Darren L. SliderThe Spinor Norm and Zassenhaus’s Theorem Darren’s Writings Home |
DARREN L. SLIDER (B. 1967)The Spinor Norm and Zassenhaus’s Theorem A Research Paper Submitted in Partial Fulfillment of the Requirements for the Master of Science Degree Department of Mathematics
ACKNOWLEDGEMENTS I would like to express my deepest gratitude to my advisor, Professor Andrew G. Earnest, for his abundant help and guidance with this paper, and for making my two years of graduate school a stimulating and enjoyable experience. I would also like to thank the other members of my committee, Professors Robert Fitzgerald and Mary Wright, for their time and effort with respect to this paper and its presentation. Finally, I express my wholehearted indebtedness and praise to God and to my wife, for reasons which transcend verbal explanation. THE SPINOR NORM AND ZASSENHAUS’S THEOREM The proof of Theorem 1 here given is due to Geoffrey Mason, which is sketched in his article “Groups, Discriminants, and the Spinor Norm” (see Bibliography). The examples following Zassenhaus’s Theorem were obtained from L. E. Dickson, We assume throughout that
V if for all v, w, x Î V, a Î F, the following properties hold:(1)
V.
V and is written d(V). d(V) is invariant under change of basis in the group F^{*}/F^{*2} and is thus represented as an element of that field (see O’Meara, pp. 85-87).Let n £ t. We introduce variables a (1 £ _{ij}i, j £ t) and the t ´ t matrices
I + A)(_{1}I + A)××××××(_{2}I + A)_{n}
J)_{n-1}D - NA = 0._{n}
It follows that
Furthermore, ( N)A =_{n}
Q.E.D.
I) = J._{n}D
Note that I = I + A - _{1}I = A. So (_{1}I - N)(E - _{1}I) = (I - N)A =_{1}
by the first result of the proof of Lemma 1. Assume that ( I) = J for some _{k -1}Dk with 2 ≤ k ≤ t. Then
by Lemma 1. Q.E.D. From now on, let dim a/_{ij}a, and set _{ii}T(v_{i}) = I + A =_{i}
T(v)×××_{2}T(v) - _{n}I) = (-2)^{n}det(a)/_{ij}a×××_{11}a._{nn}
J = _{t}DID = D. We may obtain det(D) = a/_{ij}a)_{ii}a) by -2 and (for 1 ≤ _{ij}i ≤ n) the ith row by 1/a; elementary linear algebra (see Anton, p. 65) tells us that this yields det(_{ii}D) = (-2)^{n}det(a)/_{ij}a×××_{11}a. Also from elementary linear algebra, we get det(_{nn}I - N) = 1 since I - N is upper triangular (see the proof of Proposition 1 and Anton, p. 64). Taking determinants on both sides of the result of Proposition 1 now yields
Q.E.D.
W to be v Î V :B(v,w) = 0 " w Î W}. We call rad(W) = W Ç W the ^{*} of radicalW. It is easy to prove that W and rad(^{*}W) are subspaces of W. We say that W is if rad(nondegenerateW) = {0}. In this case, V = W + W (O’Meara, p. 102).^{*}We assume henceforth that Let V onto itself. For w Î V, define T(w) : V ® V by T(w)(v) = v - (2B(v,w)w/Q(w)) for v Î V. T(w) is the .symmetry on V with respect to wLet { v} be a basis for _{n}V. Then if v Î V, v = a_{1}v + . . . _{1}a_{n}v for some _{n}a, . . . ,_{1}a Î _{n}F. Note that
under the interpretation B(v_{i},v). Indeed, in the _{j}ith row we have
and from this we can see that the choice of terminology for the symmetry on I + A shown above._{i}Let
with i = 1,...,n.
wñ. Then _{k}M has no eigenvalue equal to 1 if and only if V = W.
