Endomorphisms of the poset of idempotent

matrices ∗

Peter SemrlFaculty of Mathematics and Physics, and

Institute of Mathematics, Physics, and MechanicsUniversity of Ljubljana

Jadranska 19SI-1000 Ljubljana

Sloveniapeter.semrl@fmf.uni-lj.si

2000 Math. Subj. Class. 06A99, 15A30.

Keywords and phrases. Poset of idempotent matrices, division ring, order pre-serving map, supremum, orthomodular poset.

Abstract

Let D be a division ring, n ≥ 3 an integer, and Pn(D) the poset ofall n × n idempotent matrices over D with the partial order defined byP ≤ Q if PQ = QP = P . Let T ∈ Mn(D) be an invertible matrix andσ : D → D an endomorphism (τ : D → D an anti-endomorphism). Forany P ∈ Pn(D) we denote by Pσ ( P τ ) the idempotent matrix obtainedfrom P by applying σ (τ) entrywise. The map φ : Pn(D)→ Pn(D) definedby φ(P ) = TPσT−1, P ∈ Pn(D) (φ(P ) = T t(P τ )T−1, P ∈ Pn(D)) isan injective map preserving order in both directions. Every such map iscalled a standard map. It has been known before that if D is an EASdivision ring, then every injective order preserving map φ : Pn(D) →Pn(D) is either standard or it is of a very special degenerate form. Inthis paper we use some ideas from geometry of algebraic homogeneousspaces and elementary field theory to give examples showing that the EASassumption is indispensable. Then we define generalized standard mapsand using them we describe the general form of injective order preseverson Pn(D) for an arbitrary division ring D. Our proof is shorter than theoriginal one for the special case of EAS division rings. Under somewhatstronger assumptions we get two characterizations of standard maps. Allthe results are optimal as shown by counterexamples.

∗The author acknowledges the financial support from the Slovenian Research Agency, GrantNo. P1-0288.

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1 Introduction

Let D be a division ring and n a positive integer. We denote by Mn(D) the setof all n×n matrices over D. By Pn(D) we denote the set of all n×n idempotentmatrices, Pn(D) = {P ∈ Mn(D) : P 2 = P}. It is well-known that the set ofall n × n idempotent matrices is a poset (partially ordered set) with P ≤ Q ifPQ = QP = P . A map φ : Pn(D) → Pn(D) preserves order if φ(P ) ≤ φ(Q)whenever P ≤ Q, P,Q ∈ Pn(D). It preserves order in both directions if forevery pair P,Q ∈ Pn(D) we have φ(P ) ≤ φ(Q) if and only if P ≤ Q.

If A = [aij ] ∈ Mn(D) is any matrix and σ : D → D an endomorphism oran anti-endomorphism (an additive map satisfying σ(λµ) = σ(µ)σ(λ), λ, µ ∈ D,and σ(1) = 1) of division ring D, then we denote by Aσ the matrix obtained fromA by applying σ entrywise, Aσ = [aij ]

σ = [σ(aij)]. Clearly, for every invertiblematrix T ∈Mn(D) and every endomorphism σ of D the map φ : Pn(D)→ Pn(D)defined by

φ(P ) = TPσT−1, P ∈ Pn(D), (1)

is an injective map preserving order in both directions. If σ is an automorphismof D, then φ is an automorphism of the poset Pn(D), that is, a bijective mappreserving order in both directions.

Similarly, for every invertible matrix T ∈Mn(D) and every anti-endomorphismτ of D the map φ : Pn(D)→ Pn(D) defined by

φ(P ) = T t(P τ )T−1, P ∈ Pn(D), (2)

is an injective map preserving order in both directions. Here, tA denotes thetranspose of a matrix A. Every map of the form (1) or (2) will be called astandard map on Pn(D).

Assume now that D is an infinte division ring. For any positive integer j,1 ≤ j < n, we denote by P jn(D) the set of all idempotents of rank j, and defineφ : Pn(D)→ Pn(D) in the following way. Set φ(0) = 0 and φ(I) = I. Let ϕj beany injective map from P jn(D) into D, j = 1, . . . , n− 1. Put

φ(P ) =

1 ϕ1(P ) 0 0 . . . 00 0 0 0 . . . 00 0 0 0 . . . 00 0 0 0 . . . 0...

......

.... . .

...0 0 0 0 . . . 0

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for every P ∈ P 1n(D) and

φ(P ) =

1 0 0 0 . . . 00 1 0 0 . . . 00 ϕ2(P ) 0 0 . . . 00 0 0 0 . . . 0...

......

.... . .

...0 0 0 0 . . . 0

for every P ∈ P 2

n(D). We continue in a similar way. We denote by Eij , 1 ≤i, j ≤ n, the n × n matrix whose all entries are zero but the (i, j)-entry whichequals 1. For every P ∈ P 3

n(D) we set φ(P ) = E11 + E22 + E33 + ϕ3(P )E34,...Put (p, q) = (n− 1, n) if n is even, and (p, q) = (n, n− 1) if n is odd. Then wehave φ(P ) = E11 + . . . + En−1,n−1 + ϕn−1(P )Ep,q, P ∈ Pn−1n (D). It is trivialto check that each such map is injective and preserves order - all we need forthe verification is the basic fact that for every pair P,Q ∈ Pn(D) satisfyingP ≤ Q and P 6= Q we have rankP < rankQ. Composing such a map withthe transposition we obtain a map which is again an injective order preservingmap. So, besides standard maps we have two additional types of injective orderpreserving maps. If we further compose any of these maps with a similaritytransformation P 7→ TPT−1, we again arrive at an injective order-preservingmap. Any such map will be called a degenerate injective order preserving map.

The problem we are interested in is to describe the general form of orderpreserving maps on Pn(D). It has been known before that there is no hopefor a reasonable result without assuming a certain regularity assumption. Forthe sake of completeness we will illustrate this by a construction of wild orderpreserving maps that has been described already in [21]. We start with anymap φ : P 1

n(D) → Pn(D) and we will show that it can be extended to anorder preserving map. We first define φ(0) = 0. Denote by P≤kn (D) ⊂ Pn(D),1 ≤ k ≤ n − 1, the subset of all idempotents of rank at most k. Assumethat we have already extended φ to a map φ : P≤kn (D) → Pn(D) and that forevery P,Q ∈ P≤kn (D) the relation P ≤ Q yields that φ(P ) ≤ φ(Q). For everyP ∈ P k+1

n (D) we can find Q ∈ Pn(D) such that φ(R) ≤ Q for every R ∈ Pn(D)satisfying R ≤ P , R 6= P . Indeed, the choice Q = I works always but in generalwe have more freedom. We complete the inductive step by defining φ(P ) = Q.

Automorphisms of the poset of real or complex n× n idempotent matrices,n ≥ 3, were characterized by Ovchinnikov [18]. Note that in the case n = 2 anymap on P2(D) with the property rankφ(P ) = rankP , P ∈ P2(D), preservesorder. We call a division ring D an EAS-division ring if every endomorphism τ :D→ D is automatically surjective. Examples are the field of real numbers, thefield of rational numbers, and the division ring of quaternions, while the complexfield is not EAS [16]. It was proved in [21] that if D is an EAS division ring andn ≥ 3 an integer, then every injective order preserving map φ : Pn(D)→ Pn(D)is either standard, or it is an injective degenerate order preserving map. How

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essential is here the EAS assumption? In [21] a rather complicated example ofan injective order preserving map φ : P3(C) → P3(C) that is neither standardnor degenerate was given. After defining φ(0) = 0 and φ(I) = I one needs totake care only of rank one idempotents and rank two idempotents. But everyrank two idempotent can be written as I − P , where P is of rank one. Sowe need two injective maps ξ1, ξ2 : P1(C) → P1(C) and then we can defineφ(P ) = ξ1(P ) and φ(I −P ) = I − ξ2(P ) for every rank one idempotent P . It isthen rather easy to see that such a map will be an injective map preserving order(in both directions) if and only if the pair of maps ξ1, ξ2 preserves orthogonality(in both directions). It is not entirely trivial, but such maps can be constructed.The problem is that this counterexample is not satisfactory. There seems to beno way to extend this idea to higher dimensions. We already know that the2 × 2 case is exceptional and it would be possible that the 3 × 3 case is alsoexceptional because of the special structure described above (besides trivialidempotents 0 and I, the only non-trivial idempotents have either rank one,or corank one). Thus, at the time [21] was written it was not entirely clearwhether the assumption that D is an EAS division ring is indispensable in theabove theorem (except in the low-dimensional cases).

The study of order preserving maps on idempotents in both, the finite-dimensional (matrix) case and the infinite-dimensional (operator) case, was ini-tially motivated by an application in physics, see the review MR 95a:46093 of[18]. There are other applications in mathematical physics. Namely, Ovchin-nikov’s result recently proved to be useful in the study of quantum mechanicalinvariance transformations. Molnar [17] used it to considerably improve theclassical Wigner’s unitary-antiunitary theorem. The problem of characterizingorder preservers on the set of idempotents is closely related to the problem ofcharacterizing maps on idempotents preserving orthogonality. Solutions of thistype of problems can be applied in the study of homomorphism and local homo-morphisms of matrix and operator semigroups, see for example [20], [21], and[22]. The description of the general form of order preserving maps on idem-potents is of fundamental importance in the theory of general (not necessarilylinear) preservers on matrix and operator spaces. In particular, some of the cru-cial ideas used to obtain the optimal version of Hua’s fundamental theorem ofgeometry of matrices [13, 19, 23] were developed in [21]. Fundamental theoremsof geometry of matrices [5, 6, 7, 8, 9, 10, 11, 12] dealing with maps on matrixspaces preserving adjacency are closely related to geometry of algebraic homo-geneous spaces [1, 2]. In particular, the notion of adjacency on n × n matricescorresponds to the notion of adjacency on the set of finite points in Grass-mann space G(2n, n). In both cases adjacency is introduced with the use of thearithmetic distance. It turns out that idempotent n × n matrices can be char-acterized as a certain type of geometric midpoints between the zero matrix andthe identity matrix with respect to the arithmetic distance [23]. Examining thegroup of motions of the grassmannian G(2n, n) and the connections describedabove we have arrived at the following surprising example of an injective map

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φ : Mn(C)→Mn(C) preserving order in both directions. There exist an endo-morphism f of the complex field and an n× n complex matrix L such that thematrix I +LP f is invertible for every P ∈ Pn(C). Then (I +L)P f (I +LP f )−1

is an idempotent for every P ∈Mn(C) and the map

P 7→ (I + L)P f (I + LP f )−1, (3)

P ∈ Pn(C), is an injective map from Pn(C) into itself preserving order in bothdirections (the verification of all these facts will be given later).

The above example, showing that in the characterization of injective orderpreservers on idempotent matrices in [21] the EAS assumption is indispensable,motivated us to get back to this problem in the absence of the EAS assumption.In order to get a (necessarily more complicated) structural result for such mapsin this more general setting we need a completely different approach. Whilethe proofs in [21] were almost completely linear algebraic, we will combine herealgebraic and geometric approach. As a byproduct we will obtain much shorterproofs for the results on injective order preservers presented in [21].

In the next section we will introduce the notation, then give two more ex-amples, and formulate our main results. The third section will be devoted tosome technical lemmas. The proofs of main theorems will be given in the lastsection.

2 Notation, examples, and statement of mainresults

We denote by Dn the set of all 1 × n matrices. We will always consider it asa left vector space over D. Correspondingly, we have the right vector spaceof all n × 1 matrices tDn. Let A ∈ Mn(D). The row space of A is the leftvector subspace of Dn generated by the rows of A and the row rank of A is thedimension of this subspace. Similarly, the column rank of A is the dimensionof the right vector space generated by the columns of A. These two ranks areequal for every matrix over D and this common value is called the rank of amatrix.

Let k be a positive integer and x1, . . . , xk ∈ Dn. Then we denote by[x1, . . . , xk] the linear span of vectors x1, . . . , xk ∈ Dn. Similarly, for ty1, . . . ,

tyk ∈tDn, the symbol [ ty1, . . . ,

tyk] stands for the linear span of vectors ty1, . . . ,tyk

in the right vector space tDn. In particular, each element of the projective spaceP(Dn) over the left vector space Dn is of the form [x] for some nonzero x ∈ Dn,and similarly, P( tDn) = {[ ty] : ty ∈ tDn \ {0}}.

Let x = [x1 x2 . . . xn] ∈ Dn and let σ : D → D be any endomorphism oranti-endomorphism. Then we write xσ = [σ(x1) σ(x2) . . . σ(xn)] and similarly,for any ty we denote by tyσ the column matrix obtained from ty by applying σentrywise.

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Let a ∈ Dn and tb ∈ tDn be any nonzero vectors. Then tba = (tb)a is amatrix of rank one and every matrix of rank one can be written in this form.Obviously, a rank one matrix tba is an idempotent if and only if a tb = 1.

Assume that A,B ∈ Mn(D). We do not have t(AB) = tB tA in general,but it is easy to check that if τ is an anti-endomorphism of D then

t[(AB)τ ] = t(Bτ ) t(Aτ ).

As usual we identify n × n matrices with linear operators acting on Dn.More precisely, each n × n matrix A gives rise to a linear operator definedby x 7→ xA, x ∈ Dn. Then the rank of the matrix A is the dimension ofImA, the image of the corresponding operator A. The kernel of an operator A,KerA = {x ∈ Dn : xA = 0}, is the set of all vectors x ∈ Dn satisfying x(ty) = 0for every ty from the column space of A. Note that n = rankA+ dim KerA.

Further, each vector tb ∈ tDn can be considered as a linear functional onthe left D-linear space Dn given by

a 7→ a tb ∈ D, a ∈ Dn.

If U ⊂ Dn and V ⊂ tDn are any subsets, then U⊥ and V ⊥ are defined by

U⊥ = { ty ∈ tDn : u ty = 0 for all u ∈ U}

andV ⊥ = {x ∈ Dn : x tv = 0 for all tv ∈ V }.

Assume that P1, . . . , Pk ∈ Pn(D) are pairwise orthogonal, that is PiPj = 0whenever i 6= j. Denote by ri the rank of Pi. Then there exists an invertiblematrix A ∈Mn(D) such that for each i, 1 ≤ i ≤ k, we have

APiA−1 = diag (0, . . . , 0, 1, . . . , 1, 0, . . . , 0)

where diag (0, . . . , 0, 1, . . . , 1, 0, . . . , 0) is the diagonal matrix in which all thediagonal entries are zero except those in (r1 + . . .+ri−1 +1)st to (r1 + . . .+ri)throws [15, p.62, Exercise 1].

Let P,Q ∈ Pn(D). If P ≤ Q then clearly, Q−P is an idempotent orthogonalto P . Thus, by the previous paragraph, we have P ≤ Q, P 6= 0, Q 6= I, andP 6= Q if and only if there exist an invertible A ∈ Mn(D) and positive integersr1, r2 such that

APA−1 =

Ir1 0 00 0 00 0 0

and AQA−1 =

Ir1 0 00 Ir2 00 0 0

and 0 < r1 < r1 + r2 < n. In particular, if we identify matrices with linearoperators, then the image of P is a subspace of the image of Q, while the kernelof Q is a subspace of the kernel of P . Conversely, if the image of P is a subspace

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of the image of Q and the kernel of Q is a subspace of the kernel of P , thenP ≤ Q. This can be easily verified using the fact that if we identify an n × nidempotent matrix R with a linear operator, then Dn = ImR ⊕ KerR and Racts like the identity on ImR.

We are now ready to start with our examples. First we need the followingsimple statement.

Proposition 2.1 Let D be any division ring, n a positive integer, and ϕ :Pn(D) → Pn(D) an injective map preserving order (preserving order in bothdirections). Assume further that L ∈ Mn(D) is a matrix such that I + Lϕ(P )is invertible for every P ∈ Pn(D). Then

• (I +L)ϕ(P )(I +Lϕ(P ))−1 is an idempotent matrix for every P ∈ Pn(D),

• the map φ : Pn(D)→ Pn(D) defined by φ(P ) = (I+L)ϕ(P )(I+Lϕ(P ))−1,P ∈ Pn(D), preservers order (preserves order in both directions), and

• Imϕ(P ) = Imφ(P ) for every P ∈ Pn(D).

This section is devoted to the statement of our results, while the proofs willbe postponed to the next two sections. We will make an exception with theabove proposition since the proof is so short and simple. To prove the firstassertion we observe that for every P ∈ Pn(D) we have (I − ϕ(P ))ϕ(P ) = 0,and therefore the matrix

φ(P ) = (I + L)ϕ(P )(I + Lϕ(P ))−1 =

= (I + Lϕ(P ))ϕ(P )(I + Lϕ(P ))−1 + L(I − ϕ(P ))ϕ(P )(I + Lϕ(P ))−1

= (I + Lϕ(P ))ϕ(P )(I + Lϕ(P ))−1

is similar to idempotent ϕ(P ).We continue with the last assertion. We know that for every x ∈ Dn we

have

x ∈ Imφ(P ) ⇐⇒ xφ(P ) = x ⇐⇒ x(I + L)ϕ(P ) = x(I + Lϕ(P )).

Hence, x belongs to Imφ(P ) if and only if xϕ(P ) = x which is equivalent tox ∈ Imϕ(P ), as desired.

To prove the remaing assertion we will assume that ϕ preserves order in bothdirections. If it preserves order in one direction only, the same proof works withthe obvious modification that equivalences are replaced by implications in onedirection only. For any P,Q ∈ Pn(D) we have P ≤ Q if and only if ϕ(P ) ≤ ϕ(Q)which is futher equivalent to

Imϕ(P ) ⊂ Imϕ(Q) and Kerϕ(Q) ⊂ Kerϕ(P ).

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The first inclusion is equivalent to Imφ(P ) ⊂ Imφ(Q). Note that 0 ≤ E11 ≤E11 + E22 ≤ . . . ≤ I, and since ϕ preserves order and is injective we concludethat rankϕ(E11 + . . .+ Ekk) = k, k = 1, . . . , n, and in particular, ϕ(I) = I. Itfollows that I +L = I +Lϕ(I) is invertible. It is then straightforward to checkthat Ker ((I + L)ϕ(Q)) = (Kerϕ(Q))(I + L)−1. Hence, the second inclusionabove is equivalent to

Kerφ(Q) = Ker ((I + L)ϕ(Q)) = (Kerϕ(Q))(I + L)−1

⊂ (Kerϕ(P ))(I + L)−1 = Ker ((I + L)ϕ(P )) = Kerφ(P ).

This completes the proof.We know [16] that there exists an endomorphism f of the complex field such

that f(C) is a proper subfield of C. Take any complex number c satisfyingc 6∈ f(C). Set L = cE11. For any idempotent P = [pij ] ∈ Pn(C) the matrix

I + LP f =

1 + cf(p11) cf(p12) . . . cf(p1n)

0 1 . . . 0...

.... . .

...0 0 . . . 1

is invertible. Indeed, all we need to chack is that 1 + cf(p11) 6= 0. If 1 + cf(p11)was zero, then c would belong to f(C), a contradiction. It is clear that P 7→ P f

is an injective map from Pn(C) into itself that preserves order in both directions.Proposition 2.1 yields that any map of the form (3) is an injective map on Pn(C)preserving order in both directions and we have just verified that such mapsexist.

