A Banach Algebraic Approach to the Borsuk-Ulam Theorem

Abstract

Using methods from the theory of commutative graded Banach algebras, we obtain a generalization of the two-dimensional Borsuk-Ulam theorem as follows. Let 𝜙∶𝑆²→𝑆² be a homeomorphism of order 𝑛, and let 𝜆≠1 be an 𝑛th root of the unity, then, for every complex valued continuous function 𝑓 on 𝑆², the function ∑^𝑛−1_𝑖=0𝜆^𝑖𝑓(𝜙^𝑖(𝑥)) must vanish at some point of 𝑆². We also discuss some noncommutative versions of the Borsuk-Ulam theorem.

“To my Children Tarannom and Pouya”

1. 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 exists a point 𝑥∈𝑆^𝑛 such that 𝑓(𝑥)=0.

We consider the case 𝑛=2 and identify ℝ² with the complex numbers ℂ. Let 𝐴=𝐶(𝑆²) be the Banach algebra of all continuous complex-valued functions on 𝑆² with the 𝑍₂-graded structure: 𝐴=𝐴_ev𝐴_odd, (1.1) 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 nonzero 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 cannot 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(𝑆²) has a logarithm, then it lies in the same connected component of the identity. This shows that an odd element of 𝐶(𝑆²) cannot be invertible, so it would give 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 𝑆₃-graded structure for 𝐶^∗_red(𝐹₂), 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 show 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 arises from our main result; see Question 5 at the end of the paper.

2. Preliminaries

Let 𝐴 be a unital complex Banach algebra, and let 𝐺 be a finite group with neutral element 𝑒. A G-graded structure for 𝐴 is a decomposition 𝐴=_𝑔∈𝐺𝐴_𝑔, (2.1) where each 𝐴_𝑔 is a Banach subspace of 𝐴 and 𝐴_𝑔𝐴_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=𝑍₂ and 𝐴=𝐴₀⨁𝐴₁, an element of 𝐴₁ 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 let 𝐻 be a normal subgroup of 𝐺. Then there is an induced 𝐺/𝐻-graded structure for 𝐴 with 𝐴_𝑔/𝐻=_ℎ∈𝐻𝐴_𝑔ℎ. (2.2) The following proposition is used in the proof of our main theorem.

Proposition 2.1. Let 𝐺 be Abelian, and let 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.

Proof. This can be proved by induction on order of 𝐺 as follows. The first step of the induction is obvious since the only group of order 2 is 𝑍₂. 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/𝐻-graded structure for 𝐴. Now the order of 𝐺/𝐻 is strictly less than the order of 𝐺. So an induction argument on order of 𝐺 completes the proof.

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 𝐴=^𝑛−1_𝑖=0ker𝑇−𝜆^𝑖 (2.3) is a 𝑍_𝑛-graded structure. Conversely for the grading 𝐴=⨁^𝑛−1_𝑖=0𝐴_𝑖, the following multiplicative operator 𝑇 satisfies 𝑇^𝑛=𝐼𝑑∶𝑇^𝑛−1_𝑖=0𝑎_𝑖=^𝑛−1_𝑖=0𝜆^𝑖𝑎_𝑖. (2.4)

Example 2.2. Let 𝑋 be a compact topological space and 𝜙 be an order𝑛homeomorphism of 𝑋, that is 𝜙^𝑛=𝐼𝑑. Define 𝑇∶𝐶(𝑋)→𝐶(𝑋) with 𝑇(𝑓)=𝑓∘𝜙, where 𝐶(𝑋) is the space of all continuous functions on 𝑋. Then 𝑇 is a bounded multiplicative operator on Banach algebra 𝐴=𝐶(𝑋) with 𝑇^𝑛=𝐼𝑑, so we would havea𝑍_𝑛-graded structure for 𝐴. As a particular example, the rotation of circle by 2𝜋/𝑛 is an order 𝑛 homeomorphism of circle. Then we naturally obtain a 𝑍_𝑛-graded structure for 𝐶(𝑆¹).

