Saturation

Michael Kapovich

Recall the geometric interpretation of the saturation theorem for GL_{n}: Suppose that λ, μ, ν ∈ P_{+}^{∨} are coweights such that λ + μ + ν ∈ Q^{∨} and there exists a triangle in the Euclidean building X for G(ℚ_{p}) = GL_{n}(ℚ_{p}) with side-lengths λ, μ, ν. Then

(V (λ) ⊗ V (μ) ⊗ V (ν))^{G^{∨}(ℂ)} ≠ 0.

Here G^{∨}(ℂ) = GL_{n}(ℂ).

Below is an example of failure of the saturation theorem for Sp(4). Take λ = μ = ν = ϖ_{2}, the longer fundamental coroot. We can form a triangle in the tree embedded in X, so that the side-lengths are λ, μ, ν. Project this triangle to the positive chamber Δ = P_{+}^{∨} ⊗ℝ_{+}. The result is a folded triangle with two geodesic side-lengths (drawn in red) and one positively folded path p (in blue). However the path p is not an LS path: the maximality condition fails. Indeed, by using a non-affine wall through the break-point p(1/2) of p, one obtains a longer Bruhat chain from p′(1/2−) to p′(1/2+).

0 ω_{1}

ω_{2} B_{2} ≅ C_{2}

non-special positively folded, but maximality condition fails, so not an LS path τ

reflection

nonaffine wall

longer Bruhat chain

Building

apt

ω_{2}

0

\frac{1}{2}ω_{2}

ω̃_{2}

half apt

highest weights (λ, μ, ν) = (ω_{2}, ω_{2}, ω_{2}) c_{λμν} = 0 ω_{2} ω_{2} ω_{2}

ν^{∗} λ  μ  Δ LS path c_{λμν} ≠ 0  ν^{∗} = −w_{0}(ν) = ν

wall LS path

One observes (by inspection) that the path p is the only positively folded path (in Δ) of length μ connecting ϖ_{2} to itself. Thus,

c_{λμν} = 0.

However, by scaling the side-lengths by 2 one obtains an honest LS path (in Δ) of length 2ϖ_{2} connecting 2ϖ_{2} to itself. The corresponding folded triangle is obtained by taking the original one and scaling it by 2. Therefore

c_{2λ,2μ,2ν} ≠ 0.

0

2ω_{2}

ω_{2}

honest LS path: positively folded

ω_{2} vertex special maximal chain in Bruhat order (don’t introduce any non-affine walls)

(2ω_{2}, 2ω_{2}, 2ω_{2})

(2λ, 2μ, 2ν) ∈ C_{Rep} c_{2λ,2μ,2ν} ≠ 0

The goal of this talk is to generalize this argument to other groups (root systems).

How do we generalize the number 2 (the multiplication factor)? Given any root system R, there is a k_{R} such that for all vertices v in an apartment, k_{R} · v is a special vertex. Recall that a vertex is special if and only if it lies in the weight lattice P^{∨}. The number k_{R} can be computed from the Dynkin diagram: If

θ = ∑_{i=1}^{ℓ} m_{i}α_{i},

then

k_{R} = LCM(m_{1}, …, m_{ℓ}).

Now, given G^{∨}, compute its root system R and the factor k_{R}.

Theorem (Kapovich, Millson). For each triple (λ, μ, ν) ∈ (P_{+}^{∨})^{3} such that λ + μ + ν ∈ Q^{∨}, if there is an N ∈ ℕ = ℤ^{+} such that c_{Nλ,Nμ,Nν} ≠ 0, then c_{k_{R}^{2}λ,k_{R}^{2}μ,k_{R}^{2}ν} ≠ 0.

Remark. If R = A_{ℓ}, then k_{R} = 1, and we obtain saturation “on the nose.”

Conjecture (Kapovich, Millson). We can drop the squares.

The proof of the above theorem breaks into two parts:

1. analysis (responsible for the first stretching of λ, μ, ν by the factor of k_{R})

2. combinatorics (responsible for the second stretching by the factor of k_{R})

Consider the combinatorics part of the proof; i.e., assume that we have already stretched by a factor of k_{R} and have a triangle in the Euclidean building X with special vertices. Suppose we are given a triangle T = T (x, y, z) in the Euclidean (Bruhat-Tits) building corresponding to G(𝕂) (𝕂 = ℚ_{p} or 𝕂 = ℂ((z))) such that the vertices of Δ are special and have side-lengths λ⃗, μ⃗, ν⃗ ∈ P^{∨} with λ⃗ + μ⃗ + ν⃗ ∈ Q^{∨}. We want to show that c_{k_{R}λ,k_{R}μ,k_{R}ν} ≠ 0.