Lemma 2.2 There exist a field F, a proper subfield K ⊂ F, and two differentendomorphisms f, g : F→ F such that

• for every pair of positive integers n, r, 1 ≤ r < n, and for every n-tuple oflinearly independent vectors x1, . . . , xn ∈ Fn the n-tuple of vectors

xf1 , . . . , xfr , x

gr+1, . . . , x

gn

is also linearly independent, and

• f(F) ⊂ K and g(F) ⊂ K.

This technical lemma whose proof will be given in the next section is neededfor our next example. But first we need some more notation. Let F be a fieldand f : F → F a field endomorphism. We will denote by δf : Fn → Fn a mapdefined by δf (x) = xf . If f(F) is a proper subfield of F and {0} 6= U ⊂ Fn avector subspace, then δf (U) ⊂ Fn is not an F-linear subspace of Fn. By ∆f (U)we will denote the F-linear span of δf (U) in Fn. It is easy to verify that if

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dimU = r and the vectors x1, . . . , xr form a basis of U , then the set of vectorsxf1 , . . . , x

fr is a basis of ∆f (U).

For every pair of subspaces U, V ⊂ Fn we have

U ⊂ V ⇐⇒ ∆f (U) ⊂ ∆f (V ).

Let us verify just the fact that U 6⊂ V yields that ∆f (U) 6⊂ ∆f (V ), since theother direction is trivial. If U 6⊂ V , then

U = U1 ⊕ (U ∩ V ) and V = V1 ⊕ (U ∩ V )

for some subspaces U1, V1 with U1 6= {0}. Let x1, . . . , xp be a basis of U1,y1, . . . , yq a basis of U ∩ V , and z1, . . . , zr a basis of V1 with q, r nonnegativeintegers and p > 0. Then we know that

xf1 , . . . , xfp , y

f1 , . . . , y

fq , z

f1 , . . . , z

fr

are linearly independent and xf1 , . . . , xfp , y

f1 , . . . , y

fq is a basis of ∆f (U), while

yf1 , . . . , yfq , z

f1 , . . . , z

fr is a basis of ∆f (V ). Consequently, ∆f (U) 6⊂ ∆f (V ), as

desired.It follows that U 7→ ∆f (U) is an injective map from the set of all linear

subspaces of Fn into itselfLet P ∈ Pn(F) be any idempotent. Identifying the matrix P with linear

operator P on the vector space Fn we know that P is uniquely determined bytwo subspaces ImP and KerP satisfying Fn = ImP ⊕KerP .

For an arbitrary pair of complemented subspaces U, V ⊂ Fn, that is, Fn =U ⊕V , we denote by P (U, V ) the unique idempotent satisfying ImP (U, V ) = Uand KerP (U, V ) = V .

Example 2.3 Let a field F and field endomorphisms f, g : F → F be as inLemma 2.2. Let n be a positive integer. Then φ : Pn(F)→ Pn(F) defined by

φ(P (U, V )) = P (∆f (U),∆g(V )), P (U, V ) ∈ Pn(F),

is an injective map preserving order in both directions. Indeed, Lemma 2.2 yieldsthat φ is well-defined, that is, if we have Fn = U ⊕ V for a pair of subspacesU and V , then ∆f (U) and ∆g(V ) are complemented subspaces, too. We haveP (U1, V1) ≤ P (U2, V2) if and only if U1 ⊂ U2 and V2 ⊂ V1 and therefore theverification that φ is an injective map preserving order in both directions followstrivially from the above mentioned properties of the maps ∆f and ∆g.

So far we have presented two methods for constructing injective order pre-servers on Pn(F) that are neither standard nor degenerate. Combining them wewill arrive at our final example showing that our main result (Theorem 2.6) isthe best possible description of the general form of injective order preservers onthe poset of all n× n idempotents over an arbitrary division ring.

First we need an additional technical lemma.

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Lemma 2.4 Let F and φ : Pn(F)→ Pn(F) be as in Example 2.3. Then

φ(Pn(F)) ⊂ Pn(K).

Example 2.5 Let F and φ : Pn(F)→ Pn(F) be as in Example 2.3. Choose anyc ∈ F \K and set L = cE11. Finally, choose an invertible matrix A ∈ Mn(F).Then the map

P 7→ A(I + L)φ(P )(I + Lφ(P ))−1A−1, P ∈ Pn(F), (4)

is an injective map from Pn(F) into itself preserving order in both directions.Indeed, by the previous lemma we know that for each P ∈ Pn(F) all entries ofφ(P ) belong to K. Since c 6∈ K we conclude as in example (3) that I+Lφ(P ) isinvertible for every P ∈ Pn(F). Applying Proposition 2.1 and the fact that themap P 7→ APA−1 is an automorphism of Pn(F) we complete the verification ofthe desired properties of the map (4).

Motivated by the above examples we will now extentd the notion of a stan-dard map on Pn(D). So far we have considered elements of Pn(D) as matrices,then as linear operators, and finally as ordered pairs of complemented subspaces.For our purposes the best representation of an idempotent will be with a helpof a pair of dual r-tuples of vectors. Let r be a positive integer, 1 ≤ r < n,and x1, . . . , xr ∈ Dn and ty1, . . . ,

tyr ∈ tDn two linearly independent r-tuplesof vectors. Then we will denote by

P (x1, . . . , xr ; ty1, . . . ,tyr)

the idempotent whose image is the linear span of x1, . . . , xr and whose kernelis the set

[ ty1, . . . ,tyr]⊥ = {z ∈ Dn : z typ = 0 for all p = 1, . . . , r}. (5)

Of course, not all pairs of linearly independent r-tuples x1, . . . , xr ∈ Dn andty1, . . . ,

tyr ∈ tDn determine idempotents. If we want such a pair to correspondto some idempotent, then the linear span of x1, . . . , xr and the set (5) must becomplemented subspaces in Dn. Clearly, this happens if and only if any of thefollowing two (and hence, both) equivalent conditions is satisfied:

• For every α1, . . . , αr ∈ D not all zero there exists j ∈ {1, 2, . . . , r} suchthat (

r∑i=1

αixi

)tyj 6= 0.

• The linear functionals tyj : [x1, . . . , xr] → D, j = 1, . . . , r, are linearlyindependent.

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If x1, . . . , xr ∈ Dn and ty1, . . . ,tyr ∈ tDn are linearly independent r-tuples of

vectors satisfying the above two conditions, we say that they are dual r-tuplesof vectors. Hence, every idempotent P ∈ Pn(D) of rank r can be representedby a dual r-tuples of vectors.

Let A,B ∈Mn(D) be invertible matrices and τ, σ : D→ D endomorphisms.Assume that for every integer r, 1 ≤ r < n, and for every dual r-tuples ofvectors x1, . . . , xr ∈ Dn and ty1, . . . ,

tyr ∈ tDn the r-tuples

xτ1A, . . . , xτrA ∈ Dn and B tyσ1 , . . . , B

tyσr

are dual. Then it is easy to see that the map φ : Pn(D) → Pn(D) given byφ(0) = 0, φ(I) = I, and

φ(P (x1, . . . , xr ; ty1, . . . ,

tyr))

=

= P (xτ1A, . . . , xτrA ; B tyσ1 , . . . , B

tyσr )

is well-defined. Indeed, if

[x1, . . . , xr ] = [u1, . . . , ur ] and [ ty1, . . . ,tyr ]⊥ = [ tw1, . . . ,

twr ]⊥,

then it is straightforward to check that

[xτ1A, . . . , xτrA ] = [uτ1A, . . . , u

τrA ]

and[ B tyσ1 , . . . , B

tyσr ]⊥ = [ B twσ1 , . . . , Btwσr ]⊥.

Clearly, φ is injective and it preserves order in both directions.Similarly, if A,B ∈ Mn(D) are invertible matrices and τ, σ : D → D anti-

endomorphisms such that for every integer r, 1 ≤ r < n, and for every dualr-tuples of vectors x1, . . . , xr ∈ Dn and ty1, . . . ,

tyr ∈ tDn the r-tuples

yτ1A, . . . , yτrA ∈ Dn and B txσ1 , . . . , B

txσr

are dual, then the map φ : Pn(D)→ Pn(D) given by φ(0) = 0, φ(I) = I, and

φ(P (x1, . . . , xr ; ty1, . . . ,

tyr))

=

= P (yτ1A, . . . , yτrA ; B txσ1 , . . . , B

txσr )

is well-defined, injective and it preserves order in both directions. Each mapthat is of one of the above two forms will be called a generalized standard mapon Pn(D). Now we are ready to formulate our main result.

Theorem 2.6 Let D be any division ring, n ≥ 3, and assume that φ : Pn(D)→Pn(D) is an injective order preserving map. Then either φ is a generalizedstandard map, or φ is an injective degenerate order preserving map.

11

In particular, if φ : Pn(D) → Pn(D) is an injective order preserving map,then either it is degenerate, or it prerserves order in both directions.

We have several almost straightforward consequences. The first one is themain theorem from [21]:

Corollary 2.7 Let D be any EAS division ring, n ≥ 3, and assume thatφ : Pn(D) → Pn(D) is an injective order preserving map. Then either φ isa standard map, or φ is an injective degenerate order preserving map.

If we want to characterize standard maps we need stronger assumptionsthan in our main theorem. There are two natural possibilities. The first one isthat we replace order preserving property by a stronger supremum preservingcondition.

Corollary 2.8 Let D be any division ring, n ≥ 3, and let φ : Pn(D) → Pn(D)be an injective map. Assume that for every pair P,Q ∈ Pn(D) the existence ofthe supremum P ∨Q implies that the supremum φ(P ) ∨ φ(Q) exists and

φ(P ) ∨ φ(Q) = φ(P ∨Q).

Then φ is a standard map.

Of course, the assumption that supremum is preserved is stronger than theassumption that order is preserved. So, it is natural to ask if the injectivityassumption is indispensable in the last statement. The answer is positive as thefollowing example shows.

