Manuscript VII-A@1 · VII-A.tex · section · 11da3fe6

An operator-level second-order audit

Plateau covariance and the second-order intertwiner › An operator-level second-order audit · lines 1295–1352

VII-A@1#eq:Videntity, VII-A@1#eq:Widentity

It is useful to verify that the second-order assertion contains the correct density terms. Write

V = V_{g} [h], W = W_{g} [h, k], V^{'} = V_{g} [h^{'}], W^{'} = W_{g} [h^{'}, k^{'}] .

Set

X = X_{ξ}, T_{2} = X_{ζ} + \frac{1}{2} X_{ξ}^{2}, D = D_{ξ}, Δ_{2} = D_{ζ} + \frac{1}{2} D_{ξ}^{2} .

Expanding Lemma lem:plateaucov gives

V^{'} W^{'} + V^{'} X + P_{m} T_{2} = V + D P_{m} - P_{m} X, = W + D V + Δ_{2} P_{m} .

These are identities of differential operators, not only identities on free solutions. In particular, they retain the terms proportional to $P_{m}$ .

To verify the second scalar transformation directly, put

a = ϕ_{1, g}, b = ϕ_{2, g}, b_{cand} = b + X a + T_{2} ϕ_{0} .

Since $P_{m} a = - V ϕ_{0}$ and $P_{m} b = - V a - W ϕ_{0}$ , equations (eq:Videntity)–(eq:Widentity) imply

P_{m} b_{cand} = - V a - W ϕ_{0} + P_{m} X a + P_{m} T_{2} ϕ_{0} = - V a - W ϕ_{0} - D V ϕ_{0} + (V - V^{'}) a + W ϕ_{0} + D V ϕ_{0} - W^{'} ϕ_{0} - V^{'} X ϕ_{0} = - V^{'} (a + X ϕ_{0}) - W^{'} ϕ_{0} .

The candidate has zero Cauchy data. Uniqueness proves (eq:phi2transform) independently of any informal cancellation of switch terms.

Displayed formulas in this passage

VII-A@1#eq:Videntity · line 1314
$V^{'} W^{'} + V^{'} X + P_{m} T_{2} = V + D P_{m} - P_{m} X, = W + D V + Δ_{2} P_{m} .$
VII-A@1#eq:Widentity · line 1314
$V^{'} W^{'} + V^{'} X + P_{m} T_{2} = V + D P_{m} - P_{m} X, = W + D V + Δ_{2} P_{m} .$
unlabeled · line 1337
$P_{m} b_{cand} = - V a - W ϕ_{0} + P_{m} X a + P_{m} T_{2} ϕ_{0} = - V a - W ϕ_{0} - D V ϕ_{0} + (V - V^{'}) a + W ϕ_{0} + D V ϕ_{0} - W^{'} ϕ_{0} - V^{'} X ϕ_{0} = - V^{'} (a + X ϕ_{0}) - W^{'} ϕ_{0} .$
unlabeled · line 1299
$V = V_{g} [h], W = W_{g} [h, k], V^{'} = V_{g} [h^{'}], W^{'} = W_{g} [h^{'}, k^{'}] .$
unlabeled · line 1306
$X = X_{ξ}, T_{2} = X_{ζ} + \frac{1}{2} X_{ξ}^{2}, D = D_{ξ}, Δ_{2} = D_{ζ} + \frac{1}{2} D_{ξ}^{2} .$
unlabeled · line 1329
$a = ϕ_{1, g}, b = ϕ_{2, g}, b_{cand} = b + X a + T_{2} ϕ_{0} .$