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Topology, , Analysis—Relations and Missing Links Beno Eckmann

his is largely, but not entirely, a histor- The second part is about the of the ical survey. It puts various matters to- unitary groups. It leads to a different linearization gether that are usually considered in phenomenon stating that maps of a into a separate contexts. Moreover, it leads to unitary are homotopic in the infinite unitary open, probably quite difficult problems group to a linear (in the situation of the first andT has analogues in contemporary mathematics. part it is, in general, not true that any continuous There are three parts. solution is homotopic to a linear one). Again, no di- The first part is about a certain range of clas- rect proof is known—by differential , by sical results in concerning con- approximation, or by optimization. This lin- tinuous real functions and maps, vector fields, earization result is due to the close relation be- etc., that can be stated in a very simple way: just tween the classical Hurwitz-Radon matrices and replace “continuous” by “linear”. They thus seem Bott periodicity. Since the latter is the source of to be reduced to problems of algebra, essentially topological K-theory and general theo- , where the solution is relatively easy. ries, a direct approach would establish a link be- The proofs of all these statements, however, do not tween an important old matrix problem and refined use such a reduction principle. They are beautiful methods of algebraic topology. and in general quite difficult, using elaborate ideas The third part is a short outlook concerning top- and techniques of topology such as ics of contemporary interest. It is about recent operations, spectral , K-theory, and so trends in homology theories for spaces with infi- on. nite G using In these cases the absence of a reduction prin- methods (Hilbert-G-modules, `2-homology, and ciple from continuity to linearity is thus a missing `2-Betti numbers). The presentation, necessarily a link between two areas. Of course, such a link is little technical, turns around the von Neumann al- not necessary since the results are proved. Still, it gebra of G considered just as an algebra, forget- might be interesting to have, at least in special ting the analysis behind it. We first recall the (weak) cases, a direct reduction to linearity which could Bass (1976), which has become a theo- throw new light onto old and new mathematics. rem for several classes of groups but is still open We begin, in that first part, with a very ele- in general, and give it a simple von Neumann al- mentary example. It contains in a nutshell the gebra formulation. Application to algebraic topol- problem to be discussed later for analogous but ogy and group homology again lead, in a very spe- more complex phenomena. cial example, to linearity. Beno Eckmann is professor emeritus of mathematics and We probably all agree that eventually reducing retired founding director of the Institute for Mathemati- a difficult problem to a “nice” situation is at the cal Research at Eidgenössische Technische Hochschule heart of mathematics. What I present here is a (ETH) in Zürich. modest attempt to list topics where a direct link

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to linearity or to easy algebra probably exists but determined by x and v(x) and move the is not known. If found, it might, even in different point x, in the v(x)-direction, to its antipode −x, contexts, have some general significance. But in for all x ∈ Sn−1. This is a homotopy (a continuous most other areas, nice situations certainly are of deformation) between the identity map and the an- a quite different nature; there is no need to list ex- tipodal map of Sn−1. amples. At this point the homological concept of degree comes into play. Its value is 1 for the identity and Vector Fields and Vector Functions (−1)n for the antipodal map. Homotopic maps Two preliminary remarks