Manifolds with complete metrics of positive scalar

Shmuel Weinberger Joint work with Stanley Chang and Guoliang Yu

May 5–14, 2008 Classical background.

Fact n If Sp is the scalar curvature at a point p in a M , then

n+2 n VolM (B (p)) = VolR (B) − Sp · C + ··· If f : M → R is a function with f (p) < 0 for some p ∈ M, then there is a metric g so that f (p) is the scalar curvature of (M, g) at p. If there is a metric g with positive scalar curvature, then any function f : M → R is the scalar curvature of some metric.

Classical background.

Fact n If Sp is the scalar curvature at a point p in a manifold M , then

n+2 n VolM (B (p)) = VolR (B) − Sp · C + ···

Remark (Kazhdan-Warner) Suppose M is a compact n-manifold with n > 2. If there is a metric g with positive scalar curvature, then any function f : M → R is the scalar curvature of some metric.

Classical background.

Fact n If Sp is the scalar curvature at a point p in a manifold M , then

n+2 n VolM (B (p)) = VolR (B) − Sp · C + ···

Remark (Kazhdan-Warner) Suppose M is a compact n-manifold with n > 2.

If f : M → R is a function with f (p) < 0 for some p ∈ M, then there is a metric g so that f (p) is the scalar curvature of (M, g) at p. Classical background.

Fact n If Sp is the scalar curvature at a point p in a manifold M , then

n+2 n VolM (B (p)) = VolR (B) − Sp · C + ···

Remark (Kazhdan-Warner) Suppose M is a compact n-manifold with n > 2.

If f : M → R is a function with f (p) < 0 for some p ∈ M, then there is a metric g so that f (p) is the scalar curvature of (M, g) at p. If there is a metric g with positive scalar curvature, then any function f : M → R is the scalar curvature of some metric. Classical background.

Fact n If Sp is the scalar curvature at a point p in a manifold M , then

n+2 n VolM (B (p)) = VolR (B) − Sp · C + ···

Remark (Gauss-Bonnet) If a surface Σ2 has a complete metric of positive scalar curvature, 2 ∼ 2 2 ∼ 2 then Σ = S or Σ = RP . Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0. Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

Example () K3 is spin. sign K3 = 16, so hAˆ(K3), [K3]i= 6 0 K3 has no metric of positive scalar curvature. Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

2 Example (CP ) 2 CP has a metric with positive scalar curvature, 2 sign CP = 1, 2 CP is not spin. Since M is spin, M has a Dirac operatorD / .

D/ ?D/ = ∆ + Scal.

If Scal > 0, then ∆ + Scal > 0, then kerD / ? = 0 and kerD / = 0, then indD / = kerD / − kerD / ? = 0, so hAˆ(M), [M]i = 0 by Atiyah-Singer.

Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

Idea of Proof: D/ ?D/ = ∆ + Scal.

If Scal > 0, then ∆ + Scal > 0, then kerD / ? = 0 and kerD / = 0, then indD / = kerD / − kerD / ? = 0, so hAˆ(M), [M]i = 0 by Atiyah-Singer.

Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

Idea of Proof: Since M is spin, M has a Dirac operatorD / . If Scal > 0, then ∆ + Scal > 0, then kerD / ? = 0 and kerD / = 0, then indD / = kerD / − kerD / ? = 0, so hAˆ(M), [M]i = 0 by Atiyah-Singer.

Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

Idea of Proof: Since M is spin, M has a Dirac operatorD / .

D/ ?D/ = ∆ + Scal. then kerD / ? = 0 and kerD / = 0, then indD / = kerD / − kerD / ? = 0, so hAˆ(M), [M]i = 0 by Atiyah-Singer.

Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

Idea of Proof: Since M is spin, M has a Dirac operatorD / .

D/ ?D/ = ∆ + Scal.

If Scal > 0, then ∆ + Scal > 0, then indD / = kerD / − kerD / ? = 0, so hAˆ(M), [M]i = 0 by Atiyah-Singer.

Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

Idea of Proof: Since M is spin, M has a Dirac operatorD / .

D/ ?D/ = ∆ + Scal.

