A k n o t w i l l b e a s m o o t h o r i e n t e d s u b m a n i f o l d ~ n o f S n + 2 , w h e r e n i s r e q u i r e d t o h a v e t h e h o m o t o p y t y p e o f t h e n - s p h e r e s n . Two knots ~no,~i n C sn+2 are cobordant if there exists an oriented smooth submanifold V C I x S n+2 of dimension n+l such that (i) bV = V ~ (bl x S n+2) = 7- 1 U (-[o) , where }-i' i = O, i, is regarded as a submanifold of {i} X S n+2, and -Z 0 is Z 0 with reversed orientation; (2) The inclusions ~o ~ ¥' ~i ~ V are homotopy equivalences; (3) V meets bI x S n+2 orthogonally, i.e. the intersection of V with a neighbourhood of bl x S n+2 is [0,~) x~ ° U (l-G, 1] x~ 1 for some small ~ > 0.

Cobordism is an equivalence relation between knots of the same dimension. Transitivity is guaranteed by condition (3).

Let C n be the set of cobordism classes of knots 7-n c S n+2. It is easy to see that the ambiant connected sum of knots induces an addition of cobordism classes which turns C n into an abelian group. (For details, see [i], Chap. Ill.) The ambiant connected sum is defined as follows: Given two knots ~n i c S n+2, i = 0, i, let h I. : (D n+2, D n) > (S n+2' X~) be two embeddings such that ho is orientation preserving and h I orientation reversing on both D n+2 and Dn ? Form the disjoint union (S n+2 - %(0)) U ( S n+2 - hl(0)) and identify ho(tX ) with hl((l-t)x ) for 0 ~ t < 1 and x 6 S n+l = bD n+2. This construction yields an embedding of the connected sum ZO~ ~i into S n+2 = sn+2 ::~Sn+2, whose cobordism class is by definition the sum of the cobordism classes of ~o and ~-i" The standard embedding S n ~ S n+2 represents the zero element of the resulting group C n .

THEOREM i. C2k = 0 for k ~ 0, i.e. eyeEy even dimensional knot is cobordant to the standardly embedded S 2kC S 2k+2.

However, the groups C2k_l are not finitely generated. We shall give the purely algebraic description of C2k_l due to J.Levine [3]. It turns out that for k ~ 3, C2k_l depends only on the parity of k, so that C n is periodic of period 4 for n >= 4. For k >= 3, the group C2k_l is contained in an infinite unrestricted direct sum of cyclic groups Z, Z/2Z and Z/4Z. There are elements of each of these orders (~ , 4 and elements of order 2 not divisible by 2) occuring in C2k_ 1 ?

i. Seifert surfaces

It is well known that a smooth closed curve ~-I C S 3 bounds an orientable surface embedded in S 3, called a Seifert surface of the knot. This fact generalizes to all dimensions: Ever Z ~n C S n+2 i__~s the boundary of an oriented smooth submanifold ¥n+l C S n+2 .

The proof is easy. Observe first that ~n has trivial normal bundle in S n+2. Then, take an embedding ~n x D 2 ~ sn+2 extending the given knot ~n x [ 0} = ~-n C Sn+2o Now, it suffices to show that ~n x {Xo} , where x O ~ S 1 = bD 2, bounds a submanifold V in the complement X = S n+2 _ int [ ~n x D2). Let ~ : ~n x S 1 ~ S 1 be the projection on the second factor. Then, In x ~Xo} = ~-l(xo) and if we can extend ~ to a map ~ : X )S 1 regular at Xo, then ~-l(xo) will be the desired submanifold. The extension ~ exists by obstruction theory, since, at least for n >I, Hq+I(x, ZxSI;~q(SI))= 'for all q. In the sequel we only retain from ~-l(xo) the connected component V containing the given knot as its boundary. Observe that 0

Using the Seifert surface, we first sketch the proof of theorem i. Let ~2k ~ 82k+2 be a knot and V 2k+l a Seifert surface for this knot, i.e. V C S 2k+2 and bV = X. By surgery theory (see [2], § 6.), we know there exists a stably parallelizable (2k+2)-manifold W with corners, so that bW = V U (I xZ) U V o, where V ° is contractible and (W, V) has a handle decomposition with handles of types ~ k+l only. (For k = 1 see the argument in [i], p.265. ) By a theorem of M.Hirsch (or the direct argument in [i]), the embedding V c S 2k+2 extends to an immersion W C~ D 2k+3. Now, since [k+l) + (k+l) < 2k+3, any possible intersection of the handles of (W, V) can be removed by a (small) regular homotopy. Thus we get an embedding W C D 2k+3 extending V C S 2k+2. Its restriction to the part [I x~) U V o of the boundary bW yields a contractible submanifold of D 2k+3 with boundary the given knot ~2k. From this it easily follows that the cobordism class of ~2k C S 2k+2 is zero.

