1. Introduction

One of the fundamental problems in algebraic geometry is to study the geography of projective varieties, i.e., determining which Chern numbers occur for a complex smooth projective variety M. When M is a minimal surface of general type, we have Noether’s inequalities [1]:

\( {p_{g}}(M)={h^{0}}(M,{ω_{M}}) \)

\( K_{M}^{2}≥2{p_{g}}(M)-4. \)

This implies

\( 5c_{1}^{2}(M)≥{c_{2}}(M)-36. \)

While on the other hand, we have the Miyaoka-Yau inequality:

\( c_{1}^{2}(M)≤3{c_{2}}(M). \)

Hence \( \frac{{c_{2}}(M)}{c_{1}^{2}(M)} \) is bounded. When M is a threefold of general type with ample canonical divisor, Yau’s famous inequality in [2] says

\( 8{c_{1}}(M){c_{2}}(M)≤3c_{1}^{3}(M). \)

Hunt studied the geography of threefolds in [3]. Later, Chang and Lopez proved in [4] that the region described by the Chern ratios \( (\frac{c_{1}^{3}(M)}{{c_{1}}(M){c_{2}}(M)},\frac{{c_{3}}(M)}{{c_{1}}(M){c_{2}}(M)}) \) of threefolds with ample canonical divisor is bounded. Sheng, Xu and Zhang gave the inequalities of Chern numbers of complete intersection threefolds with ample canonical divisor in [5]:

\( 86c_{1}^{3}(M)≤{c_{3}}(M)≤\frac{c_{1}^{3}(M)}{6}. \)

The theorem of Chang-Lopez has been generalized to higher dimensional case by Du and Sun in [6].

Theorem 1.1. Let X be a nonsingular projective variety of dimension n over an algebraic closed field κ with any characteristic. Suppose K_X or −K_X is ample. If the characteristic of κ is 0 or the characteristic of κ is positive and O_X(K_X)(O_X(−K_X), respectively) is globally generated, then

\( (\frac{{c_{2,{1^{n-2}}}}}{c_{1}^{n}},\frac{{c_{2,2,{1^{n-4}}}}}{c_{1}^{n}},⋯,\frac{{c_{n}}}{c_{1}^{n}})∈{A^{p(n)}} \)

is contained in a convex polyhedron in A^p(n) depending on the dimension of X only, where p(n) is the partition number and the elements in the parentheses arranged from small to big in terms of the alphabet order of the lower indices of the numerators.

In this paper, we study the inequalities of Chern numbers of complete intersection threefolds in products of \( {P^{1}} \) . Throughout this paper, we always let \( {π_{i}}:\underset{n+3copies}{\underbrace{{P^{1}}×{P^{1}}×⋯×{P^{1}}}}\overset{}{→}{P^{1}} \) be the i-th projection, and \( {Q_{i}}=π_{i}^{*}(P) \) , where \( P \) is a point of \( {P^{1}} \) . Take H_i be a general divisor in the linear system \( |\sum _{t=1}^{n+3} {d_{it}}{Q_{t}}| \) , where d_it is a positive integer for 1 ≤ i ≤ n,1 ≤ t ≤ n+3. By the Bertini theorem, one can assume that H_i is a smooth hypersurface for i = 1, 2, ···, n, and \( X={H_{1}}∩{H_{2}}∩⋯∩{H_{n}} \) is a smooth threefold.

Our main result is

Theorem1.2. If \( {d_{ij}}≥4 \) for any \( 1≤i≤n,1≤j≤n+3, \) then we have \( \frac{1}{2} \lt \frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)} \lt \frac{2}{(4n-2{)^{2}}}+\frac{2}{4n-2}+1 \) . If \( {d_{ij}}≥6 \) for any \( 1≤i≤n,1≤j≤n+3 \) , then \( \frac{{c_{1}}(\bar{X}){c_{2}}(X)}{c_{1}^{3}(X)}-\frac{1}{2} \lt \frac{{c_{3}}(X)}{c_{1}^{3}(X)} \lt \frac{7}{12}. \)

In Section 2, In section 2, we recall the basic definitions and properties of Chern classes. In section 3, we will compute the Chern numbers of X. In section 4, we study the upper and lower bounds of \( \frac{{c_{3}}(X)}{c_{1}^{3}(X)}and\frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)} \) .

2. Chern classes

In this section, we introduce the definition of Chern classes.

