ISOMETRY AND PHASE-ISOMETRY OF NON-ARCHIMEDEAN NORMED SPACES∗

(王瑞东) (姚文婷)

College of Science, Tianjin University of Technology, Tianjin 300384, China

E-mail: wangruidong@tjut.edu.cn; yaowenting103@163.com

Abstract In this paper,we study isometries and phase-isometries of non-Archimedean normed spaces.We show that every isometry f : Sr(X) →Sr(X),where X is a finitedimensional non-Archimedean normed space and Sr(X) is a sphere with radius r ∈‖X‖,is surjective if and only if K is spherically complete and k is finite.Moreover,we prove that if X and Y are non-Archimedean normed spaces over non-trivially non-Archimedean valued fields with|2|=1,any phase-isometry f :X →Y is phase equivalent to an isometric operator.

Key words non-Archimedean normed spaces;isometry extension;Wigner’s theorem

1 Introduction

The study of non-Archimedean functional analysis is essential to many branches of mathematics and physics;it can be traced back to Monna,a Dutch mathematician of the 1940s (see[1–3]).

There has recently been increasing interest in the isometries on non-Archimedean normed spaces.In [4],Moslehian and Sadeghi tried to establish a Mazur-Ulam type theorem for the strictly convex non-Archimedean normed spaces.Shortly thereafter,it was proved that there are almost no strictly convex non-Archimedean spaces by Kubzdela in [5].Thus,the Mazur-Ulam theorem is invalid in non-Archimedean normed spaces.Surprisingly,Sánchez and Garmendia[6]exhibited a fractal structure for the group of isometries of an ultrametric normed space.So,it is far from being linear.

Additionally,as was shown by Schikhof [7],every isometry on a non-Archimedean valued field K is surjective if and only if K is spherically complete and the residue class fieldkis finite.Kubzdela further proved in [5]that the isometryf:X →X,whereXis a finite-dimensional non-Archimedean normed space,is surjective if and only if K is spherically complete andkis finite.Extending the result,we get an identical conclusion when the isometry is on the sphere with a radiusr ∈‖X‖(Theorem 3.10).

Mazur-Ulam’s main theorem in [8]stated,in short,that if the mappingfbetween a real normed space is a surjective isometry,thenfis affine,i.e.,fis a linear mapping up to translation.In 1987,Tingley [9]raised the question: if we letXandYrepresent Banach spaces,andSXandSYare the unit spheres ofXandY,and we assume thatf:SX →SYis surjective isometry,then,doesfhave a linear isometry extensionT:X →Ysuch thatT|SX=f? For more details regarding Tingley’s problem,we refer the reader to the survey [10–14].

Wigner’s Theorem [15–23],a milestone in the mathematical foundations of quantum mechanics in physics,shows that a mappingf:H →Ksatisfies that

for anyx,y ∈H,and is phase equivalent to a linear isometry or an anti-linear as long asHandKare real or complex inner product spaces.

It is natural to ask whether Wigner’s theorem would still be true ifHandKwere normed spaces.Unfortunately,there is no inner product in general normed spaces.However,in the real inner product space the equation (1.1) is equivalent to the equality

for anyx,y ∈H.We therefore call the operatorfbetween two real normed spaces a phaseisometry iffsatisfies the equality (1.2).Several papers have studied Wigner’s theorem on real normed spaces (see [24–28]).

Questions regarding Wigner’s type of result are amenable for non-Archimedean normed spaces,but seem to have received almost no attention in the literature.We prove in Theorem 4.2 that ifXandYare non-Archimedean normed spaces over a non-trivially non-Archimedean valued field K,and if the mappingf:X →Ysatisfies (1.2),there will exist a phase functionε:X →{1,-1} such thatεfis an isometry.

2 Preparatory Knowledge

In order to systematically study non-Archimedean normed spaces,we introduce the following definitions:

Definition 2.1A non-Archimedean valued field is a field K equipped with a valuation|·|:K→[0,∞) such that

1.|λ|=0 if and only ifλ=0,

2.|λµ|=|λ||µ|,

3.we have the strong triangle inequality|λ+µ|≤max{|λ|,|µ|} λ,µ∈K.

We often write K instead of(K,|·|).This is not isomorphic(algebraically and topologically)to either R or C.K is said to be trivially valued if the valuation take every nonzeroλinto 1 and|0|=0.

The unit diskBK:={λ ∈K:|λ|≤1} is a commutative ring with identity.The open unit disk:={λ ∈K:|λ|<1} is a maximal ideal ofBK.Thus,BK/is a field,that called the residue class field of K,which is customarily denoted byk.

A setB(λn,rn):={µ∈K:|µ-λn|≤rn}is called a closed ball in K.A non-Archimedean valued field K is spherically complete if every sequence of balls,satisfingB(λn,rn)⊃B(λn+1,rn+1) for everyn ∈N,has a nonempty intersection.