The attempt to carry the same proof idea for ~2k-1 C s2k+l fails in two places. First, one may not be able to reach a contractible manifold V O2k by performing framed surgery [in dimensions ~ k) on a Seifert surface V 2k of the given knot. Secondly, even if one could, one still runs into the more serious trouble that the (k+l)-handles of the resulting manifold pair [W, V) will in general have nonremovable intersections in D 2k+2 To measure these obstructions one introduces an algebraically defined cobordism group of bilinear forms. ~2. Cobordism of bilinear forms

Given S = ~ i, we define algebraically (following [3]) a group C£(Z) depending on S . Eventually, it will turn out that C2k_l -CZ(Z), with ~ = (-i) k for k ~ 3. (For results in lower dimensions, see [31

Gonsider integral valued bilinear forms on finitely generated free Z-modules. If A is such a form on the free Z-module H, denote by A ' t h e t r a n s p o s e o f A d e f i n e d b y A ' ( x , y ) = A ( y , x ) f o r a l l x , y ~ H . We s h a l l s a y t h a t A i s a n £ - f o r m i f t h e f o r m A + ~ A ' i s u n i m o d u l a r . ( S e e t h e f i r s t f e w l i n e s o f ~ 4 b e l o w f o r t h e d e f i n i t i o n . )

DEFINITION. An ~ -form A defined on H is null-cobordant if there exists a Z-submodule H o of H such that ( 1 ) 2 . r a n k ( H o ) = r a n k ( H ) ; (2) A vanishes on Ho, i.e. A(x, y) = 0 for all x, y ~ H o.

Remarks. Observe that such an H o can then be taken to be a direct summand of H. Observe also that if A is an ~ -form on H, then the rank of H must be even. Indeed, the form S = A + E A' induces a symmetric, unimodular form S m on the F2-space H/2H such that St(x, x) : 0 for all x e H/2H. Take a maximal F2-subspace U of H/2H such that ~(u, v) = 0 for all u, v e U. Let V be the orthogonal complement of U, i.e. V = { v E H/2H I S'(u, v) = 0 for all u ~ U I. Then, dim U + dim V = dim H/2H = rank (H), since S ~ is unimodular. On the other hand, the maximality of U implies U = V. Indeed, if v g V, then S~(v, v) = 0 together with Sm(u, v) = 0 for all u ~ U imply that Sm(Wl, w2) = 0 for all Wl, w 2 ~ F2v + U. Since U is a maximal subspace with this property, it follows that v E U. Thus rank (H) = 2.dim U.

If A is an ~-form on G and B an Z-form on H, we denote by A ~ B the ~ -form on G ? H given by

(A ? B) (x ? u, y ? v) = A(x, y) + B(u, v), where x, y ~ G and u, v ~ H.

DEFINITION. Let A, B be two E -forms defined on G and H respectively. We say that A and B are cobordant if the E-form A ~ (-B) on G ~ ~ is null-cobordant.

Cobordism of E-forms is an equivalence relation. Reflexivity and symmetry are both trivial. Transitivity follows from a cancellation lemma:

Proof of the lemma. Let A be defined on G and B on H. By assumption there exists L ~ G @ H so that 2.rank L = rank G + rank H and A ~ B vanishes on L. Also, there is H o C H with 2.rank H o = rank H and B vanishes on H O. Set L ° = L ~ (G ~ Ho) and let G O be the projection of L ° on G. If x, y ~ Go, there exist u, v g H ° such that x ~ U, y ~ v ~ L. Then,

A(x, y) = A(x, y) + B(u, v) = ( A S B) ( x ~ u, y ~ v) = o. It remains to prove that G o has the right rank. Take H o to be a direct summand of H and write H = H ° ~ H I. Projecting L on H i gives an exact sequence

0 ? L ° ~ L ~ L 1 *0.

Observe that L ~ H ° and ~o ~ L1 are orthogonal under B +~ B'. For L ~ H O and H ° this is obvious ? Let u 6 L ~ H ° and w ~ L I. There exists x ~ v 6 G ~ H ° such that x ~ [v + w) ~ L. Then, ( m ~ m ) ( u , w~ = (B+~B,)(u, v + w) = (A+EA')(0, x) + (B+EB')(u, v + w) = CA @ B)Cu, x+v+w) + Z . ( A e B) Cx+v+w, u) = 0 since ~ is also in L. Therefore, since B + £B' is unimodular, rank (L ~ Ho) + rank (H O ~ LI) ~ rank H.

