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Mass-shell polynomial rigidity

Statement

Let $(V, g, C^{+})$ be a four-dimensional time-oriented Lorentz vector space of signature $(-, +, +, +)$ , with future cone $C^{+}$ as in Time orientation, future and past timelike/null/causal vectors, and the closed cones. Let $M > 0$ and let $A$ be a symmetric bilinear form on $V$ and $b$ a covector. If $A (P, P) + b (P) = 0 for every P \in C^{+} with g (P, P) = - M^{2},$ then $A = 0$ and $b = 0$ . If instead this identity holds for every nonzero future null vector $P \in \overline{C^{+}}$ , then $b = 0$ and $A = λ g$ for some $λ \in R$ . The second clause is the exact massless conformal exception.

Facts & Assumptions

Given: The time-oriented Lorentz vector space and the symmetric form and covector in the Statement.

[F1]

There is a basis with Lorentz matrix $diag (- 1, 1, 1, 1)$ (Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1)). Replacing its timelike basis vector by its negative when needed gives a future orthonormal basis because the time orientation selects one of the two timelike components (Time orientation, future and past timelike/null/causal vectors, and the closed cones, The timelike set has exactly two convex components, described in orthonormal coordinates).

[F2]

In this basis $g (P, P) = - (P^{0})^{2} + \sum_{i = 1}^{3} (P^{i})^{2}$ , and the future massive and null shells have respectively $P^{0} = \sqrt{M^{2} + ∣ q ∣^{2}}$ , $P^{0} > 0$ , and $P^{0} = ∣ q ∣ > 0$ (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, Timelike, null, spacelike and causal vectors of a Lorentz vector space).

Proof

1.1givenF1F2algebra

In the future massive shell write $P = (E, r n)$ , where $E = \sqrt{M^{2} + r^{2}}$ , $r > 0$ and $n \in S^{2}$ . Apply the assumed identity to the two shell vectors $(E, r n)$ and $(E, - r n)$ and subtract. Dividing the result by $2 r > 0$ gives $2 E A_{0 i} n_{i} + b_{i} n_{i} = 0$ for every $E > M$ and every unit $n$ . For fixed $n$ , the left side is affine in $E$ and vanishes on an interval, so $A_{0 i} n_{i} = b_{i} n_{i} = 0$ ; varying $n$ yields $A_{0 i} = b_{i} = 0$ .

1.2givenF2algebra

On the future null cone write $P = ρ (1, n)$ with $ρ > 0$ and $n \in S^{2}$ . The identity becomes $ρ^{2} A ((1, n), (1, n)) + ρ b (1, n) = 0$ for every $ρ > 0$ ; comparing the two polynomial coefficients gives $A ((1, n), (1, n)) = b (1, n) = 0$ . Replacing $n$ by $- n$ in the covector equation gives $b_{0} = 0$ and $b_{i} n_{i} = 0$ for every $n$ , so $b = 0$ .

2.1step 1.1F2algebra

The remaining massive-shell identity is $A_{00} (M^{2} + r^{2}) + r^{2} n^{T} A_{s p} n + b_{0} \sqrt{M^{2} + r^{2}} = 0$ , where $A_{s p} = (A_{i j})$ . Varying $n$ shows $n^{T} A_{s p} n$ is a constant $a$ on $S^{2}$ . Taking $n = e_{i}$ gives $A_{i i} = a$ ; taking $n = (e_{i} + e_{j}) / \sqrt{2}$ gives $A_{i j} = 0$ for $i \neq j$ . With $z = \sqrt{M^{2} + r^{2}}$ the remaining identity is $(A_{00} + a) z^{2} + b_{0} z - a M^{2} = 0$ for every $z > M$ . A polynomial vanishing on an interval has all coefficients zero; since $M > 0$ , this yields $a = A_{00} = b_{0} = 0$ , hence $A = b = 0$ .

3.1step 1.2algebra∎

In the quadratic null-cone equation, comparing $n$ and $- n$ gives $A_{0 i} = 0$ . Its value on $e_{i}$ yields $A_{i i} = - A_{00}$ , and its value on $(e_{i} + e_{j}) / \sqrt{2}$ yields $A_{i j} = 0$ for $i \neq j$ . Thus $A = diag (A_{00}, - A_{00}, - A_{00}, - A_{00}) = λ g$ with $λ = - A_{00}$ . This proves the null clause and completes the lemma.

Mass-shell polynomial rigidity

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources