A metric on $(1+n)$-dimensional spacetime: a function assigning to each spacetime point $x$ a matrix of components $m_{\mu\nu}(x)$ for $\mu,\nu \in \{0,1,\dots,n\}$.
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An inverse metric on $(1+n)$-dimensional spacetime: a function assigning to each spacetime point $x$ a matrix of components $(m^{-1})^{\mu\nu}(x)$ for $\mu,\nu \in \{0,1,\dots,n\}$.
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A metric Lagrangian $\mathcal{L}(\phi, \nabla\phi, m)$: a function of a scalar value, a gradient covector, and a metric matrix, returning a real number. This is the type of Lagrangian considered in CM18 Section 2 (coordinate-invariant Lagrangians).
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Evaluate a metric Lagrangian on a scalar field $\phi$ and metric $m$ at a spacetime point $x$, producing $\mathcal{L}(\phi(x), \nabla\phi(x), m(x))$.
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A one-parameter flow on spacetime generated by a smooth vector field $Y$. The map $F_\epsilon : \text{Spacetime} \to \text{Spacetime}$ satisfies $F_0 = \mathrm{id}$, $F_{-\epsilon} \circ F_\epsilon = \mathrm{id}$, and $\frac{d}{ds}F_s(x) = Y(F_s(x))$. This corresponds to the diffeomorphism family of CM18 Proposition 2.0.1.
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The $(\mu,\nu)$-component of the inverse-Jacobian matrix $(M^{-1})^\mu_{\ \nu} = \partial x^\mu / \partial \widetilde{x}^\nu$, defined as the Jacobian of $F_{-\epsilon}$ at $\widetilde{x}$.
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The transformed scalar field $\widetilde{\phi}(\widetilde{x}) := \phi(F_{-\epsilon}(\widetilde{x}))$ from CM18 Definition 2.0.6.
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The transformed gradient $\widetilde{\nabla}_\mu \widetilde{\phi}(\widetilde{x}) := (M^{-1})^\alpha_{\ \mu} \nabla_\alpha \phi(F_{-\epsilon}(\widetilde{x}))$ from CM18 Definition 2.0.6.
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The transformed metric components $\widetilde{m}_{\mu\nu}(\widetilde{x}) := (M^{-1})^\alpha_{\ \mu} (M^{-1})^\beta_{\ \nu} m_{\alpha\beta}(F_{-\epsilon}(\widetilde{x}))$ from CM18 Definition 2.0.6.
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The transformed inverse metric components $(\widetilde{m}^{-1})^{\mu\nu}(\widetilde{x}) := M^\mu_{\ \alpha} M^\nu_{\ \beta} (m^{-1})^{\alpha\beta}(F_{-\epsilon}(\widetilde{x}))$ from CM18 Definition 2.0.6.
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A metric Lagrangian $\mathcal{L}$ is coordinate invariant (CM18 Definition 2.0.7) if for every smooth flow $\Psi$ on spacetime, $\mathcal{L}(\phi(x), \nabla\phi(x), m(x)) = \mathcal{L}(\widetilde{\phi}(\widetilde{x}), \widetilde{\nabla}\widetilde{\phi}(\widetilde{x}), \widetilde{m}(\widetilde{x}))$ where the transformed fields are as in Definition 2.0.6 and $x = F_{-\epsilon}(\widetilde{x})$.
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Chain rule along the inverse flow: for a smooth scalar function $f$, $\left.\partial_\epsilon\right|_{\epsilon=0} f(F_{-\epsilon}(\widetilde{x})) = -Y^\alpha(\widetilde{x}) \, \nabla_\alpha f(\widetilde{x})$. This is the scalar case of CM18 Proposition 2.0.2.
CM18 Proposition 2.0.2, scalar part: the $\epsilon$-derivative at $\epsilon=0$ of the transformed field equals $-Y^\alpha \nabla_\alpha \phi$.
Schwarz-type swap: for a smooth $G : \mathbb{R} \times E \to F$, the time derivative of the partial space differential equals the space differential of the partial time derivative, $\partial_t \, d_x G(t, x_0)\,v = d_x (\partial_t G)(x_0)\,v$.
The inverse Jacobian $(M^{-1})^\alpha_{\ \mu}$ depends differentiably on $\epsilon$ at $\epsilon = 0$.
The derivative of the inverse Jacobian at $\epsilon = 0$: $\left.\partial_\epsilon\right|_{\epsilon=0} (M^{-1})^\alpha_{\ \mu} = -\nabla_\mu Y^\alpha$. This expands the expansion $(M^{-1})^\mu_{\ \nu} = \delta^\mu_\nu - \epsilon \nabla_\nu Y^\mu + O(\epsilon^2)$ from CM18 Proposition 2.0.1.
HasDerivAt version of chain_rule_scalar_along_inverse_flow: for a smooth scalar $g$,
the function $\epsilon \mapsto g(F_{-\epsilon}(\widetilde{x}))$ has derivative
$-Y^\alpha \nabla_\alpha g$ at $\epsilon = 0$.
Leibniz-rule expansion of $\nabla_\mu (Y^\alpha \nabla_\alpha \phi)$ as a sum, used in proving the gradient transformation rule (CM18 Proposition 2.0.2).