Now, L ~ H o = L o ~ H and there is an exact sequence 0 ~ Lo~ H ~ L ° ~ G O ~ 0.

The two exact sequences give rank G o = rank L o - rank ~L o ~ H), rank L ° = rank L - rank LI, and so rank G O = rank L - (rank (L~Ho) + rank L1).

Pluging in the above inequality and using the hypotheses on rank L and rank Ho, one gets rank G o = ½.rank G. Of course, equality must hold since A+ZA' which is unimodular vanishes on G O . This completes the proof of the lemma.

It is clear that the direct sum of E-forms induces an addition of the cobordism classes and turns the set CZ(Z) of cobordism classes of E -forms into an abelian group. We have included reference to Z in the notation because similarly defined groups over other coefficient domains will be introduced later.

Using a Seifert surface, we shall now associate with every knot ~-2k-1 C S 2k+l an (-1)k-form. Let ¥2k be a Seifert surface for a given knot Z C S 2k+l, i.e. V 2k C S 2k+l and bV =~ . Let v be the normal vector-field to Y in S 2k+l and i+, i _ : V ~ S 2k+l - V the m a p s d e f i n e d b y i + ( P ) = P ~ . v ( P ) f o r s m a l l ~ > O, P e V.

Let H = Hk(V2k)/(torsion ). If x, y g H, the linking number L(x, i + (y)) ~ Z in S 2k+l is well defined. We set

A(X, y ) = L ( X , i + ( y ) ) .

O b s e r v i n g t h a t L ( x i + ( y ) ) - L ( x , i _ ( y ) ) = I ( x , y ) , w h e r e I d e n o t e s the intersection number in V, we have

L(y, i+(x)) = L(i_(y), x) = (-1)k+iL(x, i_(y))

= ( - l ) k + l . t L ( x , i + ( y ) ) - I ( x , y ) ) , a n d t h u s

A ( x , y) . ( - 1 ) k A ( y , x) = I ( x , y ) .

By Poincar4 duality, the intersection form I on H is unimodular, and so the form A is a (-l)k-form.

THEOREM 2. The above construction provides a well-defined map L : C 2 k _ l ~ C E ( Z ) , w i t h E = ( - i ) k .

F o r k >= 3, L i s a n i s o m o r p h i s m .

Sketch of proof. If Seifert surfaces Yl' V 2 have been chosen for 2 knots ~l' ~2' their ambiant boundary connected sum V is a Seifert surface for ZI~- 2. Moreover, the (-l)k-form associated with Y is clearly the direct sum of the [-1)k-forms associated with V 1 and V 2. Thus, in order to show L well-defined, it suffices to consider the case of a null-cobordant knot ~ 2k-1 C S 2k+l and to prove that the (-1)k-form A associated with any of its Seifert surfaces is null-cobordant. Let ~ ~ D 2k+2 be a 2k-disc with boundary T-. The union M = /k U V, where V is a Seifert surface for is a closed manifold (with corners along ~) embedded in D 2k+2. An obstruction theory argument similar to the one in § l, shows that M is the boundary of an oriented submanifold W C D 2k+2. Consider H O = Ker Ji' where j~ : HkM ~ HkW. If x, y ~ Ho, and say ~, are representative cycles, then ~ ,~ bound (k+l)-chains ~,~ in W. We can view L(x, i+(y)) as the intersection coefficient I(~, i+(~)) which is clearly zero since W and i+W are disjoint. [i+ is here the extension to W of the map defined above on V using the normal vector-field.) Thus A vanishes on H ° = Ker j . It then follows that A is null-cobordant by showing that rank H ° = ½°rank HkM. For this, write the homology exact sequence of (W, M): 0 ~H2k+l~W, M)--~H2k(M) ~ H2k(W) ....

.... Hk+I(W ) ~Hk+I(W, M) d Ker Ja ~ 0, breaking it at HkM. It yields rank Ker jm = rank Hk+l(W, M) - rank Hk+l(W ) + rank Hk+l[M ) - rank Hk+2(W , M) + ...

= (-1)k+l.½E(M) + ½.rank HkM + (-1)k.E(W), where E denotes the Euler characteristic. Since the "double" of W is a closed odd-~e~ional manifold, O = E(W U W) = 2.E(W) - E(M), and so rank Ker jm = ½.rank Hk(M ) as desired. Hence, L is a welldefined homomorphism.

