HERMITE CONSTANT AND EXTREME FORMS FOR ALGEBRAIC NUMBER FIELDS

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1 HERMITE CONSTANT AND EXTREME FORMS FOR ALGEBRAIC NUMBER FIELDS M I ICAZA 1 Introduction In this paper we consider a generalization to algebraic number fields of the classical Hermite constant γ n For this constant we extend the well-known Minkowski bound and study the notion of extreme forms in this setting Let us briefly recall the definition of Hermite s constant Let (a i ) be a positive definite symmetric real nn matrix and let f(x x n ) i a i x i x be its associated quadratic form Let us set µ( f ) min f(x) x n0 d( f ) det (a i ) γ( f ) µ( f )d( f )/n Hermite s constant is defined to be γ n sup γ( f ) f where f runs over all positive definite real quadratic forms Hermite [5] proved that γ n (43)n(n )/ and later Minkowski improved this bound to γ n 4ω /n where n ω n πn/γ(n1) denotes the volume of the n-dimensional unit sphere This constant has been widely studied and it is still an open problem to determine its exact value for n 9 Following an idea suggested by R Baeza we extend the definition of Hermite s constant to arbitrary number fields and obtain upper bounds for it that extend Minkowski s bound for γ n Based on the works of Voronoi [10] and Oesterle [9] we also generalize the notion of extreme forms We establish the existence of such forms and give a partial characterization of them We make use of Humbert s reduction theory [6 7] and some results from [9] Let K be a number field with [K: ] m r2s Let σ σ r be the real embeddings of K and σ r+ σ m be the complex embeddings with σ r+i σ r+s+i for 1 i s where denotes complex conugation We write K for the ring of algebraic integers of K We consider tuples of forms f ( f σi ) for 1 i rs where f σi (x) a(i) k k x k x is a positive definite real quadratic form in n variables for 1 i r and f σi (x) a(i) k k x k x is a positive definite complex hermitian form in n variables for r1 i rs We call such a system of rs forms a Humbert tuple For any Humbert tuple let us define µ( f ) min r f σi (x)) r+s ( f σi (x))) x n K i= i=r+ where for x (x x n )n K and σ σ σ r+s we write σ(x) (σ(x ) σ(x n )) Received 3 November 1994; revised 13 March Mathematics Subect Classification 11H06 Research supported by Fondecyt and by European Union CT1-CT J London Math Soc (2) 55 (1997) 11 22

2 12 M I ICAZA We further define the determinant of a Humbert tuple f ( f σi )tobe d(f) r det ( f σi ) r+s (det ( f σi )) (1) i= i=r+ Then we set γ( f ) µ( f )d( f )/n Hermite Humbert s constant is now defined to be γ nk sup f γ( f ) where f runs over all tuples of forms described above According to Humbert s reduction theory γ nk but the upper bound remains unspecified in Humbert s work In 2 we obtain without using Humbert s work an explicit upper bound for γ nk that depends only on K and n REMARK 1 If K is a totally real number field then γ nk can also be defined in the following way Let f be a positive definite quadratic form over K that is for each embedding σ of K into and for f(x) k a k x k x with a k K we have σ( f ) k σ(a k ) x k x is a positive definite real quadratic form We set µ( f ) min N K/ ( f(x)) d( f ) N K/ x n (det ( f )) K where N K/ denotes the norm from K to Then put γ( f ) µ( f )d( f )/n Since K is dense in [K : ] it follows that γ nk can be defined by γ nk sup f γ( f ) where f runs over all positive definite quadratic forms defined over K