Extrinsically flat submanifolds of a Riemannian manifold Let $Q$ be a $d$-dimensional Riemannian manifold. A submanifold $M$ of $Q$ is said to be extrinsically flat if $R_{M}(X,Y,Z,W) = R_{Q}(X,Y,Z,W)$ for all $X,Y,Z,W \in \mathfrak{X}(M)$, where $R_{M}$ and $R_{Q}$ are the curvature tensors of $M$ and $Q$, respectively.
It is clear that if the rank of the second fundamental form of $M$ is at most one, then by the Gauss equation $M$ is extrinsically flat.
I am wondering about the existence of such type of submanifolds. For instance, the classical axiom of planes of Cartan gives obstructions to the existence of “many” totally geodesic submanifolds, as explained below.
Fix $2 \leq r < d-1$. We say that $Q$ satisfies the axiom of $r$-planes if for every $p \in Q$ and each $r$-dimensional subspace $\varSigma$ of $T_{p}Q$ there exists a totally geodesic submanifold $S$ such that $T_{p}S = \varSigma$.
Theorem. If $Q$ satisfies the axiom of $r$-planes for some $r$, then $Q$ has constant curvature.
Question. In general, given a Riemannian manifold $Q$, are extrinsically flat submanifolds of $Q$ plentiful or rare? Is it possible for a Riemannian manifold $Q$ to satisfy the “axiom of extrinsically flat submanifolds” without necessarily having constant curvature?
 A: If you take the simplest case, in which $Q$ is a $3$-dimensional Riemannian manifold, then there are plenty of extrinsically flat surfaces $M\subset Q$.  In fact, if one chooses a 'generic' curve $\gamma$ in $Q$ and a 'generic' normal vector field $\nu$ along $\gamma$, then there will be a unique 'extinsically flat' surface $M$ containing $\gamma$ that has $\nu$ as its normal along $\gamma$.
In Cartan's language, the extrinsically flat surfaces in $Q$ depend on two functions of one variable.  This does not depend on any knowledge of the curvature of $Q$.
Addendum 1: (7/23/22)  I thought a little more about this and realized that there is a natural conjecture about the existence of 'extrinsically flat' submanifolds.  Here is the statement.
Conjecture:  For each $r\ge2$, let $D_r = r^2(r^2{-}1)/12$ be the rank of the Riemann curvature tensor in dimension $r$.  For a Riemannian manifold $Q$ of dimension $d\ge r + D_r$, the PDE for 'extrinsically flat' submanifolds of dimension $r$ is involutive.  In particular, such local submanifolds are just as 'plentiful' in $Q$ as they are in flat $\mathbb{R}^d$.
The intuition for the Conjecture is this:  'Extrinsic flatness of $M\subset Q$ is the requirement that, at each point $x\in M$, the Riemann curvature tensor of $M$ at $x$ should be equal to the restriction of the Riemann curvature tensor of $Q$ at $x$ to $T_xM$.  This is $D_r$ second order PDE on $M$ as a submanifold of $Q$.  Since the $r$-dimensional submanifolds of $Q$ can be represented locally as graphs of $d{-}r$ functions of $r$ variables, this PDE system will be formally 'determined' if $d{-}r = D_r$ and 'overdetermined' if $d{-}r<D_r$.  If the symbol of the PDE system is 'non-degenerate', in the appropriate sense, when $d{-}r = D_r$, then, at least in the real-analytic category, there will be 'plenty' of local solutions.  In fact one would expect the local generality of the solutions in this case to be $2D_r$ functions of $r{-}1$ variables.  Meanwhile, for a 'generic' $Q$ of dimension $d<r + D_r$, one would expect that it would not have any 'extrinsically flat' submanifolds of dimension $r$.
The above conjecture is easily established for $r=2$.  (The case $(r,d)=(2,3)$ is just the statement in the first paragraph.)  The case $(r,d)=(3,9)$ (note that $9 = 3 + D_3$) seems to work, but I haven't checked all of the details, as the algebra is a little tricky.
Also, note that, for the 'overdetermined' non-existence result mentioned above, it is essential that one assume that the metric on $Q$ be 'generic'.  As Cartan showed, if $d=2r$ and $Q$ has constant sectional curvature $c$, then the PDE for $r$-dimensional submanifolds of constant sectional curvature $c$ is involutive, and $2r < r + D_r$ when $r>2$.
