Tangent space on the sphere

Question

Let $S^m$ be the $m$-sphere in $\mathbf{R}^{m + 1}$. Let $p \in S^m$. I want to show that $TS^m_p = \{v \in \mathbf{R}^{m +1} : \langle v, p \rangle = 0 \}$.

That is, I want to show that for a local parametrization $f : U \to f(U) \subset M$ with $f(x) = p$ we have $df_x(\mathbf{R}^{m}) = \{v \in \mathbf{R}^{m + 1} : \langle v, p \rangle = 0 \}$. How can I do this?

I had an idea for the "$\subset$" direction: I need to show that $\langle df_x(e_i), f(x) \rangle = 0$ for all $i$, that is $\langle \partial_{i} f(x), f(x) \rangle = \frac{1}{2} \partial_i \langle f(x), f(x) \rangle = 0$, hence I need to show that $\langle f(x), f(x) \rangle$ is constant for $x \in U$. But why is that the case?

Thanks for any help!

Answer 1

You are almost there for $\subset$ since $f(x)\in S^m$, $\langle f(x),f(x)\rangle =1$ so it is constant, and for the other direction, use the fact that the dimensions of the image $df_x$ and $TS^m_p$ are equal.

Answer 2

I found a second solution using the following theorem:

Let $M \subset \mathbf{R}^n$ be a sub manifold, $p \in M$, $X \in TM_p$. Then there exists a Geodesic $c : I \to M$, $0 \in I$, with $c(0) = p$, $\dot{c}(0) = X$.

The theorem follows readily from Picard-Lindelöf. We don't really need a geodesic, but assuming this we get a nice differential equation, whose solution we know to exist.

Now taking $p \in S^m$, $X \in TS^m_p$, we get such a geodesic $c$. In particular $\langle c,c \rangle = 1$, and hence $\langle c(t), \dot{c}(t) \rangle = 0$, giving $\langle p, v \rangle = 0$ at $t = 0$.