For $\varepsilon > 0$, we say that a closed, connected and oriented immersed hypersurface $M^n$ of a riemannian manifold $(N^{n+1},g)$ is $\varepsilon$-convex whenever all principal curvatures of $M$ have the same sign and are bigger than $\varepsilon$ in absolute value.


A theorem by Eschenburg states that if $M$ is $\varepsilon$-convex and $N$ has nonnegative sectional curvature, then $M$ is the boundary of a convex body in $N$; in particular, $M$ is diffeomorphic to an $n$-dimensional sphere. 


Let now $N = \mathbb{S}^{n+1}$ with the round metric. My question is:



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> Does there exist an immersed $n$-sphere $M$ inside $\mathbb{S}^{n+1}$ such that all principal curvatures of $M$ are everywhere nonzero and two of them have different signs?
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> 
> 


My idea was to somehow deform a hypersphere to have the above properties, but I don’t know if it is possible.



Special spheres: principal curvatures with different signs
Most of Cartan's isoparametric hypersurfaces in $S^4$ have your desired properties: They have constant principal curvatures (in fact, they are homogeneous), nearly all of them have all three principal curvatures nonzero, and the ones near the minimal one (i.e., the one of maximal volume) have principal curvatures of two different signs. Moreover, since they are orbits of $\mathrm{SO}(3)$ acting irreducibly on $\mathbb{R}^5$ with finite stabilizer, they are finite quotients of the simply-connected cover of $\mathrm{SO}(3)$, which is the $3$-sphere.


Here is a bit more detail (all of this is drawn from É. Cartan's original work on isoparametric hypersurfaces): Let $V$ be the space of $3$-by-$3$ traceless symmetric matrices over the reals, a vector space of real dimension $5$. Let $A\in \mathrm{SO}(3)$ act on $V$ by $A\cdot m = AmA^{-1}$. This is an irreducible representation of $\mathrm{SO}(3)$, and the $\mathrm{SO}(3)$-action leaves invariant the (positive definite) quadratic form
$$
Q(m) = \tfrac16\mathrm{tr}(m^2)
$$
as well as the cubic form
$$
C(m) = \tfrac12\det(m).
$$
Let $S^4\subset V$ be the unit $4$-sphere, defined by $\mathrm{tr}(m^2)=6$. Since every $m\in V$ can be diagonalized by an element of $\mathrm{SO}(3)$, one easily sees that $-1\le C(m)\le 1$ for all $m\in S^4$, with equality at either end happening exactly for the matrices $m\in V$ with a double eigenvalue. The Cartan isoparametric hypersurfaces are the level sets $C(m) = r$ for $|r|<1$. They are clearly $\mathrm{SO}(3)$-orbits, since the only invariants of a symmetric matrix under the $\mathrm{SO}(3)$-action are its eigenvalues, which are completedly determined by the values of $Q(m)$ and $C(m)$ (since $\mathrm{tr}(m)=0$). 


It is easy to show that the level set $C(m)=0$ is a minimal hypersurface, with one of its principal curvatures (necessarily constant) equal to $0$ and the other two of opposite sign. Meanwhile, as Cartan shows, the level sets $C(m)=\cos 3\theta$ for $0<\theta<\pi/6$ have three nonzero principal curvatures (necessarily constant) given by
$$
\cot\left(\theta-\tfrac\pi3\right),\quad \cot\left(\theta\right),\quad \cot\left(\theta+\tfrac\pi3\right).
$$
(The first is negative and the other two are positive.)


Since each such orbit is diffeomorphic to $\mathrm{SO}(3)/D$, where $D$ is the finite group of order $4$ consisting of the diagonal matrices, and since $\mathrm{SO}(3)$ is, itself, double-covered by the $3$-sphere, it follows that the simply-connected cover of each such orbit is $8$-fold and is diffeomorphic to the $3$-sphere. (The two extreme orbits, $C(m)=\pm 1$, are copies of $\mathbb{RP}^2$.)