There are many examples of non-Archimedean valued fields.A typical one is in the field ofp-adic numbers (Qp,|·|).For any non-zero rational numberxcan be written as

wherea,bandnare integers anda,bare not divisible byp.Then|x|p=p-nifx≠ 0 and|0|p=0 defines a non-Archimedean norm on Q.We usually write Qpinstead of (Q,|·|p).

Definition 2.2LetXbe a linear space over a non-Archimedean valued field K.The function‖·‖:X →[0,∞) satisfies the following conditions:

1.‖x‖=0 if and only ifx=0,

2.‖λx‖=|λ|‖x‖for allλ ∈K,x ∈X,

3.the strong triangle inequality‖x+y‖≤max{‖x‖,‖y‖} x,y ∈X.

The‖·‖is called a non-Archimedean norm.Consequently,we call(X,‖·‖)a non-Archimedean normed space.

We frequently writeXinstead of (X,‖·‖).

Lemma 2.3(Isosceles triangle principle[30,Lemma1.1.1]) LetXbe a non-Archimedean normed space.For anyx,y ∈X,if‖x‖≠‖y‖,then‖x+y‖is equal to the largest one,i.e.,

For more background on non-Archimedean normed spaces we refer the reader to [29–32].

Since the non-Archimedean norm satisfies the strong triangle inequality,the non-Archimedean normed spaces are ultrametric metric spaces.Therefore,it makes sense to consider isometries and phase-isometries between them.

Definition 2.4LetXandYbe non-Archimedean normed spaces.A mappingf:X →Yis an isometry iffsatisfies the function equation

Definition 2.5LetXandYbe non-Archimedean normed spaces.A mappingf:X →Yis an phase-isometry iffsatisfies the function equation

Let us say that a mappingf:X →Yis phase equivalent to an isometry if there exists a phase functionε:X →K with|ε(x)|=1 such thatεfis an isometry.Namely,two mappings,f1:X →Yandf2:X →Y,are said to differ by a phase factor or be phase factor equivalent if there exists a phase function with|ε(x)|=1 such thatf1(x)=ε(x)f2(x) for anyx ∈X.

3 Isometry on the Sphere of a Non-Archimedean Normed Space

LetXbe a non-Archimedean normed space,letr ∈(0,∞)andx ∈X,call the setB(x,r):={y ∈X:‖y-x‖≤r} a closed ball inX,and callB(x,r-):={y ∈X:‖y-x‖

The following result is a straightforward modification of Sánchez and Garmendia’s result(see [6]) (the statement of [6,Proposition 2.1]remains valid if we replaceB(0,r) byX):

Theorem 3.1LetXandYbe non-Archimedean normed spaces,and take the mappingT:X →YwithT(0)=0.ThenTis an isometry if and only if,for anyr ∈‖X‖,there exists an isometric mappingfr:Sr(X)→Sr(Y) such thatT|Sr(X)=fr.

ProofAssume thatTis an isometry withT(0)=0,and for anyr ∈‖X‖letfr=T|Sr(X).Obviously,fr:Sr(X)→Sr(Y) is an isometric mapping.

Conversely,takex,y ∈Xto prove thatTis an isometry.There are two cases to consider:

•If‖x‖=‖y‖=r,we get that

•If‖x‖≠‖y‖,since‖f‖x‖‖=‖x‖and‖f‖y‖‖=‖y‖,it follows that‖T(x)‖=‖x‖and‖T(y)‖=‖y‖.Thus,by Lemma 2.3,we obtain that

Hence,Tis an isometry.

Example 3.2Let{αn}n∈Z⊂Qpbe such that|αn|p=1 for anyn ∈Z.Iff: Qp →Qpsatisfiesf(x)=αnxfor anyx ∈Qpwith|x|p=pn,thenfis an isometry of Qp.

Indeed,for anyx,y ∈Qpwith|x|p=|y|p=pn.Then

Applying Theorem 3.1,fis an isometry.

Example 3.3A mappingf: Qp →Qpis an isometry withf(x)=kxfor anyx ∈Qpandk ∈N if and only ifk=ap+r,a ∈N,r ∈{1,2,···,p-1}.

Indeed,letx,y ∈Qp.Since|k|p=|ap+r|p=max(|ap|p,|r|p)=|r|p=1,it follows that

Conversely,aiming for a contradiction,sincek ∈N,supposing thatk=ap,then we have that|k|p<1.We get that

Example 3.4Iffor anyx ∈S1(Qp).Thenf1is an isometry.

Indeed,for anyx,y ∈S1(Qp),assume thatwherea,b,c,d ∈Z but are not divisible byp.We thus obtain that|a|p=|b|p=|c|p=|d|p=1 and that

By the definition off1,we obtain that

Hence,f1is an isometry.