W is spanned by {w,...,_{1}w}, for _{k}w Î W we may find a,...,_{1}a Î _{k}F such that w = a_{1}w + ××× + _{1}a_{k}w . Assume that _{k}w′ Î Çw; then for _{i}^{*}w Î W we have
so that i = 1,...,k since w Î _{i}W; thus W^{*} Í Çw_{i}^{*}. So we have W^{*} = Çw_{i}^{*}.Let
and thus T(w)(_{k}w) = w. Therefore (M - I)(w) = 0 and if w ¹ 0, w is an eigenvector corresponding to the eigenvalue 1 of M. But by hypothesis, M has no eigenvalues equal to 1, so we must have w = 0. It follows that W^{*} = {0}. In particular, rad(W) = W Ç W^{*} = {0} so that W is nondegenerate. We conclude that V = W + W^{*} = W + {0} = W.(Ü) Assume now that wñ; since _{k}k is minimal and dim(V) = n, k = n and áw,...,_{1}wñ is a basis for _{n}V. Under the interpretation v = _{i}w, _{i}a = _{ij}B(w,_{i}w)_{j}
Since w)) ¹ 0, so det(_{j}M - I) ¹ 0. We conclude that M has no eigenvalue equal to 1.Q.E.D.
U if for each u Î U, there exists a t Î N such that M - I)(^{t}u) = 0.We may assume that the largest subspace on which u’) = 0 for some m Î N. Let u Î U be arbitrary; then for some n Î N, (M - I)(^{n}u) = 0. Let t = max{m,n}; then
so that
so that ( v = -w’ Î _{1}W. It follows that im(M - I) Í W.Now consider the operators M - I), . . . on ^{3}V. Note that M - I) Í ker(M - I) Í . . . ^{2}V. Since V is finite-dimensional, there must exist k Î N such that ker(M - I) = ker(^{k}M - I) for all ^{k+j}j Î N. Let V = ker(_{0}M - I)^{k}V = im(_{1}M - I).^{k}Let V. Since _{1}x Î V, there must exist some _{1}y Î V such that x = (M - I)(^{k}y).x Î V, we must also have (_{0}M - I)(^{k}x) = 0. Thus 0 = (M - I)(^{k}x) = M - I)((^{k}M - I)(^{k}y)) = (M - I)(^{2k}y). But ker(M - I) = ker(^{2k}M-I) by the choice of ^{k}k. It follows that y Î ker(M - I), so that we have ^{k}x = (M - I)(^{k}y) = 0.V Ç _{0}V = {_{1}0}.Since V Í _{1}W and V Í _{0}U. So
and of course dim( Q.E.D.
W) = w + rad(W) and w’ + rad(_{1}W) = w’ + rad(W), it follows that w - w,_{1}w’ - w’ _{1}W) so that B(w - w,_{1}w’) = 0, whence B(w,w’) = B(w,_{1}w’). By a similar argument, B(w,_{1}w’) = B(w,_{1}w’). Therefore _{1}B(w,w’) = B(w,_{1}w’) = B(w,_{1}w’)._{1}Given this symmetric bilinear form
so that Now suppose that x)+rad(W)=(M’-I)(^{t}x+rad(W)) = 0 + rad(W); in particular, (M - I)(^{t}x) Î rad(W). Now rad(W) Í W^{*} so that (M - I)(^{t}x) Î W^{*}; by the proof of Proposition 2, we have M((M - I)(^{t}x)) = (M - I)(^{t}x). Therefore, (M - I)(^{t+1}x) = 0. It follows that x Î U. So X Í U.We shall now prove that ( Since
Consider
so that Since Q.E.D.
If v) = 0,0 = (M - I)(^{2}v) = (M - I)((M - I)(v)) so that M - I)(v) = 0,v = 0. Proceeding inductively in this way, we may conclude that (M - I)(^{t}v) = 0 Þ v = 0 for any t Î N. It follows that U = {0}; in this case R = U Ç U^{*} = {0} Ç {0}^{*} = {0} and we are done.Now assume that the largest subspace of any nondegenerate vector space of dimension less than Now consider dim(
Let Since w) = 0 for some t Î N.M’ be as in Lemma 4, we have
whence by the maximality of Since
and certainly Q.E.D.
