A Banach algebraic Approach to the Borsuk-Ulam Theorem

Using methods from the theory of commutative graded Banach algebras, we obtain a generalization of the two dimensional Borsuk-Ulam theorem as follows: Let $\phi:S^{2} \rightarrow S^{2}$ be a homeomorphism of order n and $\lambda\neq 1$ be an nth root of the unity, then for every complex valued continuous function $f$ on $S^{2}$ the function $\sum_{i=0}^{n-1} \lambda^{i}f(\phi^{i}(x))$ must be vanished at some point of $S^{2}$. We give a generalization in term of action of compact groups. We also discuss about some noncommutative versions of the Borsuk- Ulam theorem


Introduction
The classical Borsuk-Ulam theorem states that for every continuous function g : S n → R n , there always exist a point x ∈ S n such that g(−x) = g(x). If we define f (x) = g(x) − g(−x), we obtain an equivalent statement as follows: For every odd continuous function f : S n → R n , there exist a point x ∈ S n such that f (x) = 0. We consider the case n = 2 and identify R 2 with the complex numbers C. Let A = C(S 2 ) be the Banach algebra of all continuous complex valued functions on S 2 with the Z 2 -graded structure where A ev is the space of all even functions and A odd is the space of all odd functions and the decomposition is the standard decomposition of functions to even and odd functions. Then the two dimensional Borsuk-Ulam theorem says that a homogenous element of A with non zero degree, namely an odd function, is not invertible. In this paper we are mainly interested in invertible elements of a graded unital Banach algebra which are homogenous of nontrivial degree. Some natural questions about such elements are as follows: If any such element is invertible, can it be connected to the identity in the space of invertible elements? What can be said about the relative position of their spectrum with respect to the origin? As we will see in the main theorem of this paper, for a commutative Banach algebra without nontrivial idempotent, which is graded by a finite Abelian group, a nontrivial homogenous element can not be connected to the identity. On the other hand, using an standard lifting lemma in the theory of covering spaces, we conclude that an invertible element of C(S 2 ) has a logarithm then it lies in the same connected component of the identity. This shows that an odd element of C(S 2 ) can not be invertible, so it would gives us a Banach algebraic proof of the Borsuk-Ulam theorem, in dimension two. The purpose of this paper is to translate the classical Borsuk-Ulam theorem into the language of noncommutative geometry. We also give a concrete example of an S 3 -graded structure for C * red (F 2 ), the reduced C * algebra of the free group on two generators, such that a nontrivial homogenous element lies in the same component of the identity. This would shows that the commutativity of grading group and graded algebra are necessary conditions in our main theorem. Finally we give a question as a weak version of the Kaplansky-Kadison conjecture. This question naturally arise from our main result, see question 5 at the end of the paper.

Preliminaries
Let A be a unital complex Banach algebra and G be a finite group with neutral element e. A G-graded structure for A is a decomposition where each A g is a Banach subspace of A and A g A h ⊆ A gh . An element a ∈ A g , g ∈ G is called a homogenous element, it is called nontrivial homogenous if a ∈ A g where g = e. When G = Z 2 and A = A 0 A 1 , an element of A 1 is called an odd element. A morphism α : A → A is called a graded morphism provided that α(A g ) ⊆ A g for all g ∈ G. Let A be a G-graded Banach algebra and H be a normal subgroup of G. Then there is an induced G/H-graded structure for A with Let G be Abelian and a be a nontrivial homogenous element of a G-graded Banach algebra, then for some positive integer n, there is a Z n -graded structure for A such that a is a nontrivial homogenous element of A as a Z n -graded algebra. This can be proved by induction on order of G as follows: Let G be a finite Abelian group and g ∈ G where g = e. If G is not a cyclic group, then there is a subgroup H which does not contain g, so a is a nontrivial homogenous element of the induced G/Hgraded structure for A. Now the order of G/H is strictly less than the order of G. So the assertion can be proved by induction on order of G. Note that the existence of a Z n -graded structure for a Banach algebra A is equivalent to existence of a bounded multiplicative operator T : A → A with T n = Id. For any such operator, we choose a root of unity λ = 1 and observe that the decomposition is a Z n -graded structure. Conversely for the grading A = n−1 i=0 A i , the following multiplicative operator T satisfies T n = Id For a group G, a Z n -partition for G is a partition of G to disjoint subsets G i , i = 0, 1, . . . n − 1 such that G i G j ⊆ G i+j (mod n) . This is equivalent to say that G 0 is a normal subgroup of G whose quotient is isomorphic to Z n For a discrete group G, we denote by CG, the group algebra of G with complex coefficients, namely the space of all linear combinations a g g where a g 's are complex numbers. let l 2 (G) be the Hilbert space of all γ : G → C such that |γ| 2 2 = g∈G |γ(g)| 2 < ∞. G acts on l 2 (G) with g.