Proving the surjectivity of L is straightforward. Given A so that A + (-1)kA ' is unimodular, first take a matrix representative which we denote again by A. Construct a 2k-dimensional stably parallelizahle manifold V with boundary whose intersection matrix is A + (-1)kA ' in dimension k. The manifold ¥ is obtained by attaching k-handles to D 2k with linkings of the attaching (k-1)spheres dictated by the entries in A + (-l)kA ' . (See [~ , p.256 for details.) Then, bY =~ is a homotopy (2k-1)-sphere as a consequence of the assumption that A + (-l)kA ' is unimodular.(k >= 3 is needed here to quarantee ~l ~ = 0.) Take a random embedding V C S 2k+l. Its restriction to bY yields a knot for which ¥ is a 5eifert surface, and thus to which there corresponds some (-1)k-form B of the same rank as A. Actually, B + (-l)kB ' = A + (-l)kA ' , since both are equal to the intersection form on Hk¥. One can then change B to A by readjusting the mutual linking of the handles of V.

The proof of injectivity of L proceeds in two steps: (1) One shows that every knot ~ k-I ~ S 2k+l is cobordant to a simple knot _~20k-1 C s2k+l, i0e. ~o is the boundary of a (k-l)-connected manifold V o C s2k+l; (2) It is then enough to show for a simple knot that null-cobordism of the associated (-l)k-form, constructed using a (k-1)-connected Seifert surface, implies null-cobordism of the knot.

To prove step (1), use surgery on a Seifert surface V 2k for the knot [2k-l c s2k+l to produce a manifold W with bW = ¥ U (I x( )U V o (corners along hV = {0 } x Z and Ill x 2 = bVo) , where V O is (k-1)-connected and ~W, V) has a handle decomposition with all handles of type = k. As in the proof of theorem l, it is easy to extend V ~ S 2k+l to an embedding W C D 2k+2. The problem is then to embed a (2k+2)-disc D2k+2o in D 2k+2 so that (int Do) ~ W = @, in such a way as to engulf V ° c bD ° = S~ k+l. I1" this can be done, then D 2k+2 - int D o2k+2 which is diffeomorphic to I x S 2k+l contains a cobordism between [ ~ S 2k+l and Z o = bV o C S 2ok+l. Since V o is (k-l)-connected, Z o C S o2k+l is a simple knot. The existence of an embedding D o2k+2 ~ D 2k+2 with V o ~ bD o follows from the engulfing theorem of M.Hirsch. (Theorem 2 of "Embeddings and compression of polyhedra and smooth manifolds". Topology, Vol.4 (1966), 361-369.)

Remark. It was pointed out to me that the existence of a PL-embedding of D o2k+2 with the desired properties is fairly obvious, taking for granted the relative regular neighbourhood theorem. Indeed, since V ° has k-dimensional spine, the cone over spine (Vo) embeds disjointly form int (W), observing that (W, V) has only h-handles with h <= k. Thus D o can be taken to be a regular neighbourhood of such an embedded cone.

Step (2) is an application of surgery. We have Z 2k-I ~ S 2k+l a simple knot bounding V 2k C S 2k+l, where V is (k-l)-connected. The problem is to perform ambiant surgery on V in D 2k+2 so as to produce a contractible submanifold of D 2k+2 with boundary ~2k-l. We are assuming that the associated (-l)k-form is null-cobordant, so by hypothesis there exists a basis Xl, ..., Xr, Xr+l, ..., X2r of HkV with the property that L(x~, i+(x~)) = 0 for all ~,~ ~ r. Clearly then, I(x~, xfl) = 0 for all ~,~ ~ r and since V is (k-l)-connected (and k > 3), we can take disjoint embeddings f~ : S k

= 1 ..... r representing x~. The conditions L(x~, i+(xfl)) = 0 for ~ ~ ~ then mean that the ~ 's can be extended to mutually disjoint embeddings F~ : D k+l ~ D 2k+2. Moreover, L(x~, i+(x~)) = 0 implies that F~ is extendible to an embedding F~ : D k+l x Dk---~D 2k+2 such that F~ I S k x D k is a tubular neighbourhood of f~(S k) in V. It is well known then, and easy to check that surgery on x i , ..., x r g Hk(V ) (as just shown possible) replaces V by a contractible manifold with the same boundary. Indeed, A'(x, y) = A(y, x) = S(sy, x) = 0 since sy g L. Then, A(x, y) = 0 for x ~ M, y E L follows from S = A + E A'.

Therefore, A induces a form B on M/L. Since M is a pure submodule, and hence a direct summand in H, it follows that T = B + E B' is unimodular. Thus B is an Z -form. Consider A @(-B) on H e(M/L). It vanishes on the submodule D of all (x, xa), where x ~ M and x = class of x mod L. Now, rank D = rank M = rank H - rank L since M is the orthogonal complement of L ~nder the non-singular form S. Thus rank (H + (M/L)) = rank H - rank L + rank M = 2.rank D. Hence, A is cobordant to B of strictly smaller rank.

Exercise. Define A to be reduced if A is non-singular and there is no non-zero submodule invariant by s on which A vanishes. Prove that if A and B are reduced £-forms on G and H respectively and are cobordant, then there exists an isometry f : GQ------~HQ, i.e. a Q-isomorphism tGQ = G ezQ ) such that B(fx, fy) = A(x, y). On the other hand, the example

A = ,

B = e x a m p l e , i f A i s a n E - f o r m , t h e n ~ A ( X ) = d e t ( A . X + E A ' ) i s a (well-defined) reciprocal polynomial in Z[X] called the Alexander polynomial of A. Let A be defined on H say, and set s = S-IA, where as ab,ove S = A + ~ A', i.e. s is defined by A(x, y) = S(sx, Y7 for all x, y ~ ~. Suppose that A is non-singular. Then t = 1 - s -i : H $ Q - H ? Q is an automorphism. (It is easily verified that t = -ZA-IA'.) Moreover, t is an isometry for A and S extended to H ? Q. The polynomial ~A(X) = act (A.X + ~A') is the characteristic polynomial of t (up to the non-zero factor det A).

Now, let ~ = + 1 be given together with an irreducible, reciprocal polynomial of even degree ~ 6 Z[X] such that ~(i) = (_~)m where 2m = deg~.

Define an isometric structure over Z or Q associated with ~ , to be an £-sy~m~etric unimodular hilinear form S on a Z-space (resp. Q-space 7 V together with an injection s : V ~ V satisfying (i) S(sx, y) + S(x, sy) = S(x, y), (27 ~(X) = X 2m I(I-X -I) is the minimal polynomial of s.

Note. Condition (i) is equivalent with S(tx, ty 7 = S(x, Y7 for t = i - s -i . We have stated it in the more complicated form of condition (17 because t is in general only defined over Q. Also, condition (2) simply says ~ is minimal polynomial of t.

There are obvious definitions of cobordism: An isometric structure IS, s) on V is null-cobordant if there exists an s-invariant subspave V o ~ V of half the rank of V on which S vanishes. Further, (Sl, s17 and ($2, s27 are cobordant if (S 1 @ (-$27 , s I ? s2) is null-cobordant. Trivial modifications in the proof of the lenmla in 2 show that cobordism of isometric structures (associated with the saune £ , ~ ) is an equivalence relation. The set of equivalence classes C 1Z(Z), resp. C~(Q) is a group under direct sum.

Prop. 3. There are injections

Z~ c~~ ( z ) i , c e ( z ) J ,Z~c x ( e ) , where the extreme terms are restricted direct irreducible, reciprocal ~ £ Z[X] of even degree ~ ( 1 ) : ( _ ~ ) m . sums over all 2m satisfying

Proof. Let (S, s) be an isometric structure on a free Z-module H representing an element o- in some C[(Z). Define A by A(x, y) = S(sx, y) for all x, y ~ H. Condition [i) on S and s implies that A + Z A' = S. Thus A is an E-form on H. Set i(0-) = cobordism class of A. It is immediate that we get a well-defined map i : ZA C~(Z) , CZ(Z). Next we show that i is injective. Let (Sk, Sk) 6 C~k[Z ), k = 1 ..... n, be such that A = A 1 @ ... @ A n is null-cobordant, where Ak(X , y) = Sk(SkX, y) on H k. Thus we assume that H = H 1 @ ... $ H n contains a Z-subspace N c H with A(x, y) = 0 for all x, y ~ N and rank H = 2.rank N. We can assume that N is pure. Observe first that sN ~ N, where s = s I @ ... ~ s n . Indeed, since A(x, y) = 0 and thus A' (x, y)= 0 for all x, y g N, we have S(x, y) = 0 for all x, y g N. If now x is a fixed element of N, then S(sx, N) = A(x, N) = 0. Therefore, N + Zsx is S-orthogonal to N. Since S is unimodular, rank (N + Zsx) <= rank H rank N = rank N. Thus sx ~ N, since N is pure. It follows next that the projection N k of N into H k is contained in N. (k = 1 .... , n.) To see this, let ~k = ~ ~ where ~k is the minimal polynomial that each A k is null-cobordant.