Example 2.9 Let ϕ : P 14 (R) → P 2

4 (R) be a map with the property that forevery pair of distinct rank one idempotents P,Q ∈ P 1

4 (R) we have Imϕ(P ) ∩Imϕ(Q) = {0}. Define φ : P4(R)→ P4(R) by φ(0) = 0, φ(P ) = ϕ(P ) for everyP ∈ P 1

4 (R) and φ(P ) = I for all idempotents P ∈ P4(R) of rank at least two.It is easy to verify that φ(P ) ∨ φ(Q) exists whenever P ∨ Q exists and that inthis case we have φ(P )∨ φ(Q) = φ(P ∨Q). However, φ is not a standard map.

Finally, we can consider Pn(D) as an orthomodular poset. Then we havethe following.

Corollary 2.10 Let D be any division ring, n ≥ 3, and assume that φ :Pn(D) → Pn(D) is an injective order preserving map. Suppose further thatfor every P ∈ Pn(D) we have φ(I − P ) = I − φ(P ). Then φ is a standard map.

Again we will see that the injectivity assumption is indispensable.

Example 2.11 Let D be a division ring, n ≥ 3 an integer, and Q a fixedidempotent from Pn(D). We define φ : Pn(D) → Pn(D) by φ(P ) = 0 wheneverP ≤ Q, φ(P ) = I whenever I −Q ≤ P , and φ(P ) = P otherwise. It is easy tosee that φ preserves order in one direction and φ(I − P ) = I − φ(P ) for everyP ∈ Pn(D).

12

3 Preliminary results

Our main results will be proved in the last section. In this section we will provesome technical results and two lemmas from the previous section.

Proof of Lemma 2.2. Let Q be the field of rational numbers. Choose a countableset of real numbers T = {tk : k = 1, 2, . . .} that are algebraically independentover Q. Set F = Q(T ).

Every element of F is of the form

p(t1, . . . , tm)

q(t1, . . . , tm)

for some positive integer m and some polynomials p, q ∈ Q[X1, . . . , Xm] withq 6= 0. Define f, g : F→ F by

f

(p(t1, . . . , tm)

q(t1, . . . , tm)

)=p(t2, t4, . . . , t2m)

q(t2, t4, . . . , t2m)

and

g

(p(t1, . . . , tm)

q(t1, . . . , tm)

)=p(t3, t5, . . . , t2m+1)

q(t3, t5, . . . , t2m+1).

Using the fact that T is a transcendental basis of F over Q one can easilyconclude that both maps f and g are well-defined and that they are field endo-morphisms.

Set K = Q({t2, t3, . . .}). Then clearly, f(F) ⊂ K and g(F) ⊂ K.Choose now positive integers n, r, 1 ≤ r < n, and an n-tuple of linearly

independent vectors x1, . . . , xn ∈ Fn and assume that

xf1 , . . . , xfr , x

gr+1, . . . , x

gn

are linearly dependent. In order to complete the proof we need to show thatthis leads to a contradiction.

We write x1 = [x11 x12 . . . x1n], x2 = [x21 x22 . . . x2n],..., and xn = [xn1 xn2 . . . xnn].There exist a positive integer m and polynomials pij , qij ∈ Q[X1, . . . , Xm],1 ≤ i, j ≤ n, with qij 6= 0 for all i, j, 1 ≤ i, j ≤ n, such that

xij =pij(t1, . . . , tm)

qij(t1, . . . , tm), 1 ≤ i, j ≤ n.

Our assumption that xf1 , . . . , xfr , x

gr+1, . . . , x

gn are linearly dependent is equiva-

13

lent to the fact that the determinant∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

p11(t2,...,t2m)q11(t2,...,t2m)

p12(t2,...,t2m)q12(t2,...,t2m) . . . p1n(t2,...,t2m)

q1n(t2,...,t2m)

......

. . ....

pr1(t2,...,t2m)qr1(t2,...,t2m)

pr2(t2,...,t2m)qr2(t2,...,t2m) . . . prn(t2,...,t2m)

qrn(t2,...,t2m)pr+1,1(t3,...,t2m+1)qr+1,1(t3,...,t2m+1)

pr+1,2(t3,...,t2m+1)qr+1,2(t3,...,t2m+1)

. . .pr+1,n(t3,...,t2m+1)qr+1,n(t3,...,t2m+1)

......

. . ....

pn1(t3,...,t2m+1)qn1(t3,...,t2m+1)

pn2(t3,...,t2m+1)qn2(t3,...,t2m+1)

. . . pnn(t3,...,t2m+1)qnn(t3,...,t2m+1)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣is zero.

Multiplying this equality by∏1≤i≤r,1≤j≤n

qij(t2, . . . , t2m)∏

r+1≤i≤n,1≤j≤n

qij(t3, . . . , t2m+1) 6= 0

we get a polynomial equation

F (t2, t3, . . . , t2m+1) = 0

where F ∈ Q[X1, X2, . . . , X2m]. But t2, . . . , t2m+1 are algebraically indepen-dent, and consequently, F = 0. In particular,

F (t1, t1, t2, t2, . . . , tm, tm) = 0,

which futher yields that∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

p11(t1,...,tm)q11(t1,...,tm)

p12(t1,...,tm)q12(t1,...,tm) . . . p1n(t1,...,tm)

q1n(t1,...,tm)

......

. . ....

pr1(t1,...,tm)qr1(t1,...,tm)

pr2(t1,...,tm)qr2(t1,...,tm) . . . prn(t1,...,tm)

qrn(t1,...,tm)pr+1,1(t1,...,tm)qr+1,1(t1,...,tm)

pr+1,2(t1,...,tm)qr+1,2(t1,...,tm) . . .

pr+1,n(t1,...,tm)qr+1,n(t1,...,tm)

......

. . ....

pn1(t1,...,tm)qn1(t1,...,tm)

pn2(t1,...,tm)qn2(t1,...,tm) . . . pnn(t1,...,tm)

qnn(t1,...,tm)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 0

or equivalently, x1, . . . , xn are linearly dependent, a contradiction.

2

Proof of Lemma 2.4. Let P (U, V ) be any idempotent in Pn(F). Denote Q =P (∆f (U),∆g(V )). We need to show that all entries of Q belong to K. Chooseany n-tuple of vectors x1, . . . , xr, xr+1, . . . , xn ∈ Fn such that x1, . . . , xr form abasis of U and xr+1, . . . , xn form a basis of V . Let A be the n×n matrix whosej-th row equals

• xfj , if 1 ≤ j ≤ r,

14

• xgj , if r + 1 ≤ j ≤ n.

Then clearly, A is an invertible matrix with entries in the subfield K. It followsthat A−1 ∈Mn(K), too.

Let e1, . . . , en be the standard basis of Fn. From ejA = xfj , 1 ≤ j ≤ r,

ejA = xgj , r+ 1 ≤ j ≤ n, xfjQ = xfj , 1 ≤ j ≤ r, and xgjQ = 0, r+ 1 ≤ j ≤ n, weconclude that

AQA−1 =

[Ir 00 0

],

and consequently, Q = A−1[Ir 00 0

]A ∈Mn(K).

2

The following lemma on the existence of the supremum of two idempotentmatrices might be of some independent interest.

Lemma 3.1 Let D be any division ring and P,Q ∈ Pn(D). Then P ∨Q existsif and only if one of the following two conditions is satisfied:

• KerP ∩KerQ ⊂ ImP + ImQ,

• Dn = (ImP + ImQ)⊕ (KerP ∩KerQ).

Proof. Assume first that KerP∩KerQ ⊂ ImP+ImQ and let R ∈ Pn(D) be anyidempotent such that P ≤ R and Q ≤ R. Then ImP ⊂ ImR and ImQ ⊂ ImR.Thus,

ImP + ImQ ⊂ ImR. (6)

Similarly,KerR ⊂ KerP ∩KerQ. (7)

It follows that KerR ⊂ ImR. This together with Dn = ImR ⊕ KerR yieldsthat

KerR = ImR ∩KerR = {0},

which further implies that R = I. We conclude ther the identity is the onlyidempotent that is above P and above Q, and consequently, P ∨Q = I.

We assume now that Dn = (ImP +ImQ)⊕ (KerP ∩KerQ). Let T ∈ Pn(D)be the idempotent whose image is ImP+ImQ and whose kernel is KerP∩KerQ.We claim that T = P ∨ Q. Clearly, P ≤ T and Q ≤ T . For any R satisfyingP ≤ R and Q ≤ R we have (6) and (7), and therefore, T ≤ R, as desired.

Finally, assume that P ∨ Q exists. Set R = P ∨ Q. If KerP ∩ KerQ ⊂ImP + ImQ, we are done. So assume that

KerP ∩KerQ 6⊂ ImP + ImQ. (8)

15

Hence,V = (ImP + ImQ) ∩ (KerP ∩KerQ)

is a proper subspace of KerP ∩KerQ. From

V ⊂ ImP + ImQ ⊂ ImR

we conclude that V ∩KerR = {0}. We further know that KerR ⊂ KerP∩KerQ.We claim that V = {0} and KerR = KerP ∩ KerQ. If this was not true

than it would be possible to find a subspace W ⊂ KerP ∩KerQ such that

KerP ∩KerQ = V ⊕W and W 6⊂ KerR.

Let Z be any subspace complemented to (ImP + ImQ)⊕W . Then

Dn = (ImP + ImQ)⊕W ⊕ Z.

Let T ∈ Pn(D) be the idempotent whose image is (ImP + ImQ)⊕Z and whosekernel is W . Obviously we have P ≤ T and Q ≤ T , and consequently, R ≤ Timplying that W ⊂ KerR, a contradiction.

