Heat kernel: DeWitt iterative procedure The DeWitt ansatz for the heat kernel is given by
$$K(t ; x, y ; D)=(4 \pi t)^{-n / 2} \Delta_{V V M}^{1 / 2}(x, y) \exp \left(-\frac{\sigma(x, y)}{2 t}\right) \Xi(t ; x, y ; D)$$
where $\sigma$ is the Synge's world function, $\Delta_{V V M}$ is the Van Vleck-Morte determinant. Due to the heat equation
$$(\partial_t + D)K(t ; x, y ; D) = 0$$ one obtains
$$\left(\partial_{t}+t^{-1}\left(\nabla^{\mu} \sigma\right) \nabla_{\mu}+\Delta_{V V M}^{-1 / 2} D \Delta_{V V M}^{1 / 2}\right) \Xi=0$$ where $D$ is some linear differential operator. Does anyone know how this happens ? (Ref. Eq. (4.37)-(4.39) Sec. 4.3 of https://arxiv.org/abs/hep-th/0306138)
 A: I am only getting a partial answer. Here is what and how I am getting:
$D$ is of form $-g^{\mu\nu}(\nabla_{\mu}\nabla_{\nu}+C)$ where $C$ is some constant, given in section 1 of the paper. I will take $C$ to be $0$ (zero) in the following also I will strip away all constant, which can be stripped, from $K$ for the ease of typing:
$$K(t ; x, y ; D)=\frac{1}{t^{n / 2}}\Delta^{1 / 2}\exp \left(-\frac{\sigma}{2 t}\right) \Xi$$
$$(\partial_t + D)K(t ; x, y ; D)=\partial_tK+DK$$
For the first term
$$\partial_tK=-\frac{n}{2t}\frac{1}{t^{n / 2}}\Delta^{1 / 2}\exp \left(-\frac{\sigma}{2 t}\right) \Xi+\frac{\sigma}{2t^2}\frac{1}{t^{n / 2}}\Delta^{1 / 2}\exp \left(-\frac{\sigma}{2 t}\right) \Xi+\frac{1}{t^{n / 2}}\Delta^{1 / 2}\exp \left(-\frac{\sigma}{2 t}\right)\Big(\partial_t \Xi\Big)$$
while the second term
$$DK=-\frac{1}{t^{n/2}}g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\bigg(\Big(\Delta^{1 / 2}\Xi\Big)\exp \left(-\frac{\sigma}{2 t}\right) \bigg)$$
$$=-\frac{g^{\mu\nu}}{t^{n/2}}\Bigg(\exp \left(-\frac{\sigma}{2 t}\right)\nabla_{\mu}\nabla_{\nu}\Big(\Delta^{1 / 2}\Xi\Big)+2\nabla_{\mu}\bigg[\exp \left(-\frac{\sigma}{2 t}\right)\bigg]\nabla_{\nu}\Big(\Delta^{1 / 2}\Xi\Big)+\Big(\Delta^{1 / 2}\Xi\Big)\nabla_{\mu}\nabla_{\nu}\exp \left(-\frac{\sigma}{2 t}\right)\Bigg)$$
$$=\frac{1}{t^{n/2}}\Bigg(\exp \left(-\frac{\sigma}{2 t}\right)D\Big(\Delta^{1 / 2}\Xi\Big)-
2\bigg[-\frac{1}{2t}\exp \left(-\frac{\sigma}{2 t}\right)\nabla_{\mu}\sigma\bigg]\nabla^{\mu}\Big(\Delta^{1 / 2}\Xi\Big)+\Big(\Delta^{1 / 2}\Xi\Big)D\exp \left(-\frac{\sigma}{2 t}\right)\Bigg)$$
$$=\frac{1}{t^{n/2}}\Bigg(\exp \left(-\frac{\sigma}{2 t}\right)D\Big(\Delta^{1 / 2}\Xi\Big)+\frac{1}{t}\bigg[\exp \left(-\frac{\sigma}{2 t}\right)\nabla_{\mu}\sigma\bigg]\nabla^{\mu}\Big(\Delta^{1 / 2}\Xi\Big)-\frac{1}{2t}\Big(\Delta^{1 / 2}\Xi\Big)\exp \left(-\frac{\sigma}{2 t}\right)D\sigma\Bigg)$$
The second term can be expanded using $\nabla^{\mu}\Big(\Delta^{1 / 2}\Xi\Big)=\Xi\nabla^{\mu}\Delta^{1 / 2}+\Delta^{1 / 2}\nabla^{\mu}\Xi$. Now collecting all terms and cancelling $\frac{1}{t^{n/2}}\exp(-\sigma/2t)$ we get
$$-\frac{n}{2t}\Delta^{1/2}\Xi+\frac{\sigma}{2t^2}\Delta^{1/2}\Xi+\Delta^{1/2}\partial_t\Xi+D\Big(\Delta^{1/2}\Xi\Big)+t^{-1}\nabla_{\mu}\sigma\Big(\Xi\nabla^{\mu}\Delta^{1 / 2}+\Delta^{1 / 2}\nabla^{\mu}\Xi\Big)-\frac{1}{2t}\Big(\Delta^{1/2}\Xi\Big)D\sigma=0$$
multiplying both side by $\Delta^{-1/2}$  we get
$$-\frac{n}{2t}\Xi+\frac{\sigma}{2t^2}\Xi+\partial_t\Xi+\Delta^{-1/2}D\Big(\Delta^{1/2}\Xi\Big)+t^{-1}\nabla_{\mu}\sigma\Big(\Xi\Delta^{-1/2}\nabla^{\mu}\Delta^{1 / 2}+\nabla^{\mu}\Xi\Big)-\frac{1}{2t}\Big(\Xi\Big)D\sigma=0$$
$$\implies \partial_t\Xi+\Delta^{-1/2}D\Big(\Delta^{1/2}\Xi\Big)+t^{-1}\nabla_{\mu}\sigma\nabla^{\mu}\Xi+\bigg[-\frac{n}{2t}\Xi+\frac{\sigma}{2t^2}\Xi+t^{-1}\nabla_{\mu}\sigma\Xi\Delta^{-1/2}\nabla^{\mu}\Delta^{1/2}-\frac{1}{2t}\Xi D\sigma\bigg]=0$$
$$\big(\partial_t+t^{-1}(\nabla^{\mu}\sigma)\nabla_{\mu}+\Delta^{-1/2}D\Delta^{1/2}\big)\Xi+
\bigg[-\frac{n}{2t}\Xi+\frac{\sigma}{2t^2}\Xi+\frac{1}{2t}\sigma\Xi\Delta^{-1}\nabla_{\mu}\sigma\nabla^{\mu}\Delta-\frac{1}{2t}\Xi D\sigma\bigg]=0$$
The expression in $[...]$ has to be $0$ (zero), which is something I can't prove, embarrassingly not even for Minkowski space ($\Delta=\eta^{\mu\nu}$ and $\sigma=(x-y)^2/2$). You can check this reference for some identities of $\Delta_{VVM}$ to bash with expression in $[...]$. Since I'm not well familiarized with heat kernels, I can't say with surety but $t$ has to have some relation with $\sigma$ for the second expression to drop.