In this paper we shall prove the following results THEOREM 1 Let K be a number field of degree [K: ] m r2s with r real embeddings and 2s complex embeddings Let D K be the discriminant of K Then γ nk 4r+s ω r/n n ω s/n D n K where ω n denotes the olume of the n-dimensional unit sphere THEOREM 2 There exists a Humbert tuple f ( f σi )r+s such that i= γ( f ) γ nk We call a tuple f ( f σi ) that satisfies the condition of Theorem 2 a critical tuple The more general notion of extreme tuple will be studied in 3 We also introduce there the definitions of perfect and eutactic Humbert tuples We prove that if a positive definite Humbert tuple is extreme then it is perfect and eutactic This result partially generalizes a theorem of Voronoi [9 8] 2 Proof of Theorem 1 We refer the reader to the statement of Theorem 1 given in the Introduction Proof of Theorem 1 Let σ i for 1 i r be the real embeddings of K and let σ i for r i r2s be the complex embeddings of K with σ r+i σ for 1 i s r+s+i Consider the Humbert tuple f ( f σ f σr f σr+ f σr+s ) with the first r components

3 HERMITE CONSTANT AND EXTREME FORMS 13 being positive definite real n-dimensional quadratic forms and the remaining s components being positive definite complex hermitian forms of the same dimension Suppose that each f σi has a matrix [ f σi ] A i Then from (1) d( f ) r det ( f σi ) r+s (det ( f σi )) r det A i r+s det A i i= i=r+ i= i=r+ For 1 i r there is a matrix B i n n () with det (B i ) 1 such that D i BT i A i B i A i [B i ] is a diagonal matrix D i diag (d(i) d (i) n ) Here B T denotes the transpose of a matrix B (see [4]) Then for x n K 0 we have A i [σ i (x)] C i B σ i i (x) where z zt z denotes the square of the Euclidean norm of the vector z n and C denotes the diagonal matrix C i diag (d(i) d (i) n ) Similarly for r1 i rs we diagonalize the matrices A i and obtain D i B T i A i B i A i [B i ] where B i n n () and C i diag (c(i) c (i) n ) such that c (i) c (i) d(i) for 1 n Thus A i [σ i (x)] C i B σ i i (x) where z z T z Let us define the following domain in nrns For l 0 and y (y y r z r+ z r+s ) nrns let V l y: C i B y i i l and C i B z i i l for 1 i rs The volume of V l is given by ωr vol (V l ) n lrn/ ωs lsn n (r i= det A i )/ (r+s i=r+ det A i ) ω r ωs n lmn/ n d( f )/ We impose conditions on l so that Minkowski s convex body theorem holds for V l Therefore we need vol (V l ) 2mn vol (n K ) 2mn 2 ns D K n/ (2) It will follow that for all x (x x r+s ) V l the inequality r i= f σi (x i ) r+s i=r+ ( f σ (x i i )) C d( f )/n nk holds for a certain constant C nk which we will determine below In (2) it is enough to choose l so that the equality holds Therefore it is enough to take l so that ωr ωs n lmn/ n 2mn vol (n d( f )/ K ) 2mn 2 ns D K n/ From now on we choose l so that On the other hand for x V l we have r f σi (x i ) r+s ( f σi (x i )) i= i=r+ n r i= l 2 n(r+s) ωr n ωs n D K n/ d( f )/ /mn 2nr ωr n C i B (x i i ) r+s i=r+ 2ns ωs n D K n d( f ) C i B (x i i ) n l rn lsn

