On Seifert fibered spaces bounding definite manifolds
Ahmad Issa Duncan McCoy









Department of Mathematics 

         The University of Texas At Austin 

         Austin, TX, 78712, USA
aissa@math.utexas.edu


Department of Mathematics 

         The University of Texas At Austin 

         Austin, TX, 78712, USA
d.mccoy@math.utexas.edu

 We establish an inequality which gives strong restrictions on when the standard definite plumbing intersection lattice of a Seifert fibered space over S^2 can embed into a standard diagonal lattice, and give two applications. First, we answer a question of Neumann-Zagier on the relationship between Donaldson's theorem and Fintushel-Stern's R-invariant. We also give a short proof of the characterisation of Seifert fibered spaces which smoothly bound rational homology S^1 × D^3's.


On Seifert fibered spaces bounding definite manifolds
    Duncan McCoy
    December 30, 2023
=====================================================





§ INTRODUCTION

Donaldson's diagonalization theorem <cit.> has led to many great successes in understanding several important questions in low dimensional topology, and in knot theory in particular. For example, Donaldson's theorem can often be used to answer questions concerning sliceness, unknotting number, 3-manifolds bounding rational homology balls, and surgery questions. In these cases, one typically uses Donaldson's theorem to obstruct a certain 3-manifold from bounding a certain type of smooth negative definite 4-manifold, with the obstruction taking the form of the existence of a certain map of intersection lattices. However, understanding this obstruction for large families of examples is often highly non-trivial, and can require combinatorial ingenuity.







One appealing application of Donaldson's theorem is to prove the well-known fact that the Poincaré homology sphere P = S^2(2; 2, 3/2, 5/4) does not bound a smooth integral homology 4-ball. This fact can, of course, be proved in many other ways, for example by using Rokhlin's theorem, Fintushel-Stern's R-invariant, or by using the d-invariant coming from Heegaard Floer homology. Assuming that P is oriented to bound the positive E_8 plumbing, the proof by Donaldson's theorem is as follows. If P were the boundary of a smooth integral homology W, then we could form a closed positive definite manifold by gluing -W to the positive E_8 plumbing. Donaldson's theorem would then imply that the E_8 intersection form is diagonalizable, which is, of course, untrue. In fact, as the E_8 intersection form does not embed into any diagonal lattice, this arguments shows that P does not bound any smooth negative definite 4-manifold. The purpose of this paper is to generalize this argument to other Seifert fibered spaces. We prove the following theorem.





thmsfsineq
  
  Let Y = S^2(e; p_1/q_1, …, p_k/q_k), k ≥ 3, be a Seifert fibered space over S^2 in standard form, that is, with e > 0, p_i/q_i > 1 for all i∈{1,2,…,k} and ε(Y) ≥ 0.
  Suppose that Y bounds a smooth 4-manifold W such that σ(W) = -b_2(W) and the inclusion induced map H_1(Y; ) → H_1(W; ) is injective. Then there is a partition of {1,2,…,k} into at most e classes such that for each class C,
  
    ∑_i∈ Cq_i/p_i≤ 1.



We note that the condition that ε(Y) := e - ∑_i=1^k q_i/p_i≥ 0 in Theorem <ref> guarantees that Y is oriented to bound a positive (semi-)definite plumbing 4-manifold. When Y is a rational homology sphere the map H_1(Y; ) → H_1(W; ) is automatically injective so in this case we are simply obstructing the existence of a negative definite manifold bounding Y. Although we do not discuss the details in this paper, one can easily obtain analogous results for Seifert fibered spaces over any orientable base surface. In our notation, the Poincaré homology sphere oriented to bound the positive E_8 plumbing is P=S^2(2;2,3/2, 5/4), see Figure <ref>. The reader can easily verify that Theorem <ref> obstructs P from bounding a negative definite manifold. 
Finally, we note that the converse to Theorem <ref> is not true. The integer homology sphere S^2(1; 3, 5, 13/6) passes the obstruction, but does not bound a negative definite manifold as it bounds a positive definite plumbing whose intersection form does not embed in a diagonal lattice.
 

We give two applications of Theorem <ref>. First, we prove the following theorem.


Let Y = S^2(e; p_1/q_1, …, p_k/q_k) be a Seifert fibered integral homology sphere in standard form, that is, with p_i/q_i > 1 for all i∈{1,2,…,k}, e > 0 and with Y oriented to bound a smooth positive definite plumbing 4-manifold. If Y bounds a smooth negative definite 4-manifold, then e = 1.


In the course of proving Theorem <ref>, we obtain a positive answer to the following question asked by Neumann-Zagier <cit.>.

Question: Let Y be as in Theorem <ref>. If the intersection form of the plumbing of Y is diagonalizable over , must e be equal to 1?



The motivation for this question comes from the R-invariant. 
Fintushel-Stern <cit.> used gauge theory to define an invariant R(Y) of Seifert fibered integral homology spheres with the property that if R(Y) > 0 then Y does not bound a smooth negative definite 4-manifold W with H_1(W) having no 2-torsion. 
Fintushel-Stern originally gave an expression for R(Y) as a trigonometric sum involving the Seifert invariants of Y. Neumann-Zagier <cit.> proved that these sums could be simply evaluated in terms of the central weight e of the standard positive definite plumbing bounding Y, showing that R(Y) = 2e - 3. Thus, if e > 1 then the R-invariant shows that Y does not bound a smooth negative definite 4-manifold W with H_1(W) having no 2-torsion. In this light, the positive answer to Neumann-Zagier's question implies that this result obtained from the R-invariant is also a consequence of Donaldson's theorem.



