Splitting of uniform bundles on quadrics
Xinyi Fang Duo Li Yanjie Li

§ ABSTRACT 
We show that there exist only constant morphisms from ℚ^2n+1(n≥ 1) to 𝔾(l,2n+1) if l is even (0<l<2n) and (l,2n+1) is not (2,5). As an application, we prove on ℚ^2m+1 and ℚ^2m+2(m≥ 3), any uniform bundle of rank 2m splits. 

Nodeless superconductivity and topological nodal states in molybdenum carbide
    Toni Shiroka
    
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Keywords: uniform bundle; quadric; splitting of vector bundles.


MSC: 14M15; 14M17; 14J60.






§ INTRODUCTION

In this article, we assume all the varieties are defined over ℂ. Assume that X is a rational homogeneous variety of Picard number 1 (we call
X a generalised Grassmannian for short),  then X is swepted by lines.  Let E be a vector bundle on X. We consider  its restriction  E|_L(≃𝒪(a_1)⊕⋯⊕𝒪(a_k)) to any line L⊆ X. If the splitting type (a_1,…,a_k) of   E|_L is independent of the choice of L, we call E a uniform bundle.


 Every uniform bundle on ℙ^n whose  rank is  smaller than n splits (cf. the main theorems of <cit.> and <cit.>).  On a Grassmannian Gr(k,n+1)(k≤ n+1-k), any uniform bundle whose  rank is smaller than k splits (cf. <cit.>). On some other generalised Grassmannians, there are similar results, see <cit.>, <cit.> and <cit.>.


In a recent article <cit.>, for an arbitrary generalised Grassmannian X, we prove that once the rank of a uniform bundle E is smaller than e.d.(VMRT)+1 (for the definition of  e.d.(VMRT), see <cit.>), the bundle E necessarily splits. Moreover, for most generalised Grassmannians, we classify unsplit uniform bundles of minimal ranks. The upper bound e.d.(VMRT) is optimal for many generalised Grassmannians, for example it is optimal for ℚ^n(3≤ n ≤ 6) (see <cit.>). However,  we will see that this upper bound is not optimal for higher dimensional quadrics.



In this article, we first show that there exist only constant morphisms from ℚ^2n+1(n≥ 1) to 𝔾(l,2n+1) if l is even (0<l<2n) and (l,2n+1) is not (2,5).
Combining this result with the method of <cit.>, we prove that every uniform bundle of rank 2n on ℚ^2n+1 and ℚ^2n+2(n≥ 3) splits and the upper bound 2n is optimal for ℚ^2n+1. This gives the first known example such that the optimal upper bound is bigger than e.d.(VMRT).


This article is organised as follows: in Section 2, we study morphisms from ℚ^2n+1(n≥ 1) to 𝔾(l,2n+1)  and the main result of this section is Proposition <ref>; in Section 3, we study uniform bundles on smooth quadrics and the main result of this section is Theorem <ref>. Furthermore, we also study uniform bundles on B_n/P_k and D_n/P_k, see Corollary <ref>,  Corollary <ref>, Corollary <ref> and Corollary <ref>.


§ MORPHISMS FROM QUADRICS TO GRASSMANNIANS


    There exist only constant morphisms from ℚ^2n+1(n≥ 1) to 𝔾(l,2n+1) if l is even (0<l<2n) and (l,2n+1) is not (2,5).


   We mainly follow the proof of the main theorem in <cit.>. Let H be the cohomology class of a hyperplane on ℚ^2n+1. The cohomology ring of ℚ^2n+1 is
    
    H^∙(ℚ^2n+1,ℤ)=ℤ⊕ℤH ⊕⋯⊕ℤH^n-1⊕ℤH^n/2⊕⋯⊕ℤH^2n+1/2.

    Let U (resp. Q) be the universal subbundle (resp. quotient bundle) on 𝔾(l,2n+1). Suppose that f:ℚ^2n+1→𝔾(l,2n+1) is a non-constant morphism. Let c_i and d_j be rational numbers satisfying
    
    c_i(f^*U^∨)=c_iH^i  for 1≤ i ≤ l+1    and   c_j(f^*Q)=d_jH^j  for 1≤ j ≤ 2n+1-l.

    By <cit.>, both c_i(1≤ i ≤ l+1) and d_j(1≤ j ≤ 2n+1-l) are non-negative. We note that for any 1≤ i,j<n, c_i and d_j are integers. For any i,j≥ n, 2c_i and 2d_j are integers. 
    Then from the exact sequence 0→ f^*U →𝒪_ℚ^2n+1^⊕ 2n+2→ f^*Q → 0, we get the equality of polynomials:
    
    (1-c_1t+c_2t^2+⋯+(-1)^l+1c_l+1t^l+1)(1+d_1t+⋯+d_2n+1-lt^2n+1-l)
     
            =   1+(-1)^l+1c_l+1d_2n+1-lt^2n+2.

    If c_l+1d_2n+1-l is 0, then by (<ref>) both c_1 and d_1 are zero, which implies that f is constant. So we may assume the numbers c_1,d_1,c_l+1 and d_2n+1-l are non-zero. By <cit.>, all the rational numbers c_1,c_2,…,c_l+1 and d_1,d_2,…,d_2n+1-l are positive.


