Let $G$ be a real matrix group of dimension $n$. Let $\mathfrak{g}$ denote the lie algebra of $G$. Suppose $X \subset \mathfrak{g}$ such that $e^X$ is a submanifold of $G$ of dimension $m$.
Does this imply that $X$ is a subspace of $\mathfrak{g}$ of dimension at least $m$?
Edit: Replace "subspace" by "submanifold".
Thanks!
No. Let $G$ be any connected compact group. Then $\exp: \mathfrak g \to G$ is onto. This is a nontrivial statement, but follows from the following argument. Choose a bi-invariant metric on $G$ (choose any metric, and average for both the left and right actions of $G$); its derivative at the origin is a $G$-invariant metric on $\mathfrak g$, which you can extend to a translation-invariant metric. Then $\exp$ is distance-preserving when restricted to any line through the origin of $\mathfrak g$. It follows from the compactness of $G$ that you can choose $r\in \mathbb R_{>0}$ such that, if $B_r$ denotes the ball centered at $0\in \mathfrak g$ of radius $r$, then $\exp : B_r \to G$ is onto.
So let $X = B_r \cup Y$, where $Y \subseteq \mathfrak g \smallsetminus B_r$ is some randoms subset, which is not a manifold. Then $\exp(X) = G$ is definitely a submanifold of $G$, even though $X$ is not a manifold. If you want an example where $\exp(X)$ is a proper submanifold, just embed $G$ into a larger compact group, and repeat the argument.
You could also ask the following question. Suppose that $M \looparrowright G$ is an immersed submanifold (or, perhaps, restrict to embedded submanifolds). Is the full preimage $\exp^{-1}(M)$ a submanifold of $\mathfrak g$? If $\exp$ were a submersion, the answer would be yes, since pullbacks along submersions always exist in manifolds. Unfortunately, $\exp$ is not a submersion.
To see this, consider $G = \mathrm{SU}(2) \cong S^3$, with Lie algebra $\mathfrak{su}(2) = \mathbb R^3$ with the cross product. Let $S^2_r \subset \mathbb R^3$ denote the sphere of radius $r$; then $\exp^{-1}(1)$ is a disjoint union of spheres of radius $2\pi n$ for $n \in \mathbb N$. Draw a curve through the origin. Its preimage will look like a curve unioned with a bunch of 2-spheres, and so jumps in dimension.