We are in fact able to prove a more general version of Theorem <ref> which holds for all |H_1(Y)| ∈{1,2,3,5,6,7}, see Theorem <ref> of Section <ref>. Some particular cases of Theorem <ref> are known. In their original paper, Neumann-Zagier <cit.> claimed to have proved the cases when k = 3, and when k=4 and e ≠ 3, but do not provide a proof, remarking that their proof was “clearly not the right proof”. The special case when e = k-1 follows from <cit.>.

Finally, we note that a positive answer to Neumann-Zagier's question is a special case of a more general conjecture made by Neumann <cit.>, stating that if an integral homology sphere Y is given as the boundary of a positive definite plumbing tree Γ and the intersection lattice of Γ is isomorphic to a diagonal lattice, then some vertex of Γ has weight 1. This general form of Neumann's conjecture for graph manifolds remains open.

Lidman-Tweedy <cit.> asked whether a Seifert fibered integral homology sphere with central weight different from 1 must have non-vanishing Heegaard-Floer d-invariant. As a corollary of Theorem <ref>, we answer their question positively.

corolcoroldinv Let Y be a Seifert fibered integral homology sphere, and let e∈ be the central weight in the standard definite plumbing graph for Y. If |e| ≠ 1, then d(Y) ≠ 0.


As a second application, we give a short proof of the following theorem which, in particular, gives a classification of the Seifert fibered spaces bounding rational homology S^1 × D^3's.

thmpaolosthm Let Y be a Seifert fibered space over S^2 with H_*(Y;) ≅ H_*(S^1× S^2;). The following are equivalent:
  
    
  *  Y is of the form S^2(m; p_1/q_1, p_1/p_1-q_1, …, p_m/q_m, p_m/p_m - q_m), where m ≥ 0 and p_i/q_i > 1 for all i∈{1,…,m}.
    
  *  Y = ∂ W, where W is a smooth 4-manifold with H_*(W; ) ≅ H_*(S^1 × D^3; ).
    
  *  Y is the boundary of smooth 4-manifolds W_+ and W_- such that σ(W_±) = ± b_2(W_±) and each of the inclusion-induced maps H_1(Y; ) → H_1(W_±; ) is injective.
  




Seifert fibered spaces bounding rational homology S^1× D^3's naturally arise in two contexts. First, a Seifert fibered space rational homology S^1× S^2 which embeds in S^4 necessarily bounds a rational homology S^1 × D^3. Indeed, in this context Donald <cit.> proved the implication (<ref>) implies (<ref>) of Theorem <ref>.

Second, a smoothly slice 2-component Montesinos link has double branched cover a Seifert fibered space over S^2 bounding a rational homology S^1 × D^3. Motivated by trying to determine the slice 2-component Montesinos links, Aceto <cit.> also classified Seifert fibered spaces bounding rational homology S^1× D^3's.

Much like the proofs by Donald and Aceto, our proof also proceeds by means of Donaldson's theorem. However, their proofs rely on the work of Lisca <cit.> which gives a detailed analysis on sums of linear lattices embedding in a full-rank lattice. We give a short proof of Theorem <ref> circumventing the reliance on Lisca's work. 
We obtain the additional equivalent condition (<ref>) in Theorem <ref>, since our method does not require the lattice embeddings to have full-rank.








Finally, we note that Theorem <ref> also plays a key role in a forthcoming paper <cit.>, where we analyse which Seifert fibered spaces smoothly embed in S^4, and in particular, completely determine the Seifert fibered spaces Y = S^2(e; r_1,…, r_k) with r_i ∈_>1 for all i, ε(Y) > 0 and e > k/2 which smooothly embed in S^4.












In Section <ref>, we recall some standard facts and establish notation and conventions. In Section <ref>, we prove the key technical theorem used to prove Theorem <ref>. In Section <ref>, we analyse when gluing compact 4-manifolds with boundary results in a definite 4-manifold and give a proof of Theorem <ref>. In Section <ref>, we prove Theorem <ref> answering Neumann-Zagier's question, as well as prove Corollary <ref>. Finally, in Section <ref> we prove Theorem <ref> determining the Seifert fibered spaces which bound rational homology S^1× D^3's.



 §.§ Acknowledgements

The first author would like to thank Cameron Gordon for his support and encouragement, and Josh Greene for a helpful conversation on Neumann-Zagier's question.



§ PRELIMINARIES

In this section we briefly recall some standard facts about Seifert fibered spaces and intersection lattices, as well as establish notation and conventions. See <cit.> for a more indepth treatment on Seifert fibered spaces and plumbings.

Given r ∈_>1, there is a unique (negative) continued fraction expansion

    r = [a_1, …, a_n]^- := a_1 - 1a_2 - 1⋱      
        a_n-1 - 1a_n,

where n ≥ 1 and a_i ≥ 2 are integers for all i ∈{1, …, n}. We associate to r the weighted linear graph (or linear chain) given in Figure <ref>. We call the vertex with weight labelled by a_i the ith vertex of the linear chain associated to r, so that the vertex labelled with weight a_1 is the first, or starting vertex of the linear chain.