γ(h) = γ(g −1 h). This action defines a unitary representation of G on l 2 (G). We extend this action by linearity to CG. So each element of CG can be considered as an element of B(l 2 (G)), the space of all bounded operators on l 2 (G) and every element of G can be considered as a unitary operator on l 2 (G). The reduced C* algebra of G, C * red (G), is the closure of CG, with respect to operator norm | . | op defined on B(l 2 (G)). Assume that G = i∈Zn G i is a Z n -partition for G, then CG = CG i and l 2 (G) = Let x i ∈ CG i and ε > 0 is given. We apply the definition of norm operator on operator x i , then we conclude that there is a γ = This shows that there is a j ∈ Z n such that This shows that the operator norm on CG is equivalent to direct sum norm | x i | op = |x i | op . We summarize the above discussion in the following proposition which proof would be completed provided we prove the lemma below.
Since the original norm of F is equivalent to the direct sum norm we conclude that each sequence {f i k } converges to zero. So f i = 0 for all i. From continuity of norm we have the equivalency of norm and direct sum norm on the space n i=0 F i . So this direct sum is a topological direct sum. let x ∈ E is given, there is a sequence n i=0 f k i , k ∈ N, which converges to x. This shows that each sequence {f k i } is a cauchy sequence which converges to an element f i ∈ F i . So x = n i=0 f i . This completes the proof of lemma. Remark 1. Let G and H be two groups where H is finite. Similar to above we can define an H-partition for G. It is a decomposition G = h∈H G h such that G h1 G h2 ⊆ G h1h2 . This is equivalent to say that G e is a normal subgroup of G whose quotient is isomorphic to H. In the same manner as above we can prove that an H-partition for a group G, gives an H-graded structure for C * red (G). Moreover it is clear from the definition that an element g ∈ G which is not in G e , can be considered as a nontrivial homogenous element of H-graded algebra C * red (G).
A part of the philosophy of noncommutative geometry is to translate the classical facts about compact topological spaces into language of (noncommutative) Banach or C * algebras, see [3] or [5]. According to the Gelfand -Naimark theorem, there is a natural contravariant functor from the category of compact Hausdorff topological space to the category of unital complex C * algebras. This functor assign to X, the commutative C * algebra C(X) of all complex valued continuous functions on X. It also assign to a continuous function f ; X → Y , the C* algebra morphism Conversely every unital commutative C * algebra A is isomorphic to C(X) for some compact topological space X and every morphism from C(Y ) to C(X) is equal to f * for some continuous function f : X → Y . In particular constant maps from X to Y correspond to morphism from C(Y ) to C(X) with one dimensional range. Most of the statements about topological spaces have an algebraic analogy in the world of C * algebras. For example it can be easily shown that a topological space X is connected if and only if the algebra C(X) has no nontrivial idempotent, where a nontrivial idempotent a in an algebra A is an element a ∈ A such that a 2 = a and a = 0, 1. So nonexistence of nontrivial idempotent for a (noncommutative) Banach algebra A is interpreted as "noncommutative connectedness". For a topological space X, considering the above functor, it is natural to identify C(I × X), with C(I, C(X)), where I is the unit interval. So in order to obtain a homotopy theory, for a Banach algebra A, we define AI= the Banach algebra of all continuous γ : Obviously this is a natural algebraic analogy of classical null homotopicity. For a unital Banach algebra A with unit element 1, the element λ.1 is simply shown by λ where λ ∈ C is an scalar. For an element a ∈ A, we denote sp(a) for all λ ∈ C such that a − λ is not invertible. The spectral radius of a, which is denoted by spr(a), is defined as the infimum of all r such that the disc of radius r around the origin contains sp(a).