If Scal > 0, then ∆ + Scal > 0, then kerD / ? = 0 and kerD / = 0, so hAˆ(M), [M]i = 0 by Atiyah-Singer.

Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

Idea of Proof: Since M is spin, M has a Dirac operatorD / .

D/ ?D/ = ∆ + Scal.

If Scal > 0, then ∆ + Scal > 0, then kerD / ? = 0 and kerD / = 0, then indD / = kerD / − kerD / ? = 0, Background.

Theorem (Atiyah-Singer, due to Lichnerowicz) If Mn is a compact spin manifold admitting a metric of positive scalar curvature, then

hAˆ(M), [M]i = 0.

Idea of Proof: Since M is spin, M has a Dirac operatorD / .

D/ ?D/ = ∆ + Scal.

If Scal > 0, then ∆ + Scal > 0, then kerD / ? = 0 and kerD / = 0, then indD / = kerD / − kerD / ? = 0, so hAˆ(M), [M]i = 0 by Atiyah-Singer. ind /D = 0 ∈ KOn(pt).

M is not spin, or M is spin, and

Simply connected case.

Theorem (Gromov-Lawson, Hitchin, Stolz) If Mn with n > 4 and M simply connected, then M has a metric of positive scalar curvature iff ind /D = 0 ∈ KOn(pt).

M is spin, and

Simply connected case.

Theorem (Gromov-Lawson, Hitchin, Stolz) If Mn with n > 4 and M simply connected, then M has a metric of positive scalar curvature iff M is not spin, or ind /D = 0 ∈ KOn(pt).

Simply connected case.

Theorem (Gromov-Lawson, Hitchin, Stolz) If Mn with n > 4 and M simply connected, then M has a metric of positive scalar curvature iff M is not spin, or M is spin, and hAˆ(M), [M]i = 0? Simply connected case.

Theorem (Gromov-Lawson, Hitchin, Stolz) If Mn with n > 4 and M simply connected, then M has a metric of positive scalar curvature iff M is not spin, or

M is spin, and ind /D = 0 ∈ KOn(pt). Theorem (Schoen-Yau for n ≤ 7, Gromov-Lawson for all n) T n has no metric of positive scalar curvature.

Theorem (Gromov-Lawson) K\G/Γ, a compact locally symmetric space, has no complete metric of positive scalar curvature. Additionally, noncompact hyperbolic do not have metrics of positive scalar curvature.

Proof. Following Rosenberg, same as before but using K(C ?π).

Non-simply connected case.

Question Does T n have a metric of positive scalar curvature? Theorem (Gromov-Lawson) K\G/Γ, a compact locally symmetric space, has no complete metric of positive scalar curvature. Additionally, noncompact hyperbolic manifolds do not have metrics of positive scalar curvature.

Proof. Following Rosenberg, same as before but using K(C ?π).

Non-simply connected case.

Question Does T n have a metric of positive scalar curvature?

Theorem (Schoen-Yau for n ≤ 7, Gromov-Lawson for all n) T n has no metric of positive scalar curvature. Additionally, noncompact hyperbolic manifolds do not have metrics of positive scalar curvature.

Proof. Following Rosenberg, same as before but using K(C ?π).

Non-simply connected case.

Question Does T n have a metric of positive scalar curvature?

Theorem (Schoen-Yau for n ≤ 7, Gromov-Lawson for all n) T n has no metric of positive scalar curvature.

Theorem (Gromov-Lawson) K\G/Γ, a compact locally symmetric space, has no complete metric of positive scalar curvature. Proof. Following Rosenberg, same as before but using K(C ?π).

Non-simply connected case.

Question Does T n have a metric of positive scalar curvature?

Theorem (Schoen-Yau for n ≤ 7, Gromov-Lawson for all n) T n has no metric of positive scalar curvature.

Theorem (Gromov-Lawson) K\G/Γ, a compact locally symmetric space, has no complete metric of positive scalar curvature. Additionally, noncompact hyperbolic manifolds do not have metrics of positive scalar curvature. Non-simply connected case.

Question Does T n have a metric of positive scalar curvature?

Theorem (Schoen-Yau for n ≤ 7, Gromov-Lawson for all n) T n has no metric of positive scalar curvature.