We denote by Y = S^2(e; p_1/q_1, …, p_k/q_k) the Seifert fibered space over S^2 given in Figure <ref>, where e∈, and p_i/q_i∈ is non-zero for all i ∈{1,…,k}. The generalised Euler invariant of Y is given by ε(Y) = e - ∑_i=1^k q_i/p_i. Every Seifert fibered space Y is (possibly orientation reversing) homeomorphic to one in standard form, i.e. such that ε(Y) ≥ 0 and p_i/q_i > 1 for all i ∈{1,…,k}. We henceforth assume that Y is in standard form. If ε(Y) ≠ 0 then Y is a rational homology sphere with |H_1(Y)| = |p_1⋯ p_k ε(Y)|, and if ε(Y) = 0 then Y is a rational homology S^1× S^2. 



For each i∈{1,…, k}, we have the unique continued fraction expansion q_i/p_i = [a_1^i, …, a_h_i^i]^- where h_i ≥ 1 and a_j^i ≥ 2 are integers for all j ∈{1,…,h_i}. We associate to Y = S^2(e; p_1/q_1, …, p_k/q_k) the weighted star-shaped graph in Figure <ref>. The ith leg of the star-shaped graph is the weighted linear subgraph for p_i/q_i generated by the vertices labelled with weights a_1^i, …, a_h_i^i. The degree k vertex labelled with weight e is called the central vertex.



Let Γ be either the weighted star-shaped graph for Y, or a disjoint union of weighted linear graphs. There is an oriented smooth 4-manifold X_Γ given by plumbing D^2-bundles over S^2 according to the weighted graph Γ. We denote by |Γ| the number of vertices in Γ. Let m = |Γ| and denote the vertices of Γ by v_1, v_2, …, v_m. The zero-sections of the D^2-bundles over S^2 corresponding to each of v_1, …, v_m in the plumbing together form a natural spherical basis for H_2(X_Γ). With respect to this basis, which we call the vertex basis, the intersection form of X_Γ is given by the weighted adjacency matrix Q_Γ with entries Q_ij, 1 ≤ i,j ≤ m given by


    Q_ij = w(v_i),    i = j 
    
          -1,    v_iv_j
    
          0,    ,

where w(v_i) is the weight of vertex v_i. Denoting by Q_X the intersection form of X, we call (H_2(X), Q_X) ≅ (^m, Q_Γ) the intersection lattice of X_Γ (or of Γ). We denote the intersection pairing of two elements x,y∈^m by x· y = x^T  Q_Γ  y. Now assume that Γ is the star-shaped plumbing for Y. If ε(Y) > 0 then X_Γ is a positive definite 4-manifold and Γ is the standard positive definite plumbing graph for Y. If ε(Y) = 0, then X_Γ is a positive semi-definite manifold. 

Let ι : (^m, Q_Γ) → (^r, ), r > 0, be a map of lattices, i.e. a -linear map preserving pairings, where (^r, ) is the standard positive diagonal lattice. We call ι a lattice embedding if it is injective. We adopt the following standard abuse of notation. First, for each i∈{1,…,m}, we identify the vertex v_i with the corresponding ith basis element of (^m, Q_Γ). Moreover, we shall identify an element v ∈ (^m, Q_Γ) with its image ι(v) ∈ (^r, ).



§ THE EMBEDDING INEQUALITY


In this section, we prove Theorem <ref> below, which is the key technical result of this paper. In particular, it will be used in the next section to prove Theorem <ref>. We begin with some continued fraction identities which we will need.


Let {a_i}_i≥ 1 be a sequence of integers with a_i ≥ 2 for all i, and let p_k/q_k = [a_1,…, a_k]^- for all k ≥ 1. Then we have the following identities:


  * q_n p_n-1 - p_n q_n-1=1 for all n ≥ 2.

  * [a_1, …, a_n,x]^- = xp_n-p_n-1/xq_n - q_n-1, for all n ≥ 2 and x ∈ such that both sides are well defined.

  * p_n= [ a_1  -1   0   0;  -1 a_2  -1   0;   0  -1   ⋱  -1;   0   0  -1 a_n ] and q_n= [ a_2  -1   0   0;  -1 a_3  -1   0;   0  -1   ⋱  -1;   0   0  -1 a_n ] for all n ≥ 2.



For a∈, let M_a denote the matrix
M_a=[  a -1;  1  0 ]. If q/r=[a_2,…, a_n]^-, then p_n/q_n = a_1 - r/q. In particular, we have

    [ p_n; q_n ]
    = M_a_1[ q; r ]

Thus, one can inductively show that

    [ p_n; q_n ]
    = M_a_1… M_a_n[ 1; 0 ],

and furthermore that

    [    p_n -p_n-1;    q_n -q_n-1 ]
    = M_a_1… M_a_n.

Identity (a) follows by taking determinants of (<ref>) and observing that M_a=1 for any a. Identity (b) follows from combining (<ref>) and (<ref>) to get

    M_a_1… M_a_n M_x [ 1; 0 ] = [    p_n -p_n-1;    q_n -q_n-1 ][ x; 1 ]= [  xp_n -p_n-1; xq_n - q_n-1 ].