Main Result
In the next theorem, which is the main result of this paper, k is an arbitrary positive integer, G is a finite Abelian group and A is a unital complex Banach algebra.
Main Theorem. Let A be a G-graded Banach algebra without nontrivial idempotent and a is a nontrivial homogenous element. Then 0 belongs to the convex hull of sp(a k ). If A is commutative and a is invertible, then a k can not be in the same connected component of the identity.
Proof. Without lose of generality we assume that G = Z n . So we have a multiplicative operator T : A → A with T n = Id. Let a be a nontrivial homogenous element so T (a) = λa where λ = 1 is a root of the unity. Assume for contrary that the convex hull of sp(a k ) does not contain 0. Then sp(a k ) can be included in a disc with center z 0 , for some z 0 ∈ C, which is a subset of a branch of logarithm. Using holomorphic functional calculus as described in [9, Chapter 10], we can find Thus T (b) and b are power series in a. In particular ab = ba and , for all i and j. So we have (a −1 exp b k ) k = 1. Then sp(a −1 exp b k ) is a subset of the set of all k-th roots of unity. On the other hand since A has no nontrivial idempotent , the spectrum of each element must be connected. So we can assume that sp(a −1 exp b k ) = {1}, otherwise we multiply a with an appropriate root of unity. Put q = a −1 exp b k −1. Then q is a quasinilpotent element of A, that is its spectral radius spr(q) = 0. We have where q ′ is a quasinilpotent too because aq = qa and note that spr(aq) ≤ spr(a)spr(q) for commuting elements a and q, see [9, page 302]. Moreover we have where q 1 is a quasinilpotent.
On the other hand (a+q) −1 −a −1 = −(a+q) −1 qa −1 then (a+q) −1 = a −1 +q ′′ for some quasinilpotent q ′′ . Thus we have exp − b k = a −1 + q ′′ . We obtain from (1) and (2) and the latest equation exp T (b)−b k = λ + q 2 for some quasinilpotent q 2 . Then sp(T (b)− b) is a single point {µ} different from zero, since λ = 1, so T (b) = b + µ+ y for some quasinilpotent y. Then y = T (b) − b − µ is a quasinilpotent element that can be expanded as a power series in a. Note that we emphasize on this power series expansion only for commutativity purpose. We have T (b) = b + µ + y so by induction we obtain b = T n (b) = b + nµ + y + T (y) + T 2 (y) + . . . + T n−1 (y).
Since T is multiplicative and y has a power series expansion in a, we conclude that T j (y)'s are commuting quasinilpotent elements so their sum is quasinilpotent too, see [9, page 302]. This implies that nµ is quasinilpotent which is a contradiction. This completes the first part of the theorem. Now assume that A is commutative and a is a nontrivial homogenous element such that a k is in the same connected component of the identity. There exist an element b ∈ A with exp b = a k . Obviously the same argument as above, but without needing to expansion of b as a power series in a, leads to a contradiction. So the proof is complete.
The following corollaries are immediate consequence of the above theorem: Corallary 1 . Let X be a compact locally path connected and simply connected space and φ : X → X be a homeomorphism of order n. Assume that λ = 1 is an nth root of the unity. Then for every continuous function f : X → C, there is a point x ∈ X such that p−1 i=0 λ i f (φ i (x)) = 0 Proof Let A be the commutative Banach algebra of continuous functions f : X → C with the usual structures. Since X is connected , A has no nontrivial idempotent. Define the continuous automorphism T : , then T (g) = λ n−1 g, so g is a nontrivial homogenous element of A. If g(x) = 0 for all x ∈ X then g is an invertible element of A which is not in the same connected component of the identity, by the above theorem. On the other hand, consider the covering space exp : C → C − {0}. Since X is simply connected and locally path connected, there is a lifting h of f , that is h ∈ A with exp h = g, using the standard lifting lemma in the theory of covering space, see [6, proposition 1.33]. So g is a logarithmic element and must be in the same connected component of the identity. This contradicts to the fact that g can not be connected to the identity. This completes the proof.