    Let a be √(c_l+1d_2n+1-l). We set C_i:=c_i/a^i(1≤ i ≤ l+1) and D_j:=d_j/a^j(1≤ j ≤ 2n+1-l). We note that if a,C_i and D_j are all positive integers, then by the same proof of <cit.>, we can get a contradiction. Then it suffices to show that a,C_i and D_j are integers.

    
    We firstly show that a is an integer. Since l is even, similar to the proof in <cit.>, we can show that a is c_m+1/c_m, where m is l/2. So a is rational, we may assume a=s/t, where s and t are coprime positive integers. By definition, we have a^2n+2=c_l+1d_2n+1-l. Since 2c_l+1 and 2d_2n+1-l are integers, 4a^2n+2 is an integer. So t^2n+2 divides 4, which implies that t is 1, as n is at least 1. Hence a is an integer.  
  

    We now show that C_i(=c_i/a^i) and D_j(=d_j/a^j) are integers. Let F_1(x)⋯ F_k(x) be the irreducible factorization of 1-x^2n+2 over ℤ[x] with F_l(0)=1(1≤ l ≤ k). Then F_l(x) is also irreducible over ℚ[x] by Gauss Lemma. Then
    
    (1-c_1t+c_2t^2+⋯+(-1)^l+1c_l+1t^l+1)(1+d_1t+⋯+d_2n+1-lt^2n+1-l)
     
            =   1+(-1)^l+1c_l+1d_2n+1-lt^2n+2=1-a^2n+2t^2n+2=F_1(at)⋯ F_k(at).

    Note that F_i(at) is also irreducible over ℚ[t]. We have
    
    1-c_1t+c_2t^2+⋯+(-1)^l+1c_l+1t^l+1=F_i_1(at)⋯ F_i_k_1(at) and 
       1+d_1t+⋯+d_2n+1-lt^2n+1-l=F_j_1(at)⋯ F_j_k_2(at).

    Since the coefficients of F_l(x) are integers,  C_i(=c_i/a^i) and D_j(=d_j/a^j) are integers.



§ UNIFORM BUNDLES ON ℚ^2N+1(N≥ 3)

We are going to use the method in <cit.> to show that any uniform bundle of rank 2n on ℚ^2n+1(n≥ 3) splits. We firstly fix some notations.


Let E be a uniform bundle on X(=ℚ^2n+1)(n≥ 3) of rank 2n. Assume that E does not split, by <cit.>,  we may assume that the splitting type of E is 

    (0,…,0_l+1,-1,…,-1_2n-l-1)   (l+1≥ 2n-l-1 ≥ 1).

We denote the moduli of lines on X and the corresponding universal family by:
 
    𝒰(=B_n+1/P_1,2)[r,"q"][d,"p"]   ℳ(=B_n+1/P_2)
    
            X(=B_n+1/P_1).

 where ℳ is the moduli of lines and 𝒰 is the universal family.
 The relative Harder-Narasimhan (H-N) filtration of p^*E induces an exact sequence:
 
    0→ E_1(=q^*G_1) → p^*E → E_2(=q^*G_2⊗ p^*_X(-1)) → 0,

 where G_1 (resp. G_2) is a vector bundle on ℳ of rank l+1 (resp. 2n-l-1). For each x∈ X, the restriction of relative H-N filtration to p^-1(x) induces a morphism 
 
    ψ_x:p^-1(x)(≅ℚ^2n-1)→ Gr(l+1,2n)(≅𝔾(l,2n-1)).

 
By <cit.>, we have the following description of cohomology rings:

    H^∙(X,ℚ)=ℚ[X_1]/(X_1^2n+2),
       H^∙(𝒰,ℚ)=ℚ[X_1,X_2]/(Σ_i(X_1^2,X_2^2)_n≤ i ≤ n+1),
       H^∙(ℳ,ℚ)=ℚ[X_1+X_2,X_1X_2]/(Σ_i(X_1^2,X_2^2)_n≤ i ≤ n+1).

For a  bundle F of rank r, the Chern polynomial of F is defined as

    C_F(T):=T^r-c_1(F)T^r-1+⋯+(-1)^rc_r(F).

Let E(T,X_1)=∑_k=0^2ne_kX_1^kT^2n-k(∈ℚ[X_1,T]) and S_i(T,X_1,X_2)(∈ℚ[X_1+X_2,X_1X_2,T])(i=1,2) be homogeneous polynomials representing C_p^*E(T) and C_q^*G_i(T) in the cohomology rings respectively. There are equations

    E(T,X_1)=C_p^*E(T) and S_i(T,X_1,X_2)=C_q^*G_i(T)(i=1,2).

Let R(X_1,X_2) be the polynomial Σ_n(X_1^2,X_2^2). By (<ref>), we have an equation of Chern polynomials:

    E(T,X_1)-aR(X_1,X_2)=S_1(T,X_1,X_2)S_2(T+X_1,X_1,X_2)*.

If a is 0, then both S_1(T,X_1,X_2) and S_2(T+X_1,X_1,X_2) are polynomials only in variables T and X_1. Since S_i(T,X_1,X_2)(i=1,2) are symmetrical in X_1 and X_2, we must have S_1(T,X_1,X_2)=T^l+1 and S_2(T,X_1,X_2)=T^2n-l-1. So c_1(E_1) and c_1(E_2) are 0, and ψ_x is constant for each x∈ X, which implies that E splits (see, for example, <cit.>). Therefore, we have a 0. 


 §.§ Approximate solutions

To solve the equation (<ref>), we use the concept of approximate solutions introduced in <cit.>. The following definitions and propositions are basically from <cit.> and the proofs are similar.