4 14 M I ICAZA Therefore r i= f σi (x i ) r+s i=r+ ( f σ (x i i )) 2r 2s D (d( f ))/n ωr/n ωs/n K C nk n n By Minkowski s Convex Body Theorem there exists x n K 0 so that (σ (x) σ r+s (x)) V l and thus we obtain γ nk 4r+s ω r/n ω s/n D n n K REMARK 2 in 1 It is clear that when K we recover Minkowski s bound given 3 Proof of Theorem 2 In this section we keep the notation introduced above Instead of working with Humbert tuples of forms we shall work with Humbert tuples of matrices of the form A (A i ) for 1 i rs where the first r entries are positive definite symmetric real nn matrices and the remaining s matrices are positive define nn complex hermitian matrices The definitions of µ d and γ are extended in the obvious way for tuples of matrices In the set of all tuples of matrices we introduce the following equivalence relation Let σ i be the embeddings of K arranged as before and let A (A A r+s ) B (B B r+s ) We say that A (A i ) is equivalent to B (B i ) if there exists C GL n ( K ) such that if we put C i σ i (C) then A i CT i B i C i for 1 i r and A i C T i B i C i for r1 i rs We denote by the set of classes of tuples [A] Since [K: ] m r2s we have an inclusion GL n (K) (GL n ())r(gl n ())s Therefore GL n ( K ) can be identified with a subgroup Γ (GL n ())r(gl n ())s We introduce the map ψ: (GL n ())r(gl n ())sγ defined by ψ((a A r A r+s )) [(AT A AT r A r A T r+s A r+s )] It is easily seen that ψ is a surection and we use this fact to endow with the quotient topology We also define the continuous map where [A] µ(a)d(a)/n Then φ: (0 ) χ φ ψ: (GL n ())r(gl n ())sγ (0 ) is also continuous Theorem 2 amounts to proving that the map χ attains its maximum Before we continue with the proof of Theorem 2 we need to introduce more definitions and notation Let rs (x x r ;x r+ x r+s )rs be the set of (rs)-tuples where the first r components are real and the remaining s components are complex Let rs x rs:x i 0 for 1 i rs

5 HERMITE CONSTANT AND EXTREME FORMS 15 Thus we have a map l: rs r+s which is defined by l(x x r+s ) (log x log x r log x r+ log x r+s ) If ε ε r+s is a set of independent units of K then if we define l* (1 1 r 2 2) s the set l* l(ε ) l(ε r+s )) is an -basis for r+s (later we shall take this system of units as powers of fundamental units) Hence for any x rs we have a unique representation for l(x) r+s as l(x) ξl*ξ l(ε)ξ r+s (ε r+s ) with ξ ξ ξ r+s We identify rs with m as an m (r2s)-dimensional real linear space In m we fix the following cone X X(ε ε r+s )xrs :0ξ1 0 arg x 2πk where k is the order of the group of roots of 1 contained in K If the field K is not totally imaginary then the condition 0 arg x 2πk ust means that x 0 Our next result is a generalized version of [2 Proposition 34] PROPOSITION 1 Let σ for 1 rs be the embeddings of K into and where the complex embeddings hae been arranged as before For 1 i rs1 let d i min (σ (ε i )1rs and let x (x x r+s ) X with x i 0 for 1 i r and x i 0 for r1 i rs If N(x) x x r x r+ x r+s (d d r+s ) m then x i 1 for 1 i r and x i 1 for r1 i rs In order to prove Theorem 2 it is enough to prove the following result PROPOSITION 2 For each real number ν 0 there is a compact set B (GL n ())r(gl n ())sγ such that [ν ) Im χ χ(b) Proof From Humbert s reduction theory [7 pp ] it follows that in each class [A] of positive definite forms there exists a representative A (A A r A r+ A r+s ) with A i F T i D i F i D i [F i ] for 1 i r2 where F i I J β 1 0 β n K β n 1 L is a unipotent upper triangular matrix with β i C and the diagonal matrix D i diag (t(i) t (i) n ) with t (i) + is such that t(i) t (i) + C for 1 r1 where C and C (and all the C i below) are constants depending only on the field K and on n Hence by the correspondence determined by ψ in each class [G] r GL n () r+s GL n () Γ r+ there is a representative G (G G r+s ) with each G i D i B i where B is unipotent i upper triangular with entries bounded in absolute value by constants depending only

