On rigidity of trinomial hypersurfaces and factorial trinomial varieties
aa r X i v : . [ m a t h . AG ] J u l ON RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIALTRINOMIAL VARIETIES
SERGEY GAIFULLIN
Abstract.
Trinomial varieties are affine varieties given by some special system of equa-tions consisting of polynomials with three terms. Such varieties are total coordinate spacesof normal rational varieties with torus action of complexity one. For an affine variety X we consider the subgroup SAut( X ) of the automorphism group generated by all algebraicsubgroups isomorphic to the additive group of the ground field. An affine variety X is rigidif SAut( X ) is trivial. In opposite an affine variety is flexible if SAut( X ) acts transitivelyon the regular locus. Arzhantsev proved a criterium for a factorial trinomial hypersurfaceto be rigid. We give two generalizations of Arzhantsev’s result: a criterium for an arbitrarytrinomial hypersurface to be rigid and a criterium for a factorial trinomial variety to berigid. Also a sufficient condition for a trinomial hypersurface to be flexible is obtained. Introduction
Let K be an algebraically closed field of characteristic zero and G a be its additive group.Assume X is an affine algebraic variety over K . One of approaches to investigate the groupof regular automorphisms of X is to study G a -actions on X .Let A be a commutative associative algebra over K . A linear mapping ∂ : A → A iscalled a derivation if it satisfies the Leibniz rule ∂ ( ab ) = a∂ ( b ) + b∂ ( a ) for all a, b ∈ A .A derivation is called locally nilpotent (LND) if for any a ∈ A there is a positive integer n such that ∂ n ( a ) = 0. Exponential mapping ∂
7→ { exp( t∂ ) | t ∈ K } gives a correspondencebetween LNDs and algebraic subgroups in Aut( A ) isomorphic to G a . This correspondenceprovides the technique for investigation of G a -actions.There are two opposite situations. If X does not admit non-trivial G a -action, the variety X is called rigid . A lot of examples of rigid varieties are given in [6, 10, 13], see also [12,Chapter 10]. A rigid variety has the unique maximal torus in its automorphism group [3].This fact sometimes allows to describe all automorphisms of X . For example if X is a rigidtrinomial hypersurface the automorphism group of X is a finite extension of the maximaltorus. If for every regular point x ∈ X the tangent space T x X is spanned by tangent vectorsto G a -orbits for various regular G a -actions, then X is called flexible . Flexible varieties wereinvestigated in [4]. It was proved that flexibility is equivalent to transitivity and infinitelytransitivity of the group of special automorphism on the set of regular points of X , seeTheorem 1. Flexibility of some classes of affine varieties is proved in [4, 5, 8, 19]. In somesense flexibility of a variety means that there are a lot of G a -actions on X . That is anopposite situation to rigidity of X . We call intermediate such varieties that are neitherrigid nor flexible. Mathematics Subject Classification.
Primary 14R20,14J50; Secondary 13A50, 14L30.
Key words and phrases.
Affine variety, locally nilpotent derivation, graded algebra, torus action,trinomial.The author was supported by the Foundation for the Advancement of Theoretical Physics and Mathe-matics “BASIS”.
In this paper we investigate the class of trinomial varieties . These are varieties given bysystems of polynomial equations of the form T l . . . T l n n + T l . . . T l n n + T l . . . T l n n = 0 , n ≥ , n , n ≥ , (1)see Definition 4 and [15, Construction 1.1]. Every trinomial variety admits an algebraictorus action of complexity one, i.e. the codimension of a generic orbit equals one. Theclass of trinomial varieties is interesting since every normal, rational variety X with onlyconstant invertible functions, finitely generated divisor class group and an algebraic torusaction of complexity one can be obtained as a quotient of a trinomial variety via action ofa diagonasible group, see [15, Corollary 1.9]. This trinomial variety is the total coordinatespace of X . More information on total coordinate spaces and Cox rings one can find in [2].The simplest case of a trinomial variety is a trinomial hypersurface given by a singleequation of the form (1) in K n + n + n . A trinomial hypersurface is factorial if and only ifone of the following holds • n = 0 and gcd( l i , . . . , l in i ) = 1 for i = 1 , • n ≥ d i = gcd( l i , . . . , l in i ) are pairwise coprime.Arzhantsev [1] gives a criterium for a factorial trinomial hypersurface X with n ≥ X is rigid if and only if all l ij ≥
2. It is easy to see, that for arbitrarytrinomial hypersurface X if there is l ij = 1, then X is not rigid. But the condition l ij > i and j is not sufficient for a non-factorial trinomial hypersurface to be rigid. Weprove a criterium for an arbitrary trinomial hypersurface to be rigid, see Theorem 2. Everyhypersurface, which is not rigid and for which l ij > i and j , has the following form V (cid:16) T T m . . . T m n n + T T m . . . T m n n + T l . . . T l n n (cid:17) . (2)Note that we do not use Arzhantsev’s result and thus reprove it as a particular case ofTheorem 2. In [9] and [21] all finely homogeneous (i.e. homogeneous with respect to thefinest graduation such that all T ij are homogeneous) LNDs of the algebra of regular functionson a trinomial hypersurface with n ≥ i, j ) such that l ij = 1, see Theorem 3.In the last two sections we investigate nonrigid trinomial hypersurfaces. A trinomialhypersurface X admits an action of algebraic torus T of complexity one. For nonrigidtrinomial hypersurfaces we can explicitly construct some T -homogeneous LNDs, see Lem-mas 14 and 15. This allows to prove a sufficient condition of trinomial hypersurface to beflexible, see Theorem 4. For two classes of trinomial hypersurfaces we prove that they areintermediate, see Propositions 2 and 3.The author is grateful to Ivan Arzhantsev, Roman Budylin, Alexander Perepechko, andYulia Zaitseva for useful discussions. N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 3 Derivations
Let F be an abelian group. An algebra A is called F -graded if A = M f ∈ F A f , and A f A g ⊂ A f + g . A derivation ∂ : A → A is called F -homogeneous of degree f ∈ F if forevery a ∈ A f , we have ∂ ( a ) ∈ A f + f .Let A be a finitely generated Z -graded algebra. Let ∂ be a derivation of A . It is easy tosee, that there exists a decomposition ∂ = P ki = l ∂ i into a sum of homogeneous derivations,deg ∂ i = i . The following lemma can be found in [20], see also [11]. Lemma 1.
Let ∂ be an LND and ∂ = P ki = l ∂ i , where ∂ i is a homogeneous derivation ofdegree i . Then the extreme summands ∂ l and ∂ k are LND.Remark . Further when we wright ∂ = P ki = l ∂ i , we assume that ∂ l = 0 and ∂ k = 0.Lemma 1 implies the following. Lemma 2. If A admits a nonzero LND, then A admits a nonzero Z -homogeneous LND. Let F ∼ = Z n . Assume A is a finitely generated F -graded algebra. Applying the result ofLemma 2 n times we obtain the following lemma. Lemma 3. An F -graded algebra A admits an LND if and only if A admits an F -homogeneous LND. Let A = K [ X ] be the algebra of regular functions on an affine algebraic variety X . Thegroup X ( T ) of characters of n -dimensional torus T = ( K × ) n is isomorphic to Z n . Then Z n -gradings on K [ X ] are in bijection to regular T -actions on X . If we have a X ( T )-homogeneousderivation on K [ X ], we call it a T -homogeneous derivation.Let F and S be abelian groups. Assume A = L f ∈ F A f is an F -graded algebra. Let π : F → S be a homomorphism. Then A can be considered as S -graded algebra by thefollowing way A = M s ∈ S A s , where A s = M π ( f )= s A f . The proof of the following lemma can be found in [7].
