Some complex structures on products of

2-spheres

Gangotryi Sorcar

Hebrew University of Jerusalem, Israel

June 25, 2020

Joint with Jean-Francois Lafont and Fangyang Zheng, The Ohio State University

Basic question

• How to distinguish between two ‘distinct’ complex structures

on a fixed smooth manifold M?

• We will investigate this questions in the context of Bott

manifolds.

1

Basic question

• How to distinguish between two ‘distinct’ complex structures

on a fixed smooth manifold M?

• We will investigate this questions in the context of Bott

manifolds.

1

Bott manifolds

DefinitionA Bott manifold Mn is a complex n-manifold that admits a Bott

tower, i.e. Mn = Bn and :

Mn = Bnπn−→ · · · π3−→ B2

π2−→ B1π1−→ B0 = ∗

• Here, each Bjπj−→ Bj−1 is a fiber bundle.

• Bj = P(O⊕Sj). O = trivial line bundle. Sj = any line bundle.

• Note that B1 is always P1 the complex projective line.

2

Bott manifolds

DefinitionA Bott manifold Mn is a complex n-manifold that admits a Bott

tower, i.e. Mn = Bn and :

Mn = Bnπn−→ · · · π3−→ B2

π2−→ B1π1−→ B0 = ∗

• Here, each Bjπj−→ Bj−1 is a fiber bundle.

• Bj = P(O⊕Sj). O = trivial line bundle. Sj = any line bundle.

• Note that B1 is always P1 the complex projective line.

2

Bott manifolds

DefinitionA Bott manifold Mn is a complex n-manifold that admits a Bott

tower, i.e. Mn = Bn and :

Mn = Bnπn−→ · · · π3−→ B2

π2−→ B1π1−→ B0 = ∗

• Here, each Bjπj−→ Bj−1 is a fiber bundle.

• Bj = P(O⊕Sj). O = trivial line bundle. Sj = any line bundle.

• Note that B1 is always P1 the complex projective line.

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Example: Product of spheres

Example:

• Let all Sj line bundles be trivial in the Bott tower

• This would mean all the fiber bundles Bj are products

P1 × Bj−1

• Therefore, Bn = P1 × · · · × P1 (n many copies)

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Example: Product of spheres

Example:

• Let all Sj line bundles be trivial in the Bott tower

• This would mean all the fiber bundles Bj are products

P1 × Bj−1

• Therefore, Bn = P1 × · · · × P1 (n many copies)

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Example: Product of spheres

Example:

• Let all Sj line bundles be trivial in the Bott tower

• This would mean all the fiber bundles Bj are products

P1 × Bj−1

• Therefore, Bn = P1 × · · · × P1 (n many copies)

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Example: Hirzebruch surfaces

B2 → B1 → B0 = ∗

Fm := B2 = P(O ⊕OP1(m)) (Hirzebruch surface of index m)

Line bundles on B1 = P1 are characterized by the integer m ≥ 0.

Hirzebruch showed that:

• Fm is not biholomorphic to Fn when m 6= n

• F2k are diffeomorphic to F0 = P1 × P1 (diffeo to S2 × S2)

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Z-trivial Bott manifolds

DefinitionA Bott manifold Mn is called Z-trivial iff:

• Mn is diffeomorphic to (P1)n i.e. to a product of n copies of

S2.

OR

• H∗(M;Z) is isomorphic to H∗((P1)n;Z)

The equivalence is due to a result by Masuda-Panov.

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Cohomology and Chern class of Bott manifolds

H∗(Mn;Z) = Z[x1, x2, · · · , xn]/(x21 , x

22 + x2h2, · · · , xn + xnhn)

c(M) = (1 + 2x1)(1 + 2x2 + h2)....(1 + 2xn + hn)

xj is the pullback in H2(M) of the first Chern class of OBj(1)

−hj is the pullback in H2(M) of the first Chern class of Sj

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Note:

• {x1, x2, · · · , xn} is a generating set for H∗(M;Z)

• h1 = 0

• hj is a linear combination of x1, x2, · · · , xj−1 (because hj is

pullback of 1st Chern class of Sj → Bj−1)

• If all coefficients of above linear comb. is even we say 2|hj

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Note:

• {x1, x2, · · · , xn} is a generating set for H∗(M;Z)

• h1 = 0

• hj is a linear combination of x1, x2, · · · , xj−1 (because hj is

pullback of 1st Chern class of Sj → Bj−1)

• If all coefficients of above linear comb. is even we say 2|hj

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Note:

• {x1, x2, · · · , xn} is a generating set for H∗(M;Z)

• h1 = 0

• hj is a linear combination of x1, x2, · · · , xj−1 (because hj is

pullback of 1st Chern class of Sj → Bj−1)

• If all coefficients of above linear comb. is even we say 2|hj

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Note:

• {x1, x2, · · · , xn} is a generating set for H∗(M;Z)

• h1 = 0

• hj is a linear combination of x1, x2, · · · , xj−1 (because hj is

pullback of 1st Chern class of Sj → Bj−1)

• If all coefficients of above linear comb. is even we say 2|hj

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Example: The product case

Let us denote P = (P1)n

• All Sj = O

• hj = 0 for all j

• Bn = P

• Denoting xj as yj we get:

H∗(P;Z) = Z[y1, · · · , yn]/(y21 , · · · , y2

n )

c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

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Example: The product case

Let us denote P = (P1)n

• All Sj = O• hj = 0 for all j

• Bn = P

• Denoting xj as yj we get:

H∗(P;Z) = Z[y1, · · · , yn]/(y21 , · · · , y2

n )

c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

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Example: The product case

Let us denote P = (P1)n

• All Sj = O• hj = 0 for all j

• Bn = P

• Denoting xj as yj we get:

H∗(P;Z) = Z[y1, · · · , yn]/(y21 , · · · , y2

n )

c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

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Example: The product case

Let us denote P = (P1)n

• All Sj = O• hj = 0 for all j

• Bn = P

• Denoting xj as yj we get:

H∗(P;Z) = Z[y1, · · · , yn]/(y21 , · · · , y2

n )

c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

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Example: The Hirzebruch surface case

We consider the Hirzebruch surface F2k

• S1 = O,S2 = OP1(−2k)

• h1 = 0, −h2 = π∗2(−2k[P1]) = −2kπ∗2([P1]) = −2kx1

• Because, x1 = π∗2(c1(OP1(1))) = π∗2[P1]

• So we get:

H∗(F2k ;Z) = Z[x1, x2]/(x21 , x

22 − 2kx1x2)

c(P) = (1 + 2x1)(1 + 2x2 − 2kx1)

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Example: The Hirzebruch surface case

We consider the Hirzebruch surface F2k

• S1 = O,S2 = OP1(−2k)

• h1 = 0, −h2 = π∗2(−2k[P1]) = −2kπ∗2([P1]) = −2kx1

• Because, x1 = π∗2(c1(OP1(1))) = π∗2[P1]

• So we get:

H∗(F2k ;Z) = Z[x1, x2]/(x21 , x

22 − 2kx1x2)

c(P) = (1 + 2x1)(1 + 2x2 − 2kx1)

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Example: The Hirzebruch surface case

We consider the Hirzebruch surface F2k

• S1 = O,S2 = OP1(−2k)

• h1 = 0, −h2 = π∗2(−2k[P1]) = −2kπ∗2([P1]) = −2kx1

• Because, x1 = π∗2(c1(OP1(1))) = π∗2[P1]

• So we get:

H∗(F2k ;Z) = Z[x1, x2]/(x21 , x

22 − 2kx1x2)

c(P) = (1 + 2x1)(1 + 2x2 − 2kx1)

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Example: The Hirzebruch surface case

We consider the Hirzebruch surface F2k

• S1 = O,S2 = OP1(−2k)

• h1 = 0, −h2 = π∗2(−2k[P1]) = −2kπ∗2([P1]) = −2kx1

• Because, x1 = π∗2(c1(OP1(1))) = π∗2[P1]

• So we get:

H∗(F2k ;Z) = Z[x1, x2]/(x21 , x

22 − 2kx1x2)

c(P) = (1 + 2x1)(1 + 2x2 − 2kx1)

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Chern classes of Z-trivial Bott manifolds

Theorem (L-S-Z ’18 and Kim)M is Z-trivial =⇒ there is a diffeomorphism Φ : M → P such

that Φ∗(c(P)) = c(M).

Proof.

• We showed that 2|hj and h2j = 0 using induction on the index

j

• Now write, zj = xj + 12hj

• Notice zjs generate all xjs hence are a gen. set for H∗(M;Z)

• Also, z2j = x2

j + xjhj + 14h

2j = x2

j + xjhj = 0

• Define Φ∗ by sending each yj to zj

10

Chern classes of Z-trivial Bott manifolds

Theorem (L-S-Z ’18 and Kim)M is Z-trivial =⇒ there is a diffeomorphism Φ : M → P such

that Φ∗(c(P)) = c(M).

Proof.

• We showed that 2|hj and h2j = 0 using induction on the index

j

• Now write, zj = xj + 12hj

• Notice zjs generate all xjs hence are a gen. set for H∗(M;Z)

• Also, z2j = x2

j + xjhj + 14h

2j = x2

j + xjhj = 0

• Define Φ∗ by sending each yj to zj

10

Chern classes of Z-trivial Bott manifolds

Theorem (L-S-Z ’18 and Kim)M is Z-trivial =⇒ there is a diffeomorphism Φ : M → P such

that Φ∗(c(P)) = c(M).