6 4 16 M I ICAZA on K and n and D i diag (q (i) q (i) n ) with q (i) q (i) + C for 1 i rs1n Here the equality (q(i) ) t (i) holds for 1 n 1irand the equality q(i) q (i) t(i) holds for 1 n r1 i rs For α (α α r+s )r with α i 0 it follows that χ([αg]) χ([g]) where [αg] [(α G α r+s G r+s )] By changing the class of G to the class of αg we may assume that det G i 1 and G i D i B as before i It is clear that in order to prove the proposition it is enough to consider 0 ν ν for a fixed ν that we shall choose as follows With all notation as before let 0 ν 1 be such that 1ν (d d r+s ) m where the d i are defined as in Proposition 1 and the cone X is determined by a system of squares of fundamental units of K For ν ν let [G] χ [ν ) be given by [G] [(G G r+s )] with det G i 1 and G i D (1 0 0) we get i B i as before Letting e χ([g]) min x n K r D i B i σ i (x)r+s D i B i σ i (x) r+ r D i B i σ i (e )r+s D i B i σ i (e ) t () t (r) t (r+) t(r+s) () r+ Consider the element ω (ω ω )r+s r+s defined by ω i q(i) for i 1 and ω q() ν Then N(ω) (1ν)((t() t (r) )) ((t (r+) t(r+s) )) χ([g])ν νν 1ν 1ν By [2 Lemma 32] there exists a unit η K such that if (η η r+s )(σ (η) σ r+s (η)) then ω (η ω η ω η ω η r+s ω r+s )X Since N(ω) N(ω ) we may apply Proposition 1 to get (η q() )ν 1 (η q () ) 1 (η r q (r) )1 η r+s q (r+s) 1 (3) Consider in the class [G] ofg the representative G defined by G (η G η r+s G r+s ) (G G r+s ) Then χ[g ] χ[g] and G has a decomposition G i D i B i where D i diag (q (i) q (i) n ) diag (η i q (i) η i q (i) n ) for 1 i rs and the q (i) satisfy q (i) q (i) (+) C for all 1 rs1 and all i 1 n By equation (3) (q ())ν 1 (q ()) 1 (q () ) 1 (q (r+s)) 1 On the other hand if we put C C(r+s)(n )n/ then for i 1 rs we have 1 1 r+s q () q () n r+s (q () n ) C q() n C for i νn/(q(i) = = n C) for i 2 rs Thus C/n q (i) ν/ for i 1 and C/nν/ q (i) 1 for i 2 rs The fact that the q (i) are bounded from below for all 1 i rs together with the inequalities q (i) q (i) (+) C and 1 r+s i= q (i) q (i) allow us to obtain an upper n bound for all q (i) with 2 n and 1 i rs Hence a compact subset of (GL n ())r(gl ())s n surects onto Im (χ) [ν ) We have thus proved Theorem 2 (see the Introduction)

7 HERMITE CONSTANT AND EXTREME FORMS 17 We shall call a Humbert tuple extreme if it is a maximum of the map χ Those extreme tuples which are an absolute maximum of χ will be called critical tuples In particular Theorem 2 asserts that there exist critical Humbert tuples 4 Extreme and perfect forms In this section we assume as before that K is any number field with [K: ] m r2s and we keep all the notation already introduced We shall generalize the classical notions of eutactic and perfect forms (see [8]) and we shall partially extend Voronoi s Theorem namely by showing that extreme forms are perfect and eutactic [10] Let f ( f i ) for 1 i rs be a positive definite Humbert tuple where each f i for 1 i rs is a positive definite quadratic or hermitian form of dimension n For a vector u n K we denote its class modulo units of K by [u] that is [u] ξu:ξ K We say that the vector u (u u n ) n K is a minimal ector for f if it satisfies µ( f ) r+ i= f i (u)) r+s i=r+ ( f i (u))) LEMMA 1 ectors A positie definite tuple f has only finitely many classes of minimal Proof We keep all the notation introduced in 3 Let y (y y r+s ) with y i n for 1 i r and y i n for r1 i rs Let f ( f f r+s ) be a positive