Let M be a smooth projective variety of dimension n. Let \( A(M)=⊕_{i=1}^{n}{A^{i}}(M) \) be the Chow ring of M. E is a vector bundle on M of rank r. The Chern class \( {c_{i}}(E) \) is a cycle in \( {A^{i}}(M) \) , here c₀(E) = 1. We let \( {c_{t}}(E)=1+{c_{1}}(E)t+⋯+{c_{r}}(E){t^{r}} \) be the Chern polynomial of E.

Chern class c_i(E) satisfies the properties below:

(1) If D is a divisor on M and \( E≅{O_{M}}(D) \) is a line bundle, then c₁(E)=D.

(2) If f : M′ → M is a morphism of projective varieties, then c_i(f∗E) = f∗c_i(E).

(3) If 0 → E′ → E → E′′ → 0 is a short exact sequence of a vector bundle, then

\( \begin{array}{c} {c_{t}}(E)={c_{t}}({E^{ \prime }})\cdot {c_{t}}({E^{″}}) \\ =(1+{c_{1}}({E^{ \prime }})t+⋯+{c_{{r^{ \prime }}}}({E^{ \prime }}){t^{{r^{ \prime }}}})(1+{c_{1}}({E^{″}})t+⋯+{c_{{r^{ \prime \prime }}}}({E^{″}}){t^{{r^{ \prime \prime }}}}) \\ ={c_{{r^{ \prime }}}}({E^{ \prime }})\cdot {c_{{r^{ \prime \prime }}}}({E^{ \prime \prime }}){t^{{r^{ \prime }}+{r^{ \prime \prime }}}}+⋯. \end{array} \)

Assume that rank \( {E^{ \prime }}={r^{ \prime }} \) , rank \( {E^{ \prime \prime }}={r^{ \prime \prime }} \) , so that rank E = r′ + r′′. As a result, we have \( {c_{{r^{ \prime }}+{r^{ \prime \prime }}}}(E)={c_{{r^{ \prime }}}}({E^{ \prime }}){c_{{r^{ \prime \prime }}}}({E^{ \prime \prime }}). \)

(4) Let s be a global section of E. Assume that the zero set Z(s) of s satisfies that dim Z(s) =dimM − r, then \( {c_{r}}(E)=Z(s)∈{A^{r}}(M). \)

We call \( {c_{i}}(M)={c_{i}}({T_{M}}) \) the i-th Chern class of M.

3. Chern numbers of complete intersection three- folds in products of projective spaces

In this section, we compute the Chern numbers of X.

\( M=\underset{n+3}{\underbrace{{P^{1}}×{P^{1}}×⋯×{P^{1}}}} \)

then one sees

\( \begin{array}{c} {c_{t}}(M)={c_{t}}({T_{M}})={c_{t}}(π_{1}^{*}{T_{{P^{1}}}}⊕⋯⊕π_{n+3}^{*}{T_{{P^{1}}}}) \\ =(1+2{Q_{1}}t)(1+2{Q_{2}}t)⋯(1+2{Q_{n+3}}t). \end{array} \)

From the standard exact sequence

\( 0⟶{O_{{H_{1}}}}(-{H_{1}})⟶{Ω_{M}}{|_{{H_{1}}}}⟶{Ω_{{H_{1}}}}⟶0, \)

after taking duality, we have

\( 0⟶{T_{{H_{1}}}}⟶{T_{M}}{|_{{H_{1}}}}⟶{O_{{H_{1}}}}({H_{1}})⟶0. \)

Hence, we have

\( \begin{array}{c} {c_{t}}({H_{1}})=\frac{{c_{t}}({T_{M}}{|_{{H_{1}}}})}{{c_{t}}({O_{{H_{1}}}}({H_{1}}))} \\ =\frac{(1+2{Q_{1}}t)(1+2{Q_{2}}t)⋯(1+2{Q_{n+3}}t){|_{{H_{1}}}}}{(1+{H_{1}}t){|_{{H_{1}}}}}. \end{array} \)

From the exact sequence

\( 0⟶{T_{{H_{1}}∩{H_{2}}}}⟶{T_{{H_{1}}}}{|_{{H_{1}}}}∩{H_{2}}⟶{O_{{H_{1}}∩{H_{2}}}}({H_{2}})⟶0, \)

