Proof that $S^2$ is not parallelizable I'm trying work out a proof from Arnold's Ordinary Differential Equations that the sphere $S^2$ is not parallelizable.
First, he sets as an exercise (page 298) to determine which of the manifolds below are equivalent by diffeomorphism:

*

*$SO(3)$

*$\mathbb{R}P^3$

*$T_{1}S^2$, the unit tangent vectors to a sphere embedded in $\mathbb{R}^3$

*$M = \{z_1, z_2, z_3 \in \mathbb{C} : z_1^2 + z_2^2 + z_3^2 = 0, |z_1|^2 + |z_2|^2 + |z_3|^2 = 2\}$, viewed as a submanifold of $\mathbb{R}^6$
I understand that $SO(3) \cong\mathbb{R}P^3.$
Also, I found that $M \cong T_{1}S^2$ by embedding $T_{1}S^2$ in $\mathbb{R}^6$ by identifying a tangent vector with its tangent point $(x_1, x_2, x_3)$ and its direction $(t_1, t_2, t_3)$, setting $z_k = x_k + it_k$, and showing that the conditions that $x$ and $t$ are orthogonal unit vectors correspond to the complex equations.
He then gives a further hint that $\mathbb{R}P^3 \not \cong S^{1} \times S^{2}$. I looked for a proof of this and found this question, but the answer doesn't show that no diffeomorphism exists (it shows that such a diffeomorphism would contradict that $S^2$ is not parallelizable, which is what we are trying to prove). How can I prove that $\mathbb{R}P^3 \not \cong S^{1} \times S^{2}$, hopefully using the manifold $M$?
I think that $T_{1}S^2 \cong S^{1} \times S^{2}$ since the unit tangent vectors at a point can be identified with $S^1$. Then $TS^2 = \mathbb{R} \times S^{1} \times S^{2} \not \cong \mathbb{R^2} \times S^2$, which would conclude the argument.
 A: There's a mistake in my work: $T_{1}S^2$ is not diffeomorphic to $S^1 \times S^2$ just because the unit tangent vectors at each point are diffeomorphic to $S^1$ (it's the same fallacy as thinking that all manifolds are parallelizable because the tangent space at each point is diffeomorphic to a Euclidean space). All four spaces in the question are diffeomorphic, since the first two columns of a matrix in $SO(3)$ can be used to give an embedding into $M$ in $\mathbb{R}^6$, as in Andrew Hwang's comment.
What is true is that if there were a smooth non-vanishing vector field on $S^2$, then normalizing the vectors gives a field of unit tangent vectors $x(p)$ and embedding in $\mathbb{R}^3$ and defining $y(p) = p \times x(p)$ would give a smooth frame $(x, y)$, so $S^2$ would be parallelizable. Then $TS^{2} \cong \mathbb{R}^2 \times S^2$, and the diffeomorphism restricts to $T_1 S^{2} \cong S^1 \times S^2$.
The author (probably) intended $\mathbb{R}P^3 \not \cong S^1 \times S^2$ to be taken on faith to answer the question. As in the comments, the fundamental group of $\mathbb{R}P^3$ is $\mathbb{Z}_2$ and the fundamental group of $S^1 \times S^2$ is $\mathbb{Z}$, so they are not homeomorphic. Hence $T_1 S^{2} \cong \mathbb{R}P^3 \not \cong S^1 \times S^2$, so there cannot be a continuous non-vanishing vector field on $S^2$.
