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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 $C^\infty$?

Thanks for your help!

1 Answers 1

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Given positive integers $m

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 $S^2$.