We obtain

\( {c_{t}}({H_{1}}∩{H_{2}})=\frac{(1+2{Q_{1}}t)⋯(1+2{Q_{n+3}}t){|_{{H_{1}}∩{H_{2}}}}}{(1+{H_{1}}t)(1+{H_{2}}t){|_{{H_{1}}∩{H_{2}}}}} \)

By repeating the procedure above, it can be obtained that

\( {c_{t}}(X)={c_{t}}({H_{1}}∩⋯∩{H_{n}})=\frac{(1+2{Q_{1}}t)(1+2{Q_{2}}t)⋯(1+2{Q_{n+3}}t){|_{X}}}{(1+{H_{1}}t)(1+{H_{2}}t)⋯(1+{H_{n}}t){|_{X}}} \)

It follows that

\( \begin{array}{c} (1+{c_{1}}(X)t+{c_{2}}(X){t^{2}}+{c_{3}}(X){t^{3}})(1+{H_{1}}t)(1+{H_{2}}t)⋯(1+{H_{n}}t){|_{X}} \\ =(1+2{Q_{1}}t)(1+2{Q_{2}}t)⋯(1+2{Q_{n+3}}t){|_{X}}. \end{array} \)

By considering the coefficient of t, we can get

\( \begin{array}{c} {c_{1}}(X)+{H_{1}}{|_{X}}+{H_{2}}{|_{X}}+⋯+{H_{n}}{|_{X}} \\ =2{Q_{1}}{|_{X}}+2{Q_{2}}{|_{X}}+⋯+2{Q_{n+3}}{|_{X}}. \end{array} \)

Thus,

\( \begin{array}{c} {c_{1}}(X)=(2{Q_{1}}+2{Q_{2}}+⋯+2{Q_{n+3}}-{H_{1}}-{H_{2}}-⋯-{H_{n}}){|_{X}} \\ =\sum _{i=1}^{n+3} (2-{d_{1i}}-{d_{2i}}-⋯-{d_{ni}}){Q_{i}}{|_{X}}. \end{array} \ \ \ (1) \)

As for the coefficient of t², we see that

\( \begin{array}{c} {c_{2}}(X)+{c_{1}}(X)({H_{1}}+{H_{2}}+⋯+{H_{n}}){|_{X}}+\sum _{1≤i \lt j≤n} {H_{i}}{H_{j}}{|_{X}} \\ =4\sum _{1≤i \lt j≤n+3} {Q_{i}}{Q_{j}}{|_{X}}. \end{array} \)

Since

\( \begin{array}{c} {H_{1}}+{H_{2}}+⋯+{H_{n}} \\ =\sum _{i=1}^{n} {d_{i1}}{Q_{1}}+\sum _{i=1}^{n} {d_{i2}}{Q_{2}}+⋯+\sum _{i=1}^{n} {d_{i,n+3}}{Q_{n+3}}, \end{array} \)

We obtain

\( \begin{array}{c} {c_{1}}(X)({H_{1}}+{H_{2}}+⋯+{H_{n}}) \\ =\sum _{1≤i,j≤n+3} (2-{d_{1i}}-{d_{2i}}-⋯-{d_{ni}})\sum _{k=1}^{n} {d_{kj}}{Q_{i}}{Q_{j}}. \end{array} \)

Simple computations show that

\( \begin{array}{c} {H_{i}}{H_{j}}=({d_{i1}}{Q_{1}}+{d_{i2}}{Q_{2}}+⋯+{d_{i,n+3}}{Q_{n+3}})({d_{j1}}{Q_{1}}+{d_{j2}}{Q_{2}}+⋯+{d_{j,n+3}}{Q_{n+3}}) \\ =\sum _{1≤k,l≤n+3} {d_{ik}}{d_{jl}}{Q_{k}}{Q_{l}} \end{array} \)

Hence, we have

\( \begin{array}{c} {c_{2}}(X)=4\sum _{1≤i \lt j≤n+3} {Q_{i}}{Q_{j}}{|_{X}}-\sum _{1≤k,l≤n+3} {d_{ik}}{d_{jl}}{Q_{k}}{Q_{l}}{|_{X}}- \\ \sum _{1≤i,j≤n+3} (2-{d_{1i}}-{d_{2i}}{|_{X}}-⋯-{d_{ni}})\sum _{k=1}^{n} {d_{kj}}{Q_{i}}{Q_{j}}{|_{X}}. \end{array} \ \ \ (2) \)

