Discussion carried out in Jan 5 - Mar 5, 1999, in sci.math at Usenet:
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On Jan 5, 1999, Robin Chapman, from Exeter, challenged Pertti Lounesto:
> here is one theorem on Clifford algebras: refute it within a
> month and you win, prove it to my satisfaction within a month
> and I will submit another. Neither and I win.
>
> 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 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.
>
> Robin Chapman
On Jan 6, 1999, Lounesto proved Chapman's theorem of Jan 5, 1999:
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 =/= 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}.
Pertti Lounesto
On Jan 7, 1999, Chapman accepted Lounesto's proof of Jan 6, 1999:
> Well done! This is of course the Clifford algebra
> characterization of the Arf invariant.
On Jan 19, 1999, Lounesto clarified a point in the above proof:
> On Jan 7, 1999, Robin Chapman 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 me 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 subalgebra 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 independent 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
On Jan 7, 1999, Chapman presented a second challenge problem:
> 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
On March 5, 1999, Lounesto proved Chapman's theorem of Jan 7, 1999:
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 (because it is orthogonal to m(a_1,a_2) = a),
but the precise way of writing 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_2(x_2) is not 0,
and let us set k=Q_2(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_2(x_2) belongs to J; since the
ideal of Z generated by all Q_2(x_2) is Z, we conclude that J = Z,
as desired.
Pertti Lounesto
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Summary: Chapman challenged Lounesto to prove his theorems on Jan 5,
1999, and gave one month time for each proof. Thus, the first theorem
had to be proved in a month by Feb 5, and the second theorem a month
later by March 5. Lounesto proved Chapman's challenge theorems in due
time, on Jan 7 and on March 5. In addition, Lounesto clarified on
Chapman's request the first proof on Jan 19 (in due time before Feb 7).