=0 .
SECOND STEP .
The announced statement is true when Z is a field.
PROOF . Let (a_j,b_j) be an orthogonal basis of A_j for j=1,2 .
It is clear that the map x_2 \arrow m(a_1,x_2) preserves
orthogonality, and all similar maps too; consequently we get an
orthogonal basis (a,b) of A if we set
a = m(a_1,a_2) and b = m(a_1,b_2) ,
and moreover there exists elements h and h' of Z' such that
m(b_1,a_2) = h Q_1(b_1) Q_2(a_2) b ,
m(b_1,b_2) = h' Q_1(b_1) Q_2(b_2) a ;
for the moment the former equality only means that m(b_1,a_2) is
proportional to b, but the precise way in which I write this
proportionality, will be now justified by the relations to impose to
h and h', so that m actually satisfies the requirement stated in the
hypotheses; indeed there are two relations to impose to h and h' :
h+h'=0 and h^2 Q_1(a_1)Q_1(b_1)Q_2(a_2)Q_2(b_2) = 1 ;
these relations emerge from straightforward calculations.
Let f_1 be the linear map from Cl_0(A_1,Q_1) into Cl_0(A,Q)
defined in this way:
f_1(1)=1 and f_1(a_1b_1) = h Q_1(b_1) ab ;
it is easy to check that f_1 is an algebra morphism; similarly we
define f_2 in this way:
f_2(1)=1 and f_2(a_2b_2) = ab /Q_1(a_1) ;
f_2 is also an algebra morphism, and straightforward calculations
show that f_1 and f_2 satisfy the required condition; since all
modules are provided with orthogonal bases, all these calculations
are quite easy.
THIRD STEP .
The announced statement is true without more hypotheses.
PROOF . We already know that there are two algebra morphisms
f_j : Cl_0(A_j,Q_j) \arrow Cl'_0(A',Q')
such that the required condition is true in the Clifford algebra
Cl'(A',Q') ; it remains to prove that they take their values in
Cl_0(A,Q) ; I shall prove it for f_1 (for f_2 the proof would be
similar). Let J be the subset of all z in Z such that
z f_1(c_1) belongs to Cl_0(A,Q) for all c_1 in Cl_0(A_1,Q_1) ;
obviously J is an ideal of Z, and we have to prove that J=Z ;
let x_2 be an element of A_2 such that Q(x_2) is not 0,
and let us set k=Q(x_2) ; let g be the map x_1 \arrow
m(x_1,x_2) from A_1 into A ; obviously Q(g(x_1)) = k Q_1(x_1)
for all x_1 in A_1 ; consequently there is a unique algebra
morphism f from Cl_0(A_1,Q_1) into Cl'_0(A',Q') such that
g(c_1x_1) = f(c_1)g(x_1) for all c_1 in Cl_0(A_1,Q_1) and all x_1
in A_1 ; we also know that kf(c_1) belongs to Cl_0(A,Q) for all
c_1 in Cl_0(A_1,Q_1) ; now since f is unique, it must coincide with
f_1 , and all this proves that Q(x_2) belongs to J ; since the
ideal of Z generated by all Q(x_2) is Z, we conclude that J=Z,
as desired.
Pertti Lounesto