Theorem (Gromov-Lawson) K\G/Γ, a compact locally symmetric space, has no complete metric of positive scalar curvature. Additionally, noncompact hyperbolic manifolds do not have metrics of positive scalar curvature.

Proof. Following Rosenberg, same as before but using K(C ?π). ∞ The boundary has a π1 M. ∞ This suggests an obstruction in KO(Bπ1M, Bπ1 M). Assembly map for pairs into L-theory. But in the C ?-algebra setting only works really well if the fundamental group injects. One mystery of the Baum-Conjecture is the functoriality aspect (even in the torsion free case.) Why should there be functoriality associated to homomorphisms?

What happens for interiors of manifolds with boundary?

M ∂M

M is the interior of a manifold with boundary. ∞ This suggests an obstruction in KO(Bπ1M, Bπ1 M). Assembly map for pairs into L-theory. But in the C ?-algebra setting only works really well if the fundamental group injects. One mystery of the Baum-Conjecture is the functoriality aspect (even in the torsion free case.) Why should there be functoriality associated to homomorphisms?

What happens for interiors of manifolds with boundary?

M ∂M

M is the interior of a manifold with boundary. ∞ The boundary has a fundamental group π1 M. Assembly map for pairs into L-theory. But in the C ?-algebra setting only works really well if the fundamental group injects. One mystery of the Baum-Conjecture is the functoriality aspect (even in the torsion free case.) Why should there be functoriality associated to homomorphisms?

What happens for interiors of manifolds with boundary?

M ∂M

M is the interior of a manifold with boundary. ∞ The boundary has a fundamental group π1 M. ∞ This suggests an obstruction in KO(Bπ1M, Bπ1 M). One mystery of the Baum-Conjecture is the functoriality aspect (even in the torsion free case.) Why should there be functoriality associated to homomorphisms?

What happens for interiors of manifolds with boundary?

M ∂M

M is the interior of a manifold with boundary. ∞ The boundary has a fundamental group π1 M. ∞ This suggests an obstruction in KO(Bπ1M, Bπ1 M). Assembly map for pairs into L-theory. But in the C ?-algebra setting only works really well if the fundamental group injects. What happens for interiors of manifolds with boundary?

M ∂M

M is the interior of a manifold with boundary. ∞ The boundary has a fundamental group π1 M. ∞ This suggests an obstruction in KO(Bπ1M, Bπ1 M). Assembly map for pairs into L-theory. But in the C ?-algebra setting only works really well if the fundamental group injects. One mystery of the Baum-Conjecture is the functoriality aspect (even in the torsion free case.) Why should there be functoriality associated to homomorphisms? Theorem (Browder-Livesay-Levine) Mn, n > 5, is the interior of a manifold with simply connected boundary iff M has finitely generated homology and M is simply connected at infinity.

Proper homotopy equivalence of non-compact manifolds.

Definition M is simply connected at infinity if every compact K ⊂ M is contained in a larger compact C ⊃ K, so that M − C is simply connected. Proper homotopy equivalence of non-compact manifolds.

Definition M is simply connected at infinity if every compact K ⊂ M is contained in a larger compact C ⊃ K, so that M − C is simply connected.

Theorem (Browder-Livesay-Levine) Mn, n > 5, is the interior of a manifold with simply connected boundary iff M has finitely generated homology and M is simply connected at infinity. Theorem (Chang) K\G/Γ never has a complete metric of positive scalar curvature in the obvious QI class.

Idea of Proof: Marry Novikov idea to Roe’s partitioned manifold index theorem. We will discuss it in more detail later.

The case of lattices.

Theorem (Block-W) K\G/Γ has a complete metric of positive scalar curvature iff Q-rk(Γ) > 2. Marry Novikov idea to Roe’s partitioned manifold index theorem. We will discuss it in more detail later.

The case of lattices.

Theorem (Block-W) K\G/Γ has a complete metric of positive scalar curvature iff Q-rk(Γ) > 2.

Theorem (Chang) K\G/Γ never has a complete metric of positive scalar curvature in the obvious QI class.