    A non-zero homogeneous polynomial P(T,X_1) with rational coefficients in variables T and X_1 of degree 2n is called an approximate solution if P(T,X_1)-R(X_1,X_2) has a proper divisor S(T,X_1,X_2) which is symmetrical in X_1 and X_2. We call such a divisor a symmetrical divisor. 



    By the the equation (<ref>), both 1/aE(T,X_1) and 1/aE(T-X_1,X_1) are approximate solutions. We call them approximate solutions associated with (<ref>).


In the following lemma, there are some restrictions on the coefficient of X_1^2n for an arbitrary approximate solution P(T,X_1). 

    Let P(T,X_1)=∑_k=0^2np_kX_1^kT^2n-k be an approximate solution. Then one of the followings holds. 

    (1) Any symmetrical divisor of P(T,X_1)-R(X_1,X_2) is of degree one.

    (2) The coefficient p_2n is 0 and the zero set of P(0,1)-R(1,z) is ({z|z^2n+2-1=0})\{1,-1}.

    (3) The coefficient p_2n is 1 and the zero set of P(0,1)-R(1,z) is ({z|z^2n-1=0}∪{0})\{1,-1}.


    Let S(T,X_1,X_2) be a symmetrical divisor of P(T,X_1)-R(X_1,X_2). Then
    S(0,X_1,X_2) divides p_2nX_1^2n-R(X_1,X_2). As S(T,X_1,X_2) is symmetrical in X_1 and X_2, S(0,X_1,X_2) divides p_2nX_2^2n-R(X_1,X_2).
    Therefore S(0,X_1,X_2) divides p_2n(X_1^2n-X_2^2n).
    
    By the equation R(X_1,X_2)(X_1^2-X_2^2)=X_1^2n+2-X_2^2n+2, we have
    
    (X_1^2-X_2^2)(p_2nX_1^2n-R(X_1,X_2))+X_2^2(X_1^2n-X_2^2n)=(p_2n-1)X_1^2n(X_1^2-X_2^2).


    
    If p_2n is not 0 and 1, we have S(0,X_1,X_2)| X_1^2n-X_2^2n and S(0,X_1,X_2)| X_1^2n(X_1-X_2)(X_1+X_2). Since S(0,X_1,X_2) is symmetrical in X_1 and X_2, S(0,X_1,X_2) is c(X_1+X_2)  for some c∈ℚ. In particular,  (S) is 1.


    If p_2n is 0, then P(0,1)-R(1,z) is -z^2n+2-1/z^2-1(=-(z^2n+z^2n-2+⋯+1)). If p_2n is 1, then P(0,1)-R(1,z) is -z^2z^2n-1/z^2-1=(-(z^2n+z^2n-2+⋯+z^2)). Then the assertions follow immediately.
If p_2n is 0. Let F(T,X_1,X_2) be the polynomial satisfying
    
    P(T,X_1)-R(X_1,X_2)=S(T,X_1,X_2)F(T,X_1,X_2).

    Expand S(T,X_1,X_2) and F(T,X_1,X_2) as polynomials in the variable T, we get
    
    S(T,X_1,X_2)=s_mT^m+⋯+s_1(X_1,X_2)T+s_0(X_1,X_2)
       F(T,X_1,X_2)=f_2n-mT^2n-m+⋯+f_1(X_1,X_2)T+f_0(X_1,X_2).

    By substituting T=0 in (<ref>), we have s_0(X_1,X_2)f_0(X_1,X_2)=-R(X_1,X_2). Since both s_0(X_1,X_2) and R(X_1,X_2) are symmetrical in X_1,X_2, f_0(X_1,X_2) is also symmetrical in X_1,X_2. As z^2n+2-1=0 has no multiple roots, we also have (s_0,f_0)=1. Consider the coefficients of T in both sides of (<ref>), we have s_1(X_1,X_2)f_0(X_1,X_2)+s_0(X_1,X_2)f_1(X_1,X_2)=p_2n-1X_1^2n-1. If p_2n-1 is 0, we have s_0| s_1f_0. From (s_0,f_0)=1, we get s_0| s_1



    We call an approximate solution P(T,X_1)=∑_k=0^2np_kX_1^kT^2n-k a primitive approximate solution if p_2n∈{0,1} and P(T,X_1)-R(X_1,X_2) has a symmetrical divisor S_0(T,X_1,X_2) such that there is a  2(n-p_2n+1)-th primitive unit root y_0 satisfying S_0(0,1,y_0)=0.

As in <cit.>, we have the following classifications of primitive approximate solutions.

    Let P(T,X_1) be a primitive approximate solutions. If p_2n is 0, we have P(T,X_1)=bT^2n for some b∈ℚ. If p_2n is 1, we have P(T,X_1)=Σ_n(bT^2,X_1^2) for some b∈ℚ.

The proof of Proposition <ref> is similar to that of <cit.>, we leave them in the Appendix (see Propositions <ref> and <ref>).


 §.§ The case l is even

We firstly aim to show that  l is not  even and we will prove it by contradiction. Now suppose that l is even. If l is smaller than 2n-2 and (l,2n-1) is not (2,5), the morphism ψ_x (for the definition of ψ_x, see (<ref>)) is constant for each x∈ X according to Proposition <ref>. Then E splits. We exclude the remaining cases l=2n-2 and (l,2n-1)=(2,5) by calculations.