Theorem 3.5f: Qp →Qpis an additive isometry if and only if there existsα ∈Qpsuch that|α|p=1 andf(x)=αxfor anyx ∈Qp.

Proof(): Sincefis additive,it follows thatf(0)=0.Letf(1)=α ∈Qp.Then

Sincefis additive,we conclude thatf(x)=αxfor anyx ∈Qp.

Hence,fis an isometry.Suppose thatx,y ∈Qp.Then we obtain that

and conclude thatfis additive.

Example 3.6Define a map

The mappingfis a surjective isometry but is not additive.

Indeed,for anyx,y ∈Qp,

1.if|x|p=|y|p=1,then since|xy|p=1,it follows that

3.if|x|p≠1 and|y|p≠1,we have that

Hence,fis an isometry.Obviously,fis surjective.To obtain thatfis not additive,we suppose that|x|p=1,|y|p<1 andy≠0,so we have that|x+y|p=1.Hence,

The following result was given in [6,Proposition 3.4]:

Theorem 3.7LetXandYbe non-Archimedean normed spaces over a non-trivially non-Archimedean field K,and letr ∈‖X‖,f:Sr(X)→Sr(Y) be a surjective isometric mapping.Then there exists an isometric mappingT:X →Ysuch thatT|Sr(X)=f.

Theorem 3.8IfXis a finite-dimensional normed space,f:S(X)→S(X)is an isometric mapping.Thenfis surjective.

ProofAssume the contrary and suppose thatx0∈S(X)f(S(X)).Sincef(S(X)) is closed,it follows thatd(x0,f(S(X)))=d0>0.Takem,n ∈N such thatm≠n.We may as well assume thatn>m,so that

This is a clear contradiction of the fact that{fn(x0)}⊂S(X) and thatS(X) is compact.

[7,Theorem 2]says that if card (k) is infinite,then there exists a non surjective isometryT: K→K.With minor modifications,we can get the following theorem which will be used later:

Lemma 3.9Let K be a complete non-trivially non-Archimedean valued field.Then K is spherically complete and the residue class fieldkis finite if and only if each isometryT:B(K)→B(K) is surjective.Moreover,if the residue class fieldkis infinite,then for anyα0∈K with|α0|=1,there is an isometryT:B(K)→B(K) such thatα0T(B(K)).

ProofWe just have to specify part of the theorem.We suppose thatkis infinite.Then letj ↔Bj(j ∈k) be the correspondence between the elements ofkand the additive cosets of{λ ∈K:|λ|<1} inB(K).Fixα0∈K with|α0|=1,and let

Then there is bijectionτ:k →k×.For eachj ∈k,letσjbe an isometry (translation) ofBjontoBτ(j).DefineTas follows:

It is easy to see thatTis an isometry and thatα0∉T(B(K)).

Theorem 3.10LetXbe a finite-dimensional non-Archimedean normed space,r ∈‖X‖.Every isometryf:Sr(X)→Sr(X) is surjective if and only if K is spherically complete andkis finite.

ProofBy Theorem 3.7,there exists an isometric mappingTofXsuch thatT|Sr(X)=f.We conclude from [5,Theorem 12]thatTis surjective,which implies thatfis surjective.

Conversely,aiming for a contradiction,we first suppose that K is non-spherically complete.Applying [32,4.A.],Xis non-spherically complete and there exists a sequence of closed balls inXwith an empty intersection{B(xn,‖xn-xn+1‖)}n∈N.We can assume thatrn>rn+1,wherern=‖xn-xn+1‖for anyn ∈N.Letα ∈K such thatα≠0 and

Then{B(αxn,|α|‖xn-xn+1‖)}n∈Nis a sequence of closed balls inXwith an empty intersection.Fix anx0∈Sr(X).Then,for anyn ∈N andy ∈B(αxn,|α|‖xn-xn+1‖),since

it follows from Lemma 2.3 that‖x0+y‖=r.Define a map

Obviously,f:Sr(X)→Sr(X) is well defined andx0∉f(Sr(X)).We will show thatfis an isometry.For anyx,y ∈Sr(X),we assume thatf(x)=x-αxiand thatf(y)=y-αxjfor somei,j ∈N.Ifi=j,then we are done.

Ifi≠j,we assume thati

and

Obviously,‖x-αxi-x0‖>|α|riandy ∈x0+B(αxi,|α|ri).Then we obtain that

Now assume that K is spherically complete,so thatXhas an orthogonal base(see [32,Lemma 5.5]).Lettingr ∈‖X‖,we assume that‖x1‖=r.As a consequence of Lemma 3.9,if card(k)is infinite,then there exists a non-surjective isometryT:B(K)→B(K).The functionfso defined satisfies thatWe get thatfis a non-surjective isometry ofSr(X),which completes the proof.

