ag.algebraic geometry - Lie bracket on the unshifted tangent complex?

optimum transportation – Transport of touchstone Answer

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optimum transportation – Transport of touchstone

Let’s decompose $mu$ and $nu$, two possibilities on $mathbb{R}^{d}$ , in response to
pi_{okay} (x_{1},…,x_{d}) = (x_{okay},…,x_{d})

We get a household of measures and every touchstone $mu_{okay,d}^{+} =: mu^{+}$ (resp $nu^{+}_{okay,d} =: nu^{+}$) is focused on $mathbb{R}^{k-1} occasions {x_{okay} } occasions … occasions { x_{d} }$ (resp $mathbb{R}^{k-1} occasions {y_{okay} } occasions … occasions { y_{d} }$).

For every $(x_{+},y_{+}) := (x_{okay},…,x_{d},y_{okay},…,y_{d})$ we deem $gamma^{+,+}$ the Knothe exaltation touchstone between $mu^{+}$ and $nu^{+}$. We occupy it’s realized by a map $T$, i.e.
gamma^{+,+} = (textual content{id},T)_{#} mu^{+}

with $T_{#} mu^{+} = nu^{+}$

Finally let’s deem $gamma = gamma^{+,+} otimes eta in mathcal{P}(mathbb{R}^{2nd})$ with $eta in mathcal{P}(mathbb{R}^{d-k+1} occasions mathbb{R}^{d-k+1})$ and

gamma(A) = int{gamma^{++}(A) deta}

In this definition we’ve one thing to show,that is $gamma$ is certainly a touchstone. The solely factor to show is that

(x_{+},y_{+}) rightarrow gamma^{++}(A)

is measurable.

But I do not know how you can show it, do somebody know the way to try this ? I might breathe relieved to learn it.

Thanks and regards.

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