The identities in (c) can easily be proven by induction using the observation that

    [ a_1  -1   0;  -1   ⋱  -1;   0  -1 a_n ] = a_1 [ a_2  -1   0;  -1   ⋱  -1;   0  -1 a_n ] -
    [ a_3  -1   0;  -1   ⋱  -1;   0  -1 a_n ].




If a matrix A can be written as a product M M^T, then 

    A ≥ 0



Let v 0 be an eigenvector of A with eigenvalue λ. We have

    v^T A v = λv= M'^Tv≥ 0

This implies that λ≥ 0. Since A is a product of eigenvalues, this implies that A ≥ 0 as required.





Suppose we have an embedding of disjoint linear chains such that the first vertices each share a common unit vector. Then

    ∑_i=1^n q_i/p_i≤ 1.

Moreover we have equality only if there is a unit vector that has pairing ± 1 with the first vertex in each chain pairs trivially with every other vertex.



The following theorem is the key technical result of this paper.
thmlatticeineq
  
  Let ι : (^|Γ|, Q_Γ) → (^m, ) be a lattice embedding, where m > 0 and Γ is a disjoint union of weighted linear chains representing fractions p_1/q_1, …, p_n/q_n∈_>1. Suppose that there is a unit vector w ∈ (^m, ) which pairs non-trivially with (the image of) the starting vertex of each linear chain. Then
  
    ∑_i=1^n q_i/p_i≤ 1.

 Moreover, if we have equality then w has pairing ± 1 with the starting vertex of each linear chain.



 Let {e_1,…, e_m} denote the orthonormal basis of coordinates vectors of (^m, ). Since the unit vectors in (^m, ) are precisely those vectors of the form ± e_i where i∈{1,…,m}, by a change of basis if necessary, we may assume that w = e_1 ∈ (^m, ). Write ι : (^|Γ|, Q_Γ) → (^m, ) as an integer matrix with respect to the vertex basis of (^|Γ|, Q_Γ), and let M be the transpose of this matrix. Since ι preserves intersection pairings we have, u^T Q_Γ v = ι(u)^T ι(v) = (M^T u)^T (M^T v) = u^T MM^T v for all u, v ∈ (^|Γ|, Q_Γ). Thus,

    MM^T= Q_Γ =
    [ A_1   0   0;   0   ⋱   0;   0   0 A_n ],

where for each k ∈{1,…,n}, A_k on the diagonal represents a block matrix of the form

    A_k = [ a_1  -1   0   0;  -1 a_2  -1   0;   0  -1   ⋱  -1;   0   0  -1 a_l ]

where [a_1, …, a_l]^- is the continued fraction expansion for p_k/q_k. If a matrix A can be written as a product M' M'^T, then[Let v 0 be an eigenvector of A with eigenvalue λ. We have
v^T A v = λv= M'^Tv≥ 0, thus λ≥ 0. Since A is a product of eigenvalues, this implies that A ≥ 0 as required.]

    A ≥ 0.

We will prove the theorem by applying (<ref>) to a matrix of the form A=M'M'^T, where M' is a suitable modification of M.

We may write M in the form

    M=
    [ M_1;   ⋮; M_n ]

where for all k∈{1,…,n}, M_k is a matrix such that M_k M_k^T = A_k. By the assumption that e_1 pairs non-trivially with each of the starting vertices of the linear chains, we may assume that each matrix M_k is non-zero in its top left entry. For each k∈{1,…,n}, let p_k/q_k = [a_1,…, a_l]^- be the standard continued fraction expansion and choose M_k' to be the submatrix of M_k obtained by taking the first l_k rows, where l_k is chosen so that the first column w_k of M_k' takes one of the two forms:


  *  w_k = [ u 0 ⋯ 0 v ]^T,
 where l_k > 2 or a_1 > 2, and v = 0 only if M_k' = M_k.

  *  w_k = ±[  1 -1  ⋯  0  v ]^T, where a_1 = 2, and v = 0 only if M_k' = M_k.


Let M' be the matrix

    M'=
    [ 1   0  ⋯  0;        M_1';           ⋮;        M_n' ].

Then the product A= M' M'^T takes the form of the block matrix

    M'M'^T=
    [     1 w_1^T       w_n^T;   w_1  A_1'           0;                 ⋱      ;   w_n     0        A_n' ].



A can be written in the form

    A = (P_1 … P_n) (1- ∑_i=1^n Q_i/P_i),

where P_k =  A_k' and
Q_k=-[     0 w_k^T;   w_k  A_k' ] is a quantity depending only on A_k and w_k. 


By the multi-linearity of the determinant in both rows and columns we have

    [     1 w_1^T       w_n^T;   w_1  A_1'           0;     ⋮           ⋱      ;   w_n     0        A_n' ]
    =
    [    1    0         0;    0 A_1'         0;    ⋮         ⋱     ;    0    0      A_n' ]
    + ∑_1≤ i,j ≤ n B_ij,

where B_ij is the matrix

    B_ij=
    [     0     … w_j^T     …     0;     ⋮  A_1'                  ;   w_i           ⋱     0      ;     ⋮           0     ⋱      ;     0                    A_n' ].