Hence, V = {0} and KerR = KerP ∩ KerQ, and because ImP + ImQ ⊂ImR we have

Dn = (ImP + ImQ)⊕ (KerP ∩KerQ)⊕ U

andImR = (ImP + ImQ)⊕ U

for some subspace U ∈ Dn. If U was a nonzero subspace, then because KerP ∩KerQ is also nonzero (see (8)) we would be able to find a pair of nonzero vectorsu ∈ U and v ∈ KerP∩KerQ. Obviously, u+v 6∈ (ImP+ImQ)⊕(KerP∩KerQ)and therefore there would exist a subspace U1 ∈ Dn such that u+ v ∈ U1 and

Dn = (ImP + ImQ)⊕ (KerP ∩KerQ)⊕ U1.

Then the idempotent S whose image is (ImP + ImQ)⊕U1 and whose kernel isKerP ∩ KerQ would satisfy P ≤ S and Q ≤ S which would imply R ≤ S andconsequently,

(ImP + ImQ)⊕ U ⊂ (ImP + ImQ)⊕ U1.

But these two subspaces are of the same dimension yielding that the aboveinclusion is actually the equality. This would mean that u+ v ∈ U1 ⊂ (ImP +ImQ)⊕ U , a contradiction. This completes the proof.

2

We continue with three simple linear algebra statements.

16

Lemma 3.2 Let n, k be integers, n ≥ 3 and 1 ≤ k < n− 1, and P,Q ∈ Pn(D)two idempotents of rank n − 1 such that ImP = ImQ. Then there exist twoidempotents R1, R2 ∈ Pn(D) of rank k such that R1 6= R2 and R1, R2 ≤ P andR1, R2 ≤ Q.

Proof. There is nothing to prove if P = Q. So, assume that P 6= Q. After re-placing P and Q by TPT−1 and TQT−1, respectively, where T is an appropriateinvertible n× n matrix, we can assume with no loss of generality that

P =

[I 00 0

]and Q =

[I 0

en−1 0

],

where I stands for the (n−1)× (n−1) identity matrix and en−1 = [0 . . . 0 1] ∈Dn−1.

We can take R1 = En−k−1,n−k−1 + En−k,n−k + . . . + En−2,n−2 and R1 =En−k−1,n−k−1 +En−k,n−k + . . .+En−2,n−2 +En−2,n−1 to complete the proof.

2

Lemma 3.3 Let n, k be integers, n ≥ 3 and 1 ≤ k < n, and P,Q ∈ Pn(D)two idempotents of rank n − 1 such that KerP = KerQ. Then there exist twoidempotents R1, R2 ∈ Pn(D) of rank k such that R1 6= R2 and R1, R2 ≤ P andR1, R2 ≤ Q.

Proof. The proof is a slight modification of the proof of the previous lemma.

2

Let P,Q ∈ Pn(D) be two idempotents of rank n − 1. We will write P ∼ Qif ImP = ImQ or KerP = KerQ.

Lemma 3.4 Let P,Q ∈ Pn(D) be two idempotents of rank n − 1. Then thereexist idempotents R,S of rank n− 1 such that P ∼ R, R ∼ S, and S ∼ Q.

Proof. Since ImP ∪ ImQ 6= Dn we can find a nonzero vector u ∈ Dn such thatu 6∈ ImP ∪ ImQ. It is clear that idempotents R and S defined by ImP = ImRand KerR = [u] and ImQ = ImS and KerS = [u] have the desired properties.

2

The following lemma has been known before. As the proof is very short wewill repeat it here for the sake of completeness.

Lemma 3.5 Let D be a division ring, n an integer ≥ 3, and φ : Pn(D) →Pn(D) an injective order preserving map. Then for every P ∈ Pn(D) we haverankφ(P ) = rankP . In particular, φ(0) = 0 and φ(I) = I.

17

Proof. Let rankP = r. Then we can find idempotents P0, P1, . . . , Pn withP0 = 0 ≤ P1 ≤ . . . ≤ Pn−1 ≤ Pn = I, rankPk = k, k = 0, 1, . . . , n, andP = Pr. From φ(0) ≤ φ(P1) ≤ . . . ≤ φ(Pn) and injectivity of φ we concludethat rankφ(P ) = r, as desired.

2

Lemma 3.6 Let D be an EAS division ring and n an integer, n ≥ 3. Thenevery generalized standard map φ : Pn(D)→ Pn(D) is a standard map.

Proof. Let φ be a generalized standard map. We have two possibilities and wewill consider just one of them. So, assume that A,B ∈ Mn(D) are invertiblematrices and τ, σ : D → D anti-endomorphisms such that for every integerr, 1 ≤ r < n, and for every dual r-tuples of vectors x1, . . . , xr ∈ Dn andty1, . . . ,

tyr ∈ tDn the r-tuples

yτ1A, . . . , yτrA ∈ Dn and B txσ1 , . . . , B

txσr

are dual and the map φ : Pn(D)→ Pn(D) is given by φ(0) = 0, φ(I) = I, and

φ(P (x1, . . . , xr ; ty1, . . . ,

tyr))

=

= P (yτ1A, . . . , yτrA ; B txσ1 , . . . , B

txσr ).

Observe first that τ2 is an endomorphism, and therefore surjective by the EASassumption. So, τ is bijective and the same is true for σ.

We know that for every pair of vectors x ∈ Dn and ty ∈ tDn satisfyingx ty = 1 we have yτAB txσ 6= 0. Then, in particular

[ 1 0 . . . 0 ] AB

1

σ(x2)...

σ(xn)

6= 0

for every x2, . . . , xn ∈ D. Applying the fact that σ is surjective we easilyconclude that all off diagonal entries in the first row of AB must be zero. Thesame holds true for other rows, and so, AB is a diagonal matrix with diagonalentries d1, . . . , dn. Now, for every λ ∈ D we have

[ 1− τ(λ) τ(λ) 0 . . . 0 ] AB

110...0

6= 0.

18

A straightforward computation then shows that d1 = d2. In the same waywe prove that d1 = d2 = . . . = dn = d. So, if x1, . . . , xn, y1, . . . , yn ∈ D arearbitrary elements satisfying x1y1 + . . .+ xnyn = 1, then

τ(y1)dσ(x1) + . . .+ τ(yn)dσ(xn) 6= 0.

Applying τ−1 on both sides of this equation and denoting τ−1 ◦ σ = η andτ−1(d) = s we get

η(x1)sy1 + . . .+ η(xn)syn 6= 0.

Choosing x1 = 1, x2 = γ, x3 = . . . = xn = 0, y1 = 1 − γδ, y2 = δ, y3 = . . . =yn = 0 we arrive at

s 6= (sγ − η(γ)s)δ

for every pair γ, δ ∈ D. Thus, η(γ) = sγs−1, γ ∈ D, or equivalently, σ(γ) =d−1τ(γ)d, γ ∈ D. It follows that for every pair of vectors x = [x1 . . . xn ] ∈Dn and ty = t [ y1 . . . yn ] ∈ tDn satisfying x ty = 1 we have

yτAB txσ = [ τ(y1) . . . τ(yn) ] dI

d−1τ(x1)d

...d−1τ(xn)d

= τ(x1y1 + . . .+ xnyn)d = d.

Therefore for every rank one idempotent tyx, x ty = 1, we have

φ( tyx) = φ(P (x ; ty)) = P (yτA ; B txσ) = B txσ(yτAB txσ)−1yτA =

= A−1dI

d−1τ(x1)d

...d−1τ(xn)d

d−1 [ τ(y1) . . . τ(yn) ]A

= A−1

τ(x1)...

τ(xn)

[ τ(y1) . . . τ(yn) ]A = A−1 t( tyx)τA.

Hence, the map Φ : Pn(D)→ Pn(D) defined by

Φ(P ) = Aφ(t(P τ

−1))

A−1, P ∈ Pn(D),

is an injective order preserving map sending every rank one idempotent intoitself. Using Lemma 3.5 it is trivial to conclude that Φ is the identity. Thus, φis standard.

2

19

The main tool in our proof will be a recently obtained non-surjective versionof the fundamental theorem of projective geometry [4]. We will present here aweaker version for the projective space over the left vector space Dn. Recallthat a map f : Dn → Dn is called semilinear if it is additive and there exists anendomorphism τ : D→ D (note that we do not assume that τ is surjective) suchthat f(λx) = τ(λ)f(x) for every pair λ ∈ D and x ∈ Dn. Let ξ : P(Dn)→ P(Dn)be a map. We say that its image is contained in a line if there exist nonzerovectors u, v ∈ Dn such that ξ([x]) ⊂ [u] + [v] for every nonzero x ∈ Dn. Thenthe non-surjective version of the fundamental theorem of projective geometry[4] reads as follows: Let ξ : P(Dn) → P(Dn) be an injective map whose imageis not contained in a line. Assume further that for all x, y, z ∈ Dn \ {0} we have

[x] ⊂ [y] + [z]⇒ ξ([x]) ⊂ ξ([y]) + ξ([z]).

Then there exists an injective semilinear map f : Dn → Dn such that

ξ([x]) = [f(x)], x ∈ Dn \ {0}.

Let A be the n × n matrix whose k-th row is f(ek), k = 1, . . . , n. Here,e1, . . . , en is the standard basis of Dn. Then we have

f(x) = xτA, x ∈ Dn.

Indeed, this is trivially true for x ∈ {e1, . . . , en} and for all other vectors x ∈ Dnthe above equality holds by the semilinearity of f . Note that A is not invertiblein general. However, the map x 7→ xτA, x ∈ Dn, is injective, that is, xτA 6= 0for all nonzero vectors x.

One can now easily formulate an analogous statement for maps on projectivespace ove the right vector space tDn.

We will need also “right-left” and “left-right” versions of the above theorem.The idea of the proof is the same as in [4]. As there are some minor changes andsince the proofs are rather short, we will, for the sake of completeness, proveone of these two statements, and then, of course, give the other one withoutproof.