The following corollary is an obvious consequence of the last part of the main theorem. In this corollary and its sequel, G(A) is the group of all invertible elements of a Banach algebra A and G(A) 0 is the connected components of the identity.
Corallary 3 . Let A be an idempotent less commutative Banach algebra which is graded by a finite Abelian group such that a nontrivial homogenous element is invertible. Then G(A) G(A)0 is an infinite group. Now we shall present a question as a pure algebraic analogy of the above corollary. For this purpose we use some elements of stable rank theory and K-theory for both Banach algebras and complex algebras. We say that a commutative Banach algebra A has topological stable rank one if G(A) is dense in A, see [8]. Let A be a commutative complex algebra. We say that a pair (a 1 , a 2 ) is invertible if there is a pair (b 1 , b 2 ) such that a 1 b 1 + a 2 b 2 = 1 . We say A has Basse stable rank one if for every invertible pair (a, b) ∈ A 2 , there is an element x ∈ A such that a + bx is invertible, see [1]. It is mentioned in [8] that for a commutative C* algebra, the Bass stable rank coincide to the topological stable rank. For a complex Banach algebra A, let GL n (A) be the space of all n by n invertible matrices with entries in A. There is a natural topology on this space and there is a natural embedding of GL n (A) into GL n+1 (A) which sends a matrices B to diag(B,1), so we have where GL(A) 0 is the connected component of the identity. This Abelian group K 1 (A) is the standard K 1 functor defined on the category of Banach algebras. For a complex algebra A, there is a pure algebraic K alg 1 functor defined as the quotient of GL(A) by its commutator. On the other hand it is well known that for an stable rank one Banach algebra A, G(A) G(A)0 is naturally isomorph to K 1 (A), see [4] So considering this isomorphism, the equality of the Bass and topological stable rank for commutative C * algebras and the above corollary, it is natural to ask the next pure algebraic question: Question 1. Let A be an idempotentless involutive and commutative complex algebra with Basse stable rank one which is graded by a finite Abelian group. Assume that a nontrivial homogenous element of A is invertible. Does this implies that K alg 1 (A) is an infinite group? In the following example we drop simultaneously the commutativity of both grading group G and the graded idempotent less algebra A in the main theorem. We observe that in this case the theorem is no longer valid.
Example. Let F 2 be the free group on two generators x and y. We shall see that there is a normal subgroup G 0 of F 2 which does not contain yxy −1 x −1 and its quotient is isomorphic to S 3 , the permutation group on three elements. Assuming the existence of such subgroup G 0 , we obtain an S 3 -graded structure for C * red (F 2 ) for which yxy −1 x −1 , as an element of C * red (F 2 ), is a nontrivial homogenous element. On the other hand it is well known that this algebra has no nontrivial idempotent and yxy −1 x −1 ∈ C * red (F 2 ), is in the same connected component of the identity, see [4]. To prove the existence of such subgroup G 0 , we first note that F 4 , the free group on four generators, can be considered as an index three subgroup of F 2 which does not contain yxy −1 x −1 . This can be proved using certain covering space of figure-8 space as illustrated in [6, p.58]. Let h : X → X be a covering space with h(q) = p. Then the induced map h * : π 1 ( X, q) → π 1 (X, p) is an injective map. h * (π 1 ( X, q)) is isomorphic to π 1 ( X, q) which index , as a subgroup of π 1 (X, p) is equal to the cardinal of a fibre of the covering space. Moreover a loop γ ∈ π 1 (X, p) lies in the rang of h * if and only if the unique lifting γ of γ with starting point q is a loop with base point q, see [6, proposition 1.31]. Now consider the 3-fold covering space which is illustrated in the figure below. The fundamental group of the total space is F 4 and the fundamental group of the base space is F 2 . The loop yxy −1 x −1 is not in the range of projecting map of the covering because its lifting with base point q ends to a different point O Put G = F 2 and b = yxy −1 x −1 . Then G has a subgroup H of index 3 which does not contain a commutator element b. We obtain a morphism β : G → S 3 with the standard action of G on the set of left cosets of H. Then ker β is a normal subgroup of G which is contained in H and does not contain b. G/ ker β is isomorph to a subgroup of S 3 , on the other hand G/ ker β is not Abelian since ker β does not contain a commutator element b. This shows that ker β is a normal subgroup of G = F 2 which quotient is isomorphic to S 3 and does not contain the commutator element b = yxy −1 x −1 . So the above construction gives an example of a non commutative Banach algebra C * red (F 2 ), without nontrivial idempotent, which is graded by a non Abelian finite group S 3 , such that a nontrivial homogenous element yxy −1 x −1 lies in the same connected component of the identity.