Now considering the coefficient of t³, we get

\( \begin{array}{c} {c_{3}}(X)+{c_{2}}(X)\sum _{i=1}^{n} {H_{i|}}X+{c_{1}}(X)\sum _{1≤i \lt j≤n} {H_{i}}{H_{j}}{|_{X}}+ \\ \sum _{1≤i \lt j \lt k≤n} {H_{i}}{H_{j}}{H_{k}}{|_{X}}=8(\sum _{1≤i \lt j \lt k≤n+3} {Q_{i}}{Q_{j}}{Q_{k}}){|_{X}}. \end{array} \)

This implies

\( \begin{array}{c} {c_{3}}(X)=\sum _{{i_{1}},⋯,{i_{n}},i,j,k} {d_{1{i_{1}}}}⋯{d_{n{i_{n}}}}(\frac{4}{3}-\frac{1}{6}\sum _{1≤r,s,t≤n,(r-s)(s-t)(t-r)≠0} {d_{ri}}{d_{sj}}{d_{tk}} \\ -\frac{1}{2}(2-\sum _{i=1}^{n} {d_{ti}})\sum _{1≤s,t≤n,s≠t} {d_{tj}}{d_{sk}}-[2-\frac{1}{2}\sum _{1≤s,t≤n,s≠t} {d_{ti}}{d_{sj}}- \\ (2-\sum _{t=1}^{n} {d_{ti}})\sum _{t=1}^{n} {d_{tj}}]\sum _{t=1}^{n} {d_{tk}}) \end{array} \ \ \ (3) \)

where \( {i_{1}},⋯,{i_{n}},i,j,k \) take all the arrangements of \( 1,2,⋯,n+3. \)

By (1), (2), (3), we can have

\( c_{1}^{3}(X)=\sum _{{i_{1}},⋯,{i_{n}},i,j,k} {d_{1{i_{1}}}}⋯{d_{n{i_{n}}}}(2-\sum _{t=1}^{n} {d_{ti}})(2-\sum _{t=1}^{n} {d_{tij}})(2-\sum _{t=1}^{n} {d_{tk}}),\ \ \ (4) \)

\( \begin{array}{c} {c_{1}}(X){c_{2}}(X)=\sum _{{i_{1}},⋯,{i_{n}},i,j,k} {d_{1{i_{1}}}}⋯{d_{n{i_{n}}}}[2-\frac{1}{2}\sum _{1≤t,s≤n,t≠s} {d_{ti}}{d_{sj}} \\ -(2-\sum _{t=1}^{n} {d_{ti}})\sum _{t=1}^{n} {d_{tj}}](2-\sum _{t=1}^{n} {d_{tk}}) \end{array} \ \ \ (5) \)

4. Inequalities of Chern numbers

In this section, we estimate the upper and lower bounds for \( \frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)} \) and \( \frac{{c_{3}}(X)}{c_{1}^{3}(X)} \) respectively. Let

\( {A_{i}}=(\sum _{t=1}^{n} {d_{ti}})-2,\ \ \ (6) \)

\( \begin{matrix}{B_{ij}} & = & \sum _{1≤s,t≤n,s≠t} {d_{ti}}{d_{sj}}, \\ \end{matrix}\ \ \ (7) \)

\( {C_{ijk}}=\sum _{1≤r,s,t≤n,(r-s)(s-t)(t-r)≠0}{d_{ri}}{d_{sj}}{d_{tk}},\ \ \ (8) \)

We have

\( \begin{array}{c} -c_{1}^{3}(X)=\sum _{{i_{1}},⋯,{i_{n}},i,j,k} {d_{1{i_{1}}}}⋯{d_{n{i_{n}}}}{A_{i}}{A_{j}}{A_{k}}, \\ -{c_{1}}(X){c_{2}}(X)=\sum _{{i_{1}},⋯,{i_{n}},i,j,k} {d_{1{i_{1}}}}⋯{d_{n{i_{n}}}}(2-\frac{1}{2}{B_{ij}}+{A_{i}}({A_{j}}+2)){A_{k}}, \\ -{c_{3}}(X)=\sum _{{i_{1}},⋯,{i_{n}},i,j,k} {d_{1{i_{1}}}}⋯{d_{n{i_{n}}}}[2-\frac{1}{2}{B_{ij}}+{A_{i}}({A_{j}}+2)]({A_{k}}+2) \\ -\frac{1}{2}{A_{i}}{B_{jk}}+\frac{1}{6}{C_{ijk}}-\frac{4}{3}. \end{array} \ \ \ (9) \)