Proof.

• We showed that 2|hj and h2j = 0 using induction on the index

j

• Now write, zj = xj + 12hj

• Notice zjs generate all xjs hence are a gen. set for H∗(M;Z)

• Also, z2j = x2

j + xjhj + 14h

2j = x2

j + xjhj = 0

• Define Φ∗ by sending each yj to zj

10

Chern classes of Z-trivial Bott manifolds

Theorem (L-S-Z ’18 and Kim)M is Z-trivial =⇒ there is a diffeomorphism Φ : M → P such

that Φ∗(c(P)) = c(M).

Proof.

• We showed that 2|hj and h2j = 0 using induction on the index

j

• Now write, zj = xj + 12hj

• Notice zjs generate all xjs hence are a gen. set for H∗(M;Z)

• Also, z2j = x2

j + xjhj + 14h

2j = x2

j + xjhj = 0

• Define Φ∗ by sending each yj to zj

10

Chern classes of Z-trivial Bott manifolds

Theorem (L-S-Z ’18 and Kim)M is Z-trivial =⇒ there is a diffeomorphism Φ : M → P such

that Φ∗(c(P)) = c(M).

Proof.

• We showed that 2|hj and h2j = 0 using induction on the index

j

• Now write, zj = xj + 12hj

• Notice zjs generate all xjs hence are a gen. set for H∗(M;Z)

• Also, z2j = x2

j + xjhj + 14h

2j = x2

j + xjhj = 0

• Define Φ∗ by sending each yj to zj

10

Proof (Cont.)

• By a result of Choi-Masuda, an isom Φ∗ is always induced by

a diffeomorphism Φ

• Recall:

• c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

• c(M) = (1 + 2x1)(1 + 2x2 + h2) · · · (1 + 2xn + hn)

• Φ∗(c(P)) = c(M) because yj 7→ xj + 12hj

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Proof (Cont.)

• By a result of Choi-Masuda, an isom Φ∗ is always induced by

a diffeomorphism Φ

• Recall:

• c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

• c(M) = (1 + 2x1)(1 + 2x2 + h2) · · · (1 + 2xn + hn)

• Φ∗(c(P)) = c(M) because yj 7→ xj + 12hj

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Proof (Cont.)

• By a result of Choi-Masuda, an isom Φ∗ is always induced by

a diffeomorphism Φ

• Recall:

• c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

• c(M) = (1 + 2x1)(1 + 2x2 + h2) · · · (1 + 2xn + hn)

• Φ∗(c(P)) = c(M) because yj 7→ xj + 12hj

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Proof (Cont.)

• By a result of Choi-Masuda, an isom Φ∗ is always induced by

a diffeomorphism Φ

• Recall:

• c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

• c(M) = (1 + 2x1)(1 + 2x2 + h2) · · · (1 + 2xn + hn)

• Φ∗(c(P)) = c(M) because yj 7→ xj + 12hj

11

Proof (Cont.)

• By a result of Choi-Masuda, an isom Φ∗ is always induced by

a diffeomorphism Φ

• Recall:

• c(P) = (1 + 2y1)(1 + 2y2) · · · (1 + 2yn)

• c(M) = (1 + 2x1)(1 + 2x2 + h2) · · · (1 + 2xn + hn)

• Φ∗(c(P)) = c(M) because yj 7→ xj + 12hj

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Summarizing so far..

• To summarize, we showed that the total Chern class of every

Bott manifold that is diffeo to the product of n-many

2-spheres, lies in the same Diffeo orbit of the cohomology

ring.

• So how do we distinguish between non biholomorphic Z-trivial

Bott manifolds?

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Summarizing so far..

• To summarize, we showed that the total Chern class of every

Bott manifold that is diffeo to the product of n-many

2-spheres, lies in the same Diffeo orbit of the cohomology

ring.

• So how do we distinguish between non biholomorphic Z-trivial

Bott manifolds?