T(w)×××_{2}T(w), we define the _{k} of spinor normM to be Θ(M) = Q(w)_{1}Q(w)×××_{2}Q(w) Î _{k}F^{*}/F^{*2}. This is a well-defined invariant (i.e., independent of the choice of product of reflections) in F^{*}/F^{*2} (see O’Meara, pp. 131-137), Furthermore, it is easily seen that Θ is a homomorphism on the group (under the operation of composition) of isometries of V onto itself.
Let dim( v)) = (M - I)(^{t}v) = 0. If t = 1, let v’ = v;t > 1, let v’ = (M - I)(^{t-1}v). Since v’ ¹ 0 (otherwise v = 0 or t is not minimal, respectively), v’ = αv for some 0 ¹ α Î F and we have 0 = (M - I)(v’) = (M - I)(αv) = α(M - I)(v). It follows that (M - I)(v) = 0, whence M(v) = v. So for any w Î V, say, w = βv,M(w) = M(βv) = βM(v) = βv = w. Therefore M = I.Since T(w) with _{k}Q(w) ¹ 0, then in particular, _{i}Q(w) ¹ 0 and _{1}w ¹ _{1}0 so that V = áwñ. Hence for _{1}w Î V,w = βw,_{1}
so that T(w). Therefore Θ(_{1}M) = Q(w)_{1}Q(w) = (_{1}Q(w))_{1} = 1 in ^{2}F^{*}/F^{*2}.Now assume that the spinor norm of any isometry acting as a unipotent operator on any vector space of dimension < Let v) = 0 for some minimal t Î N. If t = 1, then 1 is an eigenvalue of M corresponding to v. If t > 1, then 1 is an eigenvalue of M corresponding to (M - I)(^{t-1}v) (which is in this case an eigenvector of M as (M - I)((M - I)(^{t-1}v)) = (M - I)(^{t}v) = 0). In either case, M has an eigenvalue equal to 1 and by Proposition 2, V ¹ W. Then there is 0 ¹ w Î W^{*} (for otherwise W^{*} = {0} implies that V = W by the proof of Proposition 2). Consider the space Y / áwñ, where Y = áwñ + Z and Z = {v Î V : B(v, w) = 0}.Given the symmetric bilinear form w + z + á_{1}wñ = aw + z + áwñ, a’_{1}w + z’ + á_{1}wñ = a’w + z’ + áwñ, then z - z, _{1}z’ - z’ Î á_{1}wñz - z = β_{1}w for some β Î F and
since z’). By a similar argument, B(z,_{1}z’) = B(z,_{1}z’) so that _{1}B(z,z’) = B(z,_{1}z’) = B(z,_{1}z’)._{1}Define Furthermore, y) = 0. Thus we have M” - I)(y + áwñ) =M"(y + áwñ) - I(y + áwñ) = M(y) + áwñ - (y + áwñ) = (M(y) - y) + áwñ = (M - I)(y) + áwñ so that(M” - I)(^{n}y + áwñ) = (M - I)(^{n}y) + áwñ = áwñ. It follows that M" acts as a unipotent operator on Y/áwñ.Note that for wñ) = B"(w + á_{i}wñ,w + á_{i}wñ) = B(w,_{i}w) = _{i}Q(w) ¹ 0. Now let _{i}T"(w + á_{i}wñ) be the symmetry on Y/áwñ with respect to w + á_{i}wñ. For arbitrary y + áwñ Î Y/áwñ, we have
so that
It follows that wñ)×××T"(w + á_{k}wñ) is a representation of M" as a product of reflections such that Q"(w + á_{i}wñ) ¹ 0 for i = 1 to k.Note that dim(
by the induction hypothesis. Q.E.D.
