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?



Extrinsically flat submanifolds of a Riemannian manifold
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 $r>2$.