    There does not exist an unsplit uniform bundle of rank 2n whose  splitting type is (0,…,0,-1) on ℚ^2n+1(n≥ 3).


    Suppose E is unsplit, by the same calculation as in <cit.>, we have c_1(E_2)=(X_1+X_2)-X_1=X_2. Let f(X_1,X_2) be a homogeneous polynomial of degree 2n-1 which is  symmetrical in X_1 and X_2 and represents c_2n-1(E_1). By comparing the coefficients of T^0 on the left and right sides of the equation (<ref>), we get
    
    f(X_1,X_2)X_2=e_2nX_1^2n-a(X_1^2n+X_1^2n-2X_2^2+⋯+X_2^2n).

    Then we must have e_2n=a and f(X_1,X_2)=-a(X_1^2n-2X_2+X_1^2n-4X_2^3+⋯+X_2^2n-1), contradicting to the assumption that f is symmetrical in X_1 and X_2.

We now exclude the case (l,2n-1)=(2,5).

    There does not exist an unsplit uniform of rank 6 whose  splitting type is (0,0,0,-1,-1,-1) on ℚ^7.


    Suppose E is unsplit. Then a in the equation (<ref>) is not  0 and 
    1/aE(T,X_1)=1/a∑_k=0^6e_kX_1^kT^6-k is an approximate solution which has a symmetrical divisor of degree 3. By Lemma <ref>, we have 1/ae_6=0 or 1/ae_6=1. 
    Let f(X_1,X_2) be a homogeneous polynomial representing c_3(E_1).

    
    If 1/ae_6 is 0, we have f(X_1,X_2)| R(X_1,X_2)(=X_1^6+X_1^4X_2^2+X_1^2X_2^4+X_2^6). Since 
    
    the prime factorization of X_1^6+X_1^4X_2^2+X_1^2X_2^4+X_2^6 over ℚ[X_1,X_2] is (X_1^4+X_2^4)(X_1^2+X_2^2), R(X_1,X_2) has no divisor symmetrical in X_1 and X_2 of degree 3. 


    If 1/ae_6 is 1, then f(X_1,X_2) divides X_1^6-R(X_1,X_2)(=-X_2^2(X_1^4+X_1^2X_2^2+X_2^4)). The prime factorization of X_1^4+X_1^2X_2^2+X_2^4 over ℚ[X_1,X_2] is (X_1^2+X_1X_2+X_2^2)(X_1^2-X_1X_2+X_2^2). Therefore, X_1^6-R(X_1,X_2) has no divisor symmetrical in X_1 and X_2 of degree 3.


    In both cases, we get contradictions.




 §.§ The case l is odd

 Suppose that l is odd. We begin with a lemma.



    When l is odd, the equation E(t,1)-aR(1,0)=0 has no roots in ℝ.


    In the equation (<ref>): E(T,X_1)-aR(X_1,X_2)=S_1(T,X_1,X_2)S_2(T+X_1,X_1,X_2), we let  X_1 be 0 and let X_2 be 1. Then we get an equation T^2n-a=S_1(T,0,1)S_2(T,0,1). We write S_1(T,0,X_2)  and S_2(T,0,X_2)   as follows:   S_1(T,0,X_2)=∑_i=0^l+1a_iX_2^iT^l+1-i and S_2(T,0,X_2)=∑_j=0^2n-1-lb_jX_2^jT^2n-1-l-j, where a_i(0≤ i≤ l+1) and b_j(0≤ j ≤ 2n-1-l) are rational numbers. Then -a is a_l+1b_2n-1-l. We now wish to show  -a>0. To this end, for any x in X, we consider the embedding i_x:p^-1(x)(=ℚ^2n-1)↪𝒰(=B_n+1/P_1,2), which induces a morphism:

    i_x^*:H^∙(𝒰,ℚ)≅ℚ[X_1,X_2]/(Σ_n(X_1^2,X_2^2),Σ_n+1(X_1^2,X_2^2))→ℚ[X_2]/X_2^2n≅ H^∙(ℚ^2n-1,ℚ).
 Under the above identifications, we have S_1(T,0,X_2)=C_E_1|_p^-1(x)(T)=C_ψ_x^*U(T) and S_2(T,0,X_2)=C_E_2|_p^-1(x)(T)=C_ψ_x^*Q(T), where U (resp, Q) is the universal subbundle (resp. quotient bundle) on 𝔾(l,2n-1) and ψ_x is the morphism as in (<ref>). So we have
    
    a_l+1X_2^l+1=(-1)^l+1c_l+1(ψ_x^*U)=c_l+1(ψ_x^*U^∨) and 
       b_2n-1-lX_2^2n-1-l=(-1)^2n-1-lc_2n-1-l(ψ_x^*Q)=c_2n-1-l(ψ_x^*Q) (since l is odd, 2n-1-l is even).

    By <cit.>, c_l+1(ψ_x^*U^∨) and c_2n-1-l(ψ_x^*Q) are numerically non-negative, hence both a_l+1 and b_2n-1-l are non-negative. As -a is a_l+1b_2n-1-l and a is not 0, we have -a>0. So for any t∈ℝ, t^2n-a is bigger than 0. In other words, S_1(t,0,1) and S_2(t,0,1) are non-zero for any t∈ℝ. By the equations E(T,1)-aR(1,0)=S_1(T,1,0)S_2(T+1,1,0)=S_1(T,0,1)S_2(T+1,0,1), for any t∈ℝ, E(t,1)-aR(1,0) is not 0.


