Non existence of a non singular vector field on $S^2$ Prove that the unit tangent bundle of $S^2$, $T^1 S^2$, is not diffeomorphic to $S^2×S^1$ by showing that if so there exists a nowhere vanishing vector field on $S^2$
I do not know how to create that vector field.
Thank you
 A: First of all, it is clear that $p:T^1S^2\rightarrow S^2$ and $q:S^2\times S^1\rightarrow S^2$, as bundle, are not isomorphic. Indeed, if so, it would exist two diffeomorphism $f,g$ so that the following diagram :
$$\begin{array}{ccc}T^1S^2 & \underset{f}\rightarrow & S^2\times S^1 \\ \downarrow & \circlearrowright & \downarrow \\ S^2 & \underset{g}\rightarrow & S^2\end{array}$$
is commutative.
Since the map $s:S^2\rightarrow S^2\times S^1, x\mapsto (x,1)$ is a smooth section of $q$ i.e. $q\circ s=id$ and the diagram commutes, the map $s'=f^{-1}\circ s\circ g:S^2\rightarrow T^1S^2$ has to be a smooth section of $p$. So, we found a nonvanishing global vector field of the $2$-sphere, which is impossible.
Now, as manifold, $T^1S^2$ is not diffeomorphic to $S^2\times S^1$. You can see that here, you can't use the above trick because a section $s:S^2\rightarrow S^2\times S^1$ has no reason to give a section of $T^1S^2$ via a diffeomorphism from $T^1S^2$ to $S^2\times S^1$. 
Here is a easy way to understand why $T^1S^2$ is not diffeomorphic to $S^2\times S^1$. Actually, you will see that they are not even homeomorphic.
Let us see $T^1S^2$ as $$T^1S^2=\{(x,v)\in \mathbb R^3\times \mathbb R^3 \ | \ \|x\|=\|v\|=1, \langle x,v\rangle=0\}.$$
So, a point in $T^1S^2$ is a pair of unitary and orthogonal vectors of $\mathbb R^3$. Then, the map $$\begin{array}{rcl}T^1S^2 & \longrightarrow & SO(3) \\ (x,v) & \longmapsto & \text{Matrix}(x,v,x\times v) \end{array}$$
is well defined, continuous and bijective. It gives us a homeomorphism (or diffeomorphism). Hence if $T^1S^2$ is diffeomorphic or homeomorphic to $S^2\times S^1$ then so is $SO(3)$.
But their fundamental group doesn't coincide :  


*

*$\pi_1(S^2\times S^1)\simeq \pi_1(S^1)=\mathbb Z$,

*$\pi_1(SO(3))=\mathbb Z/2\mathbb Z$.


Actually, $SO(3)$ is homeomorphic to $\mathbb RP^3$. You can prove it using quaternions for example or Lie group theory on $SU(2)$ or even directly : 
Let $f$ be the map given by $$\begin{array}{rcl} D^3 & \longrightarrow & SO(3) \\  x & \longmapsto & \left\{\begin{array}{cc} \text{Rot}(x,\theta(x)) & \text{ if } x\neq 0 \\ I_3 & \text{ if } x=0 \end{array} \right.\end{array}$$
where 


*

*$D^3$ is the unit closed ball in $\mathbb R^3$

*$I_3$ is the identity matrix,

*$\text{Rot}(x,\theta(x))$ is the rotation of $\mathbb R^3$ whose axis is $\mathbb Rx$ oriented by $x$ and angle is $\|x\|\pi$.


This map is continuous, onto but no injective: $f(x)=f(-x)$ on the boundary of $D^3$. So it induces a continuous bijective map $\bar f:\mathbb RP^3\rightarrow SO(3)$ hence a homeomorphism.
Notice that since $\mathbb RP^3\simeq S^3/{\pm 1}$, the induced map $S^3\rightarrow SO(3)$ is a universal two-sheeted covering map and you get back $\pi_1(SO(3))$ is $\mathbb Z/2\mathbb Z$ thus different from $\mathbb Z$.
Edit: Here is an explanation of your last question : 

Show that if there exists a non vanishing vector field on $S^2$ then
  $T^1S^2$ is diffeomorphic to $S^1\times S^1.$

Let $s:S^2\rightarrow TS^2$ be such a vector field. Since it does not vanish, one can assume that for any $p\in S^2$, $\|s(p)\|=1$. Now, following the same trick as before, $s(p)$ is orthogonal to $p$ so $n(p)=p\times s(p)$ is also a smooth unitary vector field on $S^2$ and for any $p\in S^2$, $(s(p),n(p))$ is an orthonormal basis of $T_pS^2$.
Let $\phi:S^2\times \mathbb R^2\rightarrow TS^2$  be the map given by $\phi(p,(x,y))=xs(p)+yn(p)$.
You can easily check that $\phi$ is injective, smooth and onto. Furthermore, using the fact that $(s(p),n(p))$ is an orthonormal basis of $T_pS^2$, $\phi$ is actually a diffeomorphism. (This is a classical fact that any vector bundle of rank $n$ is trivial iff it has $n$ linearly independent global sections.) 
Finally, $$\begin{align*}\phi^{-1}(T^1S^2) & =\{(p,(x,y))\in S^2\times \mathbb R^2 \ | \ \|xs(p)+yn(p)\|=1\} \\ & = \{(p,(x,y))\in S^2\times \mathbb R^2 \ | \ \|xs(p)+yn(p)\|^2=1\} \\ & = \{(p,(x,y))\in S^2\times \mathbb R^2 \ | \ x^2\|s(p)\|^2+y^2\|n(p)\|^2=1\} \\ & = \{(p,(x,y))\in S^2\times \mathbb R^2 \ | \ x^2+y^2=1\} \\ & = S^2\times S^1.\end{align*}$$
So if there exists a non vanishing vector field of $S^2$, then $TS^2$ has to be diffeomorphic to $S^2\times \mathbb R^2$ and $T^1S^2$ to $S^2\times S^1$.
A: Just a supplement to @Bebop's great answer: We can also compute $\pi_1(T^1S^2)\cong \mathbb{Z}/2\mathbb{Z}$ using van Kampen's theorem. The closed hemispheres $U$ and  $L$ of $S^2$ allow us to break $T^1 S^2$ into circle bundles $T^1 U \sqcup_\phi T^1 L$, glued along some attaching map $\phi$. Since $T^1 U$ and $T^1 L$ are conveniently isomorphic to solid tori $S^1 \times D^2$ glued along their boundary tori $S^1 \times S^1$, we can analyze the attaching map and fundamental group in the same way we would analyze a lens space.  By inspection and van Kampen's theorem (or the fact that $T^1 S^2$ and $L(2,1)$ are homeomorphic)  we have $\pi_1(T^1 S^2) \cong\mathbb{Z}/2\mathbb{Z}$. 