By cofactor expansion in the first row then first column, it is not hard to check that B_ij = 0 for all i j. For i∈{1,…,n}, by row and column operations, we can put B_ij into the form of a diagonal block matrix with diagonal blocks [     0 w_k^T;   w_k  A_k' ], A_1', …, A_i-1', A_i+1', …, A_n' without changing the determinant. Hence, B_ii is the product of the determinants of these blocks, that is, B_ii = -(P_1 … P_n) Q_i/P_i. 

Since P_k>0 for all k, the previous claim combined with A≥ 0 (see (<ref>)) shows that

    ∑_i=1^n Q_i/P_i≤ 1.

So to prove the inequality in the theorem it suffices to show that
Q_k / P_k ≥ q_k /p_k for each k ∈{1,…,n}. To do this it suffices to consider some fixed k ∈{1,…,n}. For convenience, let P/Q=P_k/Q_k and p/q = p_k/q_k = [a_1, a_2, …, a_h]^- where a_i ≥ 2 for all i ∈{1,…, h}, and let l = l_k be the number of rows of A'_k.

Consider the following identity obtained by adding the second row to the first row, and the second column to the first column:

    [   0  -1   1   …   v;  -1   2  -1       0;   1  -1 a_2  -1    ;   ⋮      -1   ⋱  -1;   v   0      -1 a_l ]
    =
    [   0   1   0   …   v;   1   2  -1       0;   0  -1 a_2  -1    ;   ⋮      -1   ⋱  -1;   v   0      -1 a_l ].

Recall that w_k takes one of two possible forms. By applying the above identity if w_k takes the form <ref>, we see that regardless of the form that w_k takes, Q is equal to the determinant of a matrix of the following form

    Q=-[   0   u   0   …   v;   u a_1  -1       0;   0  -1 a_2  -1    ;   ⋮      -1   ⋱  -1;   v   0      -1 a_l ],

where if (u,v) = (±1,∓ 1) then either l > 2 or a_1 > 2. If w_k takes the form <ref>, we define u∈{± 1} via Equation (<ref>) by applying the identity.

For i∈{1,…,h}, let r_i/s_i denote the continued fraction [a_1, …, a_i]^-.
Note that P=r_l.


    Q= u^2 s_l + 2uv + v^2 r_l-1


 Applying cofactor expansion along the first column and first row in (<ref>) gives

    Q = u^2 C_1 + (-1)^l+1 uv C_2 + (-1)^l+1 uv C_3 + v^2 C_4,
 where
C_1 = [ a_2  -1   ⋯   0;  -1 a_3  -1    ;   ⋮  -1   ⋱  -1;   0      -1 a_l ],
C_2 = [    -1   a_2    -1     ⋯     0;     0    -1   a_3    -1      ;     ⋮          -1     ⋱    -1;                      -1 a_l-1;     0                      -1 ], C_3 = [    -1     0           ⋯     0;   a_2    -1                  ;    -1   a_3    -1            ;          -1     ⋱    -1      ;     0             a_l-1    -1 ]
and
C_4 = [   a_1    -1     ⋯     0;    -1   a_2    -1      ;     ⋮    -1     ⋱    -1;     0          -1 a_l-1 ].

Using the continued fraction identities in Lemma <ref>, we see that C_1 = s_l and C_4 = r_l-1. Finally, notice that C_2 (resp. C_3) is the determinant of an upper (resp. lower) triangular matrix with l-1 diagonal entries all of which are -1, hence C_2 = C_3 = (-1)^l-1.


We have Q/P≥q/p with equality only if u=± 1 and v=0.


Recall that if v=0 then r_l/s_l = p/q. Thus if v=0, Q/P = u^2q/p. Since u 0, we clearly have Q/P ≥ q/p with equality only if u^2=1, as required.
Thus assume that v 0. In this case, if l = h, or equivalently, p/q = r_l/s_l then P = p and Q = u^2 s_l + 2 u v + v^2 r_l-1 > s_l = q and thus Q/P > q/p. Hence, we assume that p/q r_l/s_l and, in particular, that p/q= [a_1, …, a_l,x]^- where x = [a_l+1, …, a_h]^- >1. Thus, by Lemma <ref> we have

    p/q= xr_l - r_l-1/xs_l -s_l-1.

Note that since u and v are both non-zero we have that

    Q= (s_l-1)u^2 + (u+v)^2 + (r_l-1-1)v^2 ≥ s_l + r_l-1 -ε,

where we take ε =2 if u=-v ∈{± 1} and ε=1 otherwise. Note that if r_l-1=2, then ε=1, as l=a_1=2 implies we cannot have u=-v ∈{± 1} by the condition stated immediately following Equation (<ref>). In either case we always have

    r_l-1-ε≥ 1.

Thus we obtain

    Q/P-q/p   ≥s_l+r_l-1-ε/r_l -xs_l-s_l-1/xr_l-r_l-1
       =(r_l-1-ε)(xr_l-r_l-1)+r_l s_l-1 - s_l r_l-1/r_l(xr_l - r_l-1)
       =(r_l-1-ε)(xr_l-r_l-1)-1/r_l(xr_l -r_l-1)
       ≥(xr_l-r_l-1)-1/r_l(xr_l -r_l-1)
       >0,

where we used the identity r_l s_l-1 - s_l r_l-1=-1 from Lemma <ref> to obtain the third line, r_l-1-ε≥ 1 to obtain the fourth line, and finally that xr_l-r_l-1>1 which follows by combining r_l ≥ r_l-1 + 1 and x > 1. This gives the desired inequality, proving the claim.

The claim together with (<ref>) proves that ∑_i=1^n q_i/p_i≤ 1 with equality only if w = e_1 has pairing ± 1 with each starting vertex, which completes the proof.
It remains only to check the case when we can have equality. If equality occurs, the above claim implies that for all k ∈{1,…,n}, that in whichever of the two possible forms <ref> and <ref> w_k takes, we have v = 0 which implies in particular that M_k' = M_k.

Recall that e_1 ∈ (,) is the unit vector pairing non-trivially with the starting vertex of each linear chain. If w_k takes form <ref> for all k ∈{1,…,n}, then we are done by the equality case in the above claim. Suppose that for some k ∈{1,…,n}, w_k has form <ref>, in particular w_k = ±[  1 -1  0  ⋯  0 ]^T. By a change of basis if necessary, we may assume the first vertex v_1 of the kth linear chain is embedded as e_1 - e_2, and the adjacent, second vertex v_2 satisfies v_2 · e_1 = -1 implying that v_2 · e_2 = 0.

We claim that e_2 pairs ± 1 with each starting vertex and pairs trivially with every other vertex. Let v be the starting vertex of some other linear chain. Then v must have weight ≥ 3 in order to pair trivially with v_1 and v_2. Thus, w_j takes <ref>. Therefore v · e_1 = ± 1 = v · e_2, which follows from v · v_1 = v · (e_1 - e_2) = 0. Similarly, we get that v' · e_2 = v' · e_1 = 0 for each vertex v' which pairs trivially with v_1 and e_1, which proves the claim and completes the proof of the equality case.






Suppose we have an embedding of disjoint linear chains such that the first vertices each share a common unit vector e_1. Then

    ∑_i=1^n q_i/p_i≤ 1.

Moreover, we have equality only if for each chain one of the following holds:


  * The unit vector e_1 has pairing ± 1 with the first vertex of the chain and does not appear in any other vertex in the chain, or
  
  * Each of the first two vertices v_1, v_2 of the chain has weight 2, (e_i · v_1, e_i · v_2) = (± 1, ∓ 1), and e_1 does not appear in any other vertex of the chain.




§ DEFINITE 4-MANIFOLDS AND THE SEIFERT FIBERED SPACE INEQUALITY

Now we consider when gluing two 4-manifolds can result in a closed definite 4-manifold.

Let U_1 and U_2 be 4-manifolds with ∂ U_1 = -∂ U_2 =Y. Then the closed 4-manifold X=U_1 ∪_Y U_2 is positive definite if and only if


  *  the inclusion-induced map (i_1)_* ⊕ (i_2)_*  H_1(Y; ) → H_1(U_1; )⊕ H_1(U_2; ) is injective and

  *  for i=1,2, U_i has the maximal possible signature, that is, 
    σ(U_i) = b_2(U_i) + b_1(U_i) - b_3(U_i) - b_2(Y).

  


In this proof all homology groups will be taken with rational coefficients. First, for i=1,2, consider the following segment of the long exact sequence in homology of the pair (U_i, Y):

    0→ H_3(U_i) → H_3(U_i,Y) → H_2(Y) → H_2(U_i)

By exactness and Lefschetz duality this shows that the rank of the map H_2(Y) → H_2(U_i) is b_2(Y)-b_1(U_i) +b_3(U_i). Homology classes in the image of H_2(Y) → H_2(U_i) pair trivially with all classes in H_2(U_i). This gives an upper bound on the signature of U_i:

    σ(U_i) ≤ b_2(U_i) + b_1(U_i) - b_3(U_i) - b_2(Y).

Now consider the segment of the Mayer-Vietoris sequence

    0→    H_3(U_1)⊕ H_3(U_2) → H_3(X) → H_2(Y) → H_2(U_1)⊕ H_2(U_2) →
    →   H_2(X) → H_1(Y) → H_1(U_1)⊕ H_1(U_2) → H_1(X) → 0.

The last three terms in this sequence show that

    b_1(U_1)+b_1(U_2)≤ b_1(Y)+b_1(X),

with equality if and only if the map induced by the inclusions 

    (i_1)_* ⊕ (i_2)_*  H_1(Y) → H_1(U_1)⊕ H_1(U_2)

is injective.

Since the Euler characteristic of an exact sequence is zero, (<ref>) shows that 

    b_2(X)=2b_1(X) + ∑_i=1^2 (b_2(U_i)-b_1(U_i) - b_3(U_i)),

where we also used that b_1(Y)=b_2(Y) and b_1(X)=b_3(X).

By Novikov additivity, we have that σ(X)=σ(U_1)+σ(U_2). So by summing the inequalities in (<ref>) for i=1,2 and comparing with (<ref>) we obtain

    b_2(X) ≥ 2(b_1(X) + b_2(Y) - b_1(U_1) - b_1(U_2)) + σ(X),

with equality if and only if we have equality in (<ref>) for both i=1,2.
Hence, X can be positive definite if and only if 

    b_1(U_1)+b_1(U_2)= b_2(Y)+b_1(X).

and we have equality in (<ref>) for i=1,2. However we have already seen that equality occurs in (<ref>) if and only if (i_1)_* ⊕ (i_2)_* is injective. 










