On global tightness of real forms in Hermitian symmetric

spaces

Hiroshi Iriyeh Tokyo Denki University

joint work with Takashi Sakai (OCAMI)

1. Definition of tight Lagrangian submanifolds

Definition. (M^{2n}, ω) : compact symplectic manifold M ⊃ L : Lagrangian submanifold

⟺^{def} dim_{ℝ} L = n, ω|_{L} = 0.

Assume : L is connected, closed and embedded.

Definition.(Y.-G.Oh, 1991) (G/K, ω, J) : cpt. Hermitian symmetric space

g(·, ·) = ω(·, J·)

G/K ⊃ L : Lagrangian submanifold L is said to be globally tight (tight )

⟺^{def} #(L ∩ g · L) = ∑ rank H_{∗}(L, ℤ_{2})

for any g ∈ G (close to id.) such that L transversely intersects with g · L.

Example 1. S^{2} ≅ ℂP^{1} = U(2)/U(1) × U(1) L =great circle L′ =small circle

Example 2 (Kobayashi-Takeuchi, 1968). All real forms in a cpt. Hermitian symmetric space G/K are tight.

2. Problems

Problem 1 (Y.-G. Oh). Classify all tight Lagrangian submanifolds in compact Hermitian symmetric spaces.

Theorem 1 (Oh, 1991). ℂP^{n} ⊃ L : tight Lagrangian submanifold.

⇒ If n ≥ 2, L = ℝP^{n} ⊂^{t.g.} ℂP^{n}.

If n = 1, L = great or small circle.

Theorem 2 (Sakai- I, 2008). S^{2}(1) × S^{2}(1) ⊃ L : tight Lagrangian surf. ⇒ L must be congruent to one of the following:

(i) L = {(x, −x)| x ∈ S^{2}(1)} ⊂^{t.g.} S^{2} × S^{2}.

(ii) L = S^{1}(a) × S^{1}(b), (1 ≥ a, b > 0).

• Are all tight Lagrangian submanifolds of G/K globally tight?

No! ⟵ small circle, etc.

Problem 2 . Are all real forms in compact Hermitian symmetric spaces globally tight?

Theorem 3 (Howard, 1993). ℝP^{n} ⊂^{t.g.} ℂP^{n} is globally tight.

The proof of the above theorem cannot be generalized to other case.

Theorem 4 (Sakai- I, 2008). The totally geodesic Lagrangian sphere

{(x, −x)| x ∈ S^{2}(1)} ⊂ S^{2} × S^{2}

is globally tight.

Our method \{ \begin{matrix}Symplectic geometry \\ (Arnold − Givental inequality) \\ Integral geometry \\ (Poincaré formula) \end{matrix}

Our method can be generalized to other case!

Example (Howard, 1993). L_{1}, L_{2} : Lagrangian submanifolds in ℂP^{n}

∫_{U(n+1)} ♯(L_{1} ∩ g(L_{2}))dg = Cvol(L_{1})vol(L_{2})

where

C = \frac{(n + 1)vol(U(n + 1))}{vol(ℝP^{n})^{2}} .

3. Arnold-Givental inequality

Theorem 5 (Oh). (M, ω, J) : cpt. Hermitian symmetric space τ : involutive anti-holomorphic isometry on M = G/K L = Fix τ “real form” Minimal Maslov number Σ_{L} ≥ 2.

⇒  ♯(L ∩ ϕ(L)) ≥ ∑ rank H_{∗}(L, ℤ_{2})

∀ϕ ∈ Ham (M) s.t. L and ϕ(L) intersect transversely.

Minimal Maslov number Σ_{L}

• w : (D^{2}, ∂D^{2}) → (M^{2n}, L^{n}), C^{∞} map

⇓

w^{∗}T M ≅ D^{2} × ℂ^{n} as symp. vector bundle

⇓

S^{1} ≅ ∂D^{2} → Λ(ℂ^{n}) := {Lag. plane in ℂ^{n}}.

⇓
I_{μ,L}(w) := μ(∂D^{2}) ∈ ℤ

where μ ∈ H^{1}(Λ(ℂ^{n}), ℤ) ≅ ℤ : Maslov class

Σ_{L} := the positive generator of the subgroup {I_{μ,L}(w)| w} ⊂ ℤ

Lemma 6 (Viterbo). w, w′ : (D^{2}, ∂D^{2}) → (M, L) with w|_{∂D} = w′|_{∂D}

u(z) := \{ \begin{matrix}w(z), & z ∈ D^{2}, \\ w′(z) & z ∈ \overline{D}^{2}. \end{matrix}

⇒ I_{μ,L}(w) − I_{μ,L}(w′) = 2c_{1}(u).

