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I'd be interested to know why there are exactly two unit normal vectors at each point of a hypersurface in a Riemannian manifold. The book I've been using for this area of study is the one titled Introduction to Riemannian Manifolds and written by John M. Lee. Now let me display a quote from chapter 8 which my question stems from.

Now we specialize the preceding considerations to the case in which $M$ is a hypersuface (i.e. a submanifold of codimension $1$) in $\widetilde{M}$. Throughout this section, our default assumption is that $(M,g)$ is an embedded $n$-dimensional Riemannian submanifold of an $(n+1)$-dimensional Riemannian manifold $(\widetilde{M},\widetilde{g})$.

In this situation, at each point of $M$, there are exactly two unit normal vectors. In terms of any local adapted orthonormal frame $(E_1,\ldots,E_{n+1})$, the two choices are $\pm E_{n+1}$. In a small enough neighborhood of each point of $M$, therefore, we can always choose a smooth unit normal vector field along $M$.

I must confess that I don't have any tiny ideas at hand. I'm sorry. The only thing I know is that at each $p\in M$, the tangent space $T_p\widetilde{M}$ is the direct sum of the space $T_p M$ (can be identified as a subspace of $T_p\widetilde{M}$) and the orthogonal complement $(T_p M)^\perp$. In symbols, $T_p\widetilde{M}=T_p M\oplus(T_p M)^\perp$. I don't think this would help any. The codimension of $M$ seems to play a role in my question, but I don't know how to bring it in. Does anyone have an idea? It would be even better if you told me where I can find relevant information in Lee's IRM. Thank you so much.

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As $M$ has dimension $n$ and $\widetilde{M}$ has dimension $n + 1$, we see that $\dim T_pM = n$ and $\dim T_p\widetilde{M} = n + 1$, so $\dim (T_pM)^{\perp} = 1$; in general, the dimension of $(T_pM)^{\perp}$ is equal to the codimension of $M$ in $\widetilde{M}$. In a one-dimensional normed real vector space $(V, \|\cdot\|)$, there are only two vectors of unit length. To see this, let $v \in V$ be non-zero, so $\|v\| = c > 0$. Set $w = \frac{1}{c}v$, and note that $\|w\| = 1$. As $V$ is one-dimensional, every vector is of the form $\lambda w$ for some $\lambda \in \mathbb{R}$. As $\|\lambda w\| = |\lambda|\|w\| = |\lambda|$, we see that there are only two vectors with unit length, namely $w$ and $-w$.

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  • $\begingroup$ May I assume you are employing the fact that $\dim(T_p\widetilde{M})=\dim(T_p M)+\dim((T_p M)^\perp)$ from linear algebra? $\endgroup$
    – Boar
    Commented Jan 31, 2022 at 15:34
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    $\begingroup$ That's correct. $\endgroup$ Commented Jan 31, 2022 at 15:42
  • $\begingroup$ Thank you, but I have one more question. Did you equip $(T_p M)^\perp$ with the inner product induced by $\widetilde{g}_p$ (the inner product on $T_p\widetilde{M}$)? $\endgroup$
    – Boar
    Commented Jan 31, 2022 at 15:59
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    $\begingroup$ Yes, that's correct. $\endgroup$ Commented Jan 31, 2022 at 18:16
  • $\begingroup$ Sorry, I was wrong. It is not necessary to equip $(T_p M)^\perp$ with an inner product. In fact, all we have to do is work in the inner product space $T_p\widetilde{M}$. $\endgroup$
    – Boar
    Commented Feb 1, 2022 at 4:23

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