4.1. Inequalities of \( \frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)} \)

In order to estimate \( \frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)} \) , we need to estimate

\( \frac{{d_{1{i_{1}}}}⋯{d_{n{i_{n}}}}(2-\frac{1}{2}{B_{ij}}+{A_{i}}({A_{j}}+2)){A_{k}}}{{d_{1{i_{1}}}}⋯{d_{n{i_{n}}}}{A_{i}}{A_{j}}{A_{k}}}=\frac{2-\frac{1}{2}{B_{ij}}+{A_{i}}({A_{j}}+2)}{{A_{i}}{A_{j}}} \)

for any \( 1≤i,j≤n+3 \) and \( i≠j. \)

Lemma 1. \( {If d_{ij}}≥4 for 1≤i≤n,1≤j≤n+3,{B_{ij}} \lt {A_{i}}{A_{j}}. \)

Proof. If \( {d_{ij}}≥4 \) , we have

\( \begin{array}{c} \sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{ij}}-2\sum _{t=1}^{n} {d_{ti}}-2\sum _{t=1}^{n} {d_{ij}}+4 \\ =\frac{1}{2}\sum _{t=1}^{n} {d_{ti}}{d_{tj}}-2\sum _{t=1}^{n} {d_{ti}}+\frac{1}{2}\sum _{t=1}^{n} {d_{ti}}{d_{tj}}-2\sum _{t=1}^{n} {d_{ij}}+4 \\ =\sum _{t=1}^{n} (\frac{1}{2}{d_{tj}}-2){d_{ti}}+\sum _{t=1}^{n} (\frac{1}{2}{d_{ti}}-2){d_{tj}}+4≥4. \end{array} \)

Since

\( \begin{array}{c} {A_{i}}{A_{j}}=(\sum _{t=1}^{n} {d_{ti}}-2)(\sum _{t=1}^{n} {d_{ij}}-2) \\ =\sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{ij}}-2\sum _{t=1}^{n} {d_{ti}}-2\sum _{t=1}^{n} {d_{ij}}+4 \\ ={B_{ij}}+\sum _{t=1}^{n} {d_{ti}}{d_{tj}}-2({A_{i}}+2)-2({A_{j}}+2)+4 \\ ={B_{ij}}-2{A_{i}}-2{A_{j}}-4+\sum _{t=1}^{n} {d_{ti}}{d_{tj}} \end{array} \)

One sees that

\( {A_{i}}{A_{j}}≥{B_{ij}}+4 \gt {B_{ij}}. \)

Lemma 2. When \( {d_{ij}}≥4 for 1≤i≤n,1≤j≤n+3 \) , then we have \( \frac{1}{2} \lt \frac{2-\frac{1}{2}{B_{ij}}+{A_{i}}({A_{j}}+2)}{{A_{i}}{A_{j}}} \lt \frac{2}{(4n-2{)^{2}}}+\frac{2}{4n-2}+1 \) .

Proof. As \( {d_{ij}}≥4, \) one sees \( {A_{j}}≥4n-2 \) , which means \( \frac{1}{{A_{j}}}≤\frac{1}{4n-2} \) . We can also have \( \frac{1}{{A_{i}}} \gt 0 \) . By Lemma 1, we have

\( \frac{2-\frac{1}{2}{B_{ij}}+{A_{i}}({A_{j}}+2)}{{A_{i}}{A_{j}}}=\frac{2}{{A_{i}}{A_{j}}}-\frac{\frac{1}{2}{B_{ij}}}{{A_{i}}{A_{j}}}+\frac{2}{{A_{j}}}+1 \gt 1-\frac{1}{2}=\frac{1}{2} \)

On the other hand, we have

\( \frac{2}{{A_{i}}{A_{j}}}-\frac{\frac{1}{2}{B_{ij}}}{{A_{i}}{A_{j}}}+\frac{2}{{A_{j}}}+1 \lt \frac{2}{{A_{i}}{A_{j}}}+\frac{2}{{A_{j}}}+1 \lt \frac{2}{(4n-2{)^{2}}}+\frac{2}{4n-2}+1\ \ \ (10) \)

Theorem 4.1. If \( {d_{ij}}≥4 for any 1≤i, j ≤n+3, \) then we have \( \frac{1}{2} \lt \) \( \frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)} \lt \frac{2}{(4n-2{)^{2}}}+\frac{2}{4n-2}+1. \)

Proof. The desired conclusion follows from Lemma 2.