A(x~, yK) = A(/L&d(t), / 4 ~ ( t ) y ) = A(x, t ~. / ~ ( t ) . /~C.(t)y) = O. where m = -deg ~W . This proves that each ~K is reciprocal.

The statement about the sign follows from ~A(1) = (_£)m, where 2m = d e g A A [This in t u r . because if l A(1) is the determinant of the skew-symmetric matrix A - A' . Thus it must be a square. If g = +i, then ~A(i) = det [A + A') = (_l)~]ran2k_si1gna~ture Now since A + A' is unimodular and even, the signature is divisible by 8.]

It remains to verify that ~is the minimal polynomial of t~ = t]Y~. But if not, there is an integer e, i< e ~ e~, such that W~ = ~ e-I (tq)¥g ~ 0 and X~e( tW)V~ = 0 . Then, for arbitrary x~, yw 6W~ one has

A ( ~ , y~) = A(Ae~-i t t ~ ) ~ , A~e-i tt~)v~)

= A{u -XK2e ~tY)~K ) = o , since 2e-2 ~ e. Hence, again, A would not be reduced.

It is clear that the map j is well-defined since the cobordism class of a reduced E-form A determines its isomorphism class over Q. To see that j is a homomorphism, observe that there is an obvious definition of a reduced isometric structure, i.e. (S, s) defined on H is reduced if s is injective and there is no non-zero subspace H O ~ H such that SIH O = 0 and S[Ho) ~ Ho. The reduction processes for an ~-form described in Prop.l and Prop.2 yield reduction processes for isometric structures. Hence, first reducing A 1 ? A 2 and then applying j yields the same as gotten by reduction of J(AI) ~ J[A2). We leave the details to the reader.

It is easily seen that none of the maps i nor j is surjective. So, non-degenerate bilinear form S on a finite dimensional Q-space V together with an isometry t : V , V, i.e. S(~x, ty) = S(x, y), such that ~ is the minimal polynomial of t. (The polynomial E Z[X] is given as irreducible, reciprocal and satisfying (i) = ~ i. This implies that the degree of ~ is even. The sign can then be fixed so that ~(i) = (-E) m, where 2m = deg~ .)

Prop.4. The group C ke(Q) is independent of ~ , i.e . C +AI(Q) and C~(Q) are isomorphic, where X= (-l)m~.

Suppose QSo, t) is an isometric structure on a Q-space V with S o symmetric. We associate with (So, t) the structure (SI, t) where S 1 is defined by Sl(X, y) = So(X , (t - t-l)y). I t is easy to check that S 1 is skew-symmetric. The conditions on ~ imply that t - t -I is an isomorphism, and so S 1 is non-degenerate. Clearly, t is an isometry for S I. It is immediate that the map (So, t) ~ (SI, t ) is compatible with cobordism and induces an isomorphism C~I(Q) - C~I(Q). For the balance of the paragraph we let CA(Q ) stand for C~I<Q). The study of this group can be attacked in the following way: Let K = Q[X]/(~(X)) and let T E K be the element corresponding to X. Since ~ is reciprocal, ~-i 6 K is also a root of ~ and there is a Q-automorphism of K sending into ~-i. We denote it with a bar: a : ~ a. Let F he its fixed field: x ~ F iff x = x. Now, suppose V is a (finite dimensional) Q-space with an isometric structure ~S, t) associated with (~ = +i, ~ 6 Z[X]) as above. Following J.Milnor [5] , we can equip V with a K-space structure and a hermitian form ( , ) with respect to the involution on K. To give V a K-space structure it is enough to define ~.x = t(x). (Recall we are assuming that is minimal polynomial of t.) To define the hermitian form ( , ), fix a pair x, y g V. Then S defines a Q-linear map K m Q via & ~S(ax, y). Since K/Q is separable, there exists a unique (x, y) by definition,

of K such that tracexl la-( ' Y) I = S(a , y ) , for all a ~ K. It is easy to verify that (x, y) is linear in the first variable and (y, x) = [~-,~-~. (For details, see J.Milnor [SJ.)

Conversely, given a (non-singular) hermitian K-space V, the formulae t(x) = ~.x and S(x, y) = traceK/Q(X , y) define an isometric structure (S, t) on V, viewed as a Q-space now, where t has minimal polynomial ~. Thus, if we define a hermitian K-space ¥ to be nullcobordant if it contains a K-subspace U such that dim U = ½.dim V on which the hermitian form vanishes, we get a one-to-one correspondence between the elements of CI(Q} and the cobordism classes of (non-singular) hermitian K-spaces.