This allows us to prove the main theorem.

*
 Let X be the standard positive (semi-)definite plumbing 4-manifold with ∂ X = Y, and let Z = X ∪_Y -W. It follows from Proposition <ref> that Z is positive definite. To see that condition <ref> of Proposition <ref> holds, note that from the homology long exact sequence for (Y, W) we get
  
    0 → H_1(Y;) → H_1(W;) → H_1(W,Y;) → 0,

  where we used that H_1(Y;) → H_1(W;) is injective. Taking the Euler characteristic of (<ref>) and applying Lefschetz duality shows that b_3(W) - b_1(W) + b_2(Y) = 0. Thus, Z is a smooth positive definite 4-manifold, so by Donaldson's theorem Z has standard positive diagonal intersection form. The inclusion X ⊂ Z induces a map H_2(X) → H_2(Z) which preserves the intersection pairing. Thus, there is a -linear map (H_2(X), Q_X) → (^m, ) for some m > 0, which preserves intersection pairings.

  We construct a partition of {1,2,…,k} into at most e classes as follows. Denote the orthonormal basis of coordinate vectors of (^m, ) by {e_1,…,e_m}. For v ∈ (H_2(X), Q_X), we call {e_i : 1≤ i≤ m, e_i · v ≠ 0} the support of v. Without loss of generality, we may assume that the central vertex has support {e_1,e_2,…, e_n} where n ≤ e. Let v_1,v_2,…, v_k be the vertices of the plumbing adjacent to the central vertex, so that v_i is a vertex belonging to the ith leg of the plumbing graph (with fraction p_i/q_i). For i∈{1,…, n}, let B_i = { 1 ≤ j ≤ k | v_j · e_i ≠ 0 } and define B_0 = ∅. Let C_i = B_i \∪_j < i B_j for i ∈{1,…,n}. Then C_1,…, C_n are disjoint and ∪_i C_i = {1,…,k}. Thus the non-empty classes {C_i : C_i ≠∅} form a partition of {1,2,…,k} into at most e classes. By definition for each i ∈{1,2,…, n}, the starting vertices of the linear chains indexed by C_i all have support containing the common unit vector e_i. Hence, by Theorem <ref>, we have that ∑_j ∈ C_iq_j/p_j≤ 1.



§ NEUMANN-ZAGIER'S QUESTION

We prove Theorem <ref> below which, when combined with Donaldson's theorem, immediately implies Theorem <ref>. Note that the following theorem also positively answers Neumann-Zagier's question stated in the introduction.



thmnzconjthm  Let Y = S^2(e; p_1/q_1, …, p_k/q_k), k ≥ 3, be in standard form, that is, with p_i/q_i > 1 for all i∈{1,2,…,k}, e > 0 and with Y bounding a smooth positive definite plumbing X. Suppose that |H_1(Y)| ∈{1,2,3,5,6,7} and the intersection lattice (H_2(X), Q_X) embeds into a positive standard diagonal lattice. Then e = 1. 



  

For sake of contradiction, assume that e > 1. We may apply Theorem <ref>, noting that the existence of W in the hypothesis of Theorem <ref> is only required to ensure that there is a map of lattices of (H_2(X), Q_X) into a positive standard diagonal lattice. Hence, there is a partition {C_1,…, C_n} of {1,…,k} into n ≤ e classes. Moreover, for each class C, 1 - ∑_i ∈ Cq_i/p_i≥ 0, and we call C complementary if equality occurs, and non-complementary otherwise.
  
 We have 
  
    |H_1(Y)|    = p_1⋯ p_k ·ε(Y) = p_1⋯ p_k (e - ∑_i=1^k q_i/p_i) 
       = p_1 ⋯ p_k ( (e - n) + ∑_i=1^n (1 - ∑_j ∈ C_iq_j/p_j) ) 
       = p_1 ⋯ p_k (e - n) + ∑_i=1^n a_i ∏_1 ≤ l ≤ k
     l ∉C_i p_l,

  where a_i = (∏_j ∈ C_i p_j) · (1 - ∑_l∈ C_iq_i/p_i) is an integer for all i ∈{1,…,n}.
  Notice that all terms in (<ref>) are non-negative integers. Since we are assuming that |H_1(Y)| ∈{1,2,3,5,6,7}, we must have n = e, otherwise |H_1(Y)| ≥ p_1 ⋯ p_k ≥ 2· 2· 2 = 8 since k ≥ 3.
  
We claim that |C_i| ≤ k-2 for some i ∈{1,…,e} with C_i non-complementary. To see this, we argue as follows. There are n = e ≥ 2 classes in the partition, and at least one non-complementary class since |H_1(Y)| > 0. If there are two non-complementary classes then at least one has size at most k-2 since k ≥ 3. If there is only one non-complementary class, then there is a complementary class which necessarily has size at least 2, and hence the non-complementary class satisfies the claim.