But could we give any such example with a finite Abelian group G? In the other word, can we drop the hypothesis of commutativity of Banaach algebra A from the second part of the main theorem. This is a motivation for the next question which can be considered as a noncommutative analogy of the two dimensional Borsuk -Ulam theorem: Question 2. Let A be a Banach algebra without nontrivial idempotent which is equipped with a G-graded structure, where G is a finite Abelian group. Can one prove that the connected component of the identity has null intersection with nontrivial homogenous elements? As a particular case, put A = C * red (F 2 ), with the Z 2 -graded structure correspond to the Z 2 -partition of F 2 to the union of odd and even words. Can a linear combination of odd words be connected to the identity. Is the decomposition of this algebra to even and odd words, the only Z 2 graded structure for A , up to graded isomorphism?
What is a Banach algebraic formulation of the higher dimensional Borsuk-Ulam theorem? In order to obtain a noncommutative version of this theorem , we restate the classical case as follows; Let f 1 , f 2 , · · · , f n be n odd continuous real valued functions on S n , then the function n i=1 f 2 i is not an invertible element of C(S n ) or equivalently is not in the same connected component of the identity (Sine every invertible element can be connected to the identity). In fact f i 's are self adjoint elements of C(S n ) which are odd elements of the standard Z 2 graded structure of C(S n ). Now a relevant noncommutative version of this statement can be presented as the next question. So it seems natural to ask that for what type of noncommutative spheres the answer to the next question is affirmative? Question 3. Assume that A is a Z 2 graded noncommutative n-sphere and a 1 , a 2 , . . . a n are self adjoint elements of A which are odd elements of this graded algebra. Is it true to say that n i=1 a 2 i is either non invertible or is not in the same connected component of the identity? Remark 2. A family of noncommutative spheres is a family of C* algebras A θ , θ ∈ R with some relations as a natural generalization of the algebra of continuous functions on n-sphere. For more information on noncommutative spheres, see [2] or [7] Another candidate for the noncommutative analogy of the Borsuk Ulam theorem can be presented as follows; Consider the equivalent statement of the Borsuk Ulam theorem which says that an odd continuous mapf : S n → S n is not null homotopic, namely it is not homotopic to a constant map. We try to translate this statement into the language of Banach or C * algebras. The antipodal map φ(x) = −x defines an order two automorphism T ; C(S n ) → C(S n ) with T (g) = g • φ which naturally gives a Z 2 graded structure for C(S n ). Similarly an odd map f : S n → S n defines a morphism α : C(S n ) → C(S n ) which satisfies T α = αT . This means that α is a graded morphism. So we ask: for what type of noncommutative spheres (spaces), the answer to the next question is affirmative? Question 4. Let A be a noncommutative sphere with a nontrivial Z 2 graded structure and α : A → A be a graded morphism. Is it true to say that α is not a null homotopic morphism ? .
Final remark on the main theorem. We explain that the main theorem gives us a weaker version of the Kaplansky-Kadison conjecture. This conjecture says that for a torsion free group Γ, C * red (Γ) has no nontrivial idempotent, see [10]. Now as a consequence of our theorem we present the following question as a weaker version of the Kaplansky-Kadison conjecture: Question 5 (A weak version of the Kaplansky-Kadison conjecture). Let Γ be a torsion free group and C * red (Γ) is equipped with a Z n -graded structure. Is it possible that the convex hull of the spectrum of a nontrivial homogenous element does not contain the origin?