M| are well-defined isometries. Furthermore, let the characteristic polynomial of _{U*}M be given by χ(z). Set χ(z) = (z - 1)χ^{f}_{0}(z) where (z - 1) does not divide χ(_{0}z). Then the characteristic polynomials of M|, _{U}M| are given by (_{U*}z - 1), χ^{f}(_{0}z) respectively.
u) = 0. Then (M - I)(^{t}M(u)) = M((M - I)(^{t}u)) = M(0) = 0. It follows that M(u) Î U for all u Î U. Therefore M| is well-defined and is an isometry because _{U}M is an isometry.Since U, M| is injective; since _{U}U is finite-dimensional, M must also be surjective. In particular, if we let u Î U, there exists x Î U such that M(x) = M|(_{U}x) = u. Now let u’ Î U^{*} be arbitrary. Since M is an isometry, we have B(u,M(u’)) = B(M(x),M(u’)) = B(x,u’) = 0. It follows (since u was also arbitrary) that M(u’) Î U^{*}. Therefore M| is well-defined and is an isometry because _{U*}M is an isometry.We claim that charpoly( M|). Indeed, since _{U*}V = U + U^{*}, there exists a basis {v, . . . ,_{1}v,_{m}v, . . . ,_{m+1}v} for _{n}V with v, . . . ,_{1}v Î _{m}U, v, . . . ,_{m+1}v Î _{n}U^{*}, and the corresponding matrix for M is in block-diagonal form.Let α be an eigenvalue of u be the corresponding eigenvector. Since u Î U, there exists n Î N such that (M - I)(^{n}u) = 0. Since α is an eigenvalue, M(u) = M|(_{U}u) = αu and (M - I)(u) = (α - 1)u so that
Proceeding inductively in this way, we get u) = (α - 1)^{n}u whence it follows that α = 1. Therefore 1 is the only eigenvalue of M|. Furthermore, if 1 is an eigenvalue of _{U}M, M(x) = x for some 0 ¹ x Î V whence we get (M - I)(x) = 0 and x Î U. Thus 1 is not an eigenvalue of M|. We may conclude that (_{U*}z - 1) | charpoly(^{ f}M|_{U}) in F[z].Let { v} be the basis for _{n}V mentioned above, and let E be a splitting field for charpoly(M|). Define _{U}V^{ #} = {α_{1}v + ××× + α_{1}_{n}v : α_{n} Î _{i}E} and U^{ #} = {α_{1}v + ××× + α_{1}_{m}v : α_{m} Î _{i}E}; it is not hard to see that these are vector spaces over E. Furthermore, define
on _{n}v) = α_{n}(_{1}Mv) + ××× + α_{1}(_{n}Mv) on _{n}V^{ #}; it follows routinely that M^{ #} is an isometry on V^{ #} with respect to B^{ #}. Furthermore, it is routinely verifiable that M^{ #} acts as a unipotent operator on U^{#}.Since M^{ #}| and _{U #}M| will have the same matrix representations and hence the same characteristic polynomials. But by an analogue to the above argument, since _{U}M^{ #} acts as a unipotent operator on U^{ #}, 1 is the only eigenvalue of M^{ #}|. Since _{U #}E is a splitting field for charpoly(M|) = charpoly(_{U}M^{ #}|), charpoly(_{U #}M|) factors completely in _{U}E[z] as an integer power of z - 1. Since 1 Î F, charpoly(M|) must also factor completely in _{U}F[z] as an integer power of z - 1. Therefore charpoly(M|) = (_{U}z - 1), as desired.^{ f}Finally, since ( (_{0}z) = χ(z) = charpoly(M) = charpoly(M|)charpoly(_{U}M|) = (_{U*}z - 1)charpoly(^{ f}M|), we must have charpoly(_{U*}M|) = χ_{U*}(_{0}z), which completes the proof.Q.E.D. The spinor norm is usually defined (as above) in terms of the symmetries of which the given isometry can be written as a product. Theorem 1, which we are now ready to prove, gives an intrinsic characterization of the spinor norm independent of such a factorization.
d(^{n}U^{*}).