Lemma 4.
Assume ∂ is an F -homogeneous derivation of degree f . Then ∂ is an S -homogeneous derivation of degree s = π ( f ) . An affine algebraic variety X is called rigid if the algebra K [ X ] admits no nonzero LNDs.An affine algebraic variety X is called flexible if for every regular point x ∈ X reg , the tangentspace T x X is spanned by tangent vectors to G a -orbits for various regular G a -actions. Thesubgroup of Aut( X ) generated by all algebraic subgroups isomorphic to G a is called subgroupof special automorphisms . We denote it by SAut( X ).Let a group G act on a set X . This action is called m -transitive if for every two m -tuples( a , . . . , a m ) and ( b , . . . , b m ), where a i = a j and b i = b j if i = j , there is an element g in G such that for all i we have g · a i = b i .If an action is m -transitive for every positive integer m , then it is called infinitely tran-sitive .One of the main results on flexible varieties is the following. SERGEY GAIFULLIN
Theorem 1. [4, Theorem 0.1]
For an irreducible affine variety X of dimension ≥ , thefollowing conditions are equivalent.(i) The group SAut( X ) acts transitively on X reg ,(ii) The group SAut( X ) acts infinitely transitively on X reg ,(iii) The variety X is flexible. Definition 1.
The subring ML( X ) ⊂ K [ X ], which is equal to the intersection of kernels ofall LNDs of K [ X ], is called the Makar-Limanov invariant of X .It is easy to see, that X is rigid if and only if ML( X ) = K [ X ]. From the other hand if X is flexible, then ML( X ) = K . But the condition ML( X ) = K does not imply flexibilityof X , see [17, § m -suspensions In this section we elaborate technique, which we use in the next sections. The mainconcept of this section is m -suspension, defined in [7]. Definition 2.
Fix a positive integer number m . Let X be an affine variety. Given anonconstant regular function f ∈ K [ X ] and positive integers k , . . . , k m we define a newaffine variety Y = Susp( X, f, k , . . . , k m ) = V (cid:0) y k y k . . . y k m m − f ( x ) (cid:1) ⊂ K m × X, called an m -suspension over X with weights k , . . . , k m .We have a natural linear action of m -dimensional algebraic torus ( K × ) m on m -dimensionalaffine space with coordinates y , . . . , y m . The stabilizer H of the monomial y k y k . . . y k m m is isomorphic to the direct product of the ( m − T and a finite group Z gcd( k ,...,k m ) . We have an effective action of H on Y = Susp( X, f, k , . . . , k m ).The variety Y can be reducible. For example, let X be a line, then K [ X ] = K [ x ]. Considerthe m -suspension Y = Susp( X, x , , . . . , K m +1 by the unique equation y . . . y m = x . Hence Y is reducible. During this section we assume that Y is irreducible. Lemma 5. [7, Lemma 3.4]
Let ∂ be a T -homogeneous LND on K [ Y ] . Then there is i ∈{ , , . . . , m } such that for all j = i we have ∂ ( y j ) = 0 . Consider the field L j = K ( y j ), which is the algebraic closure of K ( y j ). If there is a fieldsembedding K ⊂ L and Z is an affine algebraic variety over K , we denote by Z ( L ) the affinealgebraic variety over L given by the same equations as Z . Then we have Y ( L j ) = V (cid:0) y k . . . y k m m − f (cid:1) ⊂ L m − j × X ( L j ) . Since y j is a constant now, Y ( L j ) is a ( m − X ( L j ).Denote by Y j the affine algebraic variety over L j given by V (cid:16) y k . . . y k j − j − y k j +1 j +1 . . . y k m m − f (cid:17) ⊂ L m − j × X ( L j ) . It is easy to see, that Y j ∼ = Y ( L j ). Remark . The variety Y j can be reducible.The T -action on Y corresponds to a Z m − -grading of K [ Y ]. Often it is convenient toconsider Z -gradings, which are its coarsenings. Let us denote by h ij the Z -grading givenby deg( y i ) = − k j , deg( y j ) = k i . Degrees of all other y n and degrees of all g ∈ K [ X ] areequal to zero. N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 5
Lemma 6. [7, Lemma 3.7] If Y i is a rigid variety, then for any nonzero T -homogeneousLND ∂ : K [ Y ] → K [ Y ] we have1) ∂ ( y i ) = 0 ,2) ∀ j = i, ∂ ( y j ) = 0 ,3) deg( ∂ ) > with respect to Z -grading h ij . Lemma 7.
Let ≤ i = j ≤ m . Assume Y i and Y j are rigid varieties. Then Y is rigid.Proof. By Lemma 6, we obtain that any T -homogeneous LND on K [ Y ] is zero. So, byLemma 3 the variety Y is rigid. (cid:3) Lemma 8.
Let Y = Susp( X, f, k = 2 , k = a, k , . . . , k m ) , where a is an odd number.Suppose Y is rigid. Let ∂ be a T -homogeneous LND of K [ Y ] . Then for every g ∈ K [ X ] wehave y | ∂ ( g ) .Proof. Consider the grading h . Since ∂ is T -homogeneous, it is h -homogeneous. ByLemma 6 we have deg ∂ > ∂ ( y ) = 0, and ∂ ( y ) = 0. If deg ∂ < a , then deg( ∂ ( y )) < y | ∂ ( y ). This gives a contradiction with ∂ ( y ) = 0, see [12, Principle 5]. Ifdeg ∂ > a , then ∂ = y γ , where γ is a h -homogeneous LND of K [ Y ]. Let us change ∂ to δ = ∂y k for maximal possible k . Then deg δ = a . Since deg g = 0, we have deg( δ ( g )) = a .Since a is odd, we obtain y | δ ( g ). (cid:3) Lemma 9.
Let Y = Susp( X, f, k , k , . . . , k m ) and Z = Susp( X, f, ck , k , . . . , k m ) for some positive integer c . Suppose Z is rigid. Then for all j ≥ we have y j ∈ ML( Y ) .Proof. Let us fix 2 ≤ j ≤ m . Consider the Z -grading h j of K [ Y ]. Let ∂ be an h j -homogeneous LND of K [ Y ].Suppose deg ∂ ≤
0. Then deg( ∂ ( y )) <
0. Therefore, y | ∂ ( y ). Hence ∂ ( y ) = 0, see [12,Principle 5]. Let us put y = w c . Then ∂ induces a nonzero LND on { w ck y k . . . y k m m = f } ∼ = Z, a contradiction. Therefore, deg ∂ > δ be a nonzero LND of K [ Y ]. Consider δ = P ki = l δ i the decomposition intohomogeneous summands. Then δ l is a nonzero LND of degree l . Therefore, l >
0. Then y j | δ ( y j ). Since δ is an LND, we obtain δ ( y j ) = 0. So, y j ∈ ML( Y ). (cid:3) Lemma 10.
Let X be a nonrigid affine variety. Then Y = Susp( X, f, , k , . . . , k m ) is not rigid.Proof. Since X is not rigid, there exists a nonzero LND ∂ of K [ X ]. Let us show that thereexists a nonzero LND δ of K [ X ][ y , . . . , y m ] / ( y y k . . . y k m m − f ) = K [ Y ] . Define δ ( y ) = ∂ ( f ) ,δ ( y j ) = 0 , j ≥ ,δ ( g ) = ∂ ( g ) y k . . . y k m m , g ∈ K [ X ] . It is easy to see that δ is a nonzero LND of K [ Y ]. (cid:3) SERGEY GAIFULLIN
Lemma 11.
