Definition of first normal space

Question

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?

Answer 1

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.