CM18 Proposition 2.0.2, gradient part: for a smooth scalar field $\phi$, $\left.\partial_\epsilon\right|_{\epsilon=0} \widetilde{\nabla}_\mu \widetilde{\phi} = -\nabla_\mu(Y^\alpha \nabla_\alpha \phi)$.
Restatement of chain_rule_gradient_along_inverse_flow (CM18 Proposition 2.0.2, gradient
case): the $\epsilon$-derivative of the transformed gradient at $\epsilon = 0$ equals
$-\nabla_\mu(Y^\alpha \nabla_\alpha \phi)$.
For a smooth metric component $m_{\alpha\beta}$, the derivative of $\epsilon \mapsto m_{\alpha\beta}(F_{-\epsilon}(\widetilde{x}))$ at $\epsilon = 0$ is $-Y^\gamma \nabla_\gamma m_{\alpha\beta}$.
CM18 Proposition 2.0.2, metric part: $\left.\partial_\epsilon\right|_{\epsilon=0} \widetilde{m}_{\mu\nu} = -(m_{\alpha\nu} \nabla_\mu Y^\alpha + m_{\mu\alpha} \nabla_\nu Y^\alpha + Y^\alpha \nabla_\alpha m_{\mu\nu})$.
Restatement of chain_rule_metric_along_inverse_flow (CM18 Proposition 2.0.2, metric case):
the $\epsilon$-derivative of the transformed metric at $\epsilon=0$ equals
$-(m_{\alpha\nu} \nabla_\mu Y^\alpha + m_{\mu\alpha} \nabla_\nu Y^\alpha + Y^\alpha \nabla_\alpha m_{\mu\nu})$.
Derivative of the Jacobian along the inverse flow at the moving point: $\left.\partial_\epsilon\right|_{\epsilon=0} M^\mu_{\ \alpha}(F_{-\epsilon}(\widetilde{x})) = \nabla_\alpha Y^\mu(\widetilde{x})$. This is the key identity for differentiating the transformed inverse metric.
CM18 Proposition 2.0.2, inverse metric part: $\left.\partial_\epsilon\right|_{\epsilon=0} (\widetilde{m}^{-1})^{\mu\nu} = (m^{-1})^{\alpha\nu} \nabla_\alpha Y^\mu + (m^{-1})^{\mu\alpha} \nabla_\alpha Y^\nu - Y^\alpha \nabla_\alpha (m^{-1})^{\mu\nu}$.
Restatement of chain_rule_inverse_metric_along_flow (CM18 Proposition 2.0.2, inverse metric
case): the $\epsilon$-derivative of the transformed inverse metric at $\epsilon = 0$.
For any non-identity permutation $\sigma$ and any index $i$, there exists some $j \ne i$ that $\sigma$ does not fix. Used in the proof of the Jacobi determinant identity.
Jacobi's formula applied to the inverse-Jacobian flow at $\epsilon = 0$: $\left.\partial_\epsilon\right|_{\epsilon=0} \det M^{-1} = -\nabla_\alpha Y^\alpha$, matching the expansion $\det M^{-1} = 1 - \epsilon \nabla_\alpha Y^\alpha + O(\epsilon^2)$ in CM18 Proposition 2.0.1.
Restatement of jacobi_formula_flow: the $\epsilon$-derivative of $\det M^{-1}$ at $\epsilon=0$
equals $-\nabla_\alpha Y^\alpha$ (CM18 Proposition 2.0.1, determinant case).
The partial derivative $\partial \mathcal{L} / \partial m_{\mu\nu}$ of a metric Lagrangian with respect to the $(\mu,\nu)$-entry of the metric, evaluated at $(\phi(x), \nabla\phi(x), m(x))$.
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Decomposition of the Fréchet derivative on a product: for $h : E \times F \to G$ differentiable at $(a,b)$, $dh_{(a,b)}(da, db) = dh^{(\cdot, b)}_a(da) + dh^{(a, \cdot)}_b(db)$, i.e. the total differential splits into the two partial differentials.
Matrix version of clm_pi_decomp_chain: a continuous linear functional on the matrix space
$\mathbb{R}^{(n+1) \times (n+1)}$ decomposes as
$D(v) = \sum_{\mu,\nu} D(E_{\mu\nu})\, v_{\mu\nu}$, where $E_{\mu\nu}$ is the matrix with a single
$1$ in position $(\mu,\nu)$.
For a vector-valued function $f : \mathbb{R} \to \mathbb{R}^{n+1}$, taking the $i$-th component of $f'(x)$ equals the derivative of the scalar component function $\epsilon \mapsto f(\epsilon)_i$.
Matrix version of deriv_pi_comp_chain: for $f : \mathbb{R} \to \mathbb{R}^{(n+1)\times(n+1)}$,
the $(i,j)$ entry of $f'(x)$ equals the derivative of the scalar component
$\epsilon \mapsto f(\epsilon)_{ij}$.
