The tangent space of the sphere from curve velocities

If $v$ is the velocity of a curve $\gamma$ in $S^n$ with $\gamma (0)=p$, then differentiating $\gamma (t)\cdot \gamma (t)=1$ at $0$ gives $p\cdot v=0$.

Conversely, if $p\cdot v=0$, the normalized curve $\gamma (t):=(p+tv)\mathrm{/}\mathrm{\parallel }p+tv\mathrm{\parallel }$ lies in $S^n$, satisfies $\gamma (0)=p$, and has velocity $v$ at $0$.

Contact-equivalent curves have the same coordinate, hence ambient, velocity, so the displayed map is well defined. Conversely, choose an index $k$ with $p_k\mathrm{\ne }0$ and the standard sphere chart on the hemisphere where the sign of the $k$th coordinate is fixed, obtained by deleting that coordinate. If two curves have the same ambient derivative, their derivatives after this coordinate deletion agree, so they are contact equivalent. The ambient-derivative map is therefore injective. Steps 1.1-1.2 show that its image is exactly $p^{\perp }$, and [L1] identifies its domain with the derivation tangent space.