I have a question to the proof of the Liebmann-Süss theorem which is stated in my book as follows:
Assume $x:M^n\rightarrow\mathbb{R}^{n+1}$ is a compact and connected oriented hypersurface, if $M$ has constant mean curvature with a suitable selected origin and given a support function $S_f$ on $M$ with fixed sign (positive or negative), then $M$ surely is a $n$-dimensional Euclidean sphere; the inverse is evident.
Setting: \begin{align*} dx &= \sum_{i} \omega^{i} e_{i}\\ de_{i}&= \sum_{j}\omega^{j}_{i}e_{j}+\sum_{j,\alpha}h^{\alpha}_{ij}\omega^{j}e_{\alpha}\\ de_{\alpha}&= -\sum_{j,i}h^{\alpha}_{ij}\omega^{j}e_{i}+\sum_{\beta}\omega^{\beta}_{\alpha}e_{\beta} \end{align*} $$1\leq i,j...\leq n, \alpha,\beta=n+1$$ with $\omega^{i}$ being a one-form and $\{e_i\}$ an orthonormal frame and $\omega^\alpha_i=\sum_j h^\alpha_{ij}\omega^j$. Since $\frac{1}{n}\sum_i h_{ii}$ is mentioned in the proof to be the mean curvature, I assumed $h$ is the second fundamental form.
In the proof we use $S_i=\langle x, e_{i}\rangle$ and $S_f=S_{n+1}$. We calculate the divergence of a vector field on the manifold $M$ described by $\sum_i u_ie_i$ where $u_i=\sum_ih_{ij}S_j$ $$\sum_j u_{ij}\omega^j := du_i-\sum_ju_j\omega^j_i $$ and come to the following conclusion:
And thus $$\sum_i u_{ii}=\sum_{k,i}S_k h_{iki}+\sum_ih_{ii}+S_f\sum_{i,j}(h_{ij})^2.$$ Since $h_{ijk}-h_{ikj}=0$ and with $\sum_ih_{ii}=const.$ follows $$\sum_i u_{ii}=\sum_ih_{ii}+S_f(\sum_{i,j}h_{ij}^2) .$$
I don't get the last conclusion. As well as $h_{ijk}$.
About $h_{ijk}$ I only found this equality in my book: $$\sum_k h_{ijk}\omega^k= dh_{ij}-\sum_k h_{ik}\omega^k_j-\sum_k h_{kj}\omega^k_i + \sum h^\beta_{ij}\omega^\alpha_\beta$$ Please note, that in the book the index over which the sum runs is not mentioned in the last sum. I guess it's $\beta$ but all of this is part of my confusion with $h_{ijk}$. Since $\beta$ and $\alpha$ are the same anyway and $\omega^k_k=0$ I assumed the last part vanishes and this fits perfectly in the calculation above.
In fact, I have a hard time understanding this $h_{any~index}$ business.
Can anyone help?
Thank you very much!
The definitions you've given for $u_{ij}$ and $h_{ijk}$ are standard definitions of the covariant derivative of a tensor field. Somewhere in your book, they should have told you that $h_{ijk} = h_{ikj}$ (for all $i,j,k$) is a consequence of the Codazzi-Mainardi equations $$d\omega_i^{n+1} = \sum \omega_i^j\wedge \omega_j^{n+1},$$ as you can check by direct computation. I haven't checked the details of the calculation of the divergence, but it seems that you're asking why the term $\sum S_k h_{iki}$ vanishes. Well, $$\sum_{i,k} S_k h_{iki} = \sum_k S_k \sum_i h_{iki} = \sum_k S_k \sum_i h_{iik} = \sum_k S_k \big(\sum_i h_{ii}\big)_k = 0,$$ since mean curvature is constant.