3
$\begingroup$

I was reading Lee's IRM book, in page 328,there is a local expression for Hessian operator that I can't work it out.

Which shows in the normal coordinate (geodesics coordinate) the Hessian operator for distance function inside this neiborhood has the form:

$$\mathscr{H}_{r}=g^{i j}\left(\partial_{j} \partial_{k} r-\Gamma_{j k}^{m} \partial_{m} r\right) \partial_{i} \otimes d x^{k}$$

I try to prove it as follows but fails, first we can assume $\mathscr{H}_r = \omega^i_k \partial_i\otimes dx^k$,

then $\mathscr{H}_r(\partial_k) = \omega_k^i \partial_i$, so $$\omega^i_k g_{ij} = g(\omega_k^i\partial_i, \partial_j)=g(\mathscr{H}_r(\partial_k),\partial_j) = Hess\ r(\partial_k,\partial_j) = g(\nabla_{\partial_k}\nabla r,\partial_j) = g(\nabla_{\partial_k}((\partial^mr) \partial_m)),\partial_j) \tag{*}$$

where $\partial^mr$ denote the $m$-th component of the $\nabla r$, Now we need to expand the last term using product rule:

$$g(\partial_k(\partial^mr)\partial_m,\partial_j) + g(\partial^mr\nabla_{\partial_k}\partial_m, \partial_j) = \partial_k(\partial^mr)g(\partial_m,\partial_j) + \partial^mr\Gamma^{l}_{km}g(\partial_l,\partial_j) = \partial_k(\partial^mr)g_{mj}+\partial^mr\Gamma^{l}_{km} g_{lj}.$$

So

$$\omega^i_k g_{ij}= \partial_k(\partial^mr)g_{mj}+\partial^mr\Gamma^{l}_{km} g_{lj}$$

So

$$\omega^i_k = g^{ij}(\partial_k(\partial^mr)g_{mj}+\partial^mr\Gamma^{l}_{km} g_{lj})$$

Not the desired expression $$\omega^i_k=g^{i j}\left(\partial_{j} \partial_{k} r-\Gamma_{j k}^{m} \partial_{m} r\right)$$

I have tried many times to prove this identity, I am not sure where went wrong? To get minus sign in the second expression , it seems more reasonable to expand the last terms in (*) using metric compatibility condition. But Still seems not clear, the main problem is $g^{ij}$ and $g_{nm}$ in the expression will cancel out.

$\endgroup$
5
  • 2
    $\begingroup$ I am not sure but isn't $g_{mj}\partial^m r=\partial_jr$ and ...? $\endgroup$
    – C.F.G
    Mar 11, 2022 at 17:43
  • $\begingroup$ @C.F.G oh thank you that's where I was stuck $\endgroup$
    – yi li
    Mar 12, 2022 at 1:51
  • $\begingroup$ Thank you. I also have been thinking about this issue. Can I ask question? In the formula in $(*)$, why the fourth and fifth equalities are true? Where can we find associated proof? $\endgroup$
    – Plantation
    Nov 8, 2023 at 5:18
  • 1
    $\begingroup$ Hello @Plantation, it is a post long time ago, If I remember correctly, Petersen's Riemann geometry book contains more material talking about Hessian, and Proposition 2.2.6. may be is what you want? $\endgroup$
    – yi li
    Nov 8, 2023 at 10:44
  • $\begingroup$ @yi li : Thank you. As you informed, in the Proposition 2.2.6 in the Petersen's book, the fourth equality is derived. And why the fifth equality ( $\nabla r = (\partial^mr) \partial_m ) $ is true? $\endgroup$
    – Plantation
    Nov 9, 2023 at 2:22

1 Answer 1

1
$\begingroup$

For completeness let me write out the solution in detail.

First

$$\omega^i_k g_{ij} = g(\omega_k^i\partial_i, \partial_j)=g(\mathscr{H}_r(\partial_k),\partial_j) = Hess\ r(\partial_k,\partial_j) = g(\nabla_{\partial_k}\nabla r,\partial_j) = g(\nabla_{\partial_k}((\partial^mr) \partial_m)),\partial_j) \tag{*} $$

We apply metric compatible condition in the last term and the relation that $\partial^mr g_{mn}= \partial_nr$ $$\nabla_{\partial_k}g(\partial^mr\partial_m,\partial_j) - g(\partial^mr \partial _m,\nabla_{\partial_k}\partial_j)\\=\partial_j\partial_k r - \partial^mr \Gamma^l_{kj}g_{ml} = \partial_j\partial_k r-\partial_lr\Gamma^l_{kj}$$

therefore

$$\omega^i_k = g^{ij}(\partial_j\partial_k r - \partial_mr \Gamma^m_{kj})$$.

since the first order derivative of $g_{ij}$ vanish at point $p$, so it's $g^{ij}$,then take Taylor expansion we have $g^{ij} = \delta^{ij} + O(r^2)$, since $\Gamma^{m}_{kj}$ is continuous and vanish at $p$ ,we have $\Gamma^m_{kj} = O(r)$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .