# Why do we have exactly two unit normal vectors at each point of a hypersurface in a Riemannian manifold?

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.

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 $$-w$$.

• 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?
– Boar
Commented Jan 31, 2022 at 15:34
• That's correct. Commented Jan 31, 2022 at 15:42
• 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 $T_p\widetilde{M}$)?
– Boar
Commented Jan 31, 2022 at 15:59
• Yes, that's correct. Commented Jan 31, 2022 at 18:16
• 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 $T_p\widetilde{M}$.
– Boar
Commented Feb 1, 2022 at 4:23