seem appropriate. have the same degree, so we get 1=(−1)n; i.e., n 1) The results in the first part are all about cer- is even. tain positive integers (dimensions, number of vec- tor fields, etc.). They state that, apart from those Proposition 2. A continuous tangent unit vector n−1 − values of these integers where (multi-)linear solu- field on S exists if and only if n 1 is odd. tions exist, there are no continuous solutions. This The crucial point is now to express these facts negative statement can be turned into a positive in a different way. We write Pn for the problem: one, mainly about the existence of zeros; see Propo- Does there exist on Sn−1 a tangent unit vector sition 3. Moreover, in some of those cases where field? a linear solution exists, there are also other con- tinuous solutions of a different nature (not ho- Theorem L (“Linearization”). If Pn has a continu- motopic to linear ones). ous solution, then it also has a linear solution. 2) We, of course, do not suggest that the re- Theorem L is proved “indirectly” by using a spective topological results should have been method from algebraic topology. There is no harm proved from the outset by reduction to linearity. in doing so. By a “direct” proof we would mean a On the contrary, these problems were a source of procedure replacing a continuous field by a linear stimulation for developing interesting tools that one (for example, through a variational principle still are very important, e.g., in modern homotopy where in the space of all continuous fields there theory. The linearization phenomenon emerged would be an extremal expected to be linear). Such only a posteriori after the topological theorems had a direct proof would reduce the topological prob- been established. lem to very elementary linear algebra. Tangent on a Sphere In the more complicated situations to be de- n−1 n A tangent unit vector field v on S P⊂ R (given scribed below, a direct reduction to a transparent Rn 2 in coordinates ξ1,... ,ξn of by ξi =1) is a algebraic argument, though not necessary, might n−1 that attaches to x ∈ S a vector v(x) sat- be even more interesting. isfying Remarks. 1) The negative statement “There is, for h i (1) v(x),x =0, even n − 1, no continuous tangent unit vector field n−1 (2) |v(x)|2 = hv(x),v(x)i =1 on S ” can be turned into a nontrivial existence statement for zeros, as follows. for all x ∈ Sn−1. Here h , i is the standard scalar product in Rn. Proposition 3. Let fi(ξ1,... ,ξn), i =1,... ,n, be continuous functions satisfying Linearity implies that (2) is equivalent to X 0 ξ f (ξ ,... ,ξ )=0 (2 ) |v(x)|2 = hx, xi for all x ∈ Rn. i i 1 n

If n is even, v(x)=(−ξ2,ξ1, −ξ4,ξ3,...) is such for all x with |x| =1. If n is odd, then the fi have a field, linear in x. What about linear fields if n is a common zero. odd? Let X Otherwise the f could be normalized so as to v = a ξ i i ik k be the components of a unit tangent vector field on Sn−1. If the f are polynomials, one has an al- be the components of v(x). Then (1) means i P gebraic statement for which no algebraic proof a ξ ξ =0for all x; i.e., the matrix a is skew- ik i k ik seems to be known. For the complex analogue, symmetric. If n is odd, its determinant is 0, and however, there is an algebraic proof by van der thus there is an x ∈ Sn−1 with v(x)=0, in contra- Waerden (1954). diction to (2). 2) The above proof of Proposition 2 is by ele- Proposition 1. A linear tangent unit vector field − mentary algebraic topology. For differentiable vec- on Sn 1 exists if and only if n − 1 is odd. tor fields there are other classical proofs, using We now ask the same question for continuous analysis or geometry; they are all based on some vector fields. These are, of course, more interest- version of the concept of degree. A very different ing from the viewpoint of geometry and analysis. analytic proof, however, is due to Milnor (1978). Given such a field on Sn−1, we consider the great None of these proofs is by reduction to linearity.

