Multidimensional local skew-fields (792481), страница 15

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Then zuz −1 = u + cur z i + r(i + 1)/2c2 u2r−1 z 2i .Therefore, δ2i = δi2 . Then for any x ∈ k((u)) holds:z −1 xz = x − xδi z i + . z >2iWe put X = u−r−1 z −i , Y = u2 . ThenXY = u1−r z −i + . . . + Cur−1 z i + . . . ,C ∈ ”, C = 0Hence resi,r ([X, Y ]) = 0.An example with a(0, . . . , 0) = 0, r = 1 can be obtained likewise. (iii) Let K be askew field with αn = 1, in = ∞. We put X = u−rn z −in and Y = u. ThenXY = ξ −in u1−rn z −in + C + . . .where C = −in ξ −in +1 c = 0. Hence, resα ([X, Y ]) = 0.Remark These examples show that the Scolem-Noether theorem does not hold forskew fields defined here.Let K be the ring k((u))((∂u−1 )) of pseudodifferential operators. We have shownthat this is the only slew field such that res1,0 ([X, Y ]) = 0. Let us deduce a criterionfor two elements of this skew field to be conjugate.Let n ∈ N be a certain number. Consider the skew field K = k((t))((∂t−1 )), wheretn = u.

Then ∂t = ntn−1 ∂u , and K ⊂ K .Lemma 0.61 Let L = l−m ∂tm + . . . + l0 + l1 ∂t−1 + . . . ∈ K -be an arbitrary elementof K .L ∈ K if and only if li ∈ ti k((tn )).Proof Assume that L ∈ K. Then L = b−m ∂um + . . ., where bi ∈ k((u)) = k((tn )).Let j ∈ N. We have:∂uj = (n−1 t1−n ∂t )j , ∂u−j = (∂t−1 ntn−1 )j .67We prove first the assertion of the lemma for l−i (i > 0). For i = 1 we have∂ui = n−1 t1−n ∂t and b−1 ∂u = l−1 n−1 t1−n ∂t .

The assertion of the lemma holds, sincet1−n ∈ tk((tn )).For an arbitrary i we have∂ui =∂t −1 1−n −1 1−n i−1(n t )(n t ∂t ) + (n−1 t1−n )2 ∂t2 (n−1 t1−n ∂t )i−2 =∂t t(1 − n)(n−1 t−n )(n−1 t1−n ∂t )i−1 + (n−1 t1−n )2 ∂t2 (n−1 t1−n ∂t )i−2Since the coefficients in the expression for L in K belong to k((tn )), it is sufficient toshow that the lemma holds for ∂ui .We prove by induction that the assumption of the lemma holds for all the coefficients˜ kin (n−1 t1−n ∂t )i−1 . The same is true for (n−1 t1−n ∂t )i−2 . Let (n−1 t1−n ∂t )i−2 = i−2k=0 lk ∂ti(Let us note that there are no negative powers of ∂t in the expansion of ∂u , i > 0, andthe minimal power of∂t is equal to 1). We have:i−2i−2i−2i−2(n−1 t1−n )2 ∂t2 (l˜k ∂tk ) = (n−1 t1−n )2 (l˜k ∂tk+2 +l˜k ∂tk+1 +l˜k ∂tk )k=0k=0k=0k=0Therefore, (n−1 t1−n )2 l˜k ∈ tk+2 k((tn )), (n−1 t1−n )2 l˜k ∈ tk+1 k((tn )), (n−1 t1−n )2 l˜k ∈tk k((tn )).For i = 0 we have l0 = b0 ∈ k((tn )).Let us prove that the assertion of the lemma holds for ∂ −i , i > 0.

For i = 1 wehave: n−1 n−1 (k) −1−k −1(t ) ∂tCk .∂u−1 = n k=0˜ −k−j , l˜j ∈ t−k−j k((tn )).Assume that for k < i it is proved ∂u−k = ∞j=0 lj ∂t∂u−i=(∂t−1 ntn−1 )i= (nn−1Ck−1 (tn−1 )(k) ∂t−1−k )(∂t−1 ntn−1 )i−1 =k=0(nn−1∞l˜j ∂t−i+1−j )Ck−1 (tn−1 )(k) ∂t−1−k )(j=0k=0−1−k ˜ (p) −1−k−p. This yields the followlj ∂tFor every k ∈ {0, . .

