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The orthogonal complement of a null vector need not be positive definite

Statement refuted

Statement refuted: "for every nonzero displacement $u$ of a Lorentz vector space the restriction of $η$ to $u^{⊥}$ is positive definite" (eta-orthogonality and orthogonal complements in a Lorentz vector space).

Facts & Assumptions

Given: An inertial frame of an oriented time-oriented affine Minkowski space $(A, V, η, C^{+})$ with $η = diag (- 1, 1, 1, 1)$ in the frame's coordinates, and the displacement $u = (1, 1, 0, 0)$ .

[L1]

$η (v, v) = - (v^{0})^{2} + (v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}$ and $η (u, v) = - u^{0} v^{0} + u^{1} v^{1} + u^{2} v^{2} + u^{3} v^{3}$ in an $η$ -orthonormal basis (Inertial frames and inertial charts of the affine Minkowski model, Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space).

[L2]

$u^{⊥} = {v : η (u, v) = 0}$ is a linear subspace and $u \in u^{⊥}$ exactly when $η (u, u) = 0$ (eta-orthogonality and orthogonal complements in a Lorentz vector space).

[L3]

A vector is null when it is nonzero with $η (v, v) = 0$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L4]

A symmetric bilinear form $B$ on a subspace $W$ is positive semidefinite when $B (v, v) \geq 0$ for every $v \in W$ ; it is positive definite when $B (v, v) > 0$ for every nonzero $v \in W$ . Its kernel here means its radical, ${v \in W : B (v, w) = 0 for all w \in W}$ .

Counterexample

1.1L1L3algebra

For $u = (1, 1, 0, 0)$ one has $η (u, u) = - 1 + 1 = 0$ with $u \neq 0$ , so $u$ is null by [L3].

1.2L1L2algebra

The orthogonality conditions $η (u, v) = 0$ read $- v^{0} + v^{1} = 0$ , that is, $v^{1} = v^{0}$ ; on that hyperplane $η (v, v) = - (v^{0})^{2} + (v^{0})^{2} + (v^{2})^{2} + (v^{3})^{2} = (v^{2})^{2} + (v^{3})^{2}$ .

2.1step 1.1step 1.2L1L2L3L4algebra∎

The expression $(v^{2})^{2} + (v^{3})^{2}$ of step 1.2 is nonnegative on $u^{⊥}$ and vanishes exactly when $v^{2} = v^{3} = 0$ , and then $v^{1} = v^{0}$ is free; thus the restriction is positive semidefinite by [L4]. For $v, w \in u^{⊥}$ its bilinear pairing is $η (v, w) = v^{2} w^{2} + v^{3} w^{3}$ , since $v^{1} = v^{0}$ and $w^{1} = w^{0}$ . Testing $w = e_{2}, e_{3} \in u^{⊥}$ shows its kernel consists exactly of vectors with $v^{2} = v^{3} = 0$ , namely $v = v^{0} u$ . Therefore the kernel is the one-dimensional nonzero subspace $span (u)$ , and the restriction is not positive definite by [L4]. In particular $u$ itself is orthogonal to $u$ , by [L2] and step 1.1, and is a nonzero vector of $u^{⊥}$ with $η (u, u) = 0$ .

Remarks

The failure is exactly the one-dimensional kernel $span (u)$ found above; the statement is refuted by the single vector $u$ , which is a nonzero element of its own orthogonal complement. The timelike case is genuinely different, as the companion example Worked orthogonal complement of a timelike displacement is positive definite shows: the orthogonal complement of a timelike displacement is positive definite.

Depends on

Dependency tree · 2 levels

Sources

K. Conrad, Bilinear Forms (orthogonality, radicals, inertia)