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Vector Functions of Two (or More) Variables, Because of the “composition of quadratic forms” Multiplications in Rn given by (7), the with product rule The well-known vector cross-product x × y in R3 are also called composition algebras. In 1898 Hur- is a function of two vectors that is bilinear and ful- witz proved that such a “composition of quadratic fills forms” with bilinear functions j of the ξj and ηj, with real or complex coefficients, can exist for h × i h × i (3) x y,x = x y,y =0, n =1, 2, 4, 8 only. | × |2 | |2| |2 −h i2 (4) x y = x y x, y . Proposition 4. A bilinear multiplication in Rn with For which n does such a bilinear vector product two-sided identity and with norm product rule ex- exist in Rn? ists if and only if n =1, 2, 4, 8. We assume that it exists in Rn and imbed Rn In the same spirit as before, we consider the cor- in Rn+1 = R ⊕ Rn. We write X ∈ Rn+1 as X = ξ + x, responding problem for continuous multiplica- ξ ∈ R, x ∈ Rn, and similarly Y = η + y ∈ Rn+1, and tions. One would expect that continuity gives much put more flexibility than bilinearity. However, Adams proved in 1960 that: · −h i × (5) X Y = ξη x, y + ξy + ηx + x y. Theorem A. A continuous multiplication with two- sided identity and norm product rule exists only Then 1+0∈ Rn+1 is a two-sided identity for that for n =1, 2, 4, 8. product, and an easy computation using (3) and (4), namely, Thus, if we now write Pn for the continuous mul- tiplication problem in Rn (with the above proper- |X · Y |2 = ξ2η2 + hx, yi2 − 2ξηhx, yi + ξ2|y|2 ties), one again has Theorem L, except that linear is to be replaced by bilinear. And a “direct” proof + η2|x|2 +2ξηhx, yi + |x|2|y|2 −hx, yi2 would reduce the proof of Adams’s famous The- 2 2 2| |2 2| |2 | |2| |2 = ξ η + ξ y + η x + x y , orem A to the very old Hurwitz argument of lin- ear algebra. yields The original proof of Adams’s theorem was a real tour de force, using the whole range of meth- (6) |X · Y |2 = |X|2|Y |2 . ods of algebraic topology known at that time. A The product (5) turns Rn+1 into an “algebra”; the very simple proof became available later thanks to commutative and associative laws are not required. the development of topological K-theory and the It fulfills, however, the norm product rule (6). Atiyah-Hirzebruch integrality theorems; the proof We consider for a moment such algebras in Rn is simple, but the prerequisites are certainly not. with norm product rule. The product X · Y can eas- Here too an algebraic corollary can be men- ily be modified so as to contain a two-sided iden- tioned for which no algebraic proof is known. tity whose existence we will assume in the fol- Proposition 5. Rn is a bilinear if lowing and denote 1. A bilinear product can be and only if n =1, 2, 4, or 8. given by the multiplication table of a basis of Rn; Division algebra means a product without zero- it is convenient to have 1 as a basis element. divisors (associativity and commutativity are not The classical examples for n =1, 2, 4, and 8 required). If such a product is given, it can be are: renormalized so as to fulfill the norm product R1 = R rule, but one loses bilinearity. Theorem A then R2 = C says that n =1, 2, 4, or 8. R4 = quaternion algebra H (associative but not If we return to vector products of two vectors commutative) in Rn, the earlier arguments combined with the R8 = “Cayley numbers” or Octonion algebra Hurwitz Theorem (Proposition 4) yield (not associative, not commutative) Proposition 6. A nontrivial bilinear vector prod- We do not give the well-known multiplication ta- uct fulfilling (3) and (4) exists in R3 and in R7 and C H bles for (basis 1, i), (basis 1, i, j, k), and the in no other Rn. Octonions. We recall, however, that all these alge- bras fulfill the norm product rule (which implies Indeed, it follows that n +1must be 1, 2, 4, or that there are no zero-divisors). 8. We have to show only that such a vector prod- uct actually exists in R7. We first note that for n =3 If we write ξj for the components of X, ηj of the product (5) defines the usual quaternion mul- · Y, j of X Y, the norm product rule becomes tiplication in R4, where in X = ξ + x the R-multi- ple ξ of 1 is the “real part”, x the “imaginary part”. 2 ··· 2 2 ··· 2 2 ··· 2 (7) (ξ1 + + ξn)(η1 + + ηn)=1 + + n . Conversely, starting from the quaternion product,