. , n − 1} ∂t−1−k l˜j = ∞p=0 Cping conditions on the coefficients for fixed k and j:at ∂t−1−k−p−i+1−j , p ≥ 0, the coefficient belongs to t−1−k−i+1−j−p k((tn )).Conversely, assume that the assumptions of the lemma on the coefficients hold. Wehave obtainedthati−ji∂u = j≥0 cj ∂t , and cj ∈ ti−j k((tn )) for any i ∈ Z.68Consider the highest monomial in L:m−jml−m ∂tm = l−m c−1∂−l(cj c−1)−m0u0 ∂tj≥1−1nm−jmk((tn )). Hence, L = l−m c−1We have l−m c−10 ∈ k((t )), l−m cj c0 ∈ t0 ∂u + L1 , whereν(L1 ) > ν(L), and the the assumptions of the lemma hold for the coefficients in L1 .We complete the proof by induction.2Lemma 0.62 Let L, M ∈ K ⊂ K and ν(L) = ν(M ) = −n.

Let M = SLS −1 , whereS ∈ K . Then S ∈ K if and only ifreslν(L)+1 − mν(M )+1= 0 andlν(L)tlν(L)+1 − mν(M )+1∈ k[[t]]lν(L)Proof is similar to that of Proposition 0.57.2Theorem 0.63 Let L, M ∈ K = k((u))((∂u−1 )), ν(L) = ν(M ) < 0,−ν(M )M = mν(M ) ∂t+ . . .,−ν(L)L = lν(L) ∂t+ . . ..The following assumptions are equivalent:(i) there is an S ∈ K, ν(S) = 0, such that M = S −1 LS(ii) ν(L) = ν(M ), mν(M ) = lν(L) ,reslν(L)+1 − mν(M )+1= 0 andlν(L)tlν(L)+1 − mν(M )+1∈ k[[t]]lν(L)res(M j/(−ν(M )) ) = res(Lj/(−ν(L)) ) for all j ≥ 1 in K .Proof follows immediately from Corollary 10, Lemmas 0.61, 0.62 and the fact thatL (and M ) has precisely one nth root in K .2Theorem 0.64 Assume that L, M ∈ K = k((u))((∂u−1 )) and ν(L) = ν(M ) = 0.

Then(i) If l0 = m0 = const and l1 = m1 , then M = SLS −1 .(ii) If l0 = m0 = const, then M = SLS −1 if and only if (M − m0 )−1 = S(L − l0 )−1 S −1(see Theorem 0.63)Proof is obvious.2690.6New equations of KP-type on skew fieldsIn this section we give an answer on a question given in [22]. Namely, the classicalKP-hierarchy is constructed by means of the ring of pseudo-differential operators P =k((x))((∂ −1 )). This ring is a skew field. The point is to consider other skew fields insteadof this one.

We will study if there exist some new non-trivial generalisations of the KPhierarchy for a list of two-dimensional skew fields. In particular, we give a number ofnew partial differential equations of the KP-type.For every two-dimensional skew field from the list of theorem 1.5 we can write downa decomposition K = K+ + K− , where K− = {L ∈ K : ord(L) < 0} and K+ consists ofthe operators containing only ≥ 0 powers of z, and a ”KP-hierarchy” in the Lax form:∂L= [(Ln )+ , L],∂tnwhere L ∈ z −1 + K− ⊗ k[[. .

. , tm , . . .]]. Let L = z −1 + u1 z + u2 z 2 + . . ., where um =um (u, t1 , t2 , . . .). Further we will denote ∂/∂tn as ∂n .One can check that if the canonical automorphism α in the classification theorem1.5 is not trivial, then our ”KP-hierarchy” became trivial, i.e.