First consider the special case μ⃗ ∈ ℕω_{i}, where ω_{i} is a fundamental co-weight. Choose the apartment 𝔸 ⊂ X to contain the segment \overline{xy}. Since x is a special vertex, we can identify it with the origin 0 in 𝔸. Pick Δ ⊂ 𝔸 containing \overline{xy}.

0 = x  z y λ  μ ν X = building apt  \overline{xy} ⊂ apt 𝔸

0 = x
y

λ

z

ν μ

ν^{∗}

P_{+}^{∨} ⊗ ℝ

𝔸

Recall that we want

ν

λ

LS path?

positively folded path

Retract X to the positive chamber P_{+}^{∨} ⊗ ℝ. Namely, given P ∈ X with d⃗(0, P) = δ⃗, send P to the point ρ(P) = δ⃗ ∈ Δ.

0  • ρ(P) δ⃗ • P δ⃗  P_{+}^{∨} ⊗ ℝ apt  ρ : X → P_{+}^{∨} ⊗ ℝ = Δ is a simplicial map

This retraction is similar to the one used by Gaussent-Littelmann (but they retract from “−∞”; i.e., they move the point 0 to −∞).

Theorem. For any geodesic segment γ ∈ X, ρ(γ) is a positively folded path (i.e., a Hecke path) in Δ.

chain in Bruhat order  Δ maximality however might fail

Example. X is a tree

• 0

γ

ρ(γ)

Δ

Our assumption that μ⃗ ∈ ℕω_{i}, where ω_{i} is a fundamental co-weight implies that π := ρ(γ) travels along the edges of 𝔸.

0

Thus, all the breaks of π occur at vertices in 𝔸. After stretching π by k_{R}, all the breaks of π occur at special vertices in 𝔸. Therefore k_{R} · π satisfies maximality at every break point. Hence, k_{R} · π is an LS path.

For a general Hecke path π, of course, k_{R} · π is not an LS path.

Example. In the case of A_{2}, k_{R} = 1, but there are Hecke paths that are not LS paths.

0

Δ

Hecke but not LS

How do we get rid of our assumption that μ⃗ ∈ ℕω_{i}, where ω_{i} is a fundamental co-weight? We need to modify our path model. Rather than starting with the geodesic path connecting 0 to μ, we will take a piecewise-linear path π_{0} connecting 0 to μ. If

μ = ∑_{i=1}^{ℓ} n_{i}ω_{i},

we first travel along n_{1}ϖ_{1}, then parallel to n_{2}ϖ_{2}, etc.

0 π_{0}

μ = ∑_{i=1}^{ℓ} n_{i}ω_{i}

Δ

Apply the root operators e_{α}, f_{α} to π_{0} to get a family 𝒫_{π_{0}} of paths in 𝔸 such that all segments are edges of 𝔸. Then

Theorem (Littelmann). c_{λμν} equals the number of paths π ∈ 𝒫_{π_{0}} contained in Δ and connecting λ to ν^{∗}.

Note that the Retraction Theorem still works.

x

y λ

z

ν^{∗}

Δ

μ

chamber containing \overline{yz}

π_{μ}

Theorem. The path ρ(π_{μ}) belongs to 𝒫_{π_{0}}^{+}, the set of all positively folded paths of type π_{μ}. But ρ(π_{μ}) can fail the maximality condition. Note that ρ(π_{μ}) has breaks only at vertices on 𝔸. Stretch ρ(π_{μ}) by k_{R}. Then k_{R} · ρ(π_{μ}) has breaks only at special vertices. It follows that k_{R} · ρ(π_{μ}) satisfies the maximality condition and is still positively folded, and k_{R} · ρ(π_{μ}) ∈ 𝒫_{π_{k_{R}·μ}}. Remark. Positively-folded condition:

•

apply reflection move in positive direction alcove Thus, c_{k_{R}λ,k_{R}μ,k_{R}ν} ≠ 0.

k_{R}ν^{∗} k_{R}λ k_{R} · ρ(π_{μ})

This concludes the proof of the combinatorial part of the proof.

Sketch of the analytical part of the proof I am going to describe only some key ingredients of this part. Let X be a (nonpositively curved) symmetric space or a Euclidean building. Then ∂_{∞}X is its ideal boundary. Consider a triangle T (x_{1}, x_{2}, x_{3}) ⊂ X with the (usual) side-lengths m_{1}, m_{2}, m_{3} ∈ ℝ_{+}. Define a Gauss map

T = T (x_{1}, x_{2}, x_{3}) ↦ (m_{1}ξ_{1}, m_{2}ξ_{2}, m_{3}ξ_{3})

that sends T to a weighted configuration on ∂_{∞}X. Namely, take the first side \overline{x_{1}x_{2}} of T and extend it all way to infinity to get a point ξ_{1}. Place the mass m_{1} at ξ_{1}. Repeat the same for the two other sides.