    When l is odd and n is at least 3, the approximate solution 1/aE(T,X_1) or 1/aE(T-X_1,X_1) associated with (<ref>)  is a primitive approximate solution.


    Recall the equation 1/aE(t,1)-R(1,z)=1/aS_1(t,1,z)S_2(t+1,1,z). Note that l is odd implies l+1≥ 2n-1-l>1. So by Lemma <ref>, we have 1/ae_2n=0 or 1/ae_2n=1. And when 1/ae_2n is 0, 1/aE(0,1)-R(1,exp2iπ/2n+2) vanishes; when 1/ae_2n is 1, 1/aE(0,1)-R(1,exp2iπ/2n) vanishes.


    Suppose that n is 3. When 1/ae_2n is 0, we have 1/aS_1(0,1,z)S_2(1,1,z)=-(z^6+z^4+z^2+1). The prime factorization of z^6+z^4+z^2+1 over ℚ[z] is (z^4+1)(z^2+1). Note (S_1)≥(S_2) and (S_2)>1, we must have S_1(0,1,z)=λ(z^4+1) for some λ∈ℚ. Then exp2iπ/8 is a root of S_1(0,1,z) and hence 1/aE(T,X_1) is primitive. When 1/ae_2n is 1, we have 1/aS_1(0,1,z)S_2(1,1,z)=-(z^6+z^4+z^2)=-z^2(z^4+z^2+1)=-z^2(z^2+z+1)(z^2-z+1). Note that S_1 is symmetrical in X_1, X_2 and  (S_1) is at least (S_2), we have S_1(0,1,z)=λ(z^4+z^2+1). Then exp2iπ/6 is a root of S_1(0,1,z) and hence 1/aE(T,X_1) is also primitive.

    
    

    Suppose now n is at least 4 and that 1/aE(T,X_1) is not a primitive solution. Then we have 
    S_2(1,1,exp2iπ/2n+2)=0 or S_2(1,1,exp2iπ/2n)=0.
 We now show that 1/aE(T-X_1,X_1) is a primitive solution. First we have the following inequalities:
    
    π/2n<2π/2n+2<2π/2n<3π/2n<4π/2n+2.

    (Note that for the last inequality, we use the condition n≥ 4).
    Since the roots of S_2(0,1,z) satisfy z^2n=1 or z^2n+2=1, it suffices to show that S_2(0,1,z) has a non-zero root y_0 with argument satisfying π/2n<  y_0 <3π/2n. By the above inequalities, we have y_0=2iπ/2n+2 or y_0=2iπ/2n. Then 1/aE(T-X_1,X_1) is a primitive approximate solution.


    It is enough to show that for any t∈ℝ, S_2(t+1,1,z) as a polynomial of z has a non-zero root with argument in (π/2n,3π/2n). Denote by K the set {t∈ℝ | ∃ r(t)( 0)∈ℂ,  S_2(t+1,1,r(t))= 0  with π/2n< r(t) <3π/2n}. By (<ref>), we have 0∈ K. Note that K is open by construction, if we can show K is closed, then K is ℝ. Let t_0 be a limit point of K. Then S_2(t_0+1,1,z) has a root r(t_0) which is a limit point of {r(t) | t∈ K}. By Lemma <ref>, for any t∈ℝ, as a polynomial in z, the roots of 1/aE(t,1)-R(1,z) are non-zero. So the roots of S_2(t+1,1,z) are also non-zero. In particular, r(t_0) is not zero. Suppose that t_0∉ K, then we have r(t_0)=mπ/2n, where m is 1 or 3. We may assume r(t_0)=ρexpimπ/2n, where ρ is a positive real number. Then 1/aE(t_0,1)-R(1,ρexpimπ/2n) is 0. To get a contradiction, we show for any d∈ℝ, we have R(1,ρexpimπ/2n)+d 0. If R(1,ρexpimπ/2n)+d=0, we have the following identities:
    
    (R(1,ρexpimπ/2n)+d)(ρ^2exp2imπ/2n-1)
    
            =   R(1,ρexpimπ/2n)(ρ^2exp2imπ/2n-1)+d(ρ^2exp2imπ/2n-1)
    
            =   ρ^2n+2expim(2n+2)π/2n-1+dρ^2exp2imπ/2n-d
    
            =   (dρ^2-ρ^2n+2)expimπ/n-(d+1)=0.

    Since n is bigger than 3 and m is 1 or 3, expimπ/n is not real. We must have d+1=0 and dρ^2-ρ^2n+2=0. But it implies dρ^2-ρ^2n+2=-ρ^2-ρ^2n+2=0, which is absurd.

Now we complete the proof for the case l is odd. By Proposition <ref> and Proposition <ref>, we have the following possibilities (note the coefficient of T^2n in E(T,X_1) is 1):


 E(T,X_1) is T^2n or aΣ_n(T^2/b_1,X_1^2); E(T-X_1,X_1) is T^2n or aΣ_n(T^2/b_2,X_1^2),
 where b_i(∈ℚ) satisfy b_i^n=a(i=1,2).

   The above possibilities are all impossible.


    If E(T,X_1) is T^2n,  T^2n-aR(X_1,X_2) is S_1(T,X_1,X_2)S_2(T+X_1,X_1,X_2). However, we have
    
    X_1-exp2iπ/2n+2X_2 | R(X_1,X_2), (X_1-exp2iπ/2n+2X_2)^2 ∤ R(X_1,X_2), X_1-exp2iπ/2n+2X_2 ∤ 1.

