We have a $n$-dimensional hypersurface $\Sigma$ embedded in the Euclidean $(n+1)$-space $\mathbb{R}^{n+1}$. We know that $\Sigma$ is compact without boundary, convex (not necessarly strictly convex), and that the $k$-th symmetric polynomial in the principal curvatures is constant.
Can I deduce that $\Sigma$ is a sphere?
Note: in the case $k=1$ (maybe in the case $k=n$?) the answer is "yes".
The answer is yes. Even without convexity, a version of Alexandrov's theorem holds true for symmetric functions of the principal curvatures, i.e. if the $k$-th symmetric polynomial of the principal curvatures of a compact embedded hypersurface in $\mathbb{R}^{n+1}$ is constant, then the surface is a sphere.
See the following references:
A. Ros "Compact hypersurfaces with constant higher order mean curvatures." Rev. Mat. Iberoamericana 3 (1987), no. 3-4, 447–453.
S. Montiel and A. Ros "Compact hypersurfaces: the Alexandrov theorem for higher order mean curvatures." Pitman Monogr. Surveys Pure Appl. Math., 52, Longman Sci. Tech., Harlow, 1991.