Idea of Proof: We will discuss it in more detail later.

The case of lattices.

Theorem (Block-W) K\G/Γ has a complete metric of positive scalar curvature iff Q-rk(Γ) > 2.

Theorem (Chang) K\G/Γ never has a complete metric of positive scalar curvature in the obvious QI class.

Idea of Proof: Marry Novikov idea to Roe’s partitioned manifold index theorem. The case of lattices.

Theorem (Block-W) K\G/Γ has a complete metric of positive scalar curvature iff Q-rk(Γ) > 2.

Theorem (Chang) K\G/Γ never has a complete metric of positive scalar curvature in the obvious QI class.

Idea of Proof: Marry Novikov idea to Roe’s partitioned manifold index theorem. We will discuss it in more detail later. ∞ π1 (M) = Γ means that the pro-system

π1(M − K1) ← π1(M − K2) ← π1(M − K3) ← · · ·

is pro-equivalent to the constant system Γ ← Γ ← · · · .

Fundamental group at infinity.

Definition (Fundamental group at infinity)

K1 K2 K3 K4 K5

K1 ⊂ K2 ⊂ K3 ⊂ · · · ⊂ M is pro-equivalent to the constant system Γ ← Γ ← · · · .

Fundamental group at infinity.

Definition (Fundamental group at infinity)

K1 K2 K3 K4 K5

K1 ⊂ K2 ⊂ K3 ⊂ · · · ⊂ M ∞ π1 (M) = Γ means that the pro-system

π1(M − K1) ← π1(M − K2) ← π1(M − K3) ← · · · Fundamental group at infinity.

Definition (Fundamental group at infinity)

K1 K2 K3 K4 K5

K1 ⊂ K2 ⊂ K3 ⊂ · · · ⊂ M ∞ π1 (M) = Γ means that the pro-system

π1(M − K1) ← π1(M − K2) ← π1(M − K3) ← · · ·

is pro-equivalent to the constant system Γ ← Γ ← · · · . Takes the fear out of non-compactness when you are tame.

If tame at infinity, then there is a relative assembly theory, relative Novikov conjecture for L-classes and so on.

If not, it’s somewhat harder to describe the relevant assembly maps that enter the theory, but not impossible.

Proper homotopy equivalence of non-compact manifolds.

Theorem (Siebenmann’s thesis) The obstruction to putting a boundary on a tame manifold lies in ˜ ∞ K0(Zπ1 ). If tame at infinity, then there is a relative assembly theory, relative Novikov conjecture for L-classes and so on.

If not, it’s somewhat harder to describe the relevant assembly maps that enter the theory, but not impossible.

Proper homotopy equivalence of non-compact manifolds.

Theorem (Siebenmann’s thesis) The obstruction to putting a boundary on a tame manifold lies in ˜ ∞ K0(Zπ1 ). Takes the fear out of non-compactness when you are tame. If not, it’s somewhat harder to describe the relevant assembly maps that enter the theory, but not impossible.

Proper homotopy equivalence of non-compact manifolds.

Theorem (Siebenmann’s thesis) The obstruction to putting a boundary on a tame manifold lies in ˜ ∞ K0(Zπ1 ). Takes the fear out of non-compactness when you are tame.

If tame at infinity, then there is a relative assembly theory, relative Novikov conjecture for L-classes and so on. Proper homotopy equivalence of non-compact manifolds.

Theorem (Siebenmann’s thesis) The obstruction to putting a boundary on a tame manifold lies in ˜ ∞ K0(Zπ1 ). Takes the fear out of non-compactness when you are tame.

If tame at infinity, then there is a relative assembly theory, relative Novikov conjecture for L-classes and so on.

If not, it’s somewhat harder to describe the relevant assembly maps that enter the theory, but not impossible. To what extent does the theory of pairs capture the issues?

Question To what extent does the theory of pairs capture the issues?

K1 K2 K3 K4 K5 —but not in general. There are lim1 terms,

1 lf lf lf 0 → lim H (Ki ) → H (M) → lim H (Ki ) → 0, ?−1 ? ← ? and other terms measured off group homology’s limits.

To what extent does the theory of pairs capture the issues?