    By Eisenstein's criterion, T^2n-aR(X_1,X_2) is irreducible as the polynomial in the variable T with coefficients in ℚ[X_1,X_2]. This leads to a contradiction. Similar arguments can be applied to the case E(T-X_1,X_1) is T^2n.


    If E(T,X_1) is b_1^nΣ_n(T^2/b_1,X_1^2)(=Σ_n(T^2,b_1X_1^2)), then we have the equalities E(T,X_1)-aR(X_1,X_2)=Σ_n(T^2,b_1X_1^2)-Σ_n(b_1X_1^2,b_1X_2^2)=(T^2-b_1X_2^2)Σ_n-1(T^2,b_1X_1^2,b_1X_2^2) (for the last equality, see <cit.> for example). We will make use of the following claim, whose proof will be given in Proposition <ref>.

    
 Claim:  
 Σ_n-1(T^2,X_1^2,X_2^2) is irreducible in ℂ[T,X_1,X_2].

    
    As b_1 is not 0, Σ_n-1(T^2,b_1X_1^2,b_1X_2^2) is irreducible. We have (T^2-b_1X_2^2)Σ_n-1(T^2,b_1X_1^2,b_1X_2^2)=S_1(T,X_1,X_2)S_2(T+X_1,X_1,X_2). Since S_1(T,X_1,X_2) is symmetrical in X_1,X_2 and the coefficient of T^2n of S_1 is 1,
    then  S_1(T,X_1,X_2) is Σ_n-1(T^2,b_1X_1^2,b_1X_2^2) and hence S_2(T+X_1,X_1,X_2) is T^2-b_1X_2^2. So S_2(T,X_1,X_2) is (T-X_1)^2-b_1X_2^2, which is not symmetrical in X_1,X_2. We get a contradiction.

    
    If E(T-X_1,X_1) is b_2^nΣ_n(T^2/b_2,X_1^2)(=Σ_n(T^2,b_2X_1^2)), we have (T^2-b_2X_2^2)Σ_n-1(T^2,b_2X_1^2,b_2X_2^2)=S_1(T-X_1,X_1,X_2)S_2(T,X_1,X_2). Similarly, S_2(T,X_1,X_2) is Σ_n-1(T^2,b_2X_1^2,b_2X_2^2). But  (S_2) is at most (S_1) and n is at least 3, it is impossible. 

Now  we prove our main theorem.

    Every uniform bundle of rank 2n on ℚ^2n+1 and ℚ^2n+2(n≥ 3) splits.


    We have shown that there is no unsplit uniform bundle of rank 2n on ℚ^2n+1 for n≥ 3. Now let E' be a uniform bundle of rank 2n on ℚ^2n+2. For every smooth hyperplane section ℚ^2n+1↪ℚ^2n+2, the restriction E'|_ℚ^2n+1 splits. Then we conclude by <cit.>.


Given a generalised Grassmannian X, we  denote by μ(X) the maximum positive integer verifying that every uniform vector bundle whose rank is at most μ(X) splits. And we call  μ(X) the splitting threshold for uniform vector bundles on X. For example,  μ(ℙ^n) is n-1. The main theorem in <cit.> tells us that for most generalised Grassmannians X, μ(X) is equal to e.d.(VMRT) (see <cit.>). Note that for ℚ^2n+1 and ℚ^2n+2, their e.d.(VMRT) are all 2n-1.




  

  (1) Theorem <ref> reveals that the splitting thresholds for uniform vector bundles on ℚ^2n+1 and ℚ^2n+2(n≥ 3) are at least 2n. Actually, ℚ^2n+1 and ℚ^2n+2(n≥ 3) are the first known examples such that μ(X) is bigger than e.d.(VMRT).

  (2) The threshold μ(ℚ^2n+1) is 2n, as the tangent bundle T_ℚ^2n+1 is unsplit. Moreover, the splitting type of T_ℚ^2n+1 is (2,1,…,1,0).
    


In <cit.>, the authors classify the unsplit uniform bundles of minimal ranks on the generalised Grassmannians B_n/P_k (2≤ k < 2n/3), B_n/P_n-2, B_n/P_n-1 and D_n/P_k (2≤ k < 2n-2/3), D_n/P_n-3, D_n/P_n-2. As direct corollaries of Theorem <ref>, we can give further classification results for uniform bundles on B_n/P_k (2n/3≤ k ≤ n-3) and D_n/P_k (2n-2/3≤ k ≤ n-4).


Let X be  B_n/P_k, where k is 2n/3 and k is at least 6. Let E be a uniform vector bundle on X of rank r.

    
  * If r is smaller than k, then E is a direct sum of line bundles.
    
  * If r is k, then E is either a direct sum of line bundles or  E_λ_1⊗ L or  E_λ_1^∨⊗ L for some line bundle L, where E_λ_1 is the irreducible homogeneous bundle corresponding to the highest weight λ_1. 


  
Note that the e.d.(VMRT) of X is k-1(=2n-2k-1).  By
<cit.>, the first assertion follows. For the case r=k, since 2n-2k+1=k+1 is at least 7, E|_ℚ^2n-2k+1 splits by Theorem <ref>. Suppose E  is unsplit, then E|_ℙ^k is also unsplit. Then the second  assertion follows from  <cit.>. 
    
