I'm studying by myself Mean Curvature Flow and I'm stuck in the proof of the lemma $10.6$. I will put some notations before the lemma:
My doubts are how the terms highlighteds are derived. I couldn't do anything that helps me to derive $(1)$. My attempts for $(2)$, $(3)$ and $(4)$ are the following:
$(2)$:
$\begin{align*} H \overline{g}_{\alpha\beta} \frac{\partial^2 \nu^{\alpha}}{\partial x^i \partial x^j} \nu^{\beta} &\overset{(10.5)}{=} H \overline{g}_{\alpha\beta} \frac{\partial}{\partial x^i} \left( h_{jl} g^{lm} \frac{\partial X^{\alpha}}{\partial x^m} - \overline{\Gamma}_{\rho\sigma}^{\alpha} \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} \right) \nu^{\beta}\\ &= H \overline{g}_{\alpha\beta} \left[ \frac{\partial}{\partial x^i} \left( h_{jl} g^{lm} \right) \frac{\partial X^{\alpha}}{\partial x^m} + h_{jl} g^{lm} \frac{\partial^2 X^{\alpha}}{\partial x^i \partial x^m} - \frac{\partial \overline{\Gamma}_{\rho\sigma}^{\alpha}}{\partial x^i} \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} - \overline{\Gamma}_{\rho\sigma}^{\alpha} \frac{\partial}{\partial x^i} \left( \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} \right) \right] \nu^{\beta}\\ &= H \left( \frac{\partial}{\partial x^i} \left( h_{jl} g^{lm} \right) \frac{\partial X}{\partial x^m}, \nu^{\beta} \right)_{N^{n+1}} + H \overline{g}_{\alpha\beta} h_{jl} g^{lm} \frac{\partial^2 X^{\alpha}}{\partial x^i \partial x^m} \nu^{\beta} - H \overline{g}_{\alpha\beta} \frac{\partial \overline{\Gamma}_{\rho\sigma}^{\alpha}}{\partial x^i} \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} \nu^{\beta} - H \overline{g}_{\alpha\beta} \overline{\Gamma}_{\rho\sigma}^{\alpha} \frac{\partial}{\partial x^i} \left( \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} \right) \nu^{\beta}\\ &= H \overline{g}_{\alpha\beta} h_{jl} g^{lm} \frac{\partial^2 X^{\alpha}}{\partial x^i \partial x^m} \nu^{\beta} - H \overline{g}_{\alpha\beta} \frac{\partial \overline{\Gamma}_{\rho\sigma}^{\alpha}}{\partial x^i} \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} \nu^{\beta} - H \overline{g}_{\alpha\beta} \overline{\Gamma}_{\rho\sigma}^{\alpha} \frac{\partial}{\partial x^i} \left( \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} \right) \nu^{\beta}, \end{align*}$
but I couldn't see why $- H \overline{g}_{\alpha\beta} \overline{\Gamma}_{\rho\sigma}^{\alpha} \frac{\partial}{\partial x^i} \left( \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} \right) \nu^{\beta} = 0$.
$(3)$: It is only replace what I marked as $(2)$ in $(10.7)$ and do a reindexing appropriately.
$(4)$:
$\begin{align*} H \overline{g}_{\alpha\beta} \left( \frac{\partial \overline{\Gamma}_{i\delta}^{\alpha}}{\partial \tilde{y}^{\gamma}} \nu^{\gamma} \frac{\partial X^{\delta}}{\partial x^j} \nu^{\beta} - \frac{\overline{\Gamma}_{\rho\sigma}^{\alpha}}{\partial x^i} \frac{\partial X^{\rho}}{\partial x^j} \nu^{\sigma} \nu^{\beta} \right) &= H \overline{g}_{\alpha\beta} \left( \frac{\partial \overline{\Gamma}_{ij}^{\alpha}}{\partial \tilde{y}^{\gamma}} \nu^{\gamma} - \frac{\overline{\Gamma}_{j\sigma}^{\alpha}}{\partial x^i} \nu^{\sigma} \right) \nu^{\beta}\\ &= H \overline{g}_{\alpha\beta} \left( \frac{\partial \overline{\Gamma}_{ij}^{\alpha}}{\partial \tilde{y}^{\delta}} - \frac{\overline{\Gamma}_{j\delta}^{\alpha}}{\partial \tilde{y}^i} \right) \nu^{\delta} \nu^{\beta}\\ &= H \overline{g}_{\alpha\beta} \left( \frac{\partial \overline{\Gamma}_{ij}^{\alpha}}{\partial \tilde{y}^{\delta}} - \frac{\overline{\Gamma}_{\delta j}^{\alpha}}{\partial \tilde{y}^i} \right) \nu^{\delta} \nu^{\beta}\\ &= H \left( \overline{\nabla}_{\nu} \overline{\nabla}_{e_i} e_j - \overline{\nabla}_{e_i} \overline{\nabla}_{\nu} e_j, \nu \right)_{N^{n+1}}\\ &= H \overline{R}_{0i0j}, \end{align*}$
where I used the definition of tensor curvature given on page $6$ of this lecture notes written by a former Huisken's student.
P.S.: there is a proof of this lemma in the paper Contracting convex hypersurfaces in Riemannian manifolds by their mean curvature by Gerhard Huisken on page $469$, it's the theorem $3.4$.
Thanks in advance!