Example L = Fix τ w′ := τ ∘ w ⇒ I_{μ,L}(w) = c_{1}(u).

• L = S^{1}(1)×S^{1}(1), S^{2}(√2) ⊂ S^{2}(1)×S^{2}(1)

[u] = m([S^{2}]×pt)+n(pt×[S^{2}]) ∈ H_{2}(S^{2}×S^{2}),

c_{1}(u) = 2m + 2n.

Hence, Σ_{L} ≥ 2.

4. Poincaré formula for real forms

G/K : Riemannian homogeneous space x, y ∈ G/K, V ⊂ T_{x}(G/K), W ⊂ T_{y}(G/K)

g_{x} · o = x

g_{y} · o = y dg_{x}^{−1}V, dg_{y}^{−1}W ⊂ T_{o}(G/K)

σ_{K}(V, W) = ; ∫_{K} ∥v_{1} ∧ ⋯ ∧ v_{p} ∧ k^{−1}(w_{1} ∧ ⋯ ∧ w_{q})∥dμ(k) ; “angle”

where v_{1}, ⋯, v_{p} : o.n.b. of dg_{x}^{−1}V w_{1}, ⋯, w_{q} : o.n.b. of dg_{y}^{−1}W

Theorem 7 (Howard). G/K : Riemannian homogeneous space L, N : submanifolds of G/K with

dim L + dim N ≥ dim (G/K).

Then

∫_{G} vol(L ∩ gN)dg = ∫_{L×N} σ_{K}(T_{x}^{⊥}L, T_{y}^{⊥}N)dμ(x, y).

In general,

it is very difficult to calculate the right hand side of Howard’s formula.

But,

to prove the global tightness of real forms, it suffices to calculate Poincaré formula for L = N = a real form.

Lemma 8 (Sakai- I, 2008). S^{2}(1) × S^{2}(1) = G/K, where G = SO(3) × SO(3), K = SO(2) × SO(2)

L := {(x, −x)| x ∈ S^{2}(1)} ⊂^{t.g.} G/K

Then,

σ_{K}(T_{x}^{⊥}L, T_{y}^{⊥}L) = 2π^{2}.

Hence, we have

∫_{G} ♯(L ∩ gL)dμ(g) = 2π^{2}vol(L)^{2} = 128π^{4}.

5. Our proof for totally geodesic Lagrangian sphere

Main result (Sakai- I). The totally geodesic Lagrangian sphere

L := {(x, −x)| x ∈ S^{2}(1)} ⊂ S^{2} × S^{2}

is globally tight.

Proof. Assume that L ⊂ S^{2} × S^{2} is not globally tight. Then, there exists g_{0} ∈ G s.t. L and g_{0}L intersect transversely and

♯(L ∩ g_{0}L) ≥ ∑ rank H_{∗}(L, ℤ_{2}) + 1.

Then, there exists an open neighborhood U of g_{0} in G satisfying

♯(L ∩ hL) ≥ ∑ rank H_{∗}(L, ℤ_{2}) + 1 ⋯ (1)

for all h ∈ U.

By Lemma 8, we have

128π^{4} = ∫_{G} ♯(L ∩ gL)dμ(g)
= ∫_{G\U} ♯(L ∩ gL)dμ(g)

+ ∫_{U} ♯(L ∩ gL)dμ(g)

≥ ∫_{G\U} ∑ rank H_{∗}(L, ℤ_{2})dμ(g)

+ ∫_{U}(∑ rank H_{∗}(L, ℤ_{2}) + 1)dμ(g) (Arnold − Givental, eq(1))

= ∫_{G} ∑ rank H_{∗}(L, ℤ_{2})dμ(g) + ∫_{U} dμ(g)
= 2vol(G) + vol(U).

Since

vol(G) = vol(SO(3))^{2} = (8π^{2})^{2} = 64π^{4}, we have

128π^{4} ≥ 128π^{4} + vol(U).

This is a contradiction. Hence, L ⊂ S^{2} × S^{2} is globally tight.