  

Let X be  B_n/P_k with 2n/3< k ≤ n-3. Every uniform bundle of rank 2n-2k on X splits.


If E is a uniform bundle of rank 2n-2k on X, then E|_ℙ^k splits, as 2n-2k is smaller than k. On the other hand, since k is at most n-3 and hence 2n-2k+1 is at least 7, E|_ℚ^2n-2k+1 splits by Theorem <ref>. Because any 2-plane in X is contained in a ℙ^k or ℚ^2n-2k+1, E splits by <cit.>.
  

Similar to the proofs of the above corollaries, we can prove the following results.

Let X be  D_n/P_k, where k is 2n-2/3 and k is at least 6. Let E be a uniform vector bundle on X of rank r.

    
  * If r is smallet than k, then E is a direct sum of line bundles.
    
  * If r is k, then E is either a direct sum of line bundles or  E_λ_1⊗ L or  E_λ_1^∨⊗ L for some line bundle L, where E_λ_1 is the irreducible homogeneous bundle corresponding to the highest weight λ_1. 


  
Let X be  D_n/P_k with 2n-2/3< k ≤ n-4. Every uniform bundle of rank 2n-2k-2 on X splits.



§ CLASSIFICATION OF PRIMITIVE APPROXIMATE SOLUTIONS

We use methods in <cit.> to classify primitive approximate solutions. Let P(T,X_1)=∑_k=0^2np_kX_1^kT^2n-k be a primitive approximate solution. 

    If p_2n is 0, then P(T,X_1) is bT^2n for some rational number b.


    Let S_0(T,X_1,X_2) be a symmetrical divisor of P(T,X_1)-R(X_1,X_2) such that for a 2(n+1)-th primitive unit root y_0, we have S_0(0,1,y_0)=0. Since y_0 is a simple root of P(0,1)-R(1,z), y_0 is also a simple root of S_0(0,1,z). Therefore, by the implicit function theorem, there is a germ of holomorphic function y(x) at a neighborhood of x=0 satisfying
    
    S_0(x(1+y(x)),1,y(x))=0  and  y(0)=y_0.

    We now show by induction that for m=1,…,2n-1, we have y^(m)(0)=p_2n-m=0. As  S_0 is symmetrical in X_1 and X_2,  we have
    
    P(x(1+y(x)),1)-R(y(x),1)=0, 1
       P(x(1+y(x)),y(x))-R(y(x),1)=0. 2

    By taking the derivatives of (<ref>) and (<ref>) at x=0 and note that p_2n is 0, we obtain
    
    p_2n-1(1+y_0)-R'(y_0,1)y'(0)=0, 1'
       p_2n-1(1+y_0)y_0^2n-1-R'(y_0,1)y'(0)=0. 2'

    From R(y,1)(y^2-1)=y^2n+2-1, we have R'(y,1)(y^2-1)+R(y,1)·(2y)=(2n+2)y^2n+1. As R(y_0,1) is 0 and y_0 is not ± 1, R'(y_0,1) is (2n+2)y_0^2n+1/y_0^2-1( 0). From (<ref>) and (<ref>), we get p_2n-1(1+y_0)(y_0^2n-1-1)=0. Since y_0 is primitive, y_0^2n-1-1 and 1+y_0 are not zero. We have p_2n-1=0. By (<ref>) and note R'(y_0,1)0, y'(0) is  0.


    For the case m≥ 2, we assume by induction that y^(m')(0)=p_2n-m'=0 for m'<m. Then the m-th derivatives of (<ref>) and (<ref>) satisfy the following equations:
    
    m!p_2n-m(1+y_0)^m-R'(y_0,1)y^(m)(0)=0, 1^m
       m!p_2n-m(1+y_0)^my_0^2n-m-R'(y_0,1)y^(m)(0)=0. 2^m

    Since y_0 is primitive, y_0^2n-m-1 is not zero. We have y^(m)(0)=p_2n-m=0 as above.


    If p_2n is 1, then P(T,X_1) is Σ_n(bT^2,X_1^2) for some b∈ℚ.


    Let S_+(T,X_1,X_2) be a symmetrical divisor of P(T,X_1)-R(X_1,X_2) such that for a 2n-th primitive unit root y_0, we have S_+(0,1,y_0)=0. Note the equation Σ_n(p_2n-2T^2,X_1^2)-R(X_1,X_2)=(p_2n-2T^2-X_2^2)Σ_n-1(p_2n-2T^2,X_1^2,X_2^2) (see <cit.> for example). Let S_-(T,X_1,X_2) be Σ_n-1(p_2n-2T^2,X_1^2,X_2^2), then S_-(0,1,y_0) is also 0. Denote P(T,X_1) by P_+(T,X_1) and denote Σ_n(p_2n-2T^2,X_1^2) by P_-(T,X_1), we are going to show that P_+ equals P_-.
    Let y_±(x) be germs of holomorphic functions satisfying
     
    S_±(x(1+y_±(x)),1,y_±(x))=0  and  y_±(0)=y_0.

     As S_± is symmetrical in X_1 and X_2, we have equations:
    
    P_±(x(1+y_±(x)),1)-R(y_±(x),1)=0, 1_±
       P_±(x(1+y_±(x)),y_±(x))-R(y_±(x),1)=0. 2_±

    Let p^+_2n-m (resp. p^-_2n-m) be the coefficient of X_1^2n-mT^m in P_+(T,X_1) (resp. P_-(T,X_1)). We prove  p^+_2n-m=p^-_2n-m and y^(m)_+(0)=y^(m)_-(0) by induction for m (0≤ m ≤ 2n).