For a group 𝐺, a 𝑍_𝑛-partition for 𝐺 is a partition of 𝐺 into disjoint subsets 𝐺_𝑖, 𝑖=0,1,…𝑛−1 such that 𝐺_𝑖𝐺_𝑗⊆𝐺_{𝑖+𝑗(mod𝑛)}. This is equivalent to say that 𝐺₀ is a normal subgroup of 𝐺 whose quotient is isomorphic to 𝑍_𝑛. This is an obvious consequence of the normal property for subgroups; see [1]. So every group 𝐺 which has a normal subgroup 𝐻 such that 𝐺/𝐻 is isomorphic to 𝑍_𝑛 posses a 𝑍_𝑛-graded structure. For example, for every group 𝐾, the group 𝐺=𝐾×𝑍_𝑛 has 𝐾×{0} as a normal subgroup, and 𝐺/𝐾×{0} is isomorphic to 𝑍_𝑛. In this case the 𝑍_𝑛 partition for 𝐺 is 𝐺=∐_{𝑖∈𝑍𝑛}𝐾×{𝑖}.

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_𝑔𝑔 where 𝑎_𝑔’s are complex numbers. let 𝑙²(𝐺) be the Hilbert space of all 𝛾∶𝐺→ℂ such that |𝛾|²₂=∑_𝑔∈𝐺|𝛾(𝑔)|²<∞. 𝐺 acts on 𝑙²(𝐺) with 𝑔⋅𝛾(ℎ)=𝛾(𝑔⁻¹ℎ). This action defines a unitary representation of 𝐺 on 𝑙²(𝐺). We extend this action by linearity to ℂ𝐺. So each element of ℂ𝐺 can be considered as an element of 𝐵(𝑙²(𝐺)), the space of all bounded operators on 𝑙²(𝐺), and every element of 𝐺 can be considered as a unitary operator on 𝑙²(𝐺). The reduced 𝐶^∗ algebra of 𝐺, 𝐶^∗_red(𝐺), is the closure of ℂ𝐺, with respect to operator norm |⋅|_op defined on 𝐵(𝑙²(𝐺)).

In Proposition 2.1, we shall prove that a 𝑍_𝑛-partition for a Group 𝐺 gives a 𝑍_𝑛-graded structure for 𝐶^∗_red(𝐺). For the proof of the Proposition 2.1, we need to the following technical lemma.

Lemma 2.3. Let (𝐸,|⋅|) be a Banach space with a dense linear subspace F. Assume 𝐹=⨁^𝑛_𝑖=0𝐹_𝑖 is a direct sum of its linear subspaces 𝐹_𝑖’s. Suppose that the original norm of 𝐹 is equivalent to the direct limit norm |∑𝑓_𝑖|=∑|𝑓_𝑖|. Then 𝐸=⨁^𝑛_𝑖=0𝐹_𝑖.

Proof of Lemma. Let ∑^𝑛_𝑖=0𝑓^𝑖=0 where 𝑓^𝑖 is in 𝐹_𝑖, the closure of 𝐹_𝑖. Then there are sequences {𝑓^𝑖_𝑘}_𝑘∈ℕ of elements of 𝐹_𝑖 which converges to 𝑓^𝑖. So lim_𝑘→∞|∑^𝑛_𝑖=0𝑓^𝑖_𝑘|=0. 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=0𝐹_𝑖. So this direct sum is a topological direct sum. let 𝑥∈𝐸 be given, there is a sequence ∑^𝑛_𝑖=0𝑓^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=0𝑓_𝑖. This completes the proof of lemma.

Proposition 2.4. A 𝑍_𝑛-partition structure for a group 𝐺=∐^𝑛−1_𝑖=0𝐺_𝑖 gives a 𝑍_𝑛-graded structure for 𝐶^∗_red(𝐺)=⨁_{𝑖∈𝑍𝑛}𝐶^∗_red(𝐺_𝑖).