Hence, we can derive some information on CA(Q ) from the classification of hermitian spaces. Let Vf, f £ F, be the 1-dimensional hermitian K-space defined by (x, y) = x,y.f. If f' = fkk for some k E K, then Vf, ~ Vf. Now, every hermitian K-space is the direct sum of 1-dimensional spaces Yr. Thus, there is a surjection where NK/F is the norm from K to F and the dot means that we remove the 0-element. ~[F'/NK/FK" ] is the integral group ring of the multiplicative group F'/NK/FK'.The map is the linear extension of f ,Vf.) Actually, if we define a product in CA(Q ) by the tensor product over K of the associated hermitian K-spaces, the above map is a ring homomorphism. Its kernel can probably be described in terms of symbols. We do not pursue here and, following J.Levine shall only describe invariants for the elements of CA(Q ) obtained by embedding C~(Q) into a direct sum of cobordism groups for isometric structures over the real and p-adic completions of Q.

More precisely, let V be a hermitian space over K = Q[X]/(~(X)), where ~ E Z[X] is irreducible, reciprocal of degree 2m, and [42 , ~(i) = (-i) m. Denote again by F the fixed field of the involution on K determined by ~ ~ ~ = ~-l, where ~ is the image of X in K. We can view V as an F-space with symmetric, non-singular, F-valued bilinear form F defined by ~ (x, y) = traceK/F(x , y) and an isometry t defined by t(x) = ~.x. Set c = T + ~ ~ F. Then, t (or T) satisfies the equation t 2 - c.t + 1 = 0. We shall denote by ff the polynomial #(X) = cX + i .ow every prime in F, we extend the coefficients from F to the ~-adic comple%ion F . We get a map where C~(F~) = 0 if the polynomial ~ is reducible over ~, and otherwise C~(F~) is the eobordism group of hermitian K~-spaces (K~ being the completion of K at a prime extending~, or equivalently K~ = F~[X]/(~(X)).) Equivalently, Cr(~) can be viewed as the cobordism group of symmetric hilinear forms on F~-spaces with an isometry t satisfying/~(t) = O.

Prop.5. The map Of c o u r s e , p ( t x , tx~ = ~ ( x , x) = O. T h i s p r o v e s p r o p o s i t i o n j .

It remains to calculate C~(F~). There are 3 cases apart from tBe trivial case where ~ is reducible in F~[XJ.

Prop.6. Suppose ~ is irreduclble in the completion ~ . Then, (l) I f ~ is a real prime, C~(R) ~ Z; (II) I f ~ is a finite prime and -i is norm from K~ = ~ [ X J / ( ~ ( x ) ) t_oo~ , then ~ ( F ~ ) ~ Z/2Z ~ Z/2Z;

( I I I ) l f ~ is a finite prime and -1 is not a norm from K ~ t~o ~ , then = Z/4Z.

Observe first that given an isometric structure (p, t) on an ~ -space V, where t 2 - ct + 1 = 0, the cobordism class of the form alone determines the cobordism class of the isometric structure. II~deed, if W ' t) is reduced and ~ is null-cobordant, then V must be 0, Otherwise, there exists x E V, x ~ 0 such that ~ (x, x) = 0, and the argument at the end of the proof of Prop.5 shows that x and tx span a t-lnvariant non-zero subspace of V on which vanishes, contradicting the assumption that (p, t) was reduced.

Now, in case I, i.e. F~ = R, the cobordism class of.a symmetric bilinear form is determined by its signature. It follows that C~(R) = Z generated By Y1 in the above notation, i.e. C viewed as R-space with ~ x , y) = x~ + xy and t(x) = ~ . x , where ~ is a root o f ~ ( X ) = X2 -cX + i. (Observe that the involution defined above does coincide with complex conjugation in this case. What else could it be?)

Next, suppose t h a t ~ is a finite prime and -1 is a norm from K ~ to ~ . Since K ~ / ~ is a finite extension of local fields, [F~ : N K ~ = E K ~ : ~ ] = 2, where for simplicity N denote the norm from Kf to ~ . Let I, g be representatives of the 2 classes in /NK~ . Then, C ~ ( ~ ) = Z/2Z ~ Z/2Z generated by V 1 and Vg, where as above Yf denotes the ~ - s p a c e of dimension 2, which we identify with K~, together with the isometric structure given by ~(X, y) = trace (x.y.f) and t(x) = Tx, where the trace is from

It is clear that for every f 6 F~ , Vf ~ Vf is null-cobordant. Proof: Choose k d K so that -1 = k.k. Then U = ~(x, xk) ~ Vf x Vfl has half the dimension of Vf ~ Vf and ~ ~ vanishes on U. It is also obvious that V 1 and Vg are reduced and thus neither one is ~ull-cobordant. Finally, V 1 and Vg are not cobordant, otherwise, being reduced, they would be isomorphic. It is easy to check that V 1 and Vf are isomorphic if and only if f ~ N(K~).

