Post on 02-Mar-2021
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.
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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
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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
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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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
15
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”.
18
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.
19
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.
19
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.
19
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.
19
Thank you!
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