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one considers imaginary quaternions x, y ∈ R3 Proposition 8. Sn admits an almost-complex struc- and puts ture only for n =2and 6. x × y = x · y + hx, yi, On S2 such a structure exists, of course, since S2 can be turned into the . On S6 which is imaginary. Then a (linear) almost-complex structure can be derived from the Octonions in R8; it has been known since (x · y) · y = x · (y · y)=−|y|2x 1951 that it cannot come from a complex-analytic 2 3 (since y · y = −|y| for y ∈ R ) structure on S6. =(x × y) · y −hx, yiy A vector product of r vectors in Rn, r1 and r = n 1 there is a well- known multilinear solution: Take for v the vector Rn Proposition 7. A continuous vector product in orthogonal to the hyperplane spanned by the vec- fulfilling (3) and (4) exists if and only if n =3 tors xj ,j=1,... ,n− 1 (if they are linearly inde- or 7. pendent), with suitable orientation and suitably × − Writing Pn for the existence problem of a vec- normalized. In terms of the n (n 1)-matrix of tor product of two vectors in Rn, one has Theo- the components of the xj, the components of v rem L (with bilinear instead of linear). are given by the (n − 1) × (n − 1)-minors with the An interesting corollary of Proposition 7 con- usual signs. cerns almost-complex structures on Sn. Such a For r = n − 5 we know that (2, 7) has a bilinear structure is given by a continuous field J(x) of lin- solution. What about (3, 8)? Here again, a multi- ear transformations of the tangent space at x ∈ Sn linear vector product can be given explicitly in with J(x)2 = (minus)identity. (On a complex-ana- terms of the Octonions. A more elaborate argu- lytic , multiplication√ of complex vector ment, using Octonions again, shows that (4, 9) components by −1 is such a field. But we do not does not have any multilinear solution. On the assume that complex-analytic coordinates are other hand, a topological method (cross- of given.) a Stiefel-manifold fibering, Steenrod squares; see Given the field J on Sn, we consider x ∈ Sn, a survey [E1]) proves that even the continuous (4, 9)- unit tangent vector y(x) (i.e., two vectors problem does not have a solution. n+1 x, y ∈ R with |x| = |y| =1and hx, yi =0), and In summary: A continuous vector product of r the oriented tangent 2- determined by y and vectors in Rn exists only in the cases (1,n) with n × J(x)y. We choose x y to be the unit vector or- even, (n − 1,n), (2, 7), and (3, 8), and in these cases thogonal to y in that plane and corresponding to there is a (multi-)linear solution. the orientation. Writing P for the problem, Is there a vector × (r,n) We then have a vector product x y defined for product of r vectors in Rn? we have: |x| = |y| =1 and hx, yi =0only, but it can easily n+1 be extended to all vectors x, y ∈ R so as to ful- Theorem L. If P(r,n) admits a continuous solution, fill (3) and (4) and be continuous. Therefore n +1 then for that pair (r,n) it also has a (multi-)linear must be 3 or 7; whence n =2or 6. solution.

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Vector Functions of One Variable, Maximal gent vector fields (s-frames) on Sn−1? then we Number of Orthonormal Solutions again have Theorem L, this time with Pn replaced We return to one vector function of one variable by Pn,s. Rn n−1 in (tangent vector field problem on S ) and In the continuous case the same holds for lin- consider s orthonormal solutions. In other words, early independent fields instead of orthonormal we have s +1vector functions vj (x),j=0, 1,...,s ones, since orthonormalization does not affect Rn | | in , defined for x =1with v0(x)=x, and continuity. h i vj (x),vk(x) = δjk,j,k=0, 1,...,s. Hurwitz-Radon Matrices and Homotopy Groups We assume that the vectors v are linear func- j We begin with some remarks concerning the Hur- tions of x and write v (x)=A x where A is a real j j j witz-Radon matrix problem (8). n × n-matrix, A = E (unit matrix). Then 0 The proofs by Hurwitz and Radon were by ma- trix computations. A different proof, of a more hvj (x),v (x)i = hAj x, A xi = δ hx, xi k k jk conceptual nature, was given in 1942 by the n author using classical for all x ∈ R . This implies that all Aj are T T applied to a certain finite group Gs (generated