    When m is 0, we have p_2n^+=p_2n^-=1, y_+(0)=y_-(0)=y_0. We next show p_2n-1^+=p_2n-1^-=0 and y'_+(0)=y'_-(0)=0. Taking the derivatives of (<ref>) and (<ref>) at x=0, we get
    
    p_2n-1^±(1+y_0)-R'(y_0,1)y'_±(0)=0, 1'_±
       p_2n-1^±(1+y_0)y_0^2n-1+2ny_0^2n-1y'_±(0)-R'(y_0,1)y'_±(0)=0. 2'_±

    From R(y,1)(y^2-1)=y^2n+2-1, we have R'(y,1)(y^2-1)+R(y,1)·(2y)=(2n+2)y^2n+1. Note R(y_0,1)=1 (as y_0 is a 2n-th primitive unit root). Substituting it in the above equation, we have
    
    2ny_0-y_0^2R'(y_0,1)=-R'(y_0,1). †

    Multiplying (<ref>) by y_0^2 and using the relation (<ref>), one has
    
    p_2n-1^±(1+y_0)y_0-R'(y_0,1)y_±'(0)=0. y_0^2· 2'_±
       p_2n-1^±(1+y_0)-R'(y_0,1)y_±'(0)=0, 1'_±

   Since y_0 is primitive, |[ y_0(1+y_0) -R'(y_0,1);      1+y_0 -R'(y_0,1) ]| 
=(1-y_0)(1+y_0)R'(y_0,1) is not 0. Then one obtains p_2n-1^±=y_±'(0)=0. 


When m is at least 2, by induction, we may assume  p^+_2n-m'=p^-_2n-m' and y_+^(m')(0)=y_-^(m')(0) for m'<m. By taking the m-th derivatives of (<ref>) and (<ref>) at x=0, we have 

    m!p_2n-m^±(1+y_0)^m-R'(y_0,1)y^(m)_±(0)+T^1_±=0, 1^m_±
       m!p_2n-m^±(1+y_0)^my_0^2n-m+2ny_0^2n-1y^(m)_±(0)-R'(y_0,1)y^(m)_±(0)+T^2_±=0, 2^m_±

where T^i_± are the remaining terms satisfying T^1_-=T^1_+ and T^2_-=T^2_+. For the case m=2, by construction, we automatically have p_2n-2^-=p_2n-2=p_2n-2^+. Then one obtains y_+^(2)(0)=y_-^(2)(0) from (<ref>). Suppose now m is at least 3, multiplying (<ref>) by y_0^2 and using (<ref>), we have 

    m!p_2n-m^±(1+y_0)^my_0^2n-m+2-R'(y_0,1)y_±^(m)(0)+y_0^2T^2_±=0. y_0^2· 2^m_±

Note that |[ m!(1+y_0)^my_0^2n-m+2            -R'(y_0,1);           m!(1+y_0)^m            -R'(y_0,1) ]| 
=m!(1-y_0^2n-m+2)(1+y_0)^mR'(y_0,1) is not zero for m≥ 3. By solving the system of linear equations {(<ref>), (<ref>)} (view p_2n-m^± and y_±^(m)(0) as indeterminate), we have p_2n-m^+=p_2n-m^- and y_-^(m)(0)=y_+^(m)(0).


    Σ_n(T^2,X_1^2,X_2^2) is irreducible in ℂ[T,X_1,X_2].


    It suffices to show that the variety V:={(T,X_1,X_2)|Σ_n(T^2,X_1^2,X_2^2)=0}⊂ℂ^3 is smooth on ℂ^3\{0}. By <cit.>, the variety defined by Σ_n(T,X_1,X_2)=0 is smooth on ℂ^3\ 0. Note that the map (T,X_1,X_2)↦ (T^2,X_1^2,X_2^2) is a local isomorphism outside the locus defined by TX_1X_2=0, V is smooth on ℂ^3\{TX_1X_2=0}.


    Now suppose (t,u,v)( (0,0,0)) is a singular point of V, then one of t,u and v is 0. By symmetry, we assume that t is 0. Then there are  equations
    
    ∂/∂ TΣ_n(T^2,X_1^2,X_2^2)|_(0,u,v)=∂/∂ X_1Σ_n(T^2,X_1^2,X_2^2)|_(0,u,v)=∂/∂ X_2Σ_n(T^2,X_1^2,X_2^2)|_(0,u,v)=0.

    Furthermore, we have  equations:
    
    ∂/∂ X_1Σ_n(T^2,X_1^2,X_2^2)|_(0,u,v)=∂/∂ X_1Σ_n(X_1^2,X_2^2)|_(u,v)=0,
       ∂/∂ X_2Σ_n(T^2,X_1^2,X_2^2)|_(0,u,v)=∂/∂ X_2Σ_n(X_1^2,X_2^2)|_(u,v)=0.

    Without loss of generality, we may assume v is not 0. As Σ_n(u^2,v^2)(=Σ_n(0,u^2,v^2))  vanishes, u/v is a root of Σ_n(z^2,1)=0 with multiplicity greater than 2. 
    But Σ_n(z^2,1)=z^2n+2-1/z^2-1 has no multiple roots, which is a contradiction.


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