definite Humbert tuple For any positive real number T the set B T y(y y r+s ): f i (y i )T for 1 i rs is compact Hence there are only finitely many y n K such that f i (y)) T for 1 i rs For each y n K there exists ε K such that f i (εy)) σ i (ε) f i (y)) X where X X(ε ε r+s ) is defined as in 3 Therefore it is enough to prove that the set f i (y)) r+s i= i=r+ is finite But this easily follows from the considerations above and the fact that for a fixed positive real number T the set ω X: N(ω) T is bounded in rs where N(ω) is defined as in Proposition 1 X µ y n K : f i (y)) X and r+ ( f i (y)) µ Let [u ]l = be the set of classes of all minimal vectors of f For satisfying 1 l put f i (u )) λ i DEFINITION 1 Let f be a positive definite tuple with [u ]l = being its set of classes of minimal vectors Then f is perfect if there exists a system of representatives of [u ]l = namely u u l such that f is uniquely determined by the equations f i (u )) λ i for 1 i rs 11 REMARK 3 From the above definition it follows that a perfect tuple has at least n(n1) distinct classes of minimal vectors DEFINITION 2 Let f(x x n ) i a i x i x (respectively f(x x n ) i a i x i x ) be a positive definite real quadratic (respectively complex hermitian)

8 18 M I ICAZA form We define the dual form of f to be the form f *(x x n ) i A i x i x (respectively f *(x x n ) i A i x i x ) where (A ) is the adoint of the matrix i (a i ) Notice that f(x x n ) is positive definite (hermitian) if and only if f *(x x n ) is positive definite (hermitian) DEFINITION 3 With the same notation as in the previous definition let f * ( f f r ) be the dual tuple of f We say that f is eutactic if there exists (ξ(i) ξ (i) l ) l for 1 i rs with ξ(i) 0 for 1 1 such that f i (x) l ξ(i) k (u )x) for 1 i r k k= f i (x) l ξ(i) k (u k )x) (u k )x) for r1 i rs k= where σ i (u k )x denotes the standard dot product of the vectors σ i (u k ) and x (x x n ) REMARK 4 We say that a class of Humbert tuples [ f ] as defined in Section 2 is a perfect (eutactic) class if it contains a representative which is a perfect (eutactic) tuple The following lemma is a generalization of well-known classical result LEMMA 2 Let f be a Humbert tuple and let [u ]l = be the set of classes of its minimal ectors Then there exists a neighbourhood V of f such that the set of classes of minimal ectors of any element in V is contained in [u ]l = Proof Since for λ (λ λ r ) with λ i 0 the minimal vectors of f and of λf (λ f λ r f r ) are the same we assume that all the tuples to be considered have determinant 1 Assume that the lemma is not true Then for each ε 0 there exists a tuple f ε ( f ε f r+s ε) and a minimal vector ε n K of f ε such that ε is not a minimal vector of f We may assume as before that ( f i ε ( ε ))) X for 1 i rs Since ε is a minimal vector for f ε it follows that r f i ε ( ε )) r+s ( f i ε ( ε ))) C nk (d( f ε ))/n C nk (d( f ))/n C nk i= i=r+ where C nk is a constant depending only on n and on the field K Since we have assumed that ( f i ε ( ε ))) X we conclude that there exists a constant D such that nfk f i ε ( ε )) D Hence for all 1 i rs we have nfk σ i ( ε ) D nkf where D nkf depends only on n K and on the form f Since ε n K we conclude that there are only finitely many classes of such vectors ε Passing to a subset of the ε we may assume that [ ε ] for any ε small enough By continuity is a minimal vector of f which is a contradiction This proves the lemma We now state the analogue of a result due to Voronoi concerning a characterization of extreme positive definite tuples The proof of this result is obtained by generalizing that in the classical case