by orthogonal matrices, that Aj Aj = E where Aj denotes the transposed matrix, and that for symbols A1,... ,As ,εwith relations dictated by (8), 2 2 6 j =6 k i.e., ε =1,Aj = ε, Aj Ak = εAkAj ,j= k). Instead of looking for maximal s given n, one asks for min- T T Aj Ak + Ak Aj =0. imal n given s. Minimal n is provided by irre- ducible orthogonal representations with ε 7→ −E. T For k =0 this yields Aj + Aj =0,j=1,...,s; The advantage of this method is that it gives ex- whence plicitly all solutions and shows very simply that 2 there exist solutions with matrix entries 0, +1, (8) A = −E, Aj Ak + AkAj =0 j and −1 only; see [E2]. We note here, for use in the for j =6 k, j, k =1,...s. next section, that the same matrix problem can, of course, be formulated for unitary representations In addition the matrices have to be orthogonal and this is simpler than the orthogonal problem. or, equivalently, skew-symmetric. s s−1 Such matrices, with real or complex entries, In that case the minimal n is 2 2 for even s and 2 2 are called Hurwitz-Radon matrices. They were for odd s. All solutions are direct sums of the independently1 examined around 1920 by minimal ones. In a solution for s +1, omitting the Hurwitz and Radon; they determined for given last matrix As+1 of course yields a solution for s. n the maximum possible number s . If n = For even s, a minimal solution is obtained in this odd.16α2β,β =0, 1, 2, 3, then way from a minimal solution for s +1, since s (s+1)−1 2 2 =2 2 . In other words, the solutions for even s =8α +2β − 1. max s are not essential; they all come from s +1. The quantity ρ(n)=8α +2β is called the Radon Let A1, ..., As be a of orthogonal Hurwitz- number of n, and the relations (8) are called the Radon matrices, i.e., a solution of (8); let A0 = E; Hurwitz matrix . and let α0,α1,...,αs be real numbers with Σα2 =1. From (8) it follows easily that the n × n- Proposition 9. The maximum number of ortho- j matrix normal tangent vector fields on Sn−1 depending linearly on y ∈ Sn−1 is ρ(n) − 1. (9) f (a)=Σαj Aj Again, the continuous analogue was established is orthogonal, and this is equivalent to (8). We by Adams (1962) in his famous s s+1 write a =(α0,α1,...,αs ) ∈ S in R and con- Theorem B. The maximum number of continuous sider f as a (linear) map of Ss into the orthogonal − orthonormal tangent vector fields on Sn 1 is group O(n). Combining with the natural imbedding ρ(n) − 1. of O(n) into the infinite O (the → The proof is yet more difficult and technical than limit of the usual inclusions O(n) O(n +1)), we s → the original proof of Theorem A. So far no simpler get a map F : S O. Conversely, any linear map s → argument of algebraic topology has been found. If, F : S O of the form (9) (the image necessarily by analogy to the foregoing, we write P for the lies in some O(n)), with A0 = E, is given by or- n,s × problem, Is there a system of s orthonormal tan- thogonal Hurwitz-Radon n n-matrices. The ho- motopy class of F is an element of the homotopy 1Hurwitz died in 1919. His paper appeared in 1923. group πs O. Radon’s work was submitted in 1922 and also was pub- [Remark: The matrix (9) being orthogonal shows lished in 1923. that a solution of the Hurwitz equations (8) is

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equivalent to the composition of quadratic forms, πs U can be described in a simple way: One just generalizing (7), places the two maps Ss → U(n) over the diagonal 2 ··· 2 2 ··· 2 2 ··· 2 in U(2n). One therefore can obtain all elements of (α0 + + αs )(ξ1 +, ,ξn)=(1 + + n ), πs U, i.e., of πs U(n) for sufficiently high n, through where the i are bilinear in the αj and ξi.] unitary Hurwitz-Radon matrices. This can be ex- One can proceed in exactly the same way with pressed again as a linearization result of a differ- a unitary solution of (8). Then f and F (we use the ent nature: same letters) yield a homotopy class in π (U) 0 s Theorem L . Any continuous map Ss → U(n), where U is the infinite unitary group, the limit of s+1 s+2 n ≥ , or ≥ respectively is homotopic in U the inclusions U(n) → U(n +1). Both cases are 2 2 to a linear map. closely related to Bott periodicity (1956). Here we restrict ourselves, for simplicity, to the unitary A direct proof of that theorem would reduce Bott case; the orthogonal case can be dealt with in