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Let $M$ be a Riemannian manifold,and $UTM = \{(p,v) \in TM\mid g(v,v) = 1\}$ be the unit tangent bundle,assume we have proved that is a embedded submanifold in $TM$.Needs to prove that $UTM$ is connected if and only if $M$ is connected (provided that dim of $M$ is greater than 2 as here shows)

One direction is easy that is $p:UTM \to M$ is continuous hence if $UTM$ is connected then $M$ is connected.

I try to prove the other direction,and I hope there is some better idea:

We only need to prove that $UTM$ is path connected if $M$ is since manifold is locally path connected and locally connected.

To do this consider two point $(p,v)$ and $(q,u)$ on $UTM$,needs to find a continuous path.To do this it may be helpful to transport $v$ by some curve $\gamma$ to some other place that still preserved the length 1 that is $d\gamma (v) $ has lengith 1.

But I have no idea how to construct such a curve,I try to do like this,first there is a finite chain of the neighborhood(which is given as local trivialization domain) that connect $p$ and $q$ (by the chain characterization of "connectedness" ) denote them $U_{x_1},...,U_{x_N}$ such that exist some $y_i \in U_{x_i} \cap U_{x_{i+1}} $

Then we construct local pieces of curve from $$p\to y_1\to x_2\to y_2\to....\to x_{N} \to q$$

Where $p^{-1}(U) \cong U\times S^{n-1} $ that is locally looks like $U\times S^{n-1}$ where we can construct the desired curve If $n\ge 2$ since then it's connected.

Is there some better solution?

yi li
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    your solution looks right to me. i don't think it's overcomplicated. – hunter Jun 26 '21 at 10:47
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    This answers the question in a slightly more general setting : https://math.stackexchange.com/a/2499354/117909 – sss89 Jun 26 '21 at 15:14

1 Answers1

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Another idea:

Let $\nabla$ be a metric connection on $TM$ (for example the Levi-Civita connection). For two unit tangent vectors $X_p\in T_pM$, $Y_q\in T_qM$ let $\gamma: [0,1]\to M$ be a smooth curve connecting $p$ and $q$ and define $X(t)$ to be the parallel transport of $X_p$ along $\gamma$ w.r.t. $\nabla$. Since $\nabla$ is a metric connection, parallel transports are isometries, so $X(t)$ is a smooth curve in $UTM$. Since the unit sphere $S_q$ in $T_qM$ is path-connected ($n\geq2$), there is a continous curve in $S_q$ connecting $X(1)$ and $Y(q)$. Concatenating these two curves gives a continuous curve connecting $X_p$ and $Y_q$, so $UTM$ is path-connected and hence connected.

Claire
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