# Directed Partial Orders on Complex Numbers ad Quaternions II

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.Suppose that F is a partially ordered field with a directed partial order and K is a non-archemedean totally ordered subfield of F with K+=F+∩K. In this note, directed partial orders are constructed for complex numbers and quaternions over F. It is also shown that real quaternions cannot be made into a directed algebra over the real field with the total order.

Let T be a non-archimedean totally ordered field and CT=T+Ti be the field of complex numbers over T, where i2=−1, and HT=T+Ti+Tj+Tk be the division algebra of quaternions over T. In [4], a general method of constructing directed partial orders on CT and HT has been developed. The purpose of this note is to generalize the results in [4] to complex numbers and quaternions over non-archimedean partially ordered fields with a directed partial order.

We review a few definitions and the reader is referred to [2, 4] for undefined terminology and background on partially ordered rings and lattice-ordered rings (ℓ-rings). For a partially ordered ring R, the positive cone is defined as R+={r∈R|r≥0}. The partial order on a partially ordered ring R is called directed if each element of R can be written as a difference of two positive elements of R. A partially ordered ring (algebra) with a directed partial order is called a directed ring (algebra). In the following, we always assume that F is a partially ordered field with a directed partial order and K is a non-archimedean totally ordered subfield of F, so F+∩K=K+.

Following result will be used in the proof of main results in the paper.

Lemma 1 Suppose that R is a partially ordered ring with the property that for any r∈R, 2r≥0 implies that r≥0. Then for any x,y∈R+, and a,b∈R, −x≤a≤x and −y≤b≤y implies that −xy≤ab≤xy.

Proof From −x≤a≤x and −y≤b≤y, we have

(x+a)(y−b)≥0⇒xy+ay−xb−ab≥0, (1) and

(x−a)(y+b)≥0⇒xy−ay+xb−ab≥0. (2) Adding (1) and (2), we have 2(xy−ab)≥0, so xy−ab≥0 and xy≥ab.

From −x≤a≤x and −y≤b≤y, we also have

(x+a)(y+b)≥0⇒xy+ay+xb+ab≥0, (3) and

(x−a)(y−b)≥0⇒xy−ay−xb+ab≥0. (4) Adding (3) and (4), we have 2(xy+ab)≥0, so xy+ab≥0, that is, ab≥−xy. □

We notice that since F contains a totally ordered subfield K, for any a∈F, 2a≥0 implies that a≥0. In fact, since 0<2∈K, 0<12∈K, so a=12(2a)≥0.

Theorem 1 Take 0≤x,y≤1 in F with x≠0 or y≠0 such that x−1>0 if x≠0 and y−1>0 if y≠0, define the positive cone Px,y of CF=F+Fi as follows.

Px,y={a+bi|a∈F+,−xa≤nb≤ya in F for all positive integers n}. Then Px,y is a directed partial order on CF such that (CF,Px,y) is a partially ordered algebra over F and Px,y∩F=F+.

Proof It is clear that Px,y∩−Px,y={0}, Px,y+Px,y⊆Px,y and F+Px,y⊆Px,y. Suppose that a+bi,c+di∈Px,y. Then a,c∈F+, and for all positive integers n,

−a≤−xa≤nb≤ya≤a, −c≤−xc≤nd≤yc≤c. We show that (a+bi)(c+di)=(ac−bd)+(ad+bc)i∈Px,y.

From −a≤b≤a, −c≤d≤c and Lemma 1, we have bd≤ac, that is, ac−bd∈F+.

For any positive integer n, 3nd≤yc,3nb≤ya and a,c∈F+ implies that 3nad≤y(ac),3nbc≤y(ac). Since −a≤3b≤a and −c≤d≤c, Lemma 1 implies that 3bd≤ac; so 3ybd≤y(ac). Hence

3nad+3nbc+3ybd≤y(ac)+y(ac)+y(ac)=3y(ac). Since 3∈K+ and K is totally ordered, 13∈K+⊆F+, so 3(nad+nbc+ybd)≤3y(ac) implies that nad+nbc+ybd≤y(ac). Therefore n(ad+bc)≤y(ac−bd) for all positive integers n. Similarly −x(ac−bd)≤n(ad+bc). We have proved that (a+bi)(c+di)∈Px,y. Thus Px,y is a partial order on CF with respect to which CF is a partially ordered algebra. Clearly Px,y∩F=F+.

We verify that Px,y is a directed partial order on CF. Suppose that y≠0. A similar argument could be used in the case x≠0. Let a+bi∈CF. Since F is directed, a is a difference of two elements in F+, so a is a difference of two elements in Px,y since F+⊆Px,y. Consider bi=b1i−b2i, where b1,b2>0 in F. Since K is a non-archimedean totally ordered field, there exists z∈K+ such that n1≤z for all positive integers n. Let w=y−1b1. Then w,wz∈F+ and for all positive integers n, −x(wz)≤0≤nb1≤b1z=y(wz), that is, wz+b1i∈Px,y. Thus b1i=(wz+b1i)−wz is a difference of two positive elements in CF. Similarly b2i is a difference of two positive elements. Hence bi is a difference of two positive elements in CF, so a+bi is a difference of two positive elements. Therefore Px,y is directed. □

Remark 1 If the partial order on F is a lattice order, then 0<x<1 implies that x−1>0 since x<1 implies that x is an f-element [3, Theroem 1.20(2)].