9 HERMITE CONSTANT AND EXTREME FORMS 19 PROPOSITION 3 Let f be an extreme Humbert tuple Then f is perfect and eutactic We prove only the case when the field K is a totally real number field The general case follows in the same way after making the obvious changes From now on we assume that the field K is totally real of degree r In order to prove Proposition 3 we need the following lemma LEMMA 3 Let f ( f f r ) and g (g g r ) be two Humbert tuples Let F ti (1t) f i tg i for 0 t 1 and 1 i r Then d log (r i= det F ti ) 0 dt The proof of Lemma 3 follows from the classical case (see for instance [8 Lemma 1 Chapter 6 Section 39]) as d log ( det F ti )dt d (log det F ti )dt Proof of Proposition 3 We first prove that an extreme tuple is perfect Let f ( f f r ) be an extreme tuple which is not perfect Then there exists g (g g r ) with f g such that g has the same classes of minimal vectors as f say [u ]l = and there is set of representatives such that g i (u )) f i (u )) λ i for 1 i r and 1 l Then the values of these forms coincide for any system of representatives of minimal vectors Put f k i s(k) x i i x g k i w(k) x i i x where 1 i n and 1 k r Assume that w(k) s(k) for some k and some (i ) i i Consider the expression F kt tg k (1t) f k f k t(g k f k ) where t Let F t (F F t nt ) then F f Since the tuple f consists of positive definite forms there is an interval (t(k) t (k) ) where F is positive definite for every kt t (t(k) t (k) ) and 1 k r Choose t inf k r t (k) Then d(f t ) r det F kt 0 for t (t t ) k= Since f is extreme there exists t 0 such that µ(f t ) d(f) /n t µ( f ) d( f ) /n µ(f ) d(f ) /n for all t (t t ) We may assume that t t Therefore by Lemma 2 there exists a class of minimal vectors [u ] [(u u n )] such that d(f) /n t r k= (s(k) i t(w (k) i s (k) i )) (σ k (u i ) σ k (u )) i d( f ) /n r s(k) i (σ k (u i ) σ k (u )) i k= for t (t t ) By assumption t(w(k) i i s (k) i )(σ k (u i )σ k (u )) 0 Hence d(f) /n t r k= ( s(k) i σ k (u i ) σ k (u )) d( f ) /n r i k= s(k) i (σ k (u i ) σ k (u )) i

10 20 M I ICAZA Therefore d( f ) d(f t ) for t (t t ) Thus the map D: defined by D(t) d(f t ) has a local minimum at 0 Lemma 3 then leads to a contradiction and the tuples f ( f f r ) and g (g g r ) must be the same that is f k g k for 1 k r We now show that an extreme tuple is eutactic Because f is eutactic if and only if for λ (λ λ )r r with λ k 0 for 1 k r the tuple λf (λ f λ r f r ) is eutactic hence we may assume that det ( f k ) 1 for 1 k r For any fixed (t(k) i ) (0) with (t (k) i ) (t (k)) consider a linear half-space i Ψ (ξ(k) k i ) n(n+)/: n t(k) ξ(k) 0 and ξ(k) ξ(k) i i i i i= Suppose that Ψ k contains the elements σ k (u is ) σ k (u s ) that is tk i σ k (u is ) σ k (u s ) 0 for 1 s l i Here u is denotes the ith coordinate of the sth minimal vector Consider the forms F ρ k f k F ρ k i (s(k ) i for k k ρt(k )) x i i x Then the tuple F ρ ( f f k F ρ k f r ) is positive definite for ρ small enough Since f is extreme we may also assume that for such a ρ the inequality d(f ρ ) /n k k i µ(f ρ ) d(f ρ ) /n µ( f ) d( f ) /n holds By Lemma 2 there exists u [u ] [(u u n )] n K such that (s(k ) ρt(k i )) σ i k (u i )(u ) s(k) σ i k (u i ) σ k (u ) i d(f) /n k i s(k) σ i k (u ) σ k (u ) By assumption t(k i ) σ i k (u i )(u )0 Hence for small enough positive ρ we have (s(k ) ρt(k i )) σ i k (u i )(u ) d(f ρ ) /n k k i s(k) σ i k (u i ) σ k (u ) i d(f ρ ) /n k i s(k) σ i k (u i ) σ k (u ) After cancelling the contribution from the k k we find that det f k det (F ρ ) By k Lemma 3 d (det (F ρ ) k dρ 0 ρ = But d(d(f ρ )) k dρ ρ = d (det (F ρ k )) dρ ρ = i t(k ) i (det f k ) s i i t(k ) i (s(k ))* i Hence the point in n(n+)/ that represents the form f k lies in the interior of any linear half-space Ψ k that contains those points representing the forms (u s )x) for

11 HERMITE CONSTANT AND EXTREME FORMS 21 1 s l In particular it lies in the interior of the convex linear hull in n(n+)/ determined by the forms (u s )x) for 1 s l This proves that the tuple f is eutactic We finally give an example and some concluding remarks We shall consider the following form g(x y) xyxy which is known to be extreme over (in fact it is critical over that is it realizes the Hermite constant γ ) Let f ( f f )be the Humbert tuple over the real quadratic number field K (d) defined by f f xyxy Since µ( f ) min (uu)(uu): (0 0) (u ) K we conclude that µ( f ) 1 We want to determine the minimal vectors of f over K that is all (u ) K with N K/ (uu) 1 This equation can be written as N K/ (u) N K/ ()[(u ) (u)] 1 (4) Notice that by symmetry we may interchange the roles of u and in the last equation If N K/ () 2 then N K/ () 1 and we get a contradiction with (4) Hence we must have N K/ () N K/ (u) If u 0 then N K/ () 1 and is then a unit Hence the class of (u ) is [(u )] [(0 1)] Similarly if 0 we obtain [(u )] [(1 0)] Assume now that N K/ () N K/ (u) 1 Since we are considering classes of minimal vectors we may put u 1 Then [(u )] [(1 ε)] with ε K Equation (4) becomes N K/ (ε) N K/ (1ε)Tr K/ (ε)tr K/ (ε) 0 Let N K/ (ε) 1 Since εtr K/ (ε) εn K/ (ε) 0 we have εtr K/ (ε) ε 1 Hence Tr K/ (ε)tr K/ (ε) 2 Inserting this in the above equation we get N K/ (1ε)Tr K/ (ε)2tr K/ (ε) 0 Since N K/ (1ε) 2Tr K/ (ε) we get 2Tr K/ (ε) Tr K/ (ε) Now K is totally real so we must have Tr K/ (ε) 2 But Tr K/ (ε) 2 and N K/ (ε) 1 imply that ε 1 Hence [(1 ε)] [(1 1)] Let N K/ (ε) 1 Then εtr K/ (ε) ε 1 and Tr K/ (ε) 2Tr K/ (ε) Our equation in this case is Tr K/ (ε)tr K/ (ε)2tr K/ (ε) 0 and this is not possible We have then proved that the only classes of minimal vectors of the tuple f over the field K are [(1 0)] [(0 1)] [(1 1)] Inserting these values in the equations defining a perfect tuple (see Definition 1) we easily see that f is a perfect tuple over K The relations defining an eutactic tuple are also satisfied by f and we think that f could be an extreme tuple (or even a critical tuple) over K In general we do not expect f to be a critical tuple for any quadratic extension K since according to [1] the invariant γ nk is bounded from below by a function that depends linearly on D K / where D K is the discriminant of K Acknowledgements The author wishes to thank R Baeza and E Friedman for their helpful comments during the preparation of this work References 1 R BAEZA The volume of the space of Humbert reduced forms preprint 2 R BAEZA and M I ICAZA Decomposition of positive definite integral quadratic forms as a sum of positive definite forms Proceedings of Symposia in Pure Mathematics 582 (American Mathematical Society Providence 1995) 3 Z BOREVICH and I R SHAFAREVICH Number theory (Academic Press New York 1967)

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