K1 K2 K3 K4 K5

Answer In the findamental group tame case, pretty well There are lim1 terms,

1 lf lf lf 0 → lim H (Ki ) → H (M) → lim H (Ki ) → 0, ?−1 ? ← ? and other terms measured off group homology’s limits.

To what extent does the theory of pairs capture the issues?

K1 K2 K3 K4 K5

Answer In the findamental group tame case, pretty well—but not in general. and other terms measured off group homology’s limits.

To what extent does the theory of pairs capture the issues?

K1 K2 K3 K4 K5

Answer In the findamental group tame case, pretty well—but not in general. There are lim1 terms,

1 lf lf lf 0 → lim H (Ki ) → H (M) → lim H (Ki ) → 0, ?−1 ? ← ? To what extent does the theory of pairs capture the issues?

K1 K2 K3 K4 K5

Answer In the findamental group tame case, pretty well—but not in general. There are lim1 terms,

1 lf lf lf 0 → lim H (Ki ) → H (M) → lim H (Ki ) → 0, ?−1 ? ← ? and other terms measured off group homology’s limits. h2(S1 × D2), h3(S1 × D2), . . .

\ Whitehead = S3 − hi (S1 × D2). i

Key example: Whitehead manifold.

Example h1(S1 × D2), h3(S1 × D2), . . .

\ Whitehead = S3 − hi (S1 × D2). i

Key example: Whitehead manifold.

Example h1(S1 × D2), h2(S1 × D2), \ Whitehead = S3 − hi (S1 × D2). i

Key example: Whitehead manifold.

Example h1(S1 × D2), h2(S1 × D2), h3(S1 × D2), . . . Key example: Whitehead manifold.

··· \ Whitehead = S3 − hi (S1 × D2). i Key example: Whitehead manifold.

··· \ Whitehead = S3 − hi (S1 × D2). i

Remark ∼ 3 ∼ 4 Whitehead =6 R , but R × Whitehead = R . Key example: Whitehead manifold.

··· \ Whitehead = S3 − hi (S1 × D2). i

Remark There are uncountably many variants of this construction. A little bit like the space of Penrose tilings.

Key example: Whitehead manifold.

··· \ Whitehead = S3 − hi (S1 × D2). i

Question What does the moduli space of these manifolds look like? Key example: Whitehead manifold.

··· \ Whitehead = S3 − hi (S1 × D2). i

Question What does the moduli space of these manifolds look like? A little bit like the space of Penrose tilings. Key example: Whitehead manifold.

··· \ Whitehead = S3 − hi (S1 × D2). i

Remark No nice metric on these manifolds. (by Papakyriakopoulos’ sphere theorem, because A is the complement of a non-split link)

A is aspherical

∞ Naively construed, “π1 ” is trivial.

Observations about the Whitehead manifold.

2 T 2 T T 2 T 2 T 2

S1 D2 A A A A × (by Papakyriakopoulos’ sphere theorem, because A is the complement of a non-split link) ∞ Naively construed, “π1 ” is trivial.

Observations about the Whitehead manifold.

2 T 2 T T 2 T 2 T 2

S1 D2 A A A A ×

A is aspherical ∞ Naively construed, “π1 ” is trivial.

Observations about the Whitehead manifold.

2 T 2 T T 2 T 2 T 2

S1 D2 A A A A ×

A is aspherical (by Papakyriakopoulos’ sphere theorem, because A is the complement of a non-split link) Observations about the Whitehead manifold.

2 T 2 T T 2 T 2 T 2

S1 D2 A A A A ×

A is aspherical (by Papakyriakopoulos’ sphere theorem, because A is the complement of a non-split link) ∞ Naively construed, “π1 ” is trivial. —a contradiction.

DW = double of Whitehead manifold along T 2 DW has a positive scalar curvature metric except at A ∪ A¯ DW has positive scalar curvature metric at infinity

The Whitehead manifold and positive scalar curvature.

Theorem The Whitehead manifold has no complete metric of positive scalar curvature.

Proof: —a contradiction.

DW = double of Whitehead manifold along T 2 DW has a positive scalar curvature metric except at A ∪ A¯ DW has positive scalar curvature metric at infinity

The Whitehead manifold and positive scalar curvature.