M|) (see O’Meara, p. 139). Indeed, Θ(_{U*}M) = Θ(M|) since Θ(_{U*}M|) = 1 by Proposition 4. By Lemma 5, χ_{U}(_{0}z) is the characteristic polynomial of M|. Finally, since the largest subspace of _{U*}U^{*} upon which M| acts as a unipotent operator is {_{U*}0}, the orthogonal complement of that space is just U^{*}. Hence it suffices to prove the theorem for U^{*}.Let . Let _{V’}W and {w, . . . _{1}w} be as in Proposition 2. By Lemma 5, _{k}z - 1)M’) so that M’ has no eigenvalue equal to 1; it follows from Proposition 2 that V’ = W. Furthermore, by (2) in the proof of that result, we have
which is what we wanted. Q.E.D. As a special case of Theorem 1, we obtain a formula for the computation of the spinor norm in terms of a simple determinant. This formula was originally obtained by H. Zassenhaus as a by-product of his proof that the spinor norm is well-defined (i.e., independent of the choice of symmetries of which the given isometry is expressed as a product).
(_{0}z).We assert that v) = 0 for some v Î V. Let t be the least positive integer for which this is true for v in particular. If t = 1, then either v is an eigenvector corresponding to an eigenvalue of 1 or v = 0. Since there is no such eigenvalue, in this case v = 0. If, however, t > 1, then (-M - I)((-M - I)(^{t-1}v)) = 0 so that (-M - I)(^{t-1}v) is either an eigenvector corresponding to the eigenvalue 1 or is 0. Both are impossible, because there is no such eigenvalue and because t was chosen to be minimal; hence this case yields a contradiction. It follows that v = 0, and we have proved our assertion.Hence z). It follows that χ(1) = det(_{0}M + I).Let { x} be an orthogonal basis for _{n}V. It is almost trivial to verify that -I = T(x)×××_{1}T(x). It will now follow that Θ(-_{n}I) = Q(x)×××_{1}Q(x) = d(_{n}V). Furthermore, Θ(-M) = Θ(-I)Θ(-M) = d(V)Θ(M).Hence by Theorem 1, d( d(^{n}U^{*}) = 2d(^{n}V) so that Θ(M)det(M + I) = 2. Hence in ^{n}F^{*}/F^{*2}, multiplying both sides by 2det(^{-2n}M + I) yields Θ(M) = 2det(^{-n}M + I). By elementary linear algebra, 2det(^{-n}M + I) = det((M + I)/2), and we are done.Q.E.D.
v,_{2}v} a basis for _{3}V over Q. Define
Then Q^{*}/Q^{*2}.(2) Let v,_{2}v} a basis for _{3}V over Q. Define
Then We find that χ( z - 1) so that χ(1) = 4 = 1 in _{0}Q^{*}/Q^{*2}. Since
it follows by induction that ( (^{n}M - I) for all n Î N. Therefore u Î U if and only if M - I)(u) = 0,U = ker(M - I). By setting the matrix M - I = 0 we find that U is generated by the vector v - _{1}v - _{2}v. Therefore _{3}U^{*} =v Î V : B(v,v - _{1}v - _{2}v) = 0}. If _{3}v = α_{1}v + α_{1}_{2}v + α_{2}_{3}v Î _{3}U^{*}, we have 0 = B(v,v - _{1}v - _{2}v) = -3(α_{3} + α_{2}). In this case α_{3} = -α_{2} and it follows that _{3}v, _{1}v - _{2}v}_{3}U^{*}. By direct computation, d(U^{*}) = 8 = 2 in Q^{*}/Q^{*2}.Applying Theorem 1, we now have Θ( d(^{n}U^{*}) = (8)(2) = 16 = 1 in Q^{*}/Q^{*2}.BIBLIOGRAPHY Anton, Howard. Dickson, L. E. Mason, Geoffrey. “Groups, Discriminants and the Spinor Norm.” O'Meara, O. T. Zassenhaus, H. “On the Spinor Norm.” Author’s Note: I wrote this paper in the spring of 1991 as the research paper for a Master of Science degree in Mathematics, with which degree I graduated that May from Southern Illinois University at Carbondale. It is in the subspecialty of abstract linear algebra. |