Manuscript VI@1 · VI.tex · section · b5b60afb

Two new second-order identities

Sign, coefficient, and switching audit › Two new second-order identities · lines 2541–2574

VI@1#eq:psi2-exponential

The scalar equation is

P ϕ_{2} = - L_{h} ϕ_{1} - (L_{j} + M_{h}) ϕ_{0} .

The second relational coefficient is

ψ_{2} = ϕ_{2} - X_{w} ϕ_{1} + (w^{ν} \partial_{ν} w^{μ} - v^{μ}) \partial_{μ} ϕ_{0} + \frac{1}{2} w^{μ} w^{ν} \partial_{μ} \partial_{ν} ϕ_{0} .

Both terms involving derivatives of $w$ are necessary when this is rewritten using iterated directional derivatives.

Indeed, define

b^{μ} = v^{μ} - \frac{1}{2} w^{ν} \partial_{ν} w^{μ} .

Then

ψ_{2} = ϕ_{2} - X_{w} ϕ_{1} - X_{b} ϕ_{0} + \frac{1}{2} X_{w}^{2} ϕ_{0} .

Expanding $X_{w}^{2}$ verifies (eq:psi2-exponential). Replacing $b$ by $v$ in this last formula would lose half of the derivative-of- $w$ contribution.

Figure fig:VI-classical-descent displays the complete coordinate-composition coefficient.

Displayed formulas in this passage

VI@1#eq:psi2-exponential · line 2563
$ψ_{2} = ϕ_{2} - X_{w} ϕ_{1} - X_{b} ϕ_{0} + \frac{1}{2} X_{w}^{2} ϕ_{0} .$
unlabeled · line 2544
$P ϕ_{2} = - L_{h} ϕ_{1} - (L_{j} + M_{h}) ϕ_{0} .$
unlabeled · line 2548
$ψ_{2} = ϕ_{2} - X_{w} ϕ_{1} + (w^{ν} \partial_{ν} w^{μ} - v^{μ}) \partial_{μ} ϕ_{0} + \frac{1}{2} w^{μ} w^{ν} \partial_{μ} \partial_{ν} ϕ_{0} .$
unlabeled · line 2559
$b^{μ} = v^{μ} - \frac{1}{2} w^{ν} \partial_{ν} w^{μ} .$