It remains to handle the case where ~is a finite prime and -1 is not a norm from K~ to ~. We still have [F~ : NK~J = 2, and in this case we can take +I and -1 as representatives @of the two elements in F~/N . Obviously, V 1 ~ V_l is null-cobordant. Thus in this case V 1 suffices to generate C~(F~). Claim: Vl is precisely of order 4. Observe first that V 1 ~ V 1 is not null-cobordant. If it was, then V 1 and V 1 would be cobordant, hence isomorphic, since they are reduced. But V 1 = V_l implies that -1 is a norm. It remains to show that V 1 @ V 1 ~ V 1 ~ V 1 is null-cobordant. This relies on the fact that -I is a sum of norms from K~to ~. Assume for a moment that -1 = kl.k I + k2.k 2. An easy calculation shows that the form ~ ~ ~@~ vanishes on the subspace of V 1 ? V 1 @ V 1 ~ V 1 consisting of all (x, y, klX + k2Y , k2 x - klY), where x, y ~ K~. Hence, -1 = kl.k I + k2.k 2 implies that V 1 is of order 4 in C~(~). In order to show that -l = kl.k I + k2.k 2 we can either appeal to a general theorem stating that a regular quaternary space over a local field is universal. (See 0'Meara, [6], ~ 63C. In particular Remark 63:18.) Or, we can see this directly in the case of interest to us as follows: Let Qp be the completion of the rationals contained in F. Assume first p ~ 2. Then, -1 is a sum of squares in Qp and so afortiori a sum of norms in every quadratic extension K~/~. Sketch of proof: First, an easy counting argument shows that -1 is a sum of squares in Fp, the field with p elements. Indeed, let S be the set of non-zero squares in F P . Xf 0 ~ -1 - S, then -1 is a square and we are trough. Otherwise, -1 - S ~ F"P = S U (-S). We have - S ~ -1 - S because -1 ~ -1 - S. It follows that -i - S ~ -S because the inclusion would imply equality since the two sets have the same cardinality ½(p-l). Thus, -i - S intersects the complement S of -S, i.e. (-l-S) ~ S ~ ~. Thus there exist 2 , ~ 2 6 ~ so that -l-p 2 =~2

Now it is easy to solve -1 = x 2 + y 2 in Qp. Choose Xo' Yo ~ Z so that -1 = Xo2 + Yo2 mod p. We can actually take 0 <= Xo < P' < 0 = Yo < p" Assuming by induction that we can find s n = Xo + xlP + ... + x p n tO <= x.1 < p) and tn = Yo + Yl p + "'" + Yn p n (0 ~ Yi ~ p) such that

-i = s 2n + t n2 nod p n+l , set sn+ 1 = sn + xn+iP n+l ' tn+l = t n + Yn+l p n+l with unknown Xn+l, Yn+l" Writing 1 + S2n + t2n = kn pn+l , we find that 1 + Sn2+ 1 + t n2+l = 0 mod p n+2 if (and only if) Xn+l, Yn+l satisfy the congruence

k n + 2(XoXn+ 1 + YoYn+l ) = 0 mod p.

Solving this congruence in Xn+l, Yn+l is clearly possible since p ~ 2 and Xo, YO cannot both be 0 mod p since Xo2 + Yo2 = -i mod p. The sequences _I Snl , I tnl clearly converge in Qp to x, resp. y, satisfying, x 2 + y 2 = -i.

The case p = 2 would be more difficult. However, it does not arise here. Recall that [l-r) -I has to be integral. (~ is a root of the polynomial X 2 - cX +i defining the extension K~/~.) This implies that ~ 1 and hence c -i must be integral in ~. (Herefl2 of course.) ~ut then, using the fact that the multiplicative group of the residue class field of F has odd order (finite field of characteristic 2), one can find an integral element ~ of F such that ~2 + c-2 = 0 mode. Setting d = c 2 - 4 (the discriminant of X 2 - CX + i), we then have

( 2 ~ } 2 - d . c -2 = -1 mod 4~.

From there it is easy to find x, y E F~ such that x 2 - d.y 2 = -i using the successive approximations as in the case p ~ 2. This last equation however means that -i is a norm from Kfto ~.

Courant Institute NEW YORK, N.Y.