3
$\begingroup$

I'm reading the paper Classification of marginally trapped Lorentzian flat surfaces in $\mathbb{E}^4_2$ and its applications to biharmonic surfaces by B. Y. Chen.

Summarizing it quickly: he first proves a rigidity theorem for marginally trapped flat Lorentzian surfaces in $\Bbb R^4_2$, and then uses it to give an analogous result for marginally trapped bi-harmonic surfaces in $\Bbb R^4_2$. More precisely, the proof of Theorem 5.1 starts by proving that bi-harmonic surfaces are flat, and then calling the previous result.

However, his proof confuses me a little. He takes pseudo-orthonormal frames $(e_1,e_2)$ and $(e_3,e_4)$, tangent and normal to the surface, with all of them lightlike and $\langle e_1,e_2\rangle = \langle e_3,e_4\rangle = -1$, ok. But he assumes from the start that $\nabla_{e_i}e_j = 0$ for $i=1,2$.

Most likely I'm missing something very simple, but as far as I knew, that last condition can only be assumed once we know that the surface is flat. Without this, I cannot conclude equation (5.11) as he does. Also, it appears he has some typos in (5.9), and it wouldn't make any sense to introduce notations $\omega$, $\omega_1$ and $\omega_2$, since they all would be zero from the get go.

The proof seems circular. Help?

Of course, I'm also accepting alternative direct proofs that marginally trapped bi-harmonic surfaces in $\Bbb R^4_2$ are flat.

There seems to be a serious flaw in the proof, but I do not know how to fix it. Equations (5.9) and (5.10) in the paper should actually be $$\begin{align} e_1\gamma &= \phi_2 - \gamma\phi_1 - 2\gamma\omega_1 \\ e_2\alpha &= \phi_1-\alpha\phi_2 + 2\alpha\omega_2 \\ e_1\delta &= \delta\phi_1 - 2\delta\omega_1 \\ e_2\beta &= \beta\phi_2+2\beta\omega_2.\end{align}$$I'll give a proof on request. These reduce to (5.9) and (5.10) if $\omega = 0$, which is not supposed to be an assumption anymore. If $\delta \neq 0$ and $\beta \neq 0$, the argument the author gives using ${\rm trace}\; A_{DH} = 0$ can be fixed to give $DH = 0$ and hence $K = 0$.

However, the equation $\triangle^D H = h(e_1,A_He_2) + h(e_2,A_He_1)$ actually implies that $\delta\beta = 0$ (looking at the $e_4$-components in both sides) in all cases.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ From a quick glance I agree with your assessment. Does the proof actually use the assumption $\nabla_{e_i} e_j = 0$? Since the argument deals with nonzero $\omega$ terms, perhaps the inclusion of this assumption is just a mistake? $\endgroup$
    – Anthony Carapetis
    Feb 14, 2018 at 2:25
  • $\begingroup$ I'm exchanging emails with the author. He said that $\nabla_{e_i}e_j=0$ for $i=1,2$ is a typo that should be removed. It also appears to have another mistake in equations (5.9) and (5.10), which'll make $\omega_1$ and $\omega_2$ appear in the expressions for $DH$. Particularly, I can't see how ${\rm trace}(A_{DH})=0$ implies $DH=0$ anymore. I sent him a quick calculation trying to fix equations (5.9) and (5.10) a couple of hours ago. Let's see if we can come up with something... $\endgroup$
    – Ivo Terek
    Feb 14, 2018 at 2:31
  • $\begingroup$ (That is, despite that assumption being a typo, it appears that he actually used it. My "new" equations (5.9) and (5.10) reduce to his, in that case.) $\endgroup$
    – Ivo Terek
    Feb 14, 2018 at 2:36

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.