Lemma: Polarization of the scalar stress and the factor in the Hessian

Let $T_0$ be as in (eq:T0) and define

$T^{\mathrm{pol}}$ is the unique symmetric bilinear form with $T^{\mathrm{pol}}(\varphi ,\varphi )=T_0(\varphi )$, and $\frac{d}{ds}{T}_0(\varphi +s\psi ){|}_{s=0}=2T^{\mathrm{pol}}(\varphi ,\psi )=T^{\mathrm{sym}}(\varphi ,\psi )$. For the vertex $\mathcal{V}[\gamma ,\varphi ]=-\frac{1}{2}\int g{\gamma }_{\mu \nu }{T_0}^{\mu \nu }(\varphi )$,

Hence (eq:hess) is exactly the mixed Hessian of this vertex if XIV-J's $T^{\mu \nu }(\varphi ,\psi )$ is $T^{\mathrm{pol}}$, and is twice it if $T^{\mu \nu }(\varphi ,\psi )$ is $T^{\mathrm{sym}}$. The record does not display which.

Proof. $T_0(\varphi +s\psi )=T_0(\varphi )+2s{T}^{\mathrm{pol}}(\varphi ,\psi )+s^2{T}_0(\psi )$ by bilinearity; the $t$-derivative of $\mathcal{V}$ is linear in $\gamma$. Uniqueness of the symmetric polarization of a quadratic form is standard.