# Tangent bundle $TS^2$

Let

$$TS^2 = \{ (a,b) \in \mathbb{R}^3 \times \mathbb{R}^3: ||a||^2=1 , a\cdot b=0 \}$$

be the tangent bundle of $S^2$.

How do I see that $TS^2$ with the subspace topology of $\mathbb{R}^3 \times \mathbb{R}^3$ is a topological 4-manifold? Secondly, why there is a $C^\infty$ atlas such that the projection $\pi: TS^2 \to S^2$, $(a,b) \mapsto a$ is $C^\infty$?

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Given positive integers $m<n$, the implicit function theorem tells you that the zero set of a function $F:\mathbb{R}^n \to \mathbb{R}^m$ is an $(n-m)$-dimensional submanifold of $\mathbb{R}^n$ if the rank of $DF$ is maximal (i.e., equal to $m$) everywhere on that set. In your case $F: \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^2$ is given by $F(a,b) = (\|a\|^2 -1 , a\cdot b)$, and it is easy to check that $DF$ has rank 2 on the set where $F=0$.
The implicit function theorem also gives you charts whose regularity is the same regularity as that of $F$. Since $F$ is $C^\infty$, your charts are $C^\infty$, too. Since the projection map in the ambient space $\mathbb{R}^3 \times \mathbb{R}^3$ is $C^\infty$, you get the same regularity on your manifold.
If your definition of manifold includes connectedness, it is pretty straightforward to check, too. Any point $(a,b)$ on $TS^2$ can be connected to $(a,0)$ via $t \mapsto (a,tb)$, and the rest follows from path-connectedness of $S^2$.
Is really $TS^2$ deffeomorphic to $\{(a,b)\in {\Bbb R}^3\times {\Bbb R}^3:\|a\|^2=1,\,a⋅b=0\}$? A three dimensional cylinder in ${\Bbb R}^6$!