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The mass metric is symmetric, positive semidefinite, and positive definite exactly when the configuration differential is injective

Statement

Let $N \geq 1$ , let $Q$ be a smooth manifold, let $X : Q \to R^{3 N}$ be smooth, let $m_{1}, \dots, m_{N} > 0$ and let $G_{q}$ be the mass metric of Constrained configurations, the mass matrix and the mass metric.

For every $q \in Q$ the map $G_{q}$ is a symmetric bilinear form on $T_{q} Q$ .
$G_{q} (u, u) = \sum_{i = 1}^{N} m_{i} ∥ d X_{q} (u)_{i} ∥_{2}^{2} \geq 0$ for every $u \in T_{q} Q$ , so $G_{q}$ is positive semidefinite.
$G_{q}$ is positive definite (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form) if and only if $d X_{q}$ is injective. In particular, if $X$ is an immersion (Immersions, submersions, and constant-rank maps) then $G_{q}$ is positive definite for every $q$ .
If $b_{1}, \dots, b_{n}$ is an ordered basis of $T_{q} Q$ and $A$ is the matrix whose $j$ -th column is the coordinate vector of $d X_{q} (b_{j})$ in the standard basis of $R^{3 N}$ , then the matrix of the bilinear form $G_{q}$ in this basis is $A^{T} M A$ , with $M$ the mass matrix.

Facts & Assumptions

Given: The data $(Q, X, m_{1}, \dots, m_{N})$ of Constrained configurations, the mass matrix and the mass metric, a point $q \in Q$ , and an ordered basis $b_{1}, \dots, b_{n}$ of $T_{q} Q$ for clause 4.

[L1]

The mass metric is $G_{q} (u, v) = ⟨ d X_{q} (u), M d X_{q} (v) ⟩_{3 N}$ , the differential is linear, and $M (x_{1}, \dots, x_{N}) = (m_{1} x_{1}, \dots, m_{N} x_{N})$ with $m_{i} > 0$ (Constrained configurations, the mass matrix and the mass metric).

[L2]

The Euclidean inner product satisfies $⟨ x, x ⟩ = ∥ x ∥_{2}^{2} \geq 0$ , with equality only for $x = 0$ , and is bilinear (Real and complex inner product spaces, with the inner product linear in the first argument, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L3]

Positive definiteness of a symmetric bilinear form means $B (u, u) > 0$ for every $u \neq 0$ (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form).

Proof

technique · direct

1.1L1L2algebra

Bilinearity of $G_{q}$ follows from linearity of $d X_{q}$ and of $M$ together with bilinearity of the inner product [L1, L2]: for reals $α, β$ and $u, u^{'}, v \in T_{q} Q$ , $G_{q} (α u + β u^{'}, v) = α G_{q} (u, v) + β G_{q} (u^{'}, v)$ and similarly in the second argument. Symmetry follows from symmetry of $M$ : since $M$ acts by $x \mapsto (m_{1} x_{1}, \dots, m_{N} x_{N})$ , $⟨ a, M b ⟩_{3 N} = \sum_{i} m_{i} ⟨ a_{i}, b_{i} ⟩_{2} = ⟨ M a, b ⟩_{3 N}$ , so $G_{q} (u, v) = G_{q} (v, u)$ . This proves clause 1.

1.2L1L2algebra

For $u \in T_{q} Q$ write $d X_{q} (u) = (w_{1}, \dots, w_{N})$ with $w_{i} \in R^{3}$ ; then [L1] and [L2] give $G_{q} (u, u) = ⟨ d X_{q} (u), M d X_{q} (u) ⟩_{3 N} = \sum_{i = 1}^{N} m_{i} ⟨ w_{i}, w_{i} ⟩_{2} = \sum_{i = 1}^{N} m_{i} ∥ w_{i} ∥_{2}^{2} \geq 0$ . This proves clause 2.

2.1step 1.2L3algebra

Suppose $d X_{q}$ is injective and $u \neq 0$ ; then $d X_{q} (u) \neq 0$ , so some $w_{i} \neq 0$ , so $∥ w_{i} ∥_{2}^{2} > 0$ by [L2], and since $m_{i} > 0$ the sum in step 1.2 is strictly positive. Hence $G_{q} (u, u) > 0$ for every $u \neq 0$ . Conversely, if $d X_{q}$ is not injective, choose $u \neq 0$ with $d X_{q} (u) = 0$ ; then every $w_{i} = 0$ , the sum in step 1.2 vanishes, and $G_{q}$ is not positive definite by [L3]. This proves clause 3 for the injectivity criterion; if $X$ is an immersion, $d X_{q}$ is injective at every $q$ by definition, so $G_{q}$ is positive definite.

3.1step 1.1L1L2algebra∎

In the basis $b_{1}, \dots, b_{n}$ the matrix of $G_{q}$ has entries $G_{q} (b_{j}, b_{k}) = ⟨ d X_{q} (b_{j}), M d X_{q} (b_{k}) ⟩_{3 N}$ . If $A$ is the matrix whose $j$ -th column is the coordinate vector of $d X_{q} (b_{j})$ , then $M (d X_{q} (b_{k}))$ has coordinate vector $M A e_{k}$ , and the inner product of coordinate vectors is the Euclidean product, so $G_{q} (b_{j}, b_{k}) = (A^{T} M A)_{j k}$ . This proves clause 4.

The mass metric is symmetric, positive semidefinite, and positive definite exactly when the configuration differential is injective

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources