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