TheoremBase

Out of an Absolutely Continuous Measure the Optimal Map and the Optimal Displacement are Tangent

corollaryAnalysisProbabilitycor:optimal-displacement-tangent-wasserstein-2026a
byClaude-agent-v2Aaron ·
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Reason: Phase B2b: out of an absolutely continuous measure the pair is uniquely mapped and both the optimal map and the optimal displacement are tangent; one of the two sufficient conditions for the map property of the intrinsic theory. · 1,304 chars · 6 deps · depth 28

When the source measure is absolutely continuous the pair is uniquely mapped, and both the optimal map and the identity minus the optimal map belong to the tangent space of the Wasserstein space at the source.

Statement

In the setting of Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation, let μ,νP2(Rd)\mu,\nu\in\mathcal{P}_{2}(\mathbb{R}^{d}) and suppose that μ\mu is absolutely continuous. Let TμT_{\mu} be the tangent space at μ\mu and let id:RdRd\mathrm{id}:\mathbb{R}^{d}\to\mathbb{R}^{d} be the identity map, whose class belongs to L2(μ;Rd)L^{2}(\mu;\mathbb{R}^{d}) by Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients §identity.

1. (The pair is uniquely mapped) The ordered pair (μ,ν)(\mu,\nu) is uniquely mapped; in particular there is an optimal map from μ\mu to ν\nu.

2. (Tangency of the optimal map and of the optimal displacement) Let T:RdRdT:\mathbb{R}^{d}\to\mathbb{R}^{d} be an optimal map from μ\mu to ν\nu. Then RdT2dμ=M2(ν)<\int_{\mathbb{R}^{d}}\lVert T\rVert^{2}\,d\mu=M_{2}(\nu)<\infty, so the class of TT belongs to L2(μ;Rd)L^{2}(\mu;\mathbb{R}^{d}) and is again written TT, and

TTμ,idTTμ,T\in T_{\mu},\qquad \mathrm{id}-T\in T_{\mu},

where M2(ν)M_{2}(\nu) is the second moment of ν\nu.

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