Assume Y = Susp( X, f, k , . . . , k m ) is not rigid. Let d = gcd( k , . . . , k m ) .Then Z = Susp( X, f, d ) is not rigid.Proof. Since Y is not rigid, by Lemma 3 there is a nonzero T -homogeneous LND ∂ of K [ Y ].Let us denote h = y k d . . . y kmd m . Consider X ( T )-grading of K [ Y ]. We have K [ Y ] = K [ Y ] T = K [ X ][ h ] ∼ = K [ X ][ z ] / ( z d − f ) . If deg( y a . . . y a m m ) = deg( y b . . . y b m m ) , then there is p ∈ Z such that ∀ i : a i − b i = p k i d . Therefore, for every α ∈ X ( T ) there exist nonnegative integers s ( α ) , . . . , s m ( α ) such that K [ Y ] α = y s ( α )1 . . . y s m ( α ) m K [ Y ] . Let β = deg ∂ ∈ X ( T ). Let us define δ : K [ Y ] → K [ Y ] by the rule δ ( F ) = ∂ ( F ) y s ( β )1 . . . y s m ( β ) m , F ∈ K [ Y ] . We have δ ( F G ) = ∂ ( F G ) y s ( β )1 . . . y s m ( β ) m = ∂ ( F ) G + F ∂ ( G ) y s ( β )1 . . . y s m ( β ) m = δ ( F ) G + F δ ( G ) . Hence, δ is a derivation of K [ Y ] .Define ν ∂ : K [ Y ] \ { } → Z ≥ , where Z ≥ is the set of nonnegative integers. For nonzero g ∈ K [ Y ], let l = ν ∂ ( g ) + 1 be the least integer number such that ∂ l ( g ) = 0. By [12,Proposition 1.9], ν ∂ is a degree function on K [ Y ]. Therefore, ν ∂ ( F ) = ν ∂ ( ∂ ( F )) + 1 > ν ∂ ( ∂ ( F )) ≥ ν ∂ ∂ ( F ) y s ( β )1 . . . y s m ( β ) m ! = ν ∂ ( δ ( F )) . This implies that δ is an LND of K [ Y ] ∼ = K [ Z ]. That is Z is not rigid. (cid:3) Rigid trinomial hypersurfaces
Let us consider an affine space K n . We fix a partition n = n + n + n , where n is anonnegative integer and n , n are positive integers. The coordinates on K n we denote by T ij , i ∈ { , , } , 1 ≤ j ≤ n i . Definition 3. A trinomial hypersurface is a subvariety X in K n given by the unique equa-tion T l T l . . . T l n n + T l T l . . . T l n n + T l T l . . . T l n n = 0 , where l ij are positive integers. If n = 0, we obtain the variety V (cid:16) T l . . . T l n n + T l . . . T l n n (cid:17) . We call such a variety a trinomial hypersurface with free term . If n = 0 then X is a trinomial hypersurface without free term . N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 7
By [15, Theorem 1.2(i)] every trinomial hypersurface is irreducible and normal.We denote the monomial T l i i T l i i . . . T l ini in i by T l i i . So, trinomial hypersurface has the form V (cid:0) T l + T l + T l (cid:1) . If n i = 1 and l i = 1, then X ∼ = K n − . If for all i we have l i n i >
1, then X is factorial ifand only if the numbers d i = gcd( l i , ..., l in i ) , i = 0 , , l ij ≥
2, see [1, Theorem 1].The following theorem gives a criterium for an arbitrary trinomial hypersurface to berigid. It is a generalization of [1, Theorem 1].
Theorem 2.
A trinomial hypersurface X = V (cid:0) T l + T l + T l (cid:1) is not rigid if and only ifone of the following conditions holds1) there exist i ∈ { , , } and a ∈ { , , . . . , n i } such that l ia = 1 ,2) n = 0 and there exist i = j ∈ { , , } and a ∈ { , , . . . , n i } , b ∈ { , , . . . , n j } suchthat l ia = l jb = 2 and for all u ∈ { , , . . . , n i } , v ∈ { , , . . . , n j } the numbers l iu and l jv are even.Proof. If condition 1) or 2) holds, we can find a nonzero LND of K [ X ]. In case 1) we canassume i = 1 , a = 1. Then there is the following derivation δ on K [ X ], see [1]: δ ( T ) = ∂T ∂T , δ ( T ) = − ∂T ∂T , δ ( T pq ) = 0 if (1 , = ( p, q ) = (2 , . It is easy to check that δ ( T l + T l + T l ) = 0 and δ is an LND.In case 2) we can assume i = 0 , a = 1, j = 1 , b = 1. Denote for all u and vl u = 2 m u , l v = 2 m v . Then X is given by the equation T T m . . . T m n n + T T m . . . T m n n + T l T l . . . T l n n = 0 . For k = 0 , q T l k k = T k T m k k . . . T m knk kn k .The variety X is isomorphic to the variety Y = V (cid:0) T l − T l − T l (cid:1) . There is the following derivation δ of K [ Y ]: δ ( T ) = ∂T l ∂T · q T l T , δ ( T ) = ∂T l ∂T · q T l T ,δ ( T ) = 2 q T l T · q T l T (cid:18)q T l − q T l (cid:19) , δ ( T ij ) = 0 if j = 1 . One can check that δ (cid:0) T l − T l − T l (cid:1) = 0. Therefore, δ is a derivation of K [ Y ]. It is easyto see that δ (cid:18)q T l − q T l (cid:19) = 0 . This implies δ ( T ) = 0. Therefore, δ l +1 ( T ) = δ l +1 ( T ) = 0 . SERGEY GAIFULLIN
That is δ is an LND.Now let us assume that X is a trinomial hypersurface, for which conditions 1) and 2) donot hold. Let us prove that X is rigid by induction on n. Base of induction consists of 4particular cases. Case 1. X ∼ = V (cid:0) x a + y b + 1 (cid:1) . Since the condition 1) does not hold, we have a, b >