Example 3.11Let the map

for any (a,b)Thenfis an isometry.

Indeed,first,we show thatfis well defined.For any (a,b)∈there are several cases to consider.

•If|a|=1,|b|<1 or|a|<1,|b|=1,then|a+b|=|a-b|=max{|a|,|b|}=1.

•If|a|=|b|=1,we have that|a+b|,|a-b|≤max{|a|,|b|}=1.

Suppose that|a+b|<1.Then|a-b|=|a+b-2b|=|2b|=1.

Similarly,we can show thatfis an isometry.

4 Phase-Isometry of Non-Archimedean Normed Space

In this section,we will describe phase-isometries defined on non-Archimedean normed spaces.We start with the following simple observation:

Lemma 4.1LetXandYbe non-Archimedean normed spaces over a non-trivial nonarchimedean valued field K.Iff:X →Yis a phase-isometry,then‖f(x)‖=‖x‖for everyx ∈X.

ProofBecause of the definition of phase-isometry,which asserts that

for anyx,y ∈X,lettingy=xwe have that

Hence,‖f(x)‖=‖x‖,which completes the proof.

Theorem 4.2Let K be a non-trivially non-Archimedean valued field with|2|=1,and letXandYbe non-Archimedean normed spaces over K.Suppose thatf:X →Yis a phase-isometry.Then there exists a phase functionε:X →{-1,1} such that

for anyxandy ∈X.

ProofBy Theorem 3.1,it is enough to prove that,for anyr ∈‖X‖,there exists a phase functionε:Sr(X)→{-1,1} such that

for anyx,y ∈Sr(X).

LetA ⊂Sr(X) andε:A →{-1,1} such that

for anyx,y ∈A.There exists a non-emptyA.Indeed,we can fix anx ∈Sr(X) and letε(x)=1.Lety ∈Sr(X) andy≠x.Since

there existsε(y)∈{-1,1} such that

By Zorn’s lemma,it suffices to show that forA≠Sr(X) andy ∈Sr(X)A,there existsε(y)∈{-1,1} such that‖ε(x)f(x)-ε(y)f(y)‖=‖x-y‖for anyx ∈A.

If‖x+y‖=‖x-y‖,for anyx ∈A,we have that

Thus we can chooseε(y)∈{-1,1} such that‖ε(x)f(x)-ε(y)f(y)‖=‖x-y‖for anyx ∈A.Therefore we assume that‖x0+y‖≠‖x0-y‖for somex0∈A.We can chooseε(y)∈{-1,1}such that

Now we will show that‖ε(x)f(x)-ε(y)f(y)‖=‖x-y‖for anyx ∈A.Obviously,‖ε(x)f(x)-ε(y)f(y)‖=‖x-y‖when‖x+y‖=‖x-y‖.Thus we assume that‖x+y‖≠‖x-y‖.

The problem will be discussed by the following two cases:

•Case (I) If‖x0-x‖≠‖x0-y‖,we may assume that‖x0-x‖>‖x0-y‖.Then,by Lemma 2.3,it follows that

and by Lemma 2.3 and (4.1),we obtain that

Thus‖ε(x)f(x)-ε(y)f(y)‖=‖x-y‖.

•Case (II) If‖x0-x‖=‖x0-y‖,then

and we will show that‖ε(x)f(x)-ε(y)f(y)‖=‖x-y‖;if not,then‖ε(x)f(x)-ε(y)f(y)‖=‖x+y‖,and we observe that

1.If‖x+y‖<‖x-y‖,by (4.4),we have that‖x+y‖<‖x0-x‖,and

which contradicts that‖x0+y‖≠‖x0-y‖.

2.If‖x+y‖>‖x-y‖,we have that

By (4.5),we have that

Since‖x0-y‖≠‖x0+y‖and‖x0+y‖ ≤r,it follows that‖x0+y‖

which is a contradiction.

Thus the proof of Theorem 4.2 is complete.

Theorem 4.3Let K be a non-trivially non-Archimedean valued field with|2|=1 and letX,Ybe non-Archimedean normed spaces over K.Ifr ∈|K|,r≠ 0 and the mappingf:Sr(X)→Sr(Y) is a phase-isometry,thenfcan be extended to a phase-isometryFonX.

ProofAs was shown in Theorem 4.2,there existsε:Sr(X)→{-1,1} such that

for anyx,y ∈Sr(X);namely,εf:Sr(X)→Sr(Y) is an isometry.By Theorem 3.7,εfcan extend to an isometry:X →Y.Let

Obviously,Fis a phase-isometry mapping andF|Sr(X)=f.Thus the proof of the theorem is complete.

Conflict of InterestThe authors declare no conflict of interest.