4.2. Inequalities of \( \frac{{c_{3}}(X)}{c_{1}^{3}(X)} \)

In order to estimate the range of \( \frac{{c_{3}}(X)}{c_{1}^{3}(X)} \) , we need to estimate the range of

\( \frac{(2-\frac{1}{2}{B_{ij}}+{A_{i}}{A_{j}}+2{A_{i}})({A_{k}}+2)-\frac{1}{2}{A_{i}}{B_{jk}}+\frac{1}{6}{C_{ijk}}-\frac{4}{3}}{{A_{i}}{A_{j}}{A_{k}}} \)

Lemma 3. \( {If d_{ij}}≥6 for any i, j, then we have {A_{i}}{A_{j}}{A_{k}} \gt {C_{ijk}}. \)

Proof. One sees that

\( \begin{array}{c} {A_{i}}{A_{j}}{A_{k}}=(\sum _{t=1}^{n} {d_{ti}}-2)(\sum _{t=1}^{n} {d_{tj}}-2)(\sum _{t=1}^{n} {d_{tk}}-2) \\ =\sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{tj}}\sum _{t=1}^{n} {d_{tk}}-2\sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{tj}}-2\sum _{t=1}^{n} {d_{tj}}\sum _{t=1}^{n} {d_{tk}} \\ -2\sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{tk}}+4\sum _{t=1}^{n} ({d_{ti}}+{d_{tj}}+{d_{tk}})-8, \end{array} \)

and

\( \begin{array}{c} \sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{tj}}\sum _{t=1}^{n} {d_{tk}} \\ =\sum _{1≤r,s,t≤n} {d_{ri}}{d_{sj}}{d_{tk}} \\ ={C_{ijk}}+\sum _{1≤r≠t≤n} {d_{ri}}{d_{rj}}{d_{tk}}+\sum _{1≤r≠t≤n} {d_{ri}}{d_{tj}}{d_{tk}} \\ +\sum _{1≤r≠s≤n} {d_{ri}}{d_{sj}}{d_{rk}}+\sum _{t=1}^{n} {d_{ti}}{d_{tj}}{d_{tk}}. \end{array} \)

We can further have that

\( \begin{array}{c} {A_{i}}{A_{j}}{A_{k}} \\ ={C_{ijk}}+\sum _{1≤r≠t≤n} {d_{ri}}{d_{rj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{tj}}\sum _{t=1}^{n} {d_{tk}} \\ +\sum _{1≤r≠t≤n} {d_{ri}}{d_{tj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{tk}} \\ +\sum _{1≤r≠s≤n} {d_{ri}}{d_{sj}}{d_{rk}}-2\sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{tj}}+\sum _{t=1}^{n} {d_{ti}}{d_{tj}}{d_{tk}}. \end{array} \)

In order to see the relationship between \( {A_{i}}{A_{j}}{A_{k}} \) and \( {C_{ijk}} \) , we need to cal- culate the value of

\( \begin{array}{c} \sum _{1≤r≠t≤n} {d_{ri}}{d_{rj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{tj}}\sum _{t=1}^{n} {d_{tk}}+\sum _{1≤r≠t≤n} {d_{ri}}{d_{tj}}{d_{tk}} \\ -2\sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{tk}}+\sum _{1≤r≠s≤n} {d_{ri}}{d_{sj}}{d_{rk}}-2\sum _{t=1}^{n} {d_{ti}}\sum _{t=1}^{n} {d_{tj}}. \end{array} \)

One sees that

\( \begin{array}{c} \sum _{1≤r≠t≤n} {d_{ri}}{d_{rj}}{d_{tk}}-2∑{d_{tj}}∑{d_{tk}} \\ =\sum _{1≤r≠t≤n} {d_{ri}}{d_{rj}}{d_{tk}}-2\sum _{1≤r,t≤n} {d_{rj}}{d_{tk}} \\ =\sum _{1≤r≠t≤n} {d_{ri}}{d_{rj}}{d_{tk}}-2\sum _{1≤r≠t≤n} {d_{rj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{tj}}{d_{tk}} \\ =\sum _{1≤r≠t≤n} ({d_{ri}}-2){d_{rj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{tj}}{d_{tk}} \gt -2\sum _{t=1}^{n} {d_{tj}}{d_{tk}}. \end{array} \)