Proof:

2 T 2 T T 2 T 2 T 2

S1 D2 A A A A × —a contradiction.

DW = double of Whitehead manifold along T 2 DW has a positive scalar curvature metric except at A ∪ A¯ DW has positive scalar curvature metric at infinity

The Whitehead manifold and positive scalar curvature.

Proof:

2 T 2 T T 2 T 2 T 2

S1 D2 A A A A × —a contradiction.

DW = double of Whitehead manifold along T 2 DW has a positive scalar curvature metric except at A ∪ A¯ DW has positive scalar curvature metric at infinity

The Whitehead manifold and positive scalar curvature.

Proof:

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A —a contradiction.

DW has a positive scalar curvature metric except at A ∪ A¯ DW has positive scalar curvature metric at infinity

The Whitehead manifold and positive scalar curvature.

Proof:

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

DW = double of Whitehead manifold along T 2 —a contradiction.

DW has positive scalar curvature metric at infinity

The Whitehead manifold and positive scalar curvature.

Proof:

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

DW = double of Whitehead manifold along T 2 DW has a positive scalar curvature metric except at A ∪ A¯ —a contradiction.

The Whitehead manifold and positive scalar curvature.

Proof:

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

DW = double of Whitehead manifold along T 2 DW has a positive scalar curvature metric except at A ∪ A¯ DW has positive scalar curvature metric at infinity The Whitehead manifold and positive scalar curvature.

Proof:

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

DW = double of Whitehead manifold along T 2 DW has a positive scalar curvature metric except at A ∪ A¯ DW has positive scalar curvature metric at infinity—a contradiction. The partition defines a virtual vector bundle on the space at infinity. Only the ends of the space at infinity are independent of the quasi-isometry class of the metric.

Digression: Roe’s Partition Manifold Theorem.

Theorem (Roe) If V is spin and Z positive scalar curvature at infinity, then ind = 0.

Modern Philosophy

V

Z Only the ends of the space at infinity are independent of the quasi-isometry class of the metric.

Digression: Roe’s Partition Manifold Theorem.

Theorem (Roe) If V is spin and Z positive scalar curvature at infinity, then ind = 0.

Modern Philosophy The partition defines a virtual vector bundle on the space at infinity.

V Z Digression: Roe’s Partition Manifold Theorem.

Theorem (Roe) If V is spin and Z positive scalar curvature at infinity, then ind = 0.

Modern Philosophy The partition defines a virtual vector bundle on the space at infinity.

Only the ends of the V space at infinity are Z independent of the quasi-isometry class of the metric. In this case, V = T 2.

2 2 But [/D] ∈ K2(T ) dies on pushing forward by K2(T ) → K2(pt), soD / on T 2 does not obstruct positive scalar curvature.

2 ? 2 On the other hand, K2(T ) → K2(C (Z )) is injective.

The non-simply connected case.

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

If V is not simply connected, attractive to couple Roe’s theorem to ? C (π1V ). 2 2 But [/D] ∈ K2(T ) dies on pushing forward by K2(T ) → K2(pt), soD / on T 2 does not obstruct positive scalar curvature.

2 ? 2 On the other hand, K2(T ) → K2(C (Z )) is injective.

The non-simply connected case.

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

If V is not simply connected, attractive to couple Roe’s theorem to ? 2 C (π1V ). In this case, V = T . soD / on T 2 does not obstruct positive scalar curvature.

2 ? 2 On the other hand, K2(T ) → K2(C (Z )) is injective.

The non-simply connected case.

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

If V is not simply connected, attractive to couple Roe’s theorem to ? 2 C (π1V ). In this case, V = T . 2 2 But [/D] ∈ K2(T ) dies on pushing forward by K2(T ) → K2(pt), 2 ? 2 On the other hand, K2(T ) → K2(C (Z )) is injective.

The non-simply connected case.

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

If V is not simply connected, attractive to couple Roe’s theorem to ? 2 C (π1V ). In this case, V = T . 2 2 But [/D] ∈ K2(T ) dies on pushing forward by K2(T ) → K2(pt), soD / on T 2 does not obstruct positive scalar curvature. The non-simply connected case.