Three-argument chain rule for a Lagrangian $L(v, p, m)$ composed with three smooth families $f_1(\epsilon), f_2(\epsilon), f_3(\epsilon)$: the $\epsilon$-derivative at $0$ decomposes as $\partial_v L \cdot f_1' + \sum_\mu \partial_{p_\mu} L \cdot (f_2)_\mu'
- \sum_{\mu,\nu} \partial_{m_{\mu\nu}} L \cdot (f_3)_{\mu\nu}'$.
At $\epsilon = 0$, the transformed scalar field reduces to the original: $\widetilde{\phi}|_{\epsilon=0}(x) = \phi(x)$.
At $\epsilon = 0$, the transformed gradient reduces to the original gradient $\nabla\phi(x)$.
At $\epsilon = 0$, the transformed metric reduces to the original $m(x)$.
For smooth $\phi$, the transformed field $\epsilon \mapsto \widetilde{\phi}(x)$ is differentiable at $\epsilon = 0$.
For smooth $\phi$, the transformed gradient $\epsilon \mapsto \widetilde{\nabla}\widetilde{\phi}(x)$ is differentiable at $\epsilon = 0$.
For a smooth metric, the transformed metric $\epsilon \mapsto \widetilde{m}(x)$ is differentiable at $\epsilon = 0$.
Algebraic combination of the scalar, gradient, and metric variation formulas. Given the three flow-derivative identities (CM18 Proposition 2.0.2) as hypotheses, the $\epsilon$-derivative of $\mathcal{L}(\widetilde{\phi}, \widetilde{\nabla}\widetilde{\phi}, \widetilde{m})$ at $\epsilon=0$ expands by the three-argument chain rule into the explicit sum appearing in CM18 Corollary 2.0.3.
CM18 Corollary 2.0.3: the derivative of the Lagrangian along the flow at $\epsilon = 0$, $\left.\partial_\epsilon\right|_{\epsilon=0} \mathcal{L}(\widetilde{\phi}, \widetilde{\nabla}\widetilde{\phi}, \widetilde{m}) = -\frac{\partial \mathcal{L}}{\partial \phi} Y^\alpha \nabla_\alpha \phi - \frac{\partial \mathcal{L}}{\partial (\nabla_\mu \phi)} \nabla_\mu (Y^\alpha \nabla_\alpha \phi) - \frac{\partial \mathcal{L}}{\partial m_{\mu\nu}} (m_{\alpha\nu} \nabla_\mu Y^\alpha + m_{\mu\alpha} \nabla_\nu Y^\alpha + Y^\alpha \nabla_\alpha m_{\mu\nu})$.
The energy-momentum tensor associated to a metric Lagrangian: $T^{\mu\nu}(x) := 2 \frac{\partial \mathcal{L}}{\partial m_{\mu\nu}}
- (m^{-1})^{\mu\nu} \mathcal{L}(\phi, \nabla\phi, m)$. This is the canonical "Hilbert" energy-momentum tensor obtained by varying the action with respect to the metric.
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A scalar field $\phi$ satisfies the Euler-Lagrange equation for the Lagrangian $\mathcal{L}$ with metric $m$ if the Euler-Lagrange operator vanishes at every spacetime point.
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If $f : \mathbb{R}^{(n+1)\times(n+1)} \to \mathbb{R}$ is invariant under transpose and $M_0$ is a symmetric matrix, then the partial derivatives of $f$ at $M_0$ with respect to the $(\mu,\nu)$ and $(\nu,\mu)$ entries coincide.
If the Lagrangian $\mathcal{L}$ is invariant under transposing the metric argument and the metric $m$ is symmetric, then $\partial \mathcal{L} / \partial m_{\mu\nu}$ is symmetric in $(\mu,\nu)$.
The energy-momentum tensor is symmetric in $(\mu,\nu)$ whenever the metric, inverse metric, and Lagrangian (in its metric argument) are all symmetric: $T^{\mu\nu} = T^{\nu\mu}$.
The integration-by-parts computation underlying Noether's theorem in CM18: for a coordinate invariant Lagrangian, a solution $\phi$ of the Euler-Lagrange equation, and a compactly supported vector field $Y$, the contracted integral $\int_K (\nabla \cdot T)_\nu \, m_{\alpha\nu} \, Y^\alpha\, d^{1+n}x = 0$ vanishes.
Restatement of noether_ibp_computation as the Noether integral identity: for any compactly
supported smooth vector field $Y$,
$\int_K (\nabla \cdot T)_\nu \, m_{\alpha\nu} \, Y^\alpha\, d^{1+n}x = 0$.
A contracted du Bois-Reymond / fundamental lemma of the calculus of variations: if for every compactly supported smooth test vector field $Y$ one has $\int_K \sum_\alpha f(x) m_{\alpha\nu}(x) Y^\alpha(x)\, d^{1+n}x = 0$, then $f$ vanishes identically. Uses invertibility of $m$ via $m^{-1}$.
Noether's theorem (CM18 main result): for a coordinate-invariant Lagrangian $\mathcal{L}$ and a scalar field $\phi$ satisfying the Euler-Lagrange equation, the energy-momentum tensor is divergence-free, $\nabla_\mu T^{\mu\nu}(x) = 0$ for all $x$ and $\nu$.