Let ξ : P(Dn) → P( tDn) be a map. We say that its image is contained ina line if there exist nonzero vectors tu, tv ∈ tDn such that ξ([x]) ⊂ [ tu] + [ tv]for every nonzero x ∈ Dn.

Theorem 3.7 Let ξ : P(Dn) → P( tDn) be an injective map whose image isnot contained in a line. Assume further that for all x, y, z ∈ Dn \ {0} we have

[x] ⊂ [y] + [z]⇒ ξ([x]) ⊂ ξ([y]) + ξ([z]).

Then there exist an n×n matrix A and an anti-endomorphism σ : D→ D suchthat A txσ 6= 0 for all nonzero vectors x ∈ Dn and

ξ([x]) = [A txσ], x ∈ Dn \ {0}.

20

Proof. The following observation will be used several times. If x, y, z ∈ Dn andtu, tv, tw ∈ tDn are nonzero vectors such that

• ξ([x]) = [ tu], ξ([y]) = [ tv], ξ([z]) = [ tw],

• tu, tv, tw are linearly independent, and

• ξ([x+ y]) = [ tu+ tv], ξ([x+ z]) = [ tu+ tw],

then ξ([y + z]) = [ tv + tw] and ξ([x + y + z]) = [ tu + tv + tw]. Indeed, theseare straightforward consequences of our assumptions and relations

[x+ y + z] ⊂ ([x+ y] + [z]) ∩ ([x+ z] + [y])

and[y + z] ⊂ ([y] + [z]) ∩ ([x] + [x+ y + z]).

Whenever using the above observation we will refer to the first observation,while the second observation that we will also use several times is the followingtrivial statement. Assume that tu, tv, tw ∈ tDn are nonzero vectors such that

tu ∈ [ tv] + [ tw] and tu 6∈ [ tw] and tu 6∈ [ tv].

Then there is a unique nonzero tw0 ∈ [ tw] such that

[ tv] = [ tu+ tw0].

According to our assumptions we can find vectors a1, a2, a3 ∈ Dn such thatξ([a1]), ξ([a2]), ξ([a3]) are not contained in some line. We claim that we canfind tb1,

tb2,tb3 ∈ tDn such that ξ([ai]) = [ tbi], i = 1, 2, 3, and ξ([ai + aj ]) =

[ tbi + tbj ] whenever i 6= j. To prove this we first choose a nonzero tb1 ∈ tDnsuch that ξ([a1]) = [ tb1]. From

[a1] ⊂ [a1 + a2] + [a2]

we conclude thattb1 ∈ ξ([a1 + a2]) + ξ[a2]

and then the second observation yields the existence of a nonzero tb2 ∈ tDn suchthat ξ([a2]) = [ tb2] and ξ([a1 + a2]) = [ tb1 + tb2]. In the same way we see thatthere is a nonzero tb3 ∈ tDn such that ξ([a3]) = [ tb3] and ξ([a1 + a3]) = [ tb1 +tb3]. The first observation implies that we have also ξ([a2 + a3]) = [ tb2 + tb3],as desired.

We will now define a map f : Dn → tDn. We first set f(0) = t0. For anonzero x ∈ Dn we choose ai among a1, a2, a3 such that x and ai are linearlyindependent. We have

[ tbi] = ξ([ai]) ⊂ ξ([ai + x]) + ξ([x])

21

and by the second observation there is a unique nonzero f(x) ∈ tDn such that

ξ([x]) = [f(x)] and ξ([ai + x]) = [ tbi + f(x)].

We first need to check that f is well-defined, that is, the value of f(x)is independent of the choice of ai. We consider the case when x and ai arelinearly independent, i = 1, 2. Then there is a unique f(x) ∈ tDn such thatξ([x]) = [f(x)] and ξ([a1 + x]) = [ tb1 + f(x)]. We need to show that we havealso ξ([a2 + x]) = [ tb2 + f(x)]. If f(x) 6∈ [ tb1] + [ tb2], then the desired equalityholds by the first observation. Otherwise, we consider successively the triplesa1, a3, x and a3, a2, x.

Our next goal is to prove that f is additive, that is, f(x+ y) = f(x) + f(y),x, y ∈ Dn. There is nothing to prove if x = 0 or y = 0. So, assume they areboth nonzero vectors.

Let us first consider the case when x and y are linearly independent. Byinjectivity, ξ([x]) 6= ξ([y]). We choose tbi among tb1,

tb2,tb3 such that tbi 6∈

ξ([x]) + ξ([y]). From

ξ([ai + x]) = [ tbi + f(x)] and ξ([ai + y]) = [ tbi + f(y)]

and the first observation we conlude that

ξ([x+ y]) = [f(x) + f(y)] and ξ([ai + x+ y]) = [ tbi + f(x) + f(y)]

yielding that f(x+ y) = f(x) + f(y).If x and y are linearly dependent, then we can find z ∈ Dn such that x and

z, and hence, y and z are linearly independent. If x+ y 6= 0, then x+ y and zare linearly independent, and by the previous paragrpah we have

f(x+ y + z) = f(x+ y) + f(z).

This is obviously true also when x+ y = 0. Using the previous paragraph twicemore we get

f(x+ y + z) = f(x) + f(y + z) = f(x) + f(y) + f(z),

and consequently, f(x+ y) = f(x) + f(y) in this case as well.By definition we have ξ([x]) = [f(x)], x ∈ Dn, and therefore, for every

nonzero x ∈ Dn there exists a function σx : D→ D such that

f(λx) = f(x)σx(λ), λ ∈ D, and σx(0) = 0.

Since f is additive, each σx is additive, too. We claim that σx ≡ σ is independentof x. Indeed, if x and y are linearly independent, then, by injectivity of ξ, f(x)and f(y) are linearly independent, too. For every λ ∈ D we have

f(x)σx(λ) + f(y)σy(λ) = f(λx) + f(λy) = f(λ(x+ y))

22

= f(x+ y)σx+y(λ) = f(x)σx+y(λ) + f(y)σx+y(λ),

and therefore σx = σx+y = σy. If x and y are nonzero and linearly dependent,then we find z ∈ Dn that is linearly independent of x, and then, by the previousstep, we have σx = σz = σy.

Fromf(λx) = f(x)σ(λ), λ ∈ D, x ∈ Dn,

we get σ(1) = 1 and for any pair λ, µ ∈ D we have

f(x)σ(λµ) = f(λµx) = f(µx)σ(λ) = f(x)σ(µ)σ(λ).

Hence, σ : D→ D is an anti-endomorphism.Let A be the n× n matrix whose k-th column is f(ek), k = 1, . . . , n. Then

we havef(x) = A txσ x ∈ Dn.

This completes the proof.

2

Theorem 3.8 Let ξ : P( tDn)→ P(Dn) be an injective map whose image is notcontained in a line. Assume further that for all tx, ty, tz ∈ tDn \ {0} we have

[ tx] ⊂ [ ty] + [ tz]⇒ ξ([ tx]) ⊂ ξ([ ty]) + ξ([ tz]).

Then there exist an n×n matrix A and an anti-endomorphism σ : D→ D suchthat xσA 6= 0 for all nonzero vectors tx ∈ tDn and

ξ([ tx]) = [xσA], tx ∈ tDn \ {0}.

4 Proofs of the main results

We will first prove our main result. The special case when D is an EAS divisionring was proved in [21]. Our goal was to solve the general case but at that timewe were not aware of examples given in this paper and we were not even able toformulate a reasonable conjecture for the general case. Anyway, we formulatedthe first few steps of the proof for general division rings. So, we can start withusing Lemmas 2.3, 2.4, and 2.5 from [21]. The rest of the proof for the generalcase presented in this paper will be much shorter than the proof for the specialEAS case in [21].

We need some more notation. For nonzero vectors x ∈ Dn and ty ∈ tDn wedenote by R(x) and L( ty) the subsets of P 1

n(D) defined by

R(x) = {P (x ; tu) : tu ∈ tDn, x tu 6= 0}

andL( ty) = {P (v ; ty) : v ∈ Dn, v ty 6= 0}.

23

Clearly, if nonzero vectors x1, x2 ∈ Dn are linearly dependent, then R(x1) =R(x2), while in the case that x1 and x2 are linearly independent we have R(x1)∩R(x2) = ∅. We have an analogous statement for sets L( ty), ty ∈ tDn.

Proof of Theorem 2.6. By Lemma 3.5 we have rankφ(P ) = rankP for everyP ∈ Pn(D). Lemma 2.3 from [21] tells that for every nonzero ty ∈ tDn eitherthere exists a nonzero x ∈ Dn such that φ(L( ty)) ⊂ R(x), or there exists anonzero tw ∈ tDn such that φ(L( ty)) ⊂ L( tw).

From now we will assume that for some nonzero ty ∈ tDn we have the firstof the two possibilities above. Then by [21, Lemma 2.4], for every nonzerotw ∈ tDn we have

φ(L( tw)) ⊂ R(u)

for some nonzero u ∈ Dn. The other possibility, that is, for every nonzerotw ∈ tDn we have φ(L( tw)) ⊂ L( tz) for some tz ∈ tDn, will not be treated inthe paper because the proof goes through in exactly the same way.

We use [21, Lemma 2.5] to conclude that either for every nonzero ty ∈ tDnand every nonzero x ∈ Dn we have

φ(L( ty)) ⊂ R(u) and φ(R(x)) ⊂ L( tv) (9)

for some nonzero u ∈ Dn and some nonzero tv ∈ tDn, or there exists a nonzeroz ∈ Dn such that

φ(P 1n(D)) ⊂ R(z).

Of course, we will prove that in the first case φ is a generalized standard map,while in the second case φ is degenerate.