Proof. Assume that 𝐺=∐_{𝑖∈𝑍𝑛}𝐺_𝑖 is a 𝑍_𝑛-partition for 𝐺, then ℂ𝐺=⨁ℂ𝐺_𝑖 and 𝑙²(𝐺)=⨁𝑙²(𝐺_𝑖) where 𝑙²𝐺_𝑖=𝛾∈𝑙²(𝐺)∶𝛾∣_𝐺𝑗=0,𝑗≠𝑖. (2.5) For 𝑥_𝑖∈ℂ𝐺_𝑖 and 𝛾_𝑗∈𝑙²(𝐺_𝑗), we have 𝑥_𝑖⋅𝛾_𝑗 belongs to 𝑙²(𝐺_{𝑗+𝑖(mod𝑛)}). Assume 𝑥=∑^𝑛−1_𝑖=0𝑥_𝑖 and 𝛾=∑^𝑛−1_𝑖=0𝛾_𝑖. Then 𝑥⋅𝛾_𝑗²₂=^𝑛−1_𝑖=0𝑥_𝑖.𝛾_𝑗²₂,𝑥_𝑖⋅𝛾²₂=^𝑛−1_𝑗=0𝑥_𝑖⋅𝛾_𝑗²₂. (2.6) Let 𝑥_𝑖∈ℂ𝐺_𝑖 and 𝜀>0 be given. We apply the definition of norm operator on operator 𝑥_𝑖, then we conclude that there is a 𝛾=∑^𝑛−1_𝑖=0𝛾_𝑖 with |𝛾|²₂=∑^𝑛_𝑖=0|𝛾_𝑖|²₂=1 such that ^𝑛−1_𝑗=0𝑥_𝑖⋅𝛾_𝑗²₂=𝑥_𝑖⋅𝛾²₂≥𝑥_𝑖op−𝜀². (2.7)
This shows that there is a 𝑗∈𝑍_𝑛 such that 𝑥_𝑖op−𝜀²𝑛≤𝑥_𝑖⋅𝛾_𝑗²₂≤𝑥⋅𝛾_𝑗²≤|𝑥|²_op. (2.8)
Since |𝛾_𝑗|₂≤1, so |𝑥|_op≥|𝑥_𝑖|_op/√𝑛, for 𝑖=0,1,…𝑛−1. We conclude that the operator norm on ℂ𝐺 is equivalent to direct sum norm |∑𝑥_𝑖|_op=∑|𝑥_𝑖|_op. Now if we put 𝐸=𝐶^∗_red(𝐺), 𝐹=ℂ𝐺, 𝐹_𝑖=ℂ𝐺_𝑖 and apply the above lemma to these spaces, the proof of the proposition would be completed.

Remark 2.5. 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∈𝐻𝐺_ℎ such that 𝐺_ℎ1𝐺_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 [2] or [3]. 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 assigns to 𝑋 the commutative 𝐶^∗ algebra 𝐶(𝑋) of all complex-valued continuous functions on 𝑋. It also assigns to a continuous function 𝑓∶𝑋→𝑌, the 𝐶^∗ algebra morphism 𝑓^∗∶𝐶(𝑌)→𝐶(𝑋) by 𝑓^∗(𝜙)=𝜙∘𝑓. 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 𝑎²=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 the Banach algebra 𝐴𝐼 as follows.

Definition 2.6. Let 𝐴 be a Banach algebra, we denote by 𝐴𝐼, the Banach algebra of all continuous 𝛾∶𝐼→𝐴 with the standard operations and norm. We define 𝜋_𝑖; 𝐴𝐼→𝐴 with 𝜋_𝑖(𝛾)=𝛾(𝑖), for 𝑖=0,1.

We say that two morphisms 𝛼,𝛽∶𝐴→𝐵 are homotopic if there is a morphism Θ∶𝐴→𝐵𝐼 such that Θ₀=𝛼,Θ₁=𝛽, where Θ_𝑖=𝜋_𝑖∘Θ. A morphism 𝛼∶𝐴→𝐵 is called null homotopic if it is homotopic to a morphism with one-dimensional range. Obviously this is a natural Banach 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).

3. 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 1. Let 𝐴 be 𝑎 G-graded Banach algebra with no nontrivial idempotents. Let 𝑎∈𝐴 be a nontrivial homogenous element. Then 0 belongs to the convex hull of the spectrum sp(𝑎^𝑘).
Further, if 𝐴 is commutative and a is invertible, then 𝑎^𝑘 and 1 do not lie in the same connected component of the space of invertible elements 𝐺(𝐴).

