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Tame J-holomorphic area equals metric Dirichlet energy

Statement

Let $(M, ω, J)$ be the tame almost-complex data of Symplectic and almost-complex curve data, and let $(Σ, j)$ be a Riemann surface. A smooth Riemannian metric $h$ on $Σ$ is $j$ -compatible when $h (j ξ, j η) = h (ξ, η)$ for all tangent vectors $ξ, η$ . Assume that $h$ is $j$ -compatible. Set $g_{J}^{s y m} (v, w) = \frac{1}{2} (ω (v, J w) + ω (w, J v))$ , which is positive definite because $J$ is tame. At $z \in Σ$ , define the Hilbert--Schmidt norm of $d u_{z} : (T_{z} Σ, h_{z}) \to (T_{u (z)} M, g_{J}^{s y m})$ by $∣ d u_{z} ∣_{g_{J}^{s y m}, h}^{2} = \sum_{a = 1}^{2} g_{J}^{s y m} (d u_{z} (e_{a}), d u_{z} (e_{a})),$ where $(e_{1}, e_{2})$ is any $h_{z}$ -orthonormal basis; this sum is independent of that basis. Let $d {vol}_{h}$ be the area measure induced by $h$ . If $u : Σ \to M$ is a smooth $J$ -holomorphic map with finite symplectic area $E_{ω} (u)$ , then its Dirichlet energy $E_{D} (u) = \frac{1}{2} \int_{Σ} ∣ d u ∣_{g_{J}^{s y m}, h}^{2} d {vol}_{h}$ equals the symplectic area $E_{ω} (u) = \int_{Σ} u^{*} ω$ , and both are nonnegative. If $J$ is compatible, $g_{J}^{s y m} (v, w) = ω (v, J w)$ . For tame but noncompatible $J$ , the symmetrized form is the metric; the unsymmetrized form $ω (\cdot, J \cdot)$ is not assumed symmetric.

Statement references

Symplectic and almost-complex curve data

Facts & Assumptions

Given: The symplectic form, tame almost-complex structure, domain, map, and metrics in the statement.

[F1]

Symplectic and almost-complex curve data: $J^{2} = - I$ , $d u \circ j = J \circ d u$ , tameness means $ω (v, J v) > 0$ for $v \neq 0$ , and $g_{J}^{s y m} (v, w) = \frac{1}{2} (ω (v, J w) + ω (w, J v))$ ; $ω$ is a skew-symmetric two-form.

Proof

1.1givenF1algebra

At any point, $g_{J}^{s y m}$ is smooth and symmetric by its definition, and $g_{J}^{s y m} (v, v) = ω (v, J v) > 0$ for every nonzero $v$ by tameness. Thus it is a Riemannian metric. Choose a unit vector $e$ at the point. Since $h$ is $j$ -compatible, $h (j e, j e) = h (e, e) = 1$ and $h (e, j e) = h (j e, j^{2} e) = - h (j e, e)$ , so $h (e, j e) = 0$ . Thus $(e, j e)$ is a positively oriented $h$ -orthonormal basis of $T Σ$ . Put $v = d u (e)$ . The Cauchy–Riemann equation gives $d u (j e) = J v$ . Using $J^{2} = - I$ and skew-symmetry of $ω$ gives $g_{J}^{s y m} (v, J v) = 0$ and $g_{J}^{s y m} (J v, J v) = g_{J}^{s y m} (v, v) = ω (v, J v)$ .

2.1givenF1step 1.1algebra

The area density in this basis is $(u^{*} ω) (e, j e) = ω (v, J v)$ , while the Dirichlet density is $\frac{1}{2} (∣ d u (e) ∣^{2} + ∣ d u (j e) ∣^{2}) = g_{J}^{s y m} (v, v)$ . Thus the densities agree pointwise, including at points where $d u = 0$ ; integrating gives $E_{D} (u) = E_{ω} (u)$ . Tameness makes this common density nonnegative, and the finite-area hypothesis makes both integrals finite.

3.1F1step 1.1∎

If $J$ is compatible, $ω (\cdot, J \cdot)$ is symmetric by definition, so its symmetrization is itself. For a merely tame $J$ , only $g_{J}^{s y m}$ is used as the metric; no Riemannian metric is inferred from the unsymmetrized form.

Depends on

Symplectic and almost-complex curve data

Dependency tree · 2 levels

Sources

Chris Wendl, Lectures on Holomorphic Curves in Symplectic and Contact Geometry, version 3.2