Given an immersed submanifold $M$ of a Riemannian manifold $\overline{M}$, the *first normal space* of $M$ at a point $p \in M$ is defined as the linear subspace $N\_{p}^{1}M$ of $N\_{p}M$ spanned by the image of the second fundamental form $\alpha$ at $p$:


$$ N\_{p}^{1}M = \text{span of }\{\alpha(v,w) \in N\_{p}M \mid v,w \in T\_{p}M\}.$$


Equivalently, $N\_{p}^{1}M$ is the orthogonal complement in $N\_{p}M$ of the linear subspace of $N\_{p}M$ consisting of all normal vectors $\xi$ for which the shape operator $A\_{\xi}$ vanishes:


$$N\_{p}^{1}M = N\_{p}M \ominus \{\xi \in N\_{p}M \mid A\_{\xi}=0 \}.$$


The equivalence of these definition is claimed for example in the book "Submanifolds and Holonomy" by Berndt, Console and Olmos.


Here is my confusion: Take a smooth regular curve $\gamma$ in $\mathbb{R}^{N}$. According to the first definition, $N\_{p}^{1}\gamma$ should be the (one-dimensional) span of the acceleration vector $\overline{D}\_{t}\dot{\gamma}$. However, I cannot see why the set $ \{\xi \in N\_{p}\gamma \mid A\_{\xi}=0 \}$ should in general be $(N-1)$-dimensional. Any suggestion? 



Definition of first normal space
OK I got confused really bad yesterday. The set $ \{\xi \in N\_{t}\gamma \mid A\_{\xi}=0 \}$ is indeed $(N-1)$-dimensional. Without loss of generality, assume $\gamma$ be unit-speed, and $\overline{D}\_{t}\dot{\gamma}$ never zero. Then $\overline{D}\_{t}\dot{\gamma}(t)$ is a non-zero vector in the normal space $N\_{t}\gamma$ of $\gamma$ at $t$. Choose a smooth orthonormal frame $(\xi\_{1}, \dotsc, \xi\_{N-1})$ for the normal bundle $N\gamma$ such that $\xi\_{1} = \overline{D}\_{t}\dot{\gamma}$. Then $\overline{D}\_{t}\xi\_{k} \cdot \dot{\gamma} = 0$ for all $k = 2, \dotsc, N-1$, and the claim follows. 