Question. How do we invert the Gauss map?

Given three points ξ_{1}, ξ_{2}, ξ_{3} ∈ ∂_{∞}X and masses m_{1}, m_{2}, m_{3} ∈ ℝ_{+} placed at these points, we define the following map Φ : X → X, sending x_{1} to x_{4}. If Φ(x_{1}) = x_{1}, we obtain a triangle in X and therefore we inverted the Gauss map. In general, of course, x_{1} ≠ x_{4}.

• x_{1}  ξ_{1}, m_{1} • x_{2}  ξ_{2}, m_{2} m_{1}  • x_{3} m_{2}  ξ_{3}, m_{3} • x_{4} m_{3} ∂_{∞}X

When does Φ have a fixed point? Note that Φ : X → X is a weak contraction:

d(Φ(x), Φ(y)) ≤ d(x, y), ∀x, y ∈ X.

To see this, note that Φ is the composition of three weak contractions:

x_{1}

ξ_{1}, m_{1}

x_{2}

m_{1}

ξ_{2}, m_{2}

x_{3}

m_{3}, ξ_{3}

x′_{1} •

m_{1}

x′_{2}

In order to get a fixed point we need a bounded orbit for the system

(Φ^{n}(x_{1}))_{n∈ℕ} .

It turns out that we have a bounded orbit if and only if the weighted configuration (m_{i}ξ_{i}) is semistable in an appropriate sense. This definition coincides with Mumford’s notion of semistability provided that m_{i} ∈ ℕ and all ξ_{i}’s belong to the same flag-manifold, i.e., G-orbit in ∂_{∞}X. For a vector ξ^{0} ∈ ∂_{∞}Δ and a number m ≥ 0, we will identify mξ^{0} with a vector in Δ, whose (usual) length equals m. Then, each weighted configuration (m_{1}ξ_{1}, m_{2}ξ_{2}, m_{3}ξ_{3}) on ∂_{∞}X yields a triple of vectors in Δ:

(λ, μ, ν) = (m_{1}ξ_{1}^{0}, m_{2}ξ_{2}^{0}, m_{3}ξ_{3}^{0}),

where ξ_{i}^{0} ∈ ∂_{∞}Δ is the point of type ξ_{i}.

Question. Given a triple of vectors λ, μ, ν in Δ, when can we find a semistable configuration in ∂_{∞}X of type (λ, μ, ν)?

Theorem (Kapovich, Leeb, Millson). There exists a system of linear homogeneous inequalities on the vectors λ, μ, ν in Δ^{3} which give necessary and sufficient conditions for the existence of a weighted semistable configuration

(m_{1}ξ_{1}, m_{2}ξ_{2}, m_{3}ξ_{3})

on ∂_{∞}X of type (λ, μ, ν).

Now we can give the actual sketch:

0. Suppose that c_{Nλ,Nμ,Nν} ≠ 0 and λ + μ + ν ∈ Q^{∨}.

1. The Satake inverse then gives a triangle T_{N} in X with the Δ-side-lengths Nλ, Nμ, Nν.

2. Apply the Gauss map to this triangle. The result is a semistable weighted configuration

(Nm_{1}ξ_{1}, Nm_{2}ξ_{2}, Nm_{3}ξ_{3})

on ∂_{∞}X of type (Nλ, Nμ, Nν).

3. Then

(m_{1}ξ_{1}, m_{2}ξ_{2}, m_{3}ξ_{3})

is again semistable.

4. Hence, the corresponding map Φ has a fixed point x_{1} in X.

5. Form the associated triangle T = T (x_{1}, x_{2}, x_{3}). Then the Δ-side lengths of T are the vectors λ, μ, ν.

6. The condition λ + μ + ν ∈ Q^{∨} implies that we can choose T so that its vertices are vertices of X. However T does not necessarily have special vertices. One can “stretch” the triangle T by the factor of k_{R} so that the resulting triangle has special vertices.

How do we stretch? Retract T to ρ(T) ⊂ Δ. Then stretch ρ(T) by k_{R}. The result is a (positively) folded triangle k_{R} · ρ(T) ⊂ Δ. Its Δ-side lengths are k_{R}λ, k_{R}μ, k_{R}ν and its vertices are special. Identify it, by translation, with the triangle in Δ whose vertex is at the origin.

Now, unfold k_{R} · ρ(T) to a geodesic triangle in X. Its Δ-side lengths are k_{R}λ, k_{R}μ, k_{R}ν and its vertices are special.