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Bott Diagrams

• A Z-trivial Bott manifold M has two sets of generators:

• {x1, x2, · · · , xn} - generators from Bott tower

• {z1, z2, · · · , zn} - square zero generators

• Note that because h2j = 0 we can write hj = 2qjzσ(j), where

σ(j) < j

• The Bott diagram for M is a graph with one vertex for each

index j and one edge labeled qj from j to σ(j)

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Bott Diagrams

• A Z-trivial Bott manifold M has two sets of generators:

• {x1, x2, · · · , xn} - generators from Bott tower

• {z1, z2, · · · , zn} - square zero generators

• Note that because h2j = 0 we can write hj = 2qjzσ(j), where

σ(j) < j

• The Bott diagram for M is a graph with one vertex for each

index j and one edge labeled qj from j to σ(j)

13

Bott Diagrams

• A Z-trivial Bott manifold M has two sets of generators:

• {x1, x2, · · · , xn} - generators from Bott tower

• {z1, z2, · · · , zn} - square zero generators

• Note that because h2j = 0 we can write hj = 2qjzσ(j), where

σ(j) < j

• The Bott diagram for M is a graph with one vertex for each

index j and one edge labeled qj from j to σ(j)

13

Bott Diagrams

• A Z-trivial Bott manifold M has two sets of generators:

• {x1, x2, · · · , xn} - generators from Bott tower

• {z1, z2, · · · , zn} - square zero generators

• Note that because h2j = 0 we can write hj = 2qjzσ(j), where

σ(j) < j

• The Bott diagram for M is a graph with one vertex for each

index j and one edge labeled qj from j to σ(j)

13

Bott Diagrams

• A Z-trivial Bott manifold M has two sets of generators:

• {x1, x2, · · · , xn} - generators from Bott tower

• {z1, z2, · · · , zn} - square zero generators

• Note that because h2j = 0 we can write hj = 2qjzσ(j), where

σ(j) < j

• The Bott diagram for M is a graph with one vertex for each

index j and one edge labeled qj from j to σ(j)

13

Examples

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Examples

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Examples

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Bott diagrams and biholomorphisms

Theorem (L-S-Z)

Two Z-trivial Bott manifolds M1 and M2 are biholomorphic ⇐⇒their Bott diagrams are isomorphic as labeled rooted forests.

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Proof.

• Let Mi have Bott diagram Bi for i = 1, 2.

• ⇐ is easy because given the Bott diagram one can construct

the Bott tower.

• Need to show⇒, so we assume M1 and M2 are biholomorphic.

• We will call the new graph obtained after deleting a root v of

Bi , “the card of Bi at v”.

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Proof.

• Let Mi have Bott diagram Bi for i = 1, 2.

• ⇐ is easy because given the Bott diagram one can construct

the Bott tower.

• Need to show⇒, so we assume M1 and M2 are biholomorphic.

• We will call the new graph obtained after deleting a root v of

Bi , “the card of Bi at v”.

18

Proof.

• Let Mi have Bott diagram Bi for i = 1, 2.

• ⇐ is easy because given the Bott diagram one can construct

the Bott tower.

• Need to show⇒, so we assume M1 and M2 are biholomorphic.

• We will call the new graph obtained after deleting a root v of

Bi , “the card of Bi at v”.

18

Proof.

• Let Mi have Bott diagram Bi for i = 1, 2.

• ⇐ is easy because given the Bott diagram one can construct

the Bott tower.

• Need to show⇒, so we assume M1 and M2 are biholomorphic.

• We will call the new graph obtained after deleting a root v of

Bi , “the card of Bi at v”.

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Proof (Cont.)

• Show a bijection between roots of B1 and B2. Let’s say root

vertices ai maps to a′i .

• Show that the cards at ai and a′i are Bott diagrams of two

biholomorphic Bott manifolds.

• By induction on the number of vertices argue that these Bott

diagrams are isomorphic.

• We show when cards of B1 and B2 are isomorphic at ai and a′ifor all roots, the entire forests are isomorphic.

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Proof (Cont.)

• Show a bijection between roots of B1 and B2. Let’s say root

vertices ai maps to a′i .

• Show that the cards at ai and a′i are Bott diagrams of two

biholomorphic Bott manifolds.

• By induction on the number of vertices argue that these Bott

diagrams are isomorphic.

• We show when cards of B1 and B2 are isomorphic at ai and a′ifor all roots, the entire forests are isomorphic.

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Proof (Cont.)

• Show a bijection between roots of B1 and B2. Let’s say root

vertices ai maps to a′i .

• Show that the cards at ai and a′i are Bott diagrams of two

biholomorphic Bott manifolds.

• By induction on the number of vertices argue that these Bott

diagrams are isomorphic.

• We show when cards of B1 and B2 are isomorphic at ai and a′ifor all roots, the entire forests are isomorphic.

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Proof (Cont.)

• Show a bijection between roots of B1 and B2. Let’s say root

vertices ai maps to a′i .

• Show that the cards at ai and a′i are Bott diagrams of two

biholomorphic Bott manifolds.

• By induction on the number of vertices argue that these Bott

diagrams are isomorphic.

• We show when cards of B1 and B2 are isomorphic at ai and a′ifor all roots, the entire forests are isomorphic.

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Thank you!

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