# Every tangent vector corresponds to a line segment

For the purposes of this question, I'll take a tangent vector on a smooth manifold $M \subseteq \mathbb{R}^n$ to be defined as an equivalence class of curves $c : (-\varepsilon, \varepsilon) \to M$ with the same $c(0)$ and $c'(0)$.

Some (embeddings of) manifolds, e.g. ${\rm GL}_n(\mathbb{R}) \subseteq \mathbb{R}^{n^2}$, have the property that every tangent vector corresponds to a linear map $c(t) = m + t v$, whereas most, e.g. $S^2 \subseteq \mathbb{R}^3$, do not.

I feel like I'm missing something obvious, but is there a name for this property?

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## 2 Answers

The property is "being an open subset of an affine subspace of $\mathbb R^n$". I don't think there is a better name for it.

Proof. Let's identify every tangent space $T_pM$ with a linear subspace of $\mathbb R^n$. The property you described implies that for every $p\in M$ the manifold contains a neighborhood of $p$ in the affine subspace $p+T_pM$. Hence, $T_qM=T_pM$ for all points $q$ in this neighborhood. Therefore, the set $\{q\in M: T_qM=T_pM\}$ is open. The connectedness of $M$ implies that this set is all of $M$.

Every linear functional $f:\mathbb R^n\to\mathbb R$ that vanishes on $T_pM$ has zero gradient everywhere on $M$, hence is constant. This implies $M\subseteq p+T_pM$. Since the dimension of $T_pM$ is equal to the dimension of $M$, the latter must be an open subset of $p+T_pM$.

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It seems to me that these manifolds should just be an open set intersect a union of hyperplanes.

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