Let us start with the first case, that is, we have (9). Then obviously, themap φ induces two maps ξ : P(Dn) → P( tDn) and µ : P( tDn) → P(Dn) suchthat for every nonzero x ∈ Dn and every nonzero tv ∈ tDn we have

φ(R(x)) ⊂ L( tv) ⇐⇒ ξ([x]) = [ tv]

and for every nonzero ty ∈ tDn and every nonzero u ∈ Dn we have

φ(L( ty)) ⊂ R(u) ⇐⇒ µ([ ty]) = [u].

We claim that ξ satisfies all the assumptions of Theorem 3.7, that is, it is aninjective map whose image is not contained in a line, and for all x, y, z ∈ Dn\{0}we have

[x] ⊂ [y] + [z]⇒ ξ([x]) ⊂ ξ([y]) + ξ([z]).

Assume first that ξ is not injective. This would imply that there exist linearlyindependent vectors x, z ∈ Dn such that

φ(R(x)) ⊂ L( tv) and φ(R(z)) ⊂ L( tv)

24

for some nonzero tv ∈ tDn. As x and z are linearly independent we can findty ∈ tDn such that

x ty = z ty = 1.

By (9) there exists a nonzero u ∈ Dn such that φ(L( ty)) ⊂ R(u). It followsthat

φ( tyx) = φ( tyz) = P (u ; tv),

contradicting the injectivity of φ. Hence, ξ must be injective. In te same waywe prove that µ is injective.

Assume next that the image of ξ is contained in some line, that is, thereexist nonzero vectors tu, tv ∈ tDn such that

ξ([x]) ⊂ [ tu] + [ tv] (10)

for every nonzero x ∈ Dn. Again, we will show that this contradicts the injec-tivity of φ. Indeed, take any three linearly independent vectors x1, x2, x3 ∈ Dn.Then we can find ty1,

ty2 ∈ tDn such that

x1ty1 = 1 and x1

ty2 = 0

andx2

ty1 = 0 and x2ty2 = 1

andx3

ty1 = 0 and x3ty2 = 1.

Of course, ty1 and ty2 are linearly independent.Denote

P = P (x1, x2 ; ty1,ty2) and Q = P (x1, x3 ; ty1,

ty2).

Then P and Q are unique idempotents of rank two satisfying

P (x1 ; ty1), P (x2 ; ty2) ≤ P

andP (x1 ; ty1), P (x3 ; ty2) ≤ Q.

We choose nonzero vectors tu1,tu2,

tu3 ∈ tDn and nonzero vectors z1, z2 ∈ Dnsuch that

ξ([xi]) = [ tui], i = 1, 2, 3, and µ([ tyj ]) = [zj ], j = 1, 2.

Then

φ(P (x1 ; ty1)) = P (z1 ; tu1), φ(P (x2 ; ty2)) = P (z2 ; tu2),

andφ(P (x3 ; ty2)) = P (z2 ; tu3).

25

It follows thatP (z1 ; tu1), P (z2 ; tu2) ≤ φ(P ) (11)

andP (z1 ; tu1), P (z2 ; tu3) ≤ φ(Q). (12)

Any pair of vectors tui,tuj , i 6= j, is linearly independent and by (10) we have

span {tui, tuj} = span {tu, tv}

whenever i 6= j. Since µ is injective, the vectors z1 and z2 are linearly indepen-dent, too. Hence, (11) and (12) yield that

φ(P ) = P (z1, z2 ; tu, tv) = φ(Q),

a contradiction.Assume finally that x, y, z ∈ Dn \ {0} satisfy [x] ⊂ [y] + [z]. If y and

z are linearly dependent, then [x] = [y] = [z], and therefore, the inclusionξ([x]) ⊂ ξ([y]) + ξ([z]) is trivially true. So, assume that y and z are linearlyindependent. After choosing vectors tu, tv ∈ tDn such that

y tu = z tv = 1 and y tv = z tu = 0

we can find scalars α, β ∈ D such that

x( tuα+ tvβ) = 1.

Clearly,

P (y ; tu), P (z ; tv), P (x ; tuα+ tvβ) ≤ P (y, z ; tu, tv).

BecauseKerφ(P (y, z ; tu, tv)) ⊂ Kerφ(P (y ; tu)) = ξ([y])⊥

and similarly, Kerφ(P (y, z ; tu, tv)) ⊂ ξ([z])⊥, and since φ(P (y, z ; tu, tv)) isof rank two, we have

Kerφ(P (y, z ; tu, tv)) = ξ([y])⊥ ∩ ξ([z])⊥.

On the other hand,

Kerφ(P (y, z ; tu, tv)) ⊂ Kerφ(P (x ; tuα+ tvβ)) = ξ([x])⊥,

and therefore, ξ([x]) ⊂ ξ([y]) + ξ([z]), as desired.Hence, ξ satisfies all the assumptions of Theorem 3.7 and in the same way

we verify that µ satisfies all the assumptions of Theorem 3.8. It follows thatthere exist n×n matrices A,B and anti-endomorphisms σ, τ : D→ D such that

• A txσ 6= 0 for all nonzero vectors x ∈ Dn, and

26

• yτB 6= 0 for all nonzero vectors ty ∈ tDn, and

• ξ([x]) = [A txσ] for all nonzero x ∈ Dn, and

• µ([ ty]) = [yτB] for all nonzero ty ∈ tDn.

This further yields that for every pair of vectors x ∈ Dn and ty ∈ tDnsatisfying x ty 6= 0 we have

yτBA txσ 6= 0

andφ(P (x ; ty)) = P (yτB ; A txσ).

We will prove inductively that for every integer r, 1 ≤ r ≤ n−1, the followingis true: For every dual r-tuples of vectors x1, . . . , xr ∈ Dn and ty1, . . . ,

tyr ∈tDn the r-tuples

yτ1B, . . . , yτrB ∈ Dn and A txσ1 , . . . , A

txσr ∈ tDn

are dual andφ(P (x1, . . . , xr ; ty1, . . . ,

tyr))

=

= P (yτ1B, . . . , yτrB ; A txσ1 , . . . , A

txσr ).

We already know that this is true when r = 1.So, assume now that 2 ≤ r ≤ n − 1 and that the above statement holds

for r − 1. Let x1, . . . , xr ∈ Dn and ty1, . . . ,tyr ∈ tDn be a dual r-tuple of

vectors. We already know that this is equivalent to the fact that the linearfunctionals tyj : [x1, . . . , xr] → D, j = 1, . . . , r, are linearly independent. Inother words, we can find vectors tu1, . . . ,

tur ∈ tDn such that the linear spanof ty1, . . . ,

tyr ∈ tDn is equal to the linear span of tu1, . . . ,tur ∈ tDn and

moreover,xi

tuj = δij .

We will first prove that vectors A txσ1 , . . . , Atxσr are linearly independent. By the

induction hypothesis each subset with r− 1 elements is linearly independent. IfA txσ1 , . . . , A

txσr were linearly dependent, then the linear span of this set wouldbe of dimension (r − 1), and therefore, the linear spans of the (r − 1)-tuples

A txσ1 , Atxσ3 , . . . , A

txσr and A txσ2 , Atxσ3 , . . . , A

txσr

would coincide. Using the induction hypothesis we conclude that

φ(P (x2, x3, . . . , xr ; tu1 + tu2,tu3, . . . ,

tur))

= φ(P (x1, x3, . . . , xr ; tu1 + tu2,tu3, . . . ,

tur))

contradicting the injectivity of φ.

27

In the same way we see that both r-tuples yτ1B, . . . , yτrB and uτ1B, . . . , u

τrB

are linearly independent. Since obviously each yτjB, j = 1, . . . , r, is contained inthe linear span of uτ1B, . . . , u

τrB, the linear spans of these two r-tuples coincide.

It follows from

P (xj ; tuj) ≤ P (x1, . . . , xr ; tu1, . . . ,tur), j = 1, . . . , r,

that the image and kernel of φ(P (x1, . . . , xr ; tu1, . . . ,tur)) are

[uτ1B, . . . , uτrB] and [A txσ1 , . . . , A

txσr ]⊥,

respectively. Therefore the r-tuples uτ1B, . . . , uτrB and A txσ1 , . . . , A

txσr are dual,and consequently, the r-tuples yτ1B, . . . , y

τrB and A txσ1 , . . . , A

txσr are dual, too.Moreover, we have the desired equality

φ(P (x1, . . . , xr ; ty1, . . . ,

tyr))

= φ(P (x1, . . . , xr ; tu1, . . . ,

tur))

= P (uτ1B, . . . , uτrB ; A txσ1 , . . . , A

txσr )

= P (yτ1B, . . . , yτrB ; A txσ1 , . . . , A

txσr ).

It remains to show that A and B are both invertible. We already know thatthey are both of rank at least n− 1 and we need to show that they are of rankn. If this was not true, say if B was not invertible, then by the above equationeach idempotent P of rank n− 1 would be mapped by φ into an idempotent Qwhose image would coincide with the image of B. So, if we take two idempotentsP1 6= P2 of rank n− 1 with the same images, then φ(P1) and φ(P2) would havethe same images and the same kernels, that is, we would have φ(P1) = φ(P2),contradicting the injectivity of φ.

So far we have proved that either φ is a generalized standard map, or thereexists a nonzero z ∈ Dn such that

φ(P 1n(D)) ⊂ R(z), (13)

or there exists a nonzero tw ∈ tDn such that

φ(P 1n(D)) ⊂ L( tw).

As in the case of generalized standard maps, we will also here consider just oneof the two cases. So, we assume that we have (13). After composing φ with anappropriate similarity transformation we can assume with no loss of generalitythat φ(P 1

n(D)) ⊂ R(e1).To each injective order preserving map φ : Pn(D)→ Pn(D) we can associate

a dual map φ∗ : Pn(D)→ Pn(D) defined by

φ∗(P ) = I − φ(I − P ), P ∈ Pn(D),

28

which is again an injective order preserving map. Clearly, φ is a generalizedstandard map if and only if φ∗ is a generalized standard map.