it can be easily linearisedand solvable. We omit calculations here. So, it can be assumed that α = id. The sameis true if i > 1, because [(Ln )+ , L] = −[(Ln )− , L] = 0 mod ℘i in this case, where ℘is a maximal ideal of the first valuation in K. So, our ”KP-hierarchy” again is linearand easily solvable in this case.So, we assume i = 1, hence, r = 0 and c = 1, and there is only only one non-trivialparameter a. If a = 0, K is isomorphic to the ring P of pseudo-differential operators.Denote by u , u , . . . the subsequent derivatives by x.First for n = 1, we get∂1 u1 = u1This means that we can take t1 = x for u1 .Now we write down the first two equations for n = 2 and the first equation forn = 3.(3)∂2 u1 = u 1 + 2u2∂2 u2 = 2u3 + 2u1 u1 + u 2 + 2ax−1 u2(4)∂3 u1 = u 1 + 3u 2 + 3u3 + 6u1 u1 + 3a(x−1 u 1 − x−2 u1 )(5)Let us introduce the new notation: u = u1 (x, y, t) with y = t2 , t = t3 .

Also we use thestandart notation ut , uy , uyy , . . . for derivatives.We can eliminate u3 from equations 4 and 5 and then we get3u2y − 2ut = −6uu − 3u 2 − 2u + 6ax−1 u2 − 6ax−1 u + 6ax−2 u70(6)From 3 we findu 2 = 1/2(u y − u ), u2y = 1/2(uyy − u y )Differentiating equation 6 by x and inserting these expressions we finally get new KPequation(4ut − u − 12uu ) = 3uyy + 6a(5x−2 u − x−2 uy − 3x−1 u + x−1 uy − 4x−3 u )One can see that if a = 0, we get the usual KP-equation (see also explicite calculationsin [21]).71Chapter 1Classification of automorphisms ofa two-dimensional local field.1.1Basic results.In this chapter let K be a two-dimensional local field, K ∼= k((u))((z)); Autk (K)be a group of continuous k-automorphisms of a field K with respect to the topologygiven by fixed parametrisation, i.e.

by the parameters u and z (see [35] concerning theconnection between a topology and a parametrisation).Introduce the following notation. By Greece letters α, β, γ we will denote automorphisms of a field K. An overline will denote the residue homomorphism. As before,ν denote a valuation on the field K, ν̄ denote a valuation on the field K̄, ℘, ℘¯ arevaluation ideals of the valuations ν, ν̄, µ(k) is the group of roots of the unity, Autk (K̄)is a group of continuous k-automorphisms of the field K̄.Recall some results from chapter 1, section 3.Definition 1.1 Let K̄ be a one-dimensional local field with the residue field k,charK̄ = chark, ᾱ ∈ Autk (K̄). Putξ(ᾱ) = ᾱ(u)u−1 mod ∈ k and define i(ᾱ) ∈ N ∪ ∞ by the following:i(ᾱ) = 1 if ξ(ᾱ) ∈/ µ(k), elseni(ᾱ) = ν̄((ᾱ −Id)(u)), where n ≥ 1, ξ(ᾱ) is a primitive root of degree n, ord(ξ(ᾱ)) = n.Proposition 1.2 Let k be an arbitrary field, chark = 0.

Any automorphism ᾱ ∈Autk (k((u))) with ᾱ(u) = ξ(ᾱ)u + a2 u2 + . . . is conjugate with the automorphism β̄:β̄(u) = ξ(ᾱ)u + xui(ᾱ) + x2 yu2i(ᾱ)−1 , where x ∈ k ∗ /k ∗(i(ᾱ)−1) , y ∈ k.Two automorphisms β̄, β̄ are conjugate iff(ξ(β̄), i(β̄), x(β̄), y(β̄)) = (ξ(β̄ ), i(β̄ ), x(β̄ ), y(β̄ )).Corollary 12 1) i(ᾱ) = 1 iff ᾱ is an automorphism of infinite order and ξ(ᾱ) hasinfinite order;722) 1 < i(ᾱ) < ∞ iff ᾱ has infinite order and ξ(ᾱ) has finite order;3) i(ᾱ) = ∞ iff ᾱ has finite order.Remark. i) In the notation of proposition we have n|(i(ᾱ) − 1).ii) if k is a field of characteristic p > 0, then the following fact remains true: theautomorphism ᾱ ∈ Autk (k((u))) with ᾱ(u) = ξ(ᾱ)u + a2 u2 + .

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