Similarly, we can obtain that

\( \sum _{1≤r≠t≤n} {d_{ri}}{d_{tj}}{d_{tk}}-2∑{d_{ti}}∑{d_{tk}} \gt -2\sum _{t=1}^{n} {d_{ti}}{d_{tk}} \)

and

\( \sum _{1≤r≠s≤n} {d_{ri}}{d_{sj}}{d_{rk}}-2∑{d_{ti}}∑{d_{tj}} \gt -2\sum _{t=1}^{n} {d_{ti}}{d_{tj}}. \)

By (20), (21) and (22), we can have that

\( {A_{i}}{A_{j}}{A_{k}} \gt {C_{ijk}}-2\sum _{t=1}^{n} {d_{tj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{ti}}{d_{tk}}-2\sum _{t=1}^{n} {d_{ti}}{d_{tj}}+\sum _{t=1}^{n} {d_{ti}}{d_{tj}}{d_{tk}}+4\sum _{t=1}^{n} ({d_{ti}}+{d_{tj}}+{d_{tk}})-8 \)

One sees that

\( \begin{array}{c} \sum _{t=1}^{n} {d_{ti}}{d_{tj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{tj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{ti}}{d_{tk}}-2\sum _{t=1}^{n} {d_{ti}}{d_{tj}} \\ =(\frac{1}{3}\sum _{t=1}^{n} {d_{ti}}{d_{tj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{tj}}{d_{tk}})+(\frac{1}{3}\sum _{t=1}^{n} {d_{ti}}{d_{tj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{ti}}{d_{tk}}) \\ +(\frac{1}{3}\sum _{t=1}^{n} {d_{ti}}{d_{tj}}{d_{tk}}-2\sum _{t=1}^{n} {d_{ti}}{d_{tj}}) \\ =\sum _{t=1}^{n} (\frac{1}{3}{d_{ti}}-2){d_{tj}}{d_{tk}}+\sum _{t=1}^{n} (\frac{1}{3}{d_{tj}}-2){d_{ti}}{d_{tk}}+\sum _{t=1}^{n} (\frac{1}{3}{d_{tk}}-2){d_{ti}}{d_{tj}}. \end{array} \)

If \( {d_{ij}}≥6 \) , then we can have that

\( \sum _{t=1}^{n} (\frac{1}{3}{d_{ti}}-2){d_{tj}}{d_{tk}}+\sum _{t=1}^{n} (\frac{1}{3}{d_{tj}}-2){d_{ti}}{d_{tk}}+\sum _{t=1}^{n} (\frac{1}{3}{d_{tk}}-2){d_{ti}}{d_{tj}}≥0. \)

This implies that

\( {A_{i}}{A_{j}}{A_{k}} \gt {C_{ijk}}. \)

As a result, we have

\( 0 \lt \frac{{C_{ijk}}}{{A_{i}}{A_{j}}{A_{k}}} \lt 1. \)

Lemma 4. If \( {d_{ij}}≥6 \) for any i, j, then we have

\( \frac{\frac{8}{3}-\frac{1}{2}{B_{ij}}+{A_{i}}{A_{j}}+2{A_{i}}-\frac{1}{2}{A_{i}}{B_{jk}}+\frac{1}{6}{C_{ijk}}}{{A_{i}}{A_{j}}{A_{k}}} \gt \frac{\frac{1}{2}{B_{jk}}}{{A_{j}}{A_{k}}}-\frac{1}{2}. \)

Proof. One sees that

\( \begin{array}{c} \frac{\frac{8}{3}-\frac{1}{2}{B_{ij}}+{A_{i}}{A_{j}}+2{A_{i}}-\frac{1}{2}{A_{i}}{B_{jk}}+\frac{1}{6}{C_{ijk}}}{{A_{i}}{A_{j}}{A_{k}}} \\ =\frac{\frac{8}{3}-\frac{1}{2}{B_{ij}}+{A_{i}}{A_{j}}+2{A_{i}}+\frac{1}{6}{C_{ijk}}}{{A_{i}}{A_{j}}{A_{k}}}-\frac{\frac{1}{2}{B_{jk}}}{{A_{j}}{A_{k}}}. \end{array} \)

By Lemma 1, we have

\( {B_{ij}} \lt {A_{i}}{A_{j}},{B_{jk}} \lt {A_{j}}{A_{k}}. \)