Hence, in our case there exists a nonzero y ∈ Dn such that

φ(I − P 1n(D)) ⊂ I −R(y), (14)

or there exists a nonzero tw ∈ tDn such that

φ(I − P 1n(D)) ⊂ I − L( tw).

We will prove that φ is degenerate by induction on n. We start with thecase when n = 3.

After applying yet another similarity transformation we may assume thatthere exist two rank one idempotents P1, P2 and two rank two idempotentsQ1, Q2 such that P1, P2 ≤ Q1, Q2 and

φ(P1) =

1 0 00 0 00 0 0

and φ(P2) =

1 0 01 0 00 0 0

.Then, clearly, both φ(Q1) and φ(Q2) are of the form 1 0 0

0 1 ∗0 0 0

,and therefore, each rank one idempotent R1 and each rank two idempotent R2

are mapped into 1 0 0λ 0 0µ 0 0

and

1 0 α0 1 β0 0 0

,respectively, for some scalars λ, µ, α, β ∈ D. Furthermore, we have 1 0 0

λ 0 0µ 0 0

≤ 1 0 α

0 1 β0 0 0

if and only if µ = α = 0.

Hence, each rank one idempotent is mapped into an idempotent of the form 1 0 0∗ 0 00 0 0

,and each rank two idempotent is mapped into an idempotent of the form 1 0 0

0 1 ∗0 0 0

.29

Thus, φ is degenerate, as desired.Assume now that our statement is true for n− 1 and we want to prove it for

n. If P ∈ Pn(D) is any idempotent of rank n− 1, then both sets {Q ∈ Pn(D) :Q ≤ P} and {Q ∈ Pn(D) : Q ≤ φ(P )} can be identified with Pn−1(D). Ofcourse, for each P ∈ Pn(D) of rank n − 1 the set {Q ∈ Pn(D) : Q ≤ P} ismapped by φ injectively into {Q ∈ Pn(D) : Q ≤ φ(P )} and the restriction of φto {Q ∈ Pn(D) : Q ≤ P} considered as a map into {Q ∈ Pn(D) : Q ≤ φ(P )}can be viewed as a map from Pn−1(D) into itself.

We will need to distinguish two cases depending on whether n is even or oddand we will consider just one of them, say the case that n is even. Replacingthe map φ by a map P 7→ Tφ(P )T−1, P ∈ Pn(D), where T is an appropriateinvertible matrix, we may, and we will assume that

φ(E11 + . . .+ En−1,n−1) = E11 + . . .+ En−1,n−1.

Applying the induction hypothesis we may assume, after composing φ with yetanother similarity transformation, that φ maps the set of all rank one idempo-tents ≤ E11 + . . .+ En−1,n−1 into the set of all idempotents of the form

E11 + ∗E21, (15)

where ∗ stands for some scalar from D, that φ maps the set of all rank twoidempotents ≤ E11 + . . .+En−1,n−1 into the set of all idempotents of the form

E11 + E22 + ∗E23, (16)

where ∗ stands for some scalar from D,..., and that φ maps the set of all idem-potents ≤ E11 + . . .+En−1,n−1 of rank n− 2 into the set of all idempotents ofthe form

E11 + . . .+ En−2,n−2 + ∗En−2,n−1, (17)

where ∗ stands for some scalar from D.By induction hypothesis we know that for each P ∈ Pn(D) of rank n − 1

the restriction of φ to {Q ∈ Pn(D) : Q ≤ P} is an injective degenerate orderpreserving map into the set {Q ∈ Pn(D) : Q ≤ φ(P )}. From Lemmas 3.2 and3.3 one can easily conclude that if P and R are idempotents of rank n− 1 suchthat P ∼ R, and if φ behaves on the set {Q ∈ Pn(D) : Q ≤ P, Q 6= P} asdescribed in (15), (16),...,(17), then φ behaves on the set {Q ∈ Pn(D) : Q ≤R, Q 6= R} in the same way. Hence, by Lemma 3.4, we have (15), (16),...,(17)for all idempotents in Pn(D) of rank at most n− 2.

In order to complete the proof we need to see that all idempotents of rankn− 1 are mapped into the set of all idempotents of the form

E11 + . . .+ En−1,n−1 + ∗En,n−1,

where ∗ stands for some scalar from D. This is now rather easy and is left tothe reader.

30

2

Proof of Corollary 2.7. The assertion is a straightforward consequence of The-orem 2.6 and Lemma 3.6.

2

Proof of Corollary 2.8. It is clear that the supremum preserving property im-plies that φ preserves order. Hence, we can apply Theorem 2.6. It is straightfor-ward to verify that injective degenerate order preserves do not have supremumpreserving property. Thus, φ is a generalized standard map. We have two pos-sibilities and we will consider just one of them as the proof in the other case isalmost the same.

So, we will assume that A,B ∈ Mn(D) are invertible matrices and τ, σ :D→ D endomorphisms such that for every integer r, 1 ≤ r < n, and for everydual r-tuples of vectors x1, . . . , xr ∈ Dn and ty1, . . . ,

tyr ∈ tDn the r-tuples

xτ1A, . . . , xτrA ∈ Dn and B tyσ1 , . . . , B

tyσr

are dual, and that the map φ : Pn(D)→ Pn(D) is given by φ(0) = 0, φ(I) = I,and

φ(P (x1, . . . , xr ; ty1, . . . ,

tyr))

=

= P (xτ1A, . . . , xτrA ; B tyσ1 , . . . , B

tyσr ).

Equivalently, for any n-tuple of linearly independent vectors x1, . . . , xn andevery integer r, 1 ≤ r ≤ n− 1, we have

φ (P ([x1, . . . , xr] , [xr+1, . . . , xn])) =

= P ([xτ1A, . . . , xτrA] , [xσr+1B

−1, . . . , xσnB−1]).

Here, of course, P ([x1, . . . , xr] , [xr+1, . . . , xn]) denotes the idempotent whoseimage is the linear span of x1, . . . , xr and whose kernel is the linear span ofxr+1, . . . , xn.

Replacing φ by P 7→ Aφ(P )A−1 we may assume with no loss of generalitythat A = I (of course, when replacing A by the identity we also need to replaceB−1 by B−1A−1 = (AB)−1).

Let now x1, . . . , xn ∈ Dn be any n-tuple of linearly independent vectors. Weconsider the following two idempotents:

P = P ([x1, x4, . . . , xn] , [x2, x3]) and Q = P ([x1+x2, x4, . . . , xn] , [x2, x1−x3]).

Clearly,

KerP ∩KerQ = [x2] and ImP + ImQ = [x1, x2, x4, . . . , xn].

31

Hence, by Lemma 3.1, P ∨Q exists, and therefore, φ(P ) ∨ φ(Q) exists as well.Moreover, if P,Q ≤ R for some R ∈ Pn(D), then

KerR ⊂ KerP ∩KerQ ⊂ ImP + ImQ ⊂ ImR,

and consequently, R must be the identity matrix. It follows that φ(P )∨φ(Q) =I.

We haveKerφ(P ) ∩Kerφ(Q) = [xσ2B

−1]

andImφ(P ) + Imφ(Q) = [xτ1 , x

τ2 , x

τ4 , . . . , x

τn]

and since φ(P ) ∨ φ(Q) exists and is equal to the identity we conclude that

xσ2B−1 ∈ [xτ1 , x

τ2 , x

τ4 , . . . , x

τn].

This has to be true for every (n − 1)-tuple of linearly independent vectorsx1, x2, x4, . . . , xn and then we have a straightforwatrd consequence that for ev-ery x ∈ Dn the vector xσB−1 belongs to the linear span of xτ .

In other words, for every nonzero x ∈ Dn there exists a unique λx ∈ D suchthat

xσB−1 = λxxτ .

We will show that λx ≡ λ is independent of x. Clearly, it is enough to verifythat λx = λy for each pair of linearly independent x, y ∈ Dn. For each such pairof vectors x, y we have

λxxτ + λyy

τ = xσB−1 + yσB−1 = (x+ y)σB−1

= λx+y(x+ y)τ = λx+yxτ + λx+yy

τ ,

and therefore,λx = λx+y = λy,

as desired.Hence, for every x ∈ Dn we have xσB−1 = λxτ , and consequently,

φ (P ([x1, . . . , xr] , [xr+1, . . . , xn])) =

= P ([xτ1 , . . . , xτr ] , [λxτr+1, . . . , λx

τn])

= P ([xτ1 , . . . , xτr ] , [xτr+1, . . . , x

τn]).

Thus, φ(P ) = P τ for every P ∈ Pn(D). This completes the proof.

2

32

Proof of Corollary 2.10. We apply Theorem 2.6, and after checking the trivialfact that injective degenerate order preserves φ do not satisfy φ(I−P ) = I−φ(P )for every P ∈ Pn(D) we conclude that φ is a generalized standard map. As inthe previous proof it is enough to consider only the case when

φ (P ([x1, . . . , xr] , [xr+1, . . . , xn])) =

= P ([xτ1 , . . . , xτr ] , [xσr+1B

−1, . . . , xσnB−1]),

where B, τ , and σ are as above.We use ImP = Ker (I − P ) and φ(I − P ) = I − φ(P ) for an idempotent

P = P ([y2, . . . , yn] , [x])

to conclude that for every n-tuple of linearly independent vectors x, y2, . . . , ynwe have

P ([xτ ] , [yσ2B−1, . . . , yσnB

−1]) = I − P ([yτ2 , . . . , yτn] , [xσB−1])

= P ([xσB−1] , [yτ2 , . . . , yτn]),

which yields that for every nonzero x ∈ Dn the vector xσB−1 belongs to thelinear span of xτ . The rest of the proof is the same as in the previous case.

2

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