# A tangent vector to the unit tangent bundle $T_1S^n$ at $(x,\xi)$ can be written $(X,Z)$ with $X \cdot x = 0$ and $Z \cdot \xi = 0$.

Let $T_1S^n$ be the unit tangent bundle, $(x,\xi) \in T_1S^n$. Why can a tangent vector in $T_{(x,\xi)}T_1S^n$ be written $(X,Z)$ with $X \cdot x = 0$ and $Z \cdot \xi = 0$?

This is claimed in Cecil's Lie Sphere Geometry, p. 52 of the second edition.

## 1 Answer

Consider the unit-tangent bundle. I will assume this is the tangent bundle of $S^n$ where each tangent space is reduced to just the vectors of unit-length. So, the fibers of $T_1S^n$ are not vector spaces. A typical point in $T_1S^n$ would be a pair $(s,\xi)$ where $\xi \cdot \xi = 1$ and $s \cdot s = R^2$ if we have $S^n$ with radius $R$ centered at the origin. Often we set $R=1$ by default.

What does the tangent space of the tangent bundle at the particular point $(s,\xi)$ look like; that is, what is a typical element of $T_{(s,\xi)}T_1S^n$ ? It will be a pair $((s,\xi), v)$ where $v$ could be identified with the velocity vector of a curve through $(s,\xi)$. The natural chart at $(s,\xi)$ is formed by taking the Cartesian product of the $S^n$ chart with itself (assume $R=1$). It follows that we have a pair of $\mathbb{R}^{n-1}$ vectors for $v= (v_1,v_2)$ where $v_1$ is the vector component in the $s$-direction and $v_2$ is the vector component in the $\xi$ direction. In summary, a point in $T_{(s,\xi)}T_1S^n$ has the form $$((s,\xi), v) = ((s,\xi), (v_1,v_2))$$ where $v_1$ is tangent to $s$-direction and $v_2$ is tangent to $\xi$ direction. But, this just means $v_1 \cdot s=0$ and $v_2 \cdot \xi = 0$. If you like $\gamma: \mathbb{R} \rightarrow T_1S_n$ with $\gamma (t) = (\gamma^1(t),\gamma^2(t))$ has $\gamma^1(t) \cdot \gamma^1(t)=1$ and $\gamma^2(t) \cdot \gamma^2(t)=1$ where $\gamma^1(0)=s$ and $\gamma^2(0)=\xi$. Differentiate to obtain $2(\gamma^1)'(0) \cdot \gamma^1(0)=0$ or $v_1 \cdot x=0$ if we identify $(\gamma^1)'(0)=v_1$. Likewise, differentiate to obtain $2(\gamma^2)'(0) \cdot \gamma^2(0)=0$ or $v_2 \cdot \xi =0$ if we identify $(\gamma^2)'(0)=v_2$.

Ok so we have shown, a point in $T_{(s,\xi)}T_1S^n$ has the form $$((s,\xi), v) = ((s,\xi), (v_1,v_2))$$ where $v_1 \cdot s=0$ and $v_2 \cdot \xi = 0$. If we set-aside the point of attachment we simply have vector part $(v_1,v_2)$ with $v_1 \cdot s=0$ and $v_2 \cdot \xi = 0$. This is what you wished to be shown.