Next we consider quaternions over F. Recall that HF=F+Fi+Fj+Fk as a vector space over F with the multiplication as follows.

=(a0+a1i+a2j+a3k)(b0+b1i+b2j+b3k)(a0b0−a1b1−a2b2−a3b3)+(a0b1+a1b0+a2b3−a3b2)i+(a0b2+a2b0+a3b1−a1b3)j+(a0b3+a3b0+a1b2−a2b1)k, where ai,bi∈F.

Theorem 2 Take 0<x≤1 in F such that x−1>0, define the positive cone Px of HF as follows.

Px={a0+a1i+a2j+a3k | a0∈F+ and −xa0≤na1≤xa0,−xa0≤na2≤xa0, −xa0≤na3≤xa0 in F for all positive integers n}. Then Px is a directed partial order on HF such that (HF,Px) is a partially ordered algebra over F and Px∩F=F+.

Proof It is clear that Px∩−Px={0}, Px+Px⊆Px and F+Px⊆Px. We show that PxPx⊆Px. Suppose that a0+a1i+a2j+a3k,b0+b1i+b2j+b3k∈Px. We check that (a0+a1i+a2j+a3k)(b0+b1i+b2j+b3k)∈Px. Since a0+a1i+a2j+a3k,b0+b1i+b2j+b3k∈Px, we have for all positive integers n,

a0≥0, −xa0≤na1≤xa0, −xa0≤na2≤xa0, −xa0≤na3≤xa0, and

b0≥0, −xb0≤nb1≤xb0, −xb0≤nb2≤xb0, −xb0≤nb3≤xb0. We first check that a0b0−a1b1−a2b2−a3b3≥0 in F.

From −a0≤−xa0≤3a1≤xa0≤a0, −b0≤−xb0≤b1≤xb0≤b0 and Lemma 1, we have a0b0≥3a1b1, that is, a0b0−3a1b1≥0. Similarly, a0b0−3a2b2≥0 and a0b0−3a3b3≥0. Adding those three inequalities together, we obtain 3a0b0−3a1b1−3a2b2−3a3b3≥0. Then multiplying the both sides by 13 we get a0b0−a1b1−a2b2−a3b3≥0.

For simplicity, let w=x(a0b0−a1b1−a2b2−a3b3). We next show that for all positive integers n,

−w≤n(a0b1+a1b0+a2b3−a3b2)≤w. Consider n(a0b1+a1b0+a2b3−a3b2)≤w first. We divide the calculations into several steps. Let n be a positive integer.

Since 7nb1≤xb0 and a0≥0, we have 7na0b1≤xa0b0.

Since 7na1≤xa0 and b0≥0, we have 7na1b0≤xa0b0.

Since −a0≤7na2≤a0 and −xb0≤b3≤xb0, Lemma 1 implies that 7na2b3≤xa0b0.

Since −a0≤7na3≤a0 and −xb0≤b2≤xb0, Lemma 1 implies that 7na3b2≥−xa0b0, so −7na3b2≤xa0b0.

Since −a0≤7a1≤a0 and −b0≤b1≤b0, Lemma 1 implies that 7a1b1≤a0b0, so 7xa1b1≤xa0b0. Similarly, 7xa2b2≤xa0b0 and 7xa3b3≤xa0b0.

Adding inequalities in the above (1) to (5) together, we have

7na0b1+7na1b0+7na2b3−7na3b2+7xa1b1+7xa2b2+7xa3b3≤7xa0b0, and hence, by multiplying both sides of this inequality by 17,

na0b1+na1b0+na2b3−na3b2+xa1b1+xa2b2+xa3b3≤xa0b0. It follows that n(a0b1+a1b0+a2b3−a3b2)≤x(a0b0−a1b1−a2b2−a3b3).

Similarly we prove −w≤n(a0b1+a1b0+a2b3−a3b2). Let n be a positive integer.

Since −xb0≤7nb1 and a0≥0, we have that −xa0b0≤7na0b1.

Since −xa0≤7na1 and b0≥0, we have that −xa0b0≤7na1b0.

Since −a0≤7na2≤a0 and −xb0≤b3≤xb0, Lemma 1 implies that −xa0b0≤7na2b3.

Since −a0≤7na3≤a0 and −xb0≤b2≤xb0, Lemma 1 implies that 7na3b2≤xa0b0, so −xa0b0≤−7na3b2.