Combining the above claim with (<ref>), we see that |H_1(Y)| is a sum of integers greater than 1, and at least one of these integers is not prime. For |H_1(Y)| ∈{1,2,3,5,6,7}, this is only possible for |H_1(Y)| = 7 with decomposition 7 = 3 + 2 · 2, and for |H_1(Y)| = 6 with the two decompositions |H_1(Y)| = 2 + 2· 2 = 2· 3. We address these cases in turn.
For |H_1(Y)| = 7 = 3 + 2 · 2, comparing this decomposition with (<ref>), we see that there must exist some non-complementary C_i with |C_i| = 2 and p_j = 2 for all j ∈ C_i. However, such a C_i must be complementary since 1 - 1/2 - 1/2 = 0, a contradiction. A similar argument rules out the decomposition |H_1(Y)| = 2 + 2· 2. Finally, in the case |H_1(Y)| = 2· 3, the decomposition implies that there exists a complementary class C_i = {a,b} with p_a = 2 and p_b = 3, which is impossible.


We obtain the following corollary, answering a question of Lidman-Tweedy <cit.>.

*
 We prove the contrapositive. Assume that d(Y) = 0. Note that reversing the orientation of Y simply changes the sign of the weight of the central vertex in the definite plumbing bounding Y. Thus, by reversing the orientation of Y if necessary we assume that Y bounds a smooth negative definite plumbing X^4. Let 𝒞 = {ξ∈ H_2(X;) |ξ· v = v · v 2v∈ H_2(X;)} be the set of characteristic vectors, and let n = (H_2(X)). Elkies <cit.> proved that
  0 ≤ n + max_ξ∈𝒞ξ·ξ, with equality if and only if Q_X is diagonalizable over . However, it follows from <cit.> that n + max_ξ∈𝒞ξ·ξ≤ 4d(Y) = 0. Therefore Q_X is diagonalizable over , in particular (H_2(-X), Q_-X) embeds into a positive standard diagonal lattice. Hence, Theorem <ref> implies that |e| = 1.




§ SEIFERT FIBERED SPACES BOUNDING RATIONAL HOMOLOGY S^1 × D^3'S

In this section we prove Theorem <ref>, which in particular gives a classification of the Seifert fibered spaces which smoothly bound rational homology S^1× D^3's. We note that the implication (<ref>) implies (<ref>) was proved by Donald <cit.>, and the equivalence of (<ref>) and (<ref>) was shown by Aceto <cit.>.

*

 
  First suppose that (<ref>) holds, that is, Y = S^2(k; p_1/q_1, p_1/p_1-q_1, …, p_k/q_k, p_k/p_k - q_k), where k ≥ 0 and p_i/q_i∈_>1 for all i∈{1,…,k}. By Rolfsen twisting, Y can be put into the form S^2(0; p_1/q_1, -p_1/q_1, …, p_k/q_k, -p_k/q_k). Let M = S^2(0; p_1/q_1, p_2/q_2, …, p_k/q_k), let M^∘ be the 3-manifold with torus boundary given by removing a tubular neighbourhood of a regular fiber of M and let W = M × [0, 1]. Then ∂ W = M^∘∪_∂ -M^∘ is the double of M^∘, which is precisely Y. Finally, notice that H_*(W; ) = H_*(M^∘; ) = H_*(S^1 × D^3; ), where the last equality follows from the fact that M is a rational homology S^3 and M^∘ is obtained by removing a neighbourhood of a simple closed curve from M. This proves (<ref>).
  
  The implication (<ref>) implies (<ref>) holds by taking W_± = W and noting that H_1(Y;) → H_1(W;) is injective by the long exact sequence of the pair (Y, W).
  
  Finally assume that (<ref>) holds. Hence, Y is the boundary of smooth 4-manifolds W_+ and W_- satisfying σ(W_±) = ± b_2(W_±) and such that the inclusion induced maps H_1(Y) → H_1(W_±) are injective. Write Y as S^2(e; p_1/q_1, …, p_k/q_k) with k ≥ 3 and p_i/q_i∈_>1 for all i∈{1,…,k}. Notice that -Y = S^2(k-e; p_1/p_1 - q_1, …, p_k/p_k - q_k), and Y is of the form given in (<ref>) if and only if -Y is of this form.
 
  Thus, by reversing the orientations of both Y and W_± if necessary, we may assume that e ≥k/2.

  By Theorem <ref>, there is a partition {C_1,…, C_n} of {1,…,k} into n ≤ e classes such that for each class C, 1 - ∑_i ∈ Cq_i/p_i≥ 0. Since Y is a rational homology S^1× S^2, we thus have
    
    0    = p_1⋯ p_k ·ε(Y) = p_1⋯ p_k (e - ∑_i=1^k q_i/p_i) 
       = p_1 ⋯ p_k ( (e - n) + ∑_i=1^n (1 - ∑_j ∈ C_iq_j/p_j) ),

    where all terms in the sum are non-negative. Hence, we must have n = e and 1 - ∑_i ∈ Cq_i/p_i = 0, for all i∈{1,…,n}. This implies that |C_i| ≥ 2 for all i∈{1,…,n}. Thus, there are at least 2n = 2e legs, so e ≤k/2. However, by assumption e ≥k/2 so e = k/2 and |C_i| = 2 for all i∈{1,…,n}. Thus, C_1,…,C_n partition {1,…,k} into pairs of indices indexing pairs of fractions of the form p/q, p/p-q∈_>1, and thus (<ref>) holds.




alpha