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 𝑧₀, for some 𝑧₀∈ℂ, which is a subset of a branch of logarithm. Using holomorphic functional calculus as described in [4, Chapter 10], we can find a convergent series 𝑏=∑^∞_𝑖=0𝑐_𝑖(𝑎^𝑘−𝑧₀)^𝑖 where exp𝑏=𝑎^𝑘 where 𝑐_𝑖’s are complex numbers.
So 𝑇(𝑏)=∑^∞_𝑖=0𝑐_𝑖(𝜆^𝑘𝑎^𝑘−𝑧₀)^𝑖. Thus 𝑇(𝑏) and 𝑏 are power series in 𝑎. In particular 𝑎𝑏=𝑏𝑎 and 𝑇^𝑖(𝑏)𝑇^𝑗(𝑏)=𝑇^𝑗(𝑏)𝑇^𝑖(𝑏), for all 𝑖 and 𝑗. So we have (𝑎⁻¹exp(𝑏/𝑘))^𝑘=1. Then sp(𝑎⁻¹exp(𝑏/𝑘)) is a subset of the set of all 𝑘th 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⁻¹exp(𝑏/𝑘))={1}; otherwise we multiply 𝑎 with an appropriate root of unity. Put 𝑞=𝑎⁻¹exp(𝑏/𝑘)−1. Then 𝑞 is a quasinilpotent element of 𝐴; that is, its spectral radius spr(𝑞)=0. We have 𝑏exp𝑘=𝑎+𝑞𝑎=𝑎+𝑞^′, (3.1) 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 [4, page 302]. Moreover, we have 𝑇(𝑏)exp𝑘=𝑇(𝑎)+𝑇𝑞𝑎=𝜆𝑎+𝑇𝑞^′=𝜆𝑎+𝑞₁, (3.2) where 𝑞₁ is a quasinilpotent.
On the other hand, (𝑎+𝑞)⁻¹−𝑎⁻¹=−(𝑎+𝑞)⁻¹𝑞𝑎⁻¹, then (𝑎+𝑞)⁻¹=𝑎⁻¹+𝑞^′′ for some quasinilpotent 𝑞^′′. Thus we have exp(−𝑏/𝑘)=𝑎⁻¹+𝑞^′′. We obtain from (3.1) and (3.2) and the latest equation exp(𝑇(𝑏)−𝑏)/𝑘=𝜆+𝑞₂ for some quasinilpotent 𝑞₂. 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+𝑇𝑦+𝑇²𝑦+⋯+𝑇^𝑛−1𝑦. (3.3) 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 [4, 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 as the identity. Then there exists an element 𝑏∈𝐴 with exp𝑏=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.

Corollary 3.1. Let 𝑋 be a compact locally path connected and simply connected space, and let 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=0𝜆^𝑖𝑓(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∶𝐴↦𝐴 by 𝑇(𝑓)=𝑓∘𝜙. 𝑇 satisfies 𝑇^𝑛=1, so we have a 𝑍_𝑛-graded structure for 𝐴 in the form 𝐴=⨁^𝑛−1_𝑖=0ker(𝐴−𝜆^𝑖). Put 𝑔=∑^𝑝−1_𝑖=0𝜆^𝑖𝑓(𝜙^𝑖(𝑥)), then 𝑇(𝑔)=𝜆^𝑛−1𝑔, 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 as 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 expℎ=𝑔, using the standard lifting lemma in the theory of covering space; see [5, proposition 1.33]. So 𝑔 is a logarithmic element and must be in the same connected component as the identity. This contradicts to the fact that 𝑔 cannot be connected to the identity. This completes the proof.

Putting 𝑋=𝑆²,𝜙(𝑥)=−𝑥, and 𝜆=−1, we obtain the classical two-dimensional Borsuk Ulam theorem as follows.

Corollary 3.2 (Borsuk-Ulam Theorem). For a continuous function 𝑓∶𝑆²→ℂ, there must exist a point 𝑥∈𝑆² with 𝑓(𝑥)=𝑓(−𝑥).