the periodicity to the purely algebraic discussion of same way, though the details are a little more com- Hurwitz-Radon matrices. plicated. What Bott proved was actually more than the pe- Ω As for the πs U(n), it has been riodicity of homotopy groups: One considers U, known since around 1940 that the space of loops in U beginning and ending in 1 ∈ U. The periodicity of homotopy groups ∼ s +1 πs+2(U)=πs (U) for all s ≥ 0 is essentially the πs U(n) = πs U 2 same as a homotopy equivalence between ΩΩU s +1 and U and thus a periodicity with period 2 for all if s is odd and n ≥ , 2 iterated loop-spaces of U. For the groups of ho- motopy classes of maps of arbitrary spaces (cell ∼ s +2 complexes) X into U and into the iterated loop- πs U(n) = πs U 2 spaces of U, one therefore has the same periodic- s +2 ity. The (abelian) groups thus obtained constitute if s is even and n ≥ . a cohomology called topological K-the- 2 ory. This was the first example of an “extraordi- The isomorphisms are given by the imbedding nary” cohomology theory, and it seems interest- U(n) → U(n +1). These “stable” groups are the ho- ing that it is closely related to the unitary motopy groups πs U. Hurwitz-Radon matrices. The Bott periodicity theorem determined the Everything can also be said, mutatis mutandis, groups πs U completely: about the orthogonal (or the symplectic) Hurwitz- Radon matrices and the infinite orthogonal (or Z πs U = if s is odd, symplectic) group and the corresponding K-the- =0 ifs is even. ory; here the periodicity has period 8.

If s is even, any solution of (8) yields, as it must, Von Neumann Algebra of a Group a nullhomotopic map f : Ss → U(n), since it comes from a solution for s +1, so that f can be extended About the Bass Conjecture C to a linear map Ss+1 → U(n) and is therefore null- Here we consider the complex group algebra G homotopic (even in a linear way). of a discrete group G. Recall that it is the complex For odd s, however, one has the interesting re- having the group elements as basis, sult: with product given by the group multiplication of ∈ A minimal solution of the Hurwitz-Radon prob- the basis elements. The identity element 1 G is lem yields, for odd s, through Ss → U(n) → U given the identity for the algebra product. An idempo- tent a ∈ CG is an element fulfilling a2 = a. The by f above, a of πs U. s−1 idempotent conjecture says that the only idempo- Note that here n =2 2 , while in the usual tents in CG are 0 and 1, as in C or any division approach the generator, in the stable group , provided the group G is torsion-free, i.e., has s+1 s+1 6 πs U( 2 ) , lies in dimension n = 2 . For the no elements of finite order =1(elements of finite lowest cases s =1and 3, these n are equal, and order easily yield nontrivial idempotents of CG). the generators are easily recognized to be identi- In the following we always assume G to be torsion- cal. This is not so for s>3, so that a proof is free. needed. It makes use (see [E2]) of Bott’s theorem A strong tool to deal with this problem is the in its full topological statement. On the algebraic “canonical” trace κ(a), also called the Kaplansky side it is based on a simultaneous analysis of the trace of a ∈ CG; it is the coefficient of 1 ∈ G of a. solutions of the matrix problem (8) for all values A little more generally, let A =(aij) be an idem- of s. potent (n × n)-matrix with entries in CG and n What about the multiples of the generator? The κ(A)=Σκ(aii). The image of A in CG is a finitely group operation (addition) for two elements of generated projective CG-module P, and we write

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also κ(P) for κ(A), since κ is independent of the For the proof, however, one uses the fact that imbedding of P in some CGn. the N(G) is a finite von Neumann algebra and thus admits a center-valued trace: For the Kaplansky Theorem. κ(A) is a nonnegative real onto N(G) ⊗C P its value is precisely κ(P).identity number, equal to 0 only if A =0. G (here a deep result on finite conjugacy classes in Actually κ(A) is known to be rational, but here G following from [B], Theorem 8.1, is used). If the we will not make use of this. weak Bass conjecture holds, it is equal to We recall that the von Neumann algebra N(G) d(P).identity, which is the center-valued trace of of G can be defined