T 2 T 2 T 2 T 2 T 2 T 2

A¯ A¯ A¯ A A A A

If V is not simply connected, attractive to couple Roe’s theorem to ? 2 C (π1V ). In this case, V = T . 2 2 But [/D] ∈ K2(T ) dies on pushing forward by K2(T ) → K2(pt), soD / on T 2 does not obstruct positive scalar curvature.

2 ? 2 On the other hand, K2(T ) → K2(C (Z )) is injective. Tilt horizontal direction to the vertical.

Idea 2 2 Use π1(DW ) rather than π1(T ) = Z .

Question

Do we know strong Novikov conjecture for π1(DW )? DW aspherical.

2  H2(T ) / H2(DW ) O O

2 K2(T ) / K2(DW ) / K2(π1DW )

Theorem (Connes-Gromov-Mascovicci) Novikov conjecture holds for all 2-dimensional cohomology classes. Example S3 − cantor set = L^3#L3.

Proof: Whitehead case applies. Perelman is used—though Hamilton is probably enough.

Taming 3-manifolds.

Theorem If M3 is of finite type and has positive scalar curvature at infinity, then M is the interior of a manifold with boundary. Proof: Whitehead case applies. Perelman is used—though Hamilton is probably enough.

Taming 3-manifolds.

Theorem If M3 is of finite type and has positive scalar curvature at infinity, then M is the interior of a manifold with boundary.

Example S3 − cantor set = L^3#L3. Perelman is used—though Hamilton is probably enough.

Taming 3-manifolds.

Theorem If M3 is of finite type and has positive scalar curvature at infinity, then M is the interior of a manifold with boundary.

Example S3 − cantor set = L^3#L3.

Proof: Whitehead case applies. —though Hamilton is probably enough.

Taming 3-manifolds.

Theorem If M3 is of finite type and has positive scalar curvature at infinity, then M is the interior of a manifold with boundary.

Example S3 − cantor set = L^3#L3.

Proof: Whitehead case applies. Perelman is used Taming 3-manifolds.

Theorem If M3 is of finite type and has positive scalar curvature at infinity, then M is the interior of a manifold with boundary.

Example S3 − cantor set = L^3#L3.

Proof: Whitehead case applies. Perelman is used—though Hamilton is probably enough. Proof:

Contractible manifolds and positive scalar curvature.

Theorem For all n, there is a contractible manifold Mn having no complete metric of positive scalar curvature. 2 n = 2 R .

Contractible manifolds and positive scalar curvature.

Theorem For all n, there is a contractible manifold Mn having no complete metric of positive scalar curvature.

Proof: n = 1 R. Contractible manifolds and positive scalar curvature.

Theorem For all n, there is a contractible manifold Mn having no complete metric of positive scalar curvature.

Proof: n = 1 R. 2 n = 2 R . n = 4 The Mazur manifold. n > 4 Variations on the Mazur manifold.

Contractible manifolds and positive scalar curvature.

Theorem For all n, there is a contractible manifold Mn having no complete metric of positive scalar curvature.

Proof: n = 3 The Whitehead manifold. n > 4 Variations on the Mazur manifold.

Contractible manifolds and positive scalar curvature.

Theorem For all n, there is a contractible manifold Mn having no complete metric of positive scalar curvature.

D2 D2 Proof: ×

n = 3 The Whitehead S1 manifold. n = 4 The Mazur manifold. S1 D3 × Contractible manifolds and positive scalar curvature.

Theorem For all n, there is a contractible manifold Mn having no complete metric of positive scalar curvature.

Proof: n = 3 The Whitehead manifold. n = 4 The Mazur manifold. n > 4 Variations on the Mazur manifold. Question 4 Is R the only contractible 4-manifold with positive scalar curvature?

Dimension four?

Question Are there are interesting 4-manifolds that have complete metrics of positive scalar curvature? Dimension four?

Question Are there are interesting 4-manifolds that have complete metrics of positive scalar curvature?

Question 4 Is R the only contractible 4-manifold with positive scalar curvature?