Example 3.3. Put 𝑆²={(𝑥,𝑦,𝑧)∈ℝ³∶𝑥²+𝑦²+𝑧²=1}, and let 𝑓∶𝑆²→ℝ² be a continuous function. Then there is a point (𝑥₀,𝑦₀,𝑧₀)∈𝑆² such that 𝑓(𝑦₀,−𝑥₀,−𝑧₀)−𝑓(−𝑥₀,−𝑦₀,𝑧₀)+𝑓(−𝑦₀,𝑥₀,−𝑧₀)−𝑓(𝑥₀,𝑦₀,𝑧₀)=0. To prove this, consider the fourth-order homeomorphism 𝜙 of 𝑆² with 𝜙(𝑥,𝑦,𝑧)=(−𝑦,𝑥,−𝑧). Now apply Corollary 3.1 with 𝜆=−1.
The following corollary is an obvious consequence of the last part of the main theorem. In this corollary and its sequel, 𝐺(𝐴)₀ is the connected component of the identity.

Corollary 3.4. Let 𝐴 be an idempotentless 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.

4. Further Questions and Remarks

In this section we present some questions which naturally arise from the main theorem and the corollaries of the previous sections.

First we discuss about a pure algebraic analogy of Corollary 3.4. For this purpose, we need 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 [6]. Let 𝐴 be a commutative complex algebra. We say that a pair (𝑎₁,𝑎₂) is invertible if there is a pair (𝑏₁,𝑏₂) such that 𝑎₁𝑏₁+𝑎₂𝑏₂=1. We say 𝐴 has Bass stable rank one if, for every invertible pair (𝑎,𝑏)∈𝐴², there is an element 𝑥∈𝐴 such that 𝑎+𝑏𝑥 is invertible; see [7]. It is mentioned in [6] that, for a commutative 𝐶^∗ algebra, the Bass stable rank coincide with 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 an inductive system with GL_𝑛(𝐴)’s. Put 𝐺𝐿(𝐴)=lim_→GL_𝑛(𝐴). Then GL(𝐴) has a natural inductive limit topology and algebraic structure. Define 𝐾₁(𝐴)=GL(𝐴)/GL(𝐴)₀ where GL(𝐴)₀ is the connected component of the identity. This Abelian group 𝐾₁(𝐴) is the standard 𝐾₁ functor defined on the category of the Banach algebras. For a complex algebra 𝐴, there is a pure algebraic 𝐾^alg₁ 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 isomorphic to 𝐾₁(𝐴); see [8].

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 Bass stable rank one which is graded by a finite Abelian group. Assume that a nontrivial homogenous element of 𝐴 is invertible. Does this imply that 𝐾^alg₁(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 4.1. Let 𝐹₂ be the free group on two generators 𝑥 and 𝑦. We shall see that there is a normal subgroup 𝐺₀ of 𝐹₂ which does not contain 𝑦𝑥𝑦⁻¹𝑥⁻¹ and its quotient is isomorphic to 𝑆₃, the permutation group on three elements. Assuming the existence of such subgroup 𝐺₀, we obtain an 𝑆₃-graded structure for 𝐶^∗_red(𝐹₂) for which 𝑦𝑥𝑦⁻¹𝑥⁻¹, as an element of 𝐶^∗_red(𝐹₂), is a nontrivial homogenous element. On the other hand, it is well known that this algebra has no nontrivial idempotent and 𝑦𝑥𝑦⁻¹𝑥⁻¹∈𝐶^∗_red(𝐹₂), is in the same connected component as the identity; see [8]. To prove the existence of such subgroup 𝐺₀, we first note that 𝐹₄, the free group on four generators, can be considered as an index three subgroup of 𝐹₂ which does not contain 𝑦𝑥𝑦⁻¹𝑥⁻¹. This can be proved using certain covering space of figure-8 space as illustrated in [5, page 58]. Let ℎ∶𝑋→𝑋 be a covering space with ℎ(𝑞)=𝑝. Then the induced map ℎ_∗∶𝜋₁(𝑋,𝑞)→𝜋₁(𝑋,𝑝) is an injective map. ℎ_∗(𝜋₁(𝑋,𝑞)) is isomorphic to 𝜋₁(𝑋,𝑞) whose index, as a subgroup of 𝜋₁(𝑋,𝑝), is equal to the cardinal of a fibre of the covering space. Moreover, a loop 𝛾∈𝜋₁(𝑋,𝑝) lies in the range of ℎ_∗ if and only if the unique lifting 𝛾 of 𝛾 with starting point 𝑞 is a loop with base point 𝑞, see [5, proposition 1.31].