1. Itis proven in [10, Section 2] that X is rigid. Case 2. X ∼ = V (cid:0) x a + y b + z c (cid:1) . Since conditions 1) and 2) do not hold, we have a, b, c > X is rigid. Case 3. X ∼ = V (cid:0) x + y b + z w c (cid:1) , b = 2 , ∤ c. It is easy to see, that X is a 2-suspension over an affine plane X ∼ = Susp( K , − ( x + y b ) , , c ). Case 2 provides rigidityof V (cid:0) x + y b + w c (cid:1) . Therefore, Lemma 8 implies that for every T -homogeneous LND δ of K [ X ] we have z | δ ( x ) and z | δ ( y ).By Lemma 6, δ ( w ) = 0. Let us consider L = K ( w ). Then δ induces an LND ∂ of L [ Y ] = L [ x, y, z ] / ( x + y d + z ). We have z | ∂ ( x ) and z | ∂ ( y ).Let us consider the following Z -grading on L [ Y ]deg x = deg z = d, deg y = 2 . Consider the decomposition of ∂ into the homogeneous summands ∂ = P ki = l ∂ i . Since z ishomogeneous, we obtain z | ∂ l ( x ) and z | ∂ l ( y ). By Lemma 1, the derivation ∂ l is locallynilpotent.If d | l , then d ∤ ( l + 2). Therefore, y | ∂ l ( y ). Hence, ∂ l ( y ) = 0. Consider F = L ( y ). Then ∂ l induces an LND of F [ x, z ] / ( x + z + 1). But Case 1 implies that there are no nonzeroLNDs on F [ x, z ] / ( x + z + 1). Therefore, ∂ l is zero.If d ∤ l , then d ∤ ( l + d ). Therefore, y | ∂ l ( z ). Hence either ∂ l ( y ) = 0 or ∂ l ( z ) = 0. In bothcases by the same arguments as above ∂ l is zero.It is easy to see, that ∂ l = 0 implies δ = 0. That is X is rigid. Case 4. X ∼ = V (cid:0) x + y v b + z w c (cid:1) , ∤ b > , ∤ c > . Let us consider the following Z -grading of K [ X ]deg x = (0 , , deg y = ( − b, , deg v = (2 , , deg z = (0 , − c ) , deg w = (0 , . Suppose X is not rigid. Then by Lemma 3 there exists a Z -homogeneous derivation ∂ .Again we can consider X as 2-suspension over an affine space X ∼ = Susp( K , − ( x + y v b ) , , c ) . Case 3 provides rigidity of V (cid:0) x + y v b + w c (cid:1) . Therefore ∂ ( w ) = 0 and Lemma 8 impliesthat z | ∂ ( y ) and z | ∂ ( x ). Analogically ∂ ( v ) = 0, y | ∂ ( z ) and y | ∂ ( x ). Hence either ∂ ( z ) = 0 or ∂ ( y ) = 0. Considering the field K ( v, w ) we obtain the same situation as in Case3. So we obtain a contradiction, which implies that X is rigid. Inductive step . Let X be a a trinomial hypersurface, for which conditions 1) and 2) donot hold. Suppose X is not rigid. Note that X is an n -suspension over K n + n . Let usconsider the varieties X j = V T l + T l + T l T l j j ! , j = 1 , , . . . , n . Since the condition 1) does not hold for X , it does not hold for all X j . If for some X j thecondition 2) does not hold, then by induction hypothesis X j is rigid. By Lemma 7 if forsome i = j the varieties X i and X j are rigid, then X is rigid. Since X is not rigid, there is N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 9 i ∈ { , , . . . , n } such that for all j = i , for X j the condition 2) holds. We have analogicalsituation for the other monomials. It is easy to see that this is possible if and only if X isisomorphic to varieties from Cases 3 or 4. Theorem 2 is proved. (cid:3) Example 1.
In [13, Section 9.1] the following trinomial variety is considered X = V (cid:0) t d x a + y b + z c (cid:1) . Theorem 2 implies that X is not rigid if and only if one of the following conditions hold1) 1 ∈ { a, b, c, d } ;2) X ∼ = V (cid:0) t d x a + y + z (cid:1) ;3) X ∼ = V ( t m x + y + z c ).This coincides with the assertions of [13, Theorem 9.1] and [13, Remark 9.2].In [3] the automorphism group of a rigid trinomial hypersurface without free term isdescribed. So, using Theorem 2 we obtain description of the automorphism group of sometrinomial hypersurfaces. Example 2.
Let X = V ( x y + z u + v w ). Then X is not factorial and it is rigid byTheorem 2. Therefore, we can apply [3, Theorem 5.5] to describe automorphisms of X . Wehave Aut( X ) ∼ = S ⋌ (( Z / Z ) × ( Z / Z ) × T ) , where S permutes the monomials, T is the 4-dimensional torus acting by( t , t , t , t ) · ( x, y, z, u, v, w ) = ( t t x, t − y, t t z, t − u, t t v, t − w ) , the first copy of Z / Z acts by multiplying u by √ w by √
1. 5.
Trinomial varieties
Let us remind [15, Construction 1.1]. Fix integers r, n > m ≥
0, and q ∈ { , } . Andfix a partition n = n q + . . . + n r , n i > . For each i = q, . . . , r fix a tuple l i = ( l i , l i , . . . , l in i ) of positive integers and define amonomial T l i i = T l i i . . . T l ini in i ∈ K [ T ij , S k | q ≤ i ≤ r, ≤ j ≤ n i , ≤ k ≤ m ] . We write K [ T ij , S k ] for the above polynomial ring. Now we define a ring R ( A ) for someinput data A . Type 1. q = 1, A = ( a , . . . , a r ), a j ∈ K , if i = j , then a i = a j . Set I = { , ..., r − } andfor every i ∈ I let us define a polynomial g i = T l i i − T l i +1 i +1 − ( a i +1 − a i ) ∈ K [ T ij , S k ] . Type 2. q = 0, A = (cid:18) a a a . . . a r a a a . . . a r (cid:19) is a 2 × ( r + 1)-matrix with pairwise linearly independent columns. Set I = { , ..., r − } and define for every i ∈ I a polynomial g i = det T l i i T l i +1 i +1 T l i +2 i +2 a i a i +1 a i +2 a i a i +1 a i +2 ∈ K [ T ij , S k ] . For both Types we define R ( A ) = K [ T ij , S k ] / ( g i | i ∈ I ). Definition 4.
The variety X = Spec( R ( A )) for some A we call a trinomial variety .By [15, Theorem 1.2(i)] every trinomial variety is irreducible and normal. We need thefollowing fact. Proposition 1. [15, Theorem 1.2(iv)]
Suppose r ≥ and n i l ij > for all i, j . Then(a) in case of Type 1, R ( A ) is factorial if and only if one has gcd( l i , . . . , l in i ) = 1 for i = 1 , . . . , r ;(b) in case of Type 2, R ( A ) is factorial if and only if the numbers d i := gcd( l i , . . . , l in i ) are pairwise coprime. Now let us give a criterium for a factorial trinomial variety to be rigid.
Theorem 3.
Let X be a trinomial variety of Type 1 or a factorial trinomial variety ofType 2. Then X is not rigid if and only if one of the following holds:1) m = 0 ,2) In case of Type 1 there is b ∈ { , . . . r } such that for each i ∈ { , . . . r } \ { b } there is j ( i ) ∈ { , . . . n i } such that l ij ( i ) = 1 ,3) In case of Type 2 there are b and c in { , . . . r } such that for each i ∈ { , . . . r } \ { b, c } there is j ( i ) ∈ { , . . . n i } such that l ij ( i ) = 1 .Proof. If m = 0, we have an LND ∂∂S of R ( A ). So, in the sequel we assume m = 0.If r = 2, X is a trinomial hypersurface and the assertion of the theorem follows fromTheorem 2.Let us prove by induction on r that if one of conditions 2) and 3) holds, then X is notrigid. Base of induction r = 2 is already proved.Assume r >
2. If condition 2) holds, we can assume that b = 1. Indeed, the system ofequations of X is not symmetric with respect to groups T i . But we can replace equationsby linear combinations of them in such a way that we obtain system of equations of Type 1with permuting groups. Analogically, if the condition 3) holds, we can assume b = 0 , c = 1.Consider A = K [ T ij | q ≤ i ≤ r − , ≤ j ≤ n i ] ,Z = Spec ( A / ( g i | i ∈ I \ { r − } )) in case of Type 1 ,Z = Spec ( A / ( g i | i ∈ I \ { r − } )) in case of Type 2 . Then X = Susp( Z, f, l r , . . . , l rn r ), where f = T l r − r − + p for some p ∈ K in case of Type 1and f = p T l r − r − + p T l r − r − for some p , p ∈ K in case of Type 2. Lemma 10 implies that if Z is not rigid, then X is not rigid. This completes the inductive step.Now, let us prove that if neither 2) nor 3) holds, then X is rigid. Case of Type 1.
Suppose, condition 2) does not hold, but X is not rigid. We haveseen that X is an n r -suspension over Z . By Lemma 7 we can decrease m until m = 1.Analogically, we can decrease the numbers of variables T ij for all i . So, we obtain a trinomialvariety e X such that • e X does not satisfy the condition 2), • e X is not rigid, • for e X we have n = n = . . . = n r = 1. N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 11
We use notations e T ij and e l ij for variables and their powers in equations of e X . If e l ij = 1,then e X is isomorphic to a trinomial variety with less r . We assume that we have eliminatedall variables e T ij with e l ij = 1. So now all e l ij ≥ r ≥ Lemma 12.
Let B be a commutative K -domain and ∂ be non-zero LND of B . Suppose u, v ∈ Ker ∂ and x, y ∈ B are non-zero, and a , b are integers with a, b ≥ . Assume ux a + vy b = 0 . If ∂ ( ux a + vy b ) = 0 , then ∂ ( x ) = ∂ ( y ) = 0 . In our situation B = K [ e X ], u = 1, v = − x = e T i , y = e T i +1 , , a = e l i , b = e l i +1 , .Then ux a + vy b = e T e l i i − e T e l i +1 , i +1 , = 0 ∈ K . Hence, ∂ ( ux a + vy b ) = 0. Then, by Lemma 12, ∂ ( e T i ) = ∂ ( e T i +1 , ) = 0. Since this is true for all i , e X is rigid. Therefore, X is rigid. Case of Type 2.
We use induction on r . Base of induction r = 2 is already provedin Theorem 2.Since the condition 3) does not hold, there are a, b, c ∈ { , , . . . , r } such that for allpossible j we have l aj ≥ , l bj ≥ , l cj ≥
2. We can assume that a = 0 , b = 1 , c = 2. Againconsider Z . Since X is factorial any two of d i are coprime. Hence, Z is factorial too. In thestep of induction it is sufficiently to show that if r ≥ Z implies rigidity of X .There are two cases. Case A. n r = 1 . If l r = 1, then X ∼ = Z . So, in the sequel we assume l r ≥ Lemma 13. [13, Corollary 4.4]
Suppose R is a K -domain, g is a Z -grading of R , and f ∈ R is homogeneous of degree φ ∈ Z ( φ = 0 ). Assume that n ≥ is an integer such that B = R [ z ] / ( f + z n ) is a domain, and let h be the Z -grading of B defined by h = ( n g , φ ) . If gcd( φ, n ) = 1 , then ∂ (ML( R )) = 0 for every h -homogeneous LND ∂ of B . We would like to use this lemma in case when R = K [ Z ] , B = K [ X ] , f = − ( p T l r − r − + p T l r − r − ) , z = T r , n = l r . There are integers u i , . . . , u in i such that u i l i + . . . + u in i l in i = d i . Let us define a Z -grading g of R by the rule deg T ij = u ij d d . . . d i − d i +1 . . . d r − . Then deg f = r − Y k =1 d i . Since n r = 1, we have l r = d r . By Proposition 1,gcd( d r , deg f ) = gcd d r , r − Y k =1 d i ! = 1 . We see that all conditions of Lemma 13 are satisfied. Therefore, for every h -homogeneousLND ∂ we obtain ∂ (ML( R )) = 0. But we assume that R = K [ Z ] is rigid. That isML( R ) = R . We obtain that if i ≤ r −
1, then ∂ ( T ij ) = 0 for all j . This implies ∂ ( T r ) = 0,so ∂ ≡
0. Lemma 2 implies that X is rigid. Case B. n r ≥ . We see that X is an n r -suspension over Z . Suppose X is not rigid. ByLemma 11 the variety V = V (cid:0) f − y d r (cid:1) ⊂ Z × K is not rigid. If d r = 1 then V ∼ = Z . But Z is rigid. Therefore, d r ≥
2. Hence, V is afactorial trinomial variety, such that condition 3) does not hold. Therefore, we are in CaseA. Hence, V is rigid. This implies that X is rigid. Theorem 3 is proved. (cid:3) Example 3.
The following subvariety X in K is rigid ( x + y + z = 1; y + z + w = 1 . Indeed, it is easy to see that X is isomorphic to the variety Y given by ( T T − T = 1; T − T T = 1 . If we take A = (1 , , T l = T , T l = T T and T l = T , we obtain Y as a trinomialvariety of Type 1. Therefore, Theorem 3 implies rigidity of Y .6. Flexible trinomial hypersurfaces
In Section 4 we have obtained a criterium for trinomial hypersurface to be rigid. Now letus consider trinomial hypersurfaces which are not rigid. The opposite to rigidity situationis flexibility. Let us prove that some trinomial hypersurfaces are flexible. We introduce fivespecial types of trinomial hypersurfaces, which we denote H - H . H V (cid:0) T l + T l + T T . . . T n (cid:1) H V (cid:0) T T . . . T n + T T . . . T n + T l (cid:1) H V (cid:16) T l + T T l . . . T l n n + T T l . . . T l n n (cid:17) H V (cid:16) T T l . . . T l n n + T T m . . . T m n n + T T m . . . T m n n (cid:17) H V (cid:16) T T m . . . T m n n + T T m . . . T m n n + T T m . . . T m n n (cid:17) These types have nontrivial intersections. For example, the variety V ( T + T + T )belongs to Type H and Type H . Remark . We use the same notations as in Section 4. In particular, n ≥ n , n > V (1 + T T + T ) is of Type H , but the variety V (1 + T + T ) is not of Type H . Theorem 4.
Trinomial hypersurfaces of Types H - H are flexible. To prove this theorem we need two lemmas.
N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 13
Lemma 14.
Let X = V (cid:16) T l + T l + T T l . . . T l n n (cid:17) . Let us fix n − numbers c , . . . , c n ∈ K . Consider the subvariety X ( c , . . . , c n ) = V ( T − c , . . . , T n − c n ) ⊂ X. If for all k ∈ { , . . . , n } we have c k = 0 , then SAut( X ) acts on X ( c , . . . , c n ) transitively.Proof. For every i ∈ { , } and j ∈ { , , . . . , n i } we can define an LND γ ij of K [ X ] by γ ij ( T ) = − ∂T i ∂T ij , γ ij ( T ij ) = ∂T ∂T , γ ij ( T pq ) = 0 for all other pairs ( p, q ) . Denote τ ij ( s ) = exp( sγ ij ). Then τ ij ( s )( T ij ) = T ij + sT l . . . T l n n ,τ ij ( s )( T pq ) = T pq for all pairs ( p, q ) = (2 , p, q ) = ( i, j ).Let P, Q ∈ X ( c , . . . , c n ). Since T l . . . T l n n ( P ) = T l . . . T l n n ( Q ) = 0 , we can take s ij = T ij ( Q ) − T ij ( P ) T l . . . T l n n ( P ) . Put R = τ ij ( s ij )( P ). Then T ij ( R ) = T ij ( Q ).Starting from P we can change by τ ij ( s ij ) all T ij . As the result we get a point S ∈ X ( c , . . . , c n ) such that T ij ( S ) = T ij ( Q ) for all i = 0 , j = 1 , , . . . , n i . Then T ( S ) = − T l ( S ) + T l ( S ) T l . . . T l n n ( S ) = − T l ( Q ) + T l ( Q ) T l . . . T l n n ( Q ) = T ( Q ) . Therefore, S = Q . That is P and Q are in one SAut( X )-orbit. (cid:3) Lemma 15.
Consider X = V (cid:16) T T m . . . T m n n − T T m . . . T m n n − T l (cid:17) . Let us fix n + n − numbers: a , . . . , a n , b , . . . , b n . Consider the subvariety X ( a , . . . , a n , b , . . . , b n ) = V ( T − a , . . . , T n − a n , T − b , . . . , T n − b n ) ⊂ X. If all a u = 0 and all b v = 0 , then SAut( X ) acts on X ( a , . . . , a n , b , . . . , b n ) ∩ X reg transitively.Proof. For each 1 ≤ i ≤ n we have two LNDs δ i + and δ i − of K [ X ] given by δ i + ( T ) = ∂T l ∂T i q T l T ; δ i + ( T ) = − ∂T l ∂T i q T l T ; δ i + ( T i ) = 2 q T l q T l T T (cid:18)q T l + q T l (cid:19) ; δ i + ( T jk ) = 0 for all ( j, k ) / ∈ { (0 , , (1 , , (2 , i ) } . δ i − ( T ) = ∂T l ∂T i q T l T ; δ i − ( T ) = ∂T l ∂T i q T l T ; δ i − ( T i ) = 2 q T l q T l T T (cid:18)q T l − q T l (cid:19) ; δ i − ( T jk ) = 0 for all ( j, k ) / ∈ { (0 , , (1 , , (2 , i ) } . As in the proof of Theorem 2 one can check that δ i + and δ i − are LNDs of K [ X ].Denote α = q T l − q T l ,β = q T l + q T l . Then δ i + ( α ) = δ i − ( β ) = 2 ∂T l ∂T i q T l q T l T T , δ i + ( β ) = δ i − ( α ) = 0 . Denote ϕ i + ( s ) = exp( sδ i + ) and ϕ i − ( s ) = exp( sδ i − ). We have ϕ i ± ( s )( T i ) = T i + 2 s q T l q T l T T (cid:18)q T l ± q T l (cid:19) . Let us fix a point Q in X ( a , . . . , a n , b , . . . , b n ) ∩ X reg such, that for all i and j we have T ij ( Q ) = 0. Let P be an arbitrary point in X ( a , . . . , a n , b , . . . , b n ) ∩ X reg . If we can map P to Q by an element of SAut( X ), then X ( a , . . . , a n , b , . . . , b n ) ∩ X reg is contained in oneSAut( X )-orbit. Let us construct a sequence of automorphisms in SAut( X ), composition ofwhich maps P to Q .First of all we would like to map P by SAut( X ) to such a point R , that T i ( R ) = T i ( Q )for all i . To do this we obtain points P = P, P , . . . , P n = R ∈ ( X ( a , . . . , a n , b , . . . , b n ) ∩ X reg ) , where T ( P i ) = T ( Q ) , . . . , T i ( P i ) = T i ( Q ) and P i = β i ( P i − ) for some β i ∈ SAut( X ).We start from i = 1 and P = P . When we obtain a point P i , we increase i . The point R coincides with P n .Since all a u = 0 and all b v = 0, if T ( P i − ) = 0 or T ( P i − ) = 0, then at least one δ i + ( T i )( P i − ) or δ i − ( T i )( P i − ) is not equal to zero. Therefore, we can change T i ( P i − )by ϕ i + or by ϕ i − . Hence, there is s such that T i ( ϕ i ± ( s )( P i − )) = T i ( Q ). We can take P i = ϕ i ± ( s )( P i − ).If T ( P i − ) = T ( P i − ) = 0, then there is an index k such that T k ( P i − ) = 0. For every j < i we have T j ( P i − ) = T j ( Q ) = 0. Therefore, k ≥ i . Since P i − is a regular point, wehave l k = 1 and k is the unique index such that T k ( P i − ) = 0. Then we can consider theLND γ given by γ ( T k ) = ∂T l ∂T γ ( T ) = − ∂T l ∂T k , γ ( T pq ) = 0 for all other pairs ( p, q ) . Denote τ ( s ) = exp( sγ ). Since for all r = k it is true T r ( P i − ) = 0, we obtain γ ( T )( P i − ) = 0. Therefore, there is s ∈ K such that T ( τ ( s )( P i − )) = 0. Let us de-note P ′ i − = τ ( s )( P i − ). Since k ≥ i , we have T j ( P ′ i − ) = T j ( P i − ) = T j ( Q ) for all j < i .Now we have δ i + ( T i )( P ′ i − ) = 0. Hence, we can take P i = ϕ i + ( t )( P ′ i − )) for suitable t . N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 15
We have obtained a point R = ψ ( P ) for some ψ ∈ SAut( X ) such that for all i we have T i ( R ) = T i ( Q ), for all j ≥ T j ( R ) = T j ( Q ) = a j and T j ( R ) = T j ( Q ) = b j .Since for all i and j it is true T ij ( Q ) = 0, we have δ ( α )( R ) = 0. Hence, there is ˆ s such that α ( ϕ (ˆ s )( R )) = α ( Q ). Let us denote M = ϕ (ˆ s )( R ). Volumes of the followingfunctions at M and Q coincide: α , T j , T j , and T j for all j ≥ αβ = T l . Therefore, α ( Q ) = 0. Since volumes of all T j , T j for j ≥
2, and α at M are not equal to zero, we obtain δ − ( T )( M ) = 0. Hence, we can find t ∈ K suchthat T ( ϕ − ( t )( M )) = T ( Q ). From the other hand ϕ − ( t ) does not change T ij , j ≥ α . So, volumes of the following functions coincide at N = ϕ − ( t )( M ) and Q : T j and T j for j ≥ T i for all i , and α. Since αβ = T l and α ( Q ) = 0, we obtain β ( N ) = β ( Q ).Therefore, all variables coincide at N and Q . That is N = Q . (cid:3) Proof of Theorem 4.