Hence, we have

\( \begin{array}{c} \frac{\frac{8}{3}-\frac{1}{2}{B_{ij}}+{A_{i}}{A_{j}}+2{A_{i}}-\frac{1}{2}{A_{i}}{B_{jk}}+\frac{1}{6}{C_{ijk}}}{{A_{i}}{A_{j}}{A_{k}}} \gt \\ \frac{\frac{8}{3}+\frac{1}{2}{B_{ij}}+2{A_{i}}-\frac{1}{2}{A_{i}}{B_{jk}}+\frac{1}{6}{C_{ijk}}}{{A_{i}}{A_{j}}{A_{k}}} \gt 0. \end{array} \)

This implies

\( \frac{\frac{8}{3}-\frac{1}{2}{B_{ij}}+{A_{i}}{A_{j}}+2{A_{i}}-\frac{1}{2}{A_{i}}{B_{jk}}+\frac{1}{6}{C_{ijk}}}{{A_{i}}{A_{j}}{A_{k}}} \gt \frac{\frac{1}{2}{B_{jk}}}{{A_{j}}{A_{k}}}-\frac{1}{2}.\ \ \ (11) \)

Theorem 4.2. If \( {d_{ij}}≥6 \) for any i, j, then we have \( \frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)}-\frac{1}{2} \lt \frac{{c_{3}}(X)}{c_{1}^{3}(X)} \lt \frac{7}{12}. \)

Proof. According to Lemma 4, we have that \( -{c_{3}}(X) \gt -{c_{1}}(X){c_{2}}(X)-\frac{1}{2}c_{1}^{3}(X) \) , i.e., \( \frac{{c_{3}}(X)}{c_{1}^{3}(X)} \gt \frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)}-\frac{1}{2} \) .

Now, we consider the upper bound of \( \frac{{c_{3}}(X)}{c_{1}^{3}(X)} \)

Because \( {A_{i}}=\sum _{t=1}^{n} {d_{ti}}-2≥6n-2, \) we have

\( \begin{array}{c} \frac{\frac{8}{3}}{{A_{i}}{A_{j}}{A_{k}}}+\frac{1}{{A_{k}}}+\frac{2}{{A_{j}}{A_{k}}}+\frac{\frac{1}{6}{C_{ijk}}}{{A_{i}}{A_{j}}{A_{k}}} \\ \lt \frac{\frac{8}{3}}{(6n-2{)^{3}}}+\frac{1}{6n-2}+\frac{2}{(6n-2{)^{2}}}+\frac{1}{6} \\ ≤\frac{8}{3}+\frac{1}{4}+\frac{2}{16}+\frac{1}{6} \\ =\frac{1}{24}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6} \\ =\frac{7}{12}. \end{array} \ \ \ (12) \)

5. Conclusions

In this paper, we take \( M=\underset{n+3}{\underbrace{{P^{1}}×{P^{1}}×⋯×{P^{1}}}} \) as an example to calculate the Chern numbers of complete intersection three-folds in products of projective spaces. Thus, in our conclusion, we get its Chern number and the inequalities that it will satisfy:

If \( {d_{ij}}≥4 \) for any \( 1≤i≤n,1≤j≤n+3, \) then we have \( \frac{1}{2} \lt \frac{{c_{1}}(X){c_{2}}(X)}{c_{1}^{3}(X)} \lt \frac{2}{(4n-2{)^{2}}}+\frac{2}{4n-2}+1 \) . If \( {d_{ij}}≥6 \) for any \( 1≤i≤n,1≤j≤n+3 \) , then \( \frac{{c_{1}}(\bar{X}){c_{2}}(X)}{c_{1}^{3}(X)}-\frac{1}{2} \lt \frac{{c_{3}}(X)}{c_{1}^{3}(X)} \lt \frac{7}{12}. \)

However, those conclusions build up on an important assumption, which is the value of \( {d_{ij}} \) . This means that there is still room for exploration and explanation of those results when applying other values of \( {d_{ij}} \) .

As for the future meaning of research into this field, it may help in the field of physics. For instance, Miyaoka-Yau type inequalities are widely applied to the quantum mechanics and field theory, so we believe researches like this can be applied to more different conditions.

Acknowledgments

We would like to thank our mentor Prof. Sun Hao, who provided support on choosing the topic and revising our paper. We learned a lot about this subtle field of mathematics under his patient guidance.

Moreover, we would like to appreciate the support given by our family when writing this paper.