as the algebra of all G-equi- the above free module. And projective modules variant bounded linear operators on the Hilbert having the same center-valued trace are isomor- space `2G of square-summable complex functions phic. on G. A simple proof of the above theorem is ob- [Actually, more is true. The algebraic category tained by imbedding CG into N(G); then κ(A) can of finitely generated projective N(G)-modules is be identified with the von Neumann trace of A. Al- equivalent to the category of finitely generated though the idempotent map defined by A need not Hilbert-G-modules (and G-equivariant bounded be selfadjoint, it is equivalent to a selfadjoint one linear operators as ). And if two N(G)- (orthogonal projection). In other words, κ(A)=κ(P) projectives have the same center-valued trace, then is the von Neumann dimension of the Hilbert-G- the corresponding Hilbert-G-modules are isomet- module `2G ⊗CG P. rically G-isomorphic. The von Neumann dimension Another notion of trace is given by the aug- can be carried over to finitely generated projective mentation of Σaii. It is an integer, namely, the di- N(G)-modules; a special feature is that submod- mension of the C-vector space C ⊗CG P, which we ules of finitely generated projective N(G)-mod- write in short d(P). The Bass conjecture says that ules are again projective.] these two traces are equal: Conversely, the center-valued traces show that (11) implies the (weak) Bass conjecture for G, (10) κ(P)=d(P). κ(P)=d(P). 00 This is the weak form of the conjecture, implied Theorem L . The associativity formula (11) is by the strong one, which we do not formulate here; equivalent to the weak Bass conjecture κ(P)=d(P) C see [B]. It has been proved for several big classes for all finitely generated projective G-modules. of groups G, such as linear groups, solvable groups For many classes of groups, κ(P)=d(P); whence (of finite homological dimension), hyperbolic (11) is a theorem, and certainly not an easy one, groups, 3-manifold groups, groups of cohomo- with different proofs according to the respective logical dimension 2 (over Q). The proof of the class. Dare one ask here for a more direct ap- simple (10) is very indirect and different proach to Theorem L 00? For which groups? In what for the various classes, using methods, generality? cyclic homology of groups, homological dimen- Return to Topology: Poincaré-2-Complex sion, etc. (Bass 1976, Eckmann 1986). First, some very short technical remarks about ∈ C Note that for an idempotent a G the pro- Hilbert space methods in algebraic topology. C ⊂ C jective P is the left ideal Ga G and d(P) is In recent years homotopy invariants of a space necessarily equal to 0 or 1. Thus κ(a)=0 or X (cell-complexes of finite type with infinite fun- − κ(1 a)=0, which proves the idempotent con- damental group G) have been introduced and ap- jecture for the respective groups. plied with the help of Hilbert-G-modules: `2-ho- Projective Modules over N(G) mology modules (reduced, i.e., cycles modulo the Beyond the equality (10), i.e., the weak Bass con- of the space) and `2-Betti num- jecture, one can say more: Through the imbed- bers (their von Neumann dimension). These con- ding of CG in N(G) the projective module P be- cepts actually go back to Atiyah (1976), but were comes a finitely generated projective N(G)-module fully developed much later. N(G) ⊗CG P that turns out to be a free N(G) - In view of the category equivalence above, one module of rank equal to d(P) = dimCC ⊗CG P. This gets a purely algebraic approach to all this: G can be expressed as an isomorphism of N(G)- operates as a covering transformation group on modules the universal covering Xe of X; the chain groups of Xe are free ZG-modules, and tensoring them over ⊗ ⊗ C ⊗ N(G) CG P = N(G) C ( CG P), CG with the algebra N(G), one obtains a complex or more intuitively as an associativity formula of finitely generated free N(G)-modules. Its ho- mology groups are not projective in general, but (11) (N(G) ⊗C C) ⊗CG P = N(G) ⊗C (C ⊗CG P), finitely presented N(G)-modules. Their “projective part”, corresponding to reduced `2-homology, where everything is purely algebraic. In particular, yields the `2-Betti numbers, and they also yield the N(G) is considered just as an algebra. Novikov-Shubin invariants (Farber, Lück, 1995).