In all the items we prove that X reg is one SAut( X )-orbit, where X isthe considered trinomial hypersurface. Type H . Suppose
P, Q ∈ X such that for all 1 ≤ i ≤ n we have T i ( P ) = 0 and T i ( Q ) = 0. Then P and Q are in one SAut( X )-orbit. Indeed, by Lemma 14 there is ψ ∈ SAut( X ) such that T ( ψ ( P )) = T ( Q ) , ∀ j = 1 : T j ( ψ ( P )) = T j ( P ) . Then again by Lemma 14 there is ψ ∈ SAut( X ) such that T (( ψ ◦ ψ )( P )) = T ( Q ) , ∀ j = 2 : T j (( ψ ◦ ψ )( P )) = T j ( ψ ( P )) . And so on. Terminally we obtain, that ( ψ n ◦ . . . ◦ ψ )( P ) = Q .Now let us take R ∈ X reg . Our goal is to find ψ ∈ SAut( X ) such that for all 1 ≤ i ≤ n T i ( ψ ( R )) = 0. If there are i ∈ { , } and j ∈ { , . . . , n i } such that T l i i T ij ( R ) = 0 , then γ ij ( T )( R ) = 0. Therefore, we can find s ∈ K such that T (exp( sγ ij )( R )) = 0. Wecan choose such s that T l i i T ij (exp( sγ ij )( R )) = 0 . But T k (exp( sγ ij )( R )) = T k ( R ), k >
1. Analogically we can apply automorphisms fromSAut( X ) to get a point S with T k ( S ) = 0 for all k .If there are no i ∈ { , } and j ∈ { , . . . , n i } such that T l i i T ij ( R ) = 0 . Then for all i ∈ { , } and j ∈ { , . . . , n i } we have ∂f∂T ij = 0 , where f = T l + T l + T T . . . T n . Since R ∈ X reg , there exists 1 ≤ a ≤ n such thatfor all b = a we have T b ( R ) = 0. Applying Lemma 14 we obtain ψ ∈ SAut( X ) such that T i ( ψ ( R )) = 0 for all i . Type H . Analogically to H , using Lemma 15 we can conclude that if P, Q ∈ X reg and T i ( P ) = 0 , T i ( Q ) = 0 , T j ( P ) = 0 , T j ( Q ) = 0 for all 1 ≤ i ≤ n , 1 ≤ j ≤ n , then P and Q are in one SAut( X )-orbit. Indeed, we canfind ψ ∈ SAut( X ) such that T ( ψ ( P )) = T ( Q ) , T ( ψ ( P )) = T ( Q ) ,T i ( ψ ( P )) = T i ( P ) , T j ( ψ ( P )) = T j ( P ) for all i, j > . In this way we can find ψ ∈ SAut( X ) such that ∀ i, j : T i ( ψ )( P ) = T i ( Q ) , T j ( ψ )( P ) = T j ( Q ) , Therefore, by Lemma 15, P and Q are in one SAut( X )-orbit.Let R ∈ X reg . Suppose T ij ( R ) = 0, for some i ∈ { , } . We can assume that T ( R ) = 0.Suppose all l k >
1. Since R ∈ X reg we obtain that for all j and k we have T j ( R ) = 0 and T k ( R ) = 0. Therefore, there is s such that T ( ϕ ( s )( R )) = 0. We can take such s that T j ( ϕ ( s )( R )) = 0 for all j . So, there exists ψ ∈ SAut( X ) such that for all i and j wehave T i ( ψ ( R )) = 0 , T j ( ψ ( R )) = 0.If there is 1 ≤ k ≤ n such that l k = 1, then X is of Type H . We consider this caselater. Type H . Let
P, Q ∈ X reg . We can fix Q such that T . . . T n T . . . T n ( Q ) = 0.If T . . . T n ( P ) = 0, then using Lemma 14 we can obtain P ′ = ψ ( P ), ψ ∈ SAut( X ), T j ( P ′ ) = T j ( Q ) for all 1 ≤ j ≤ n . By Lemma 14 P ′ and Q are in one SAut( X )-orbit.Therefore P and Q are in one SAut( X )-orbit.If T . . . T n ( P ) = 0 and T . . . T n ( P ) = 0 we can swap T l and T l .If T . . . T n ( P ) = 0 and T . . . T n ( P ) = 0, then T l ( P ) = 0. Since P is regular, thereexist a ∈ { , , } and 1 ≤ j ≤ n a such that l aj = 1 and T a . . . T aj − T aj +1 . . . T an a ( P ) = 0 . Then we can swap monomials a and 1 and swap variables in such a way that T . . . T n ( P ) = 0. So, we obtain that X reg is one orbit. Type H . Let
P, Q ∈ X reg , T ij ( Q ) = 0 for all i, j . If T . . . T n ( P ) = 0, using Lemma 14we can find ψ ∈ SAut( X ) such that T ij ( ψ ( P )) = T ij ( Q ) for i = 1 ,
2. Then by Lemma 15 ψ ( P ) and Q are in one SAut( X )-orbit.If among T , . . . , T n there is only one variable T j such that T j ( P ) = 0 and l j = 1,then we can swap T and T j .If among T , . . . , T n there are at least two variables vanishing at P or there is only onevariable T j vanishing on P but l j >
1, then since P is regular, we have T . . . T n ( P ) = 0and T . . . T n ( P ) = 0. Then applying Lemma 15 we can obtain τ ∈ SAut( X ) such that T . . . T n ( τ ( P )) = 0. Type H . Let
P, Q ∈ X reg , T ij ( Q ) = 0 for all i, j . Since P is regular, at list for two i wehave T i . . . T in i ( P ) = 0. We can assume that T . . . T n ( P ) = 0 and T . . . T n ( P ) = 0.By Lemma 15 there exists ψ ∈ SAut( X ) such that for all 2 ≤ i ≤ n , 2 ≤ j ≤ n and1 ≤ k ≤ n we have T i ( ψ ( P )) = T i ( P ) , T j ( ψ ( P )) = T j ( P ) , T k ( ψ ( P )) = T k ( Q ) . Applying once more Lemma 15 we obtain ψ ∈ SAut( X ) such that for all 1 ≤ i ≤ n ,2 ≤ j ≤ n and 2 ≤ k ≤ n we have T i ( ψ ◦ ψ ( P )) = T i ( Q ) , T j (( ψ ◦ ψ ( P )) = T j ( P ) ,T k (( ψ ◦ ψ ( P )) = T k ( ψ ( P )) = T k ( Q ) . Again by Lemma 15 ψ ◦ ψ ( P ) and Q are in one SAut( X )-orbit.The proof of Theorem 4 is completed. N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 17 (cid:3)
Remark . Theorem 4 does not give a criterium for a trinomial hypersurface to be flexible,but only a sufficient condition.
Remark . Flexibility of varieties of Type H when n = 2 follows from results of [5]. Example 4.