Now consider the 3-fold covering space which is illustrated in the Figure 1. The fundamental group of the total space is 𝐹₄, and the fundamental group of the base space is 𝐹₂. The loop 𝑦𝑥𝑦⁻¹𝑥⁻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.

Figure 1 

Put 𝐺=𝐹₂ and 𝑏=𝑦𝑥𝑦⁻¹𝑥⁻¹. Then 𝐺 has a subgroup 𝐻 of index 3 which does not contain a commutator element 𝑏. We obtain a morphism 𝛽∶𝐺→𝑆₃ with the standard action of 𝐺 on the set of left cosets of 𝐻. Then ker𝛽 is a normal subgroup of 𝐺 which is contained in 𝐻 and does not contain 𝑏. 𝐺/ker𝛽 is isomorph to a subgroup of 𝑆₃; on the other hand, 𝐺/ker𝛽 is not Abelian since ker𝛽 does not contain a commutator element 𝑏. This shows that ker𝛽 is a normal subgroup of 𝐺=𝐹₂ in which quotient is isomorphic to 𝑆₃ and does not contain the commutator element 𝑏=𝑦𝑥𝑦⁻¹𝑥⁻¹. So the above construction gives an example of a noncommutative Banach algebra 𝐶^∗_red(𝐹₂), without nontrivial idempotent, which is graded by a non-Abelian finite group 𝑆₃, such that a nontrivial homogenous element 𝑦𝑥𝑦⁻¹𝑥⁻¹ lies in the same connected component as the identity.

But could we give any such example with a finite Abelian group 𝐺? In other words, can we drop the hypothesis of commutativity of Banach 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(𝐹₂), with the 𝑍₂-graded structure corresponding to the 𝑍₂-partition of 𝐹₂ 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 𝑍₂-graded structure for 𝐴, up to graded isomorphism?

There is an affirmative answer to a particular case of the second part of the above question. We thank professor Valette for his affirmative answer in this case. This affirmative answer is mentioned in the following proposition.

Proposition 4.2. Every linear combination of two odd words 𝑎 and 𝑏 in 𝐶^∗_red(𝐹₂) cannot be connected to the identity.

Proof. For a contradiction assume that 𝑟𝑎+𝑠𝑏 can be connected to the identity in the space of invertible elements, where 𝑟 and 𝑠 are two complex numbers. With a small perturbation, we can assume that |𝑟|<|𝑠|. (Note that with this perturbation the connected component does not change.) Now put 𝑧₀=𝑟/𝑠, so 𝑧₀𝑎+𝑏 lies in the same connected component as the identity with |𝑧₀|<1. Then the curve 𝑡𝑧₀𝑎+𝑏,𝑡∈[0,1] is a curve which lies in the space of invertible elements of 𝐶^∗_red(𝐹₂). Because 𝑏𝑎⁻¹ is a unitary element, its spectrum does not contain an element 𝑡𝑧₀ where 𝑡∈[0,1]. So 𝑏 as an odd element lies in the same connected component as the identity. But the only words which can be connected to the identity are members of the commutator subgroup of 𝐹₂. This contradiction shows that a linear combination of two odd words can not be connected to the identity.

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 𝑓₁,𝑓₂,…,𝑓_𝑛 be 𝑛 odd continuous real-valued functions on 𝑆^𝑛, then the function ∑^𝑛_𝑖=1𝑓²_𝑖 is not an invertible element of 𝐶(𝑆^𝑛) or equivalently is not in the same connected component as the identity (since 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 𝑍₂-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 𝑍₂-graded noncommutative 𝑛-sphere and 𝑎₁,𝑎₂,…𝑎_𝑛 are self-adjoint elements of 𝐴 which are odd elements of this graded algebra. Is it true to say that ∑^𝑛_𝑖=1𝑎²_𝑖 is either non-invertible or is not in the same connected component as the identity?

Remark 4.3. 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 [9] or [10].

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 map 𝑓∶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 𝑍₂-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 𝑇𝛼=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 𝑍₂-graded structure, and let 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 [11]. 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) 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?

Acknowledgment

The author would like to thank the referee for very valuable suggestions.