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If G fulfills the Bass conjecture, then the and Lecture Notes, vol. 6, Amer. Math. Soc., Provi- same procedure can be applied to a finitely dence, RI, 1994, pp. 23–35. dominated space X, since the chain groups of For the third part we refer to recent papers by Wolfgang Xe are finitely generated projective ZG-modules, Lück, Michael Farber, and the author, and to: [B] H. BASS, Euler characteristics and characters of discrete and the tensor product over CG yields free N(G)- groups, Invent. Math. 35 (1976), 155–196. modules. This is interesting, for example, for Poin- [E3] B. ECKMANN, Poincaré groups of dimension two caré complexes. are groups, Combinatorial A is called a Poincaré-n- and Topology, Ann. of Math. Stud., Princeton Univ. complex if it fulfills the classical Poincaré duality Press, Princeton, NJ, 1986, pp. 35–51. relations well known for closed n-— an approximation to the latter. We restrict atten- tion to a very special application. It concerns the theorem (the author et al.; cf. the survey [E3]): A Poincaré-2-complex X with infinite funda- mental group G is homotopy equivalent to a closed surface of ≥ 1. No finiteness assumptions are required; X is fi- nitely dominated. An important ingredient in the proof (we mention here only the orientable case) is to show that the first ordinary β1(X) is ≥ 2. The methods above greatly simplify the argument, as follows. The `2-Betti numbers bi(X) compute the of the space and fulfill Poincaré du- ality in the manifold- or Poincaré complex-case, ex- actly as the ordinary Betti numbers do. The Betti number b0 in case of an infinite group G is easily seen to be =0. Thus in our situation

χ(X)=β0 − β1 + β2 =2− β1 = −b1;

whence indeed β1 ≥ 2. The Poincaré complex above is aspherical; i.e., all homotopy groups in dimensions ≥ 2 are 0. It is thus a classifying space for G, and the homol- ogy of G is the same as the homology of X. (The being 2, the Bass con- jecture is fulfilled; this we have already used above implicitly.) Passing to the universal of the sur- face, one can express the result in terms of the group G, yet another linearity statement: Theorem L 000. A group whose homology fulfills Poincaré duality of dimension 2 is isomorphic to a plane group operating freely with com- pact fundamental domain on the Euclidean or hy- perbolic plane. If a group G is the fundamental group of a closed aspherical n-manifold, then its homology fulfills, of course, Poincaré duality of dimension n. Is the converse true? For dimensions n ≥ 3 this problem is still unsolved, except for partial re- sults in dimension 3.

References For the first part we refer to the survey: [E1] B. ECKMANN, Continuous solutions of linear equa- tions, Expo. Math. 9 (1991), 351–365, where the rel- evant references can be found. Similarly for the second part: [E2] ———, Hurwitz–Radon matrices revisited, The Hilton Symposium 1993 (Montreal, PQ) CRM Proceedings

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