The Score of a Penalty Pair is Determined by the Penalty

In the setting of Plans, Marginals, Vector Fields and Symmetric Matrices on the Wasserstein Space: Standing Notation, let $(\mathcal{D},\mathcal{D}_{\Sigma},\mathcal{E},\Sigma)$ be a penalty pair on $\mathcal{P}_{2}(\mathbb{R}^{d})$, let $\mu\in\mathcal{D}_{\Sigma}$, and let $T_{\mu}$, $\langle\cdot,\cdot\rangle_{\mu}$, the test functions $\psi\in C_{c}^{\infty}(\mathbb{R}^{d})$ with their gradients $\nabla\psi$, the maps $\mathrm{id}+t\,\nabla\psi$ of $\mathbb{R}^{d}$ and the push-forwards $(\mathrm{id}+t\,\nabla\psi)_{\#}\mu$ be as in that definition. For this statement only, say that $\xi\in T_{\mu}$ represents the first variation of $\mathcal{E}$ at $\mu$ if it has the property that Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §variation requires of $\Sigma(\mu)$: for every $\psi\in C_{c}^{\infty}(\mathbb{R}^{d})$ there is a real number $t_{0}>0$ such that $(\mathrm{id}+t\,\nabla\psi)_{\#}\mu\in\mathcal{D}$ for every $t$ in the open interval $(-t_{0},t_{0})$ and the function $(-t_{0},t_{0})\to\mathbb{R}$, $t\mapsto\mathcal{E}\bigl((\mathrm{id}+t\,\nabla\psi)_{\#}\mu\bigr)$, is differentiable at $0$, an interior point of that interval as recorded in that clause, with derivative $\langle\xi,\nabla\psi\rangle_{\mu}$. Then the following hold.

1. (The score is the unique representative of the first variation)¶ $\Sigma(\mu)$ is the only element of $T_{\mu}$ that represents the first variation of $\mathcal{E}$ at $\mu$.

2. (The score is determined by the penalty)¶ If $(\mathcal{D},\mathcal{D}_{\Sigma},\mathcal{E},\Sigma')$ is also a penalty pair on $\mathcal{P}_{2}(\mathbb{R}^{d})$, then $\Sigma'(\mu)=\Sigma(\mu)$.