In [13, Theorem 9.2] it is proved that ML ( V ( t x + y + z n )) = K . Thisvariety is of Type H . Hence, it is flexible by Theorem 4.7. Intermediate trinomial hypersurfaces
Theorem 2 gives a criterium for a trinomial hypersurface to be rigid. Theorem 4 givesa sufficient condition for a trinomial hypersurface to be flexible. Let us consider varieties,which are not rigid and are not isomorphic to any variety from Theorem 4. There are twotypes of them
Type A. V (cid:16) T l + T l + T . . . T k T l k +1 k +1 . . . T l n n (cid:17) , where k ≥ n − k ≥
1, all l ij ≥
2, and { l , . . . , l n } 6 = { , a , . . . , a n − } . Type B. V (cid:16) T . . . T k T m k k +1 . . . T m n n + T . . . T k T m k k +1 . . . T m n n + T l (cid:17) , where k , k ≥ n − k + n − k ≥
1, all m ij ≥
2, all l i ≥
2, and { l , . . . , l n } 6 = { , a , . . . , a n − } . Definition 5.
A variety, which is not rigid and is not flexible, we call intermediate .Let us formulate a question, which remains open.
Question 1.
Does Theorem 4 give a criterium for a trinomial hypersurface to be flexible?That is, is it true that trinomial hypersurfaces of Types A and B are intermediate?
The next two propositions give a partial answer to Question 1 when the trinomial hy-persurface is of Type A and k = 1 and when it is of Type B and k = 1. In these cases theanswer is ”yes”. This implies that there exists an intermediate trinomial hypersurface. Proposition 2.
Let X = V (cid:16) T l + T l + T T l . . . T l n n (cid:17) ,n ≥ , all l ij ≥ and { l , . . . , l n } 6 = { , a , . . . , a n − } . Then1)
ML( X ) = K [ T , . . . , T n ] ,2) generic SAut( X ) -orbits on X can be separated by functions from ML( X ) .Proof. Note that X is an n -suspension over K n + n .Consider Z = V (cid:16) T l + T l + S T l . . . T l n n (cid:17) . Theorem 2 implies that Z is rigid. Hence, Lemma 9 implies that for all 2 ≤ j ≤ n we have T j ∈ ML( X ). Therefore, K [ T , . . . , T n ] ⊂ ML( X ).In the proof of Lemma 14 we define an LND γ ij : K [ X ] → K [ X ]. We haveKer γ ij = K [ T ab | ( a, b ) = (2 , , ( a, b ) = ( i, j )] . Indeed, for any nonzero LND ∂ of K [ X ] we have a formula for transcendence degree of itskernel tr . deg Ker ∂ = tr . deg K [ X ] − n + n + n − , see for example [12, Principle 11(e)]. It is obvious that K [ T ab | ( a, b ) = (2 , , ( a, b ) = ( i, j )] ⊂ Ker γ ij . If there is a polynomial depending on T or T ij in Ker γ ij , thentr . degKer γ ij > n + n + n − , this gives a contradiction.We obtain ML( X ) ⊂ \ i =1 , j =1 ,...,n i Ker γ ij = K [ T , . . . , T n ] . follows from Lemma 14. (cid:3) Remark . SAut( X )-orbits on an arbitrary affine variety can not always be separated byML( X ). They always can be separated by rational SAut( X )-invariants FML( X ), see [4,Theorem 1.13]. For example, there are nonflexible affine threefolds with trivial Makar-Limanov invariant, see [17, § Proposition 3.
Let X = V (cid:16) T T m . . . T m n n + T . . . T k T m k k +1 . . . T m n n + T l (cid:17) , where n ≥ , ≤ k ≤ n , for all i , j , k we have l i ≥ and m jk ≥ , and { l , . . . , l n } 6 = { , a , . . . , a n − } . Then1)
ML( X ) ⊃ K [ T , . . . , T n ] ,2) If k = 1 , thena) ML( X ) = K [ T , . . . , T n , T , . . . , T n ] , b) generic SAut( X ) -orbits can be separated by ML( X ) .3) If k = n , thena) ML( X ) = K [ T , . . . , T n ] , b) generic SAut( X ) -orbits can be separated by ML( X ) .Proof. Note that X is an n -suspension over K n + n . Let us consider Z = V (cid:16) S u T m . . . T m n n + T . . . T k T m k k +1 . . . T m n n + T l (cid:17) , where u ≥
2. Theorem 2 implies that Z is rigid. It follows from Lemma 9 that for all j ≥ T j ∈ ML( X ). By 1) we have ML( X ) ⊃ K [ T , . . . , T n , T , . . . , T n ]. Lemma 15 implies that ageneric variety V ( T − a , . . . , T n − a n , T − b , . . . , T n − b n ) ∩ X reg is one SAut( X )-orbit. This implies 2(a) and 2(b). By 1) we have ML( X ) ⊃ K [ T , . . . , T n ]. Let U = { x ∈ X | ∀ i, j : T ij ( x ) = 0 } . It is clear that U ⊂ X is open. Then V = SAut( X )( U ) ⊂ X is open.Let us prove that every nonempty set W ( a , . . . , a n ) = V ( T − a , . . . , T n − a n ) ∩ V is one SAut( X )-orbit. N RIGIDITY OF TRINOMIAL HYPERSURFACES AND FACTORIAL TRINOMIAL VARIETIES 19
Since ML( X ) ⊃ K [ T , . . . , T n ], if W ( a , . . . , a n ) = ∅ , then W ( a , . . . , a n ) ∩ U = ∅ .Let P, Q ∈ W ( a , . . . , a n ). Then there are ψ P and ψ Q in SAut( X ) such that P ′ = ψ P ( P )and Q ′ = ψ Q ( Q ) are in W ( a , . . . , a n ) ∩ U .By Lemma 15 there is ψ ∈ SAut( X ) such that T ( ψ ( P ′ )) = T ( Q ′ ) , ∀ i ≥ T i ( ψ ( P ′ )) = T i ( P ′ ) . Applying again Lemma 15 we obtain ψ ∈ SAut( X ) such that T ( ψ ◦ ψ ( P ′ )) = T ( Q ′ ) , T ( ψ ◦ ψ ( P ′ )) = T ( Q ′ ) , ∀ i ≥ T i ( ψ ◦ ψ ( P ′ )) = T i ( P ′ ) . We proceed in such a way until we obtain ψ ∈ SAut( X ) such that for all i = 1 , . . . , n wehave T i ( ψ ( P ′ )) = T i ( Q ′ ). Since for all j ≥ T j ( ψ ( P ′ )) = T j ( P ′ ) = T j ( Q ′ ),Lemma 15 implies existing of τ ∈ SAut( X ) such that ( τ ◦ ψ )( P ′ ) = Q ′ . This implies that W ( a , . . . , a n ) is one SAut( X )-orbit. This proves 3(a) and 3(b). (cid:3) References [1] I. Arzhantsev.
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Lomonosov Moscow State University, Faculty of Mechanics and Mathemat-ics, Department of Higher Algebra, Leninskie Gory 1, Moscow, 119991 Russia;andNational Research University Higher School of Economics, Faculty of ComputerScience, Kochnovskiy Proezd 3, Moscow, 125319 Russia
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