Existence of a relatively flat surface of $\mathbb{S}^3$

Question

Consider $\mathbb{S}^3$ with the standard round metric.

Is there an embedded surface $S \subseteq \mathbb{S}^3$ with the following property:

$R^{\mathbb{S}^3}(X,Y)=0$ for every two tangent vector $X,Y \in TS$.

That is, I require $R^{\mathbb{S}^3}(X,Y)Z=0$ for all $X,Y \in TS$ and $Z \in T\mathbb{S}^3$.

Here $R^{\mathbb{S}^3}$ is the Riemann curvature tensor of $\mathbb{S}^3$, not the curvature tensor of the induced Riemannian manifold $S$.

Is this condition related to other geometric properties like $S$ being a totally geodesic submanifold or a minimal submanifold?

Answer 1

The answer is negative, even if we only require that $R^{\mathbb{S}^3}(X,Y)Z=0$ for all $X,Y \in TS$ and $Z \in TS$. The standard metric on $\mathbb{S}^3$ has non-zero constant sectional curvature, and so $\langle R^{\mathbb{S}^3}(X,Y)X,Y \rangle \neq 0$.