Since −a0≤7a1≤a0 and −b0≤b1≤b0, we have −a0≤7a1≤a0 and −b0≤−b1≤b0, so Lemma 1 implies that −a0b0≤−7a1b1, so −xa0b0≤−7xa1b1. Similarly, −xa0b0≤−7xa2b2 and −xa0b0≤−7xa3b3.

Adding up the above inequalities, we have

−7xa0b0≤7na0b1+7na1b0+7na2b3−7na3b2−7xa1b1−7xa2b2−7xa3b3, and hence

−xa0b0≤na0b1+na1b0+na2b3−na3b2−xa1b1−7xa2b2−xa3b3, that is, −x(a0b0−a1b1−a2b2−a3b3)≤n(a0b1+a1b0+a2b3−a3b2).

By similar calculations, we have

−w≤n(a0b2+a2b0+a3b1−a1b3)≤w, and

−w≤n(a0b3+a3b0+a1b2−a2b1)≤w. Therefore PxPx⊆Px, so (HF,Px) is a partially ordered algebra over F. It is straightforward to verify that Px∩F=F+.

Finally we show that Px is a directed partial order on HF. Take a∈F. Since F is directed, a=b−c, where b,c∈F+. Since F+⊆Px, a is a difference of two positive elements in (HF,Px). Consider ai=bi−ci. Since K is a non-archimedean totally ordered field, there exists z∈K+ such that n1≤z for all positive integers n. Let v=x−1b. Then v,vz∈F+ and for all positive integers n, nb≤bz=x(vz), that is, vz+bi∈Px. Thus bi=(vz+bi)−vz is a difference of two positive elements in (HF,Px). Similarly, ci is also a difference of two positive elements in (HF,Px), and hence ai=bi−ci is a difference of two positive elements in (HF,Px). The same argument may be used to show that aj and ak are also a difference of two positive elements in (HF,Px). It follows that each element in HF is a difference of two positive elements in (HF,Px), that is, Px is directed. This completes the proof of Theorem 2. □

In [1], Birkhoff asked if H=HR can be made into a directed algebra over R with the usual total order, where R is the field of real numbers. In the following, we show that the answer is no. More generally, for any totally ordered subfield T of R, HT cannot be a directed algebra over T.

Theorem 3 H cannot be a directed algebra over R with the usual total order.

Proof We will suppose that H is a directed algebra over R and derive a contradiction.

We first show that if w=a0+a1i+a2j+a3k>0 in H, then a0>0 in R. Note that w2−2a0w=−(a20+a21+a22+a23) by direct calculation, and that if 0<a in R and 0<z in H, then 0<az in H because H is a partially ordered algebra over R that has no divisor of zero.

Now suppose by the way of contradiction that a0≤0 in R. Then −a0≥0 in R, and thus since w>0 in H, w2−2a0w≥0 in H. That is, −(a20+a21+a22+a23)≥0 in H, and hence −(a20+a21+a22+a23)w>0 in H. But since (a20+a21+a22+a23)≥0 in R, (a20+a21+a22+a23)w≥0 in H as well, and therefore (a20+a21+a22+a23)w=0. So since w≠0 and H is a division ring, we must have a20+a21+a22+a23=0. But then a0=a1=a2=a3=0 because a0,a1,a2, and a3 are all in R, and hence w=a0+a1i+a2j+a3k=0, a contradiction. It follows that a0>0 in R.

Since the partial order on H is directed, there exists z=a+bi+cj+dk>0 in H with z∉R. For instance, i=z1−z2, where z1,z2 are positive in H. Clearly z1,z2 cannot be both in R. The argument above shows that a>0 in R. Then since R is totally ordered, a−1>0 in R and hence a−1z=1+(a−1b)i+(a−1c)j+(a−1d)k>0 in H. Suppose that a−1b=s,a−1c=t,a−1d=u. Then we have w=1+si+tj+uk>0 in H and w∉R. For simplicity, let v=si+tj+uk. Then v2=−(s2+t2+u2), so −v2∈R+. Therefore −v2w≥0 in H, and hence w3−v2w>0 in H. Since

1+3v+2v2==(1+2v)(1+v)=(1+2v+v2−v2)w(w2−v2)w=w3−v2w, we have 1+3v+2v2>0 in H. Let w1=1+3v+2v2. Then w1>0 in H, and hence (w1−2v2)w=w1w−2v2w>0 in H. Since

1+4v+3v2=(1+3v)(1+v)=(1+3v+2v2−2v2)w=(w1−2v2)w 1+4v+3v2>0 in H. If we continue this procedure, we get that for any positive integer n, (1+nv)(1+v)=1+(n+1)v+nv2>0 in H. Therefore since the real part of a positive element in H must be positive in R, we must have 0≤1+nv2 for all positive integers n, so −nv2≤1 for all positive integers n. Then −v2=0, so v2=0 since R is archimedean with respect to the total order. Hence s=t=u=0, and w∈R, a contradiction of our observation above that w∉R.

Therefore H cannot be a directed algebra over R with the total order.