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How can one find three linearly independent vector fields on $S^1\times S^2$? I know that $S^1\times S^2 \cong SO_3( \mathbb{R})$, i.e. the set of orthogonal $3 \times 3$ matrices with determinant $1$, which is a Lie group and thus parallelizable. I am, however, interested in an explicit form for three vector fields.

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    $\begingroup$ $S^1\times S^2$ is not isomorphic to $SO(3)$. One way to see this is the fundamental group of the former is $\mathbb Z$ while the fundamental group of $SO(3)$ is $\mathbb Z_2$. Maybe you thought this because there is a fibration $S^1 \to SO(3) \to S^2$. Nonetheless, $S^1\times S^2$ is parallelizable since all orientable 3-manifolds are. $\endgroup$ Apr 21 '14 at 12:06
  • $\begingroup$ I believe that there is an isomorphism - just fix the first row, then the second row, etc. As I said, I am interested in the explicit form of three vector fields. Can you help me in this regard please? $\endgroup$
    – Stanislav
    Apr 28 '14 at 1:11
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    $\begingroup$ There is not an isomorphism-- I have just proved that there isn't. $\endgroup$ Apr 28 '14 at 2:18
  • $\begingroup$ Okay. I would still appreciate an explicit form of 3 vector fields. We both consent they exist but I am struggling with producing them. $\endgroup$
    – Stanislav
    May 1 '14 at 21:00
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Draw a picture of $S^2 \times [0,1]$ as a thickened sphere in $\mathbb{R}^3$ where we think of the outside sphere as $S^2\times \{1\}$ and the inside as $S^2\times \{0\}$. We will view $S^2\times S^1$ as a quotient of this space where we identify $(p,0)$ with $(p,1)$.

Now, at each point, just draw the usual $x,y,z$ arrows. This defines a nonvanishing smooth vector field on $S^2\times [0,1]$. The identification made on the boundary maps the usual $x,y,z$ arrows to the usual $x,y,z$ arrows, so this descends to a non-vanishing vector field on $S^2\times S^1$.

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  • $\begingroup$ @user557906: I liked this description of $X$ because you can draw a picture of it - it's very hands on. Under the identification of $(p,0)$ with $(p,1)$, the $x$, $y$, and $z$ arrows are taken to themselves so produced a well defined tangent vector at each point of $X/\sim \cong S^1\times S^2$. As to why I couldn't assign "the same" vectors to each point of $S^2\times S^1$, what do you mean by "the same"? These vectors live on different manifolds, so definitely in different tangent spaces. $\endgroup$ May 1 '18 at 20:43
  • $\begingroup$ $X$ is a subset of $\mathbb{R}^3$, so there is a canonical way of identifying $T_p X$ with $\mathbb{R}^3$. But $S^1\times S^2$ can't be embedded in $\mathbb{R}^3$ (though it can be embedded in $\mathbb{R}^4$.) Thus, while the tangent space at each point in $S^2\times S^1$ is isomorphic to $\mathbb{R}^3$, there may not be a canonical such choice. In other words, for each $p\in X$ or $q\in S^1\times S^2$, we have to choose an isomorphism $T_p X\cong \mathbb{R}^3$ and $T_q (S^1\times S^2)\cong \mathbb{R}^3$. Can these choices of isomorphisms be made in a continuous fashion? On $X$, they .... $\endgroup$ May 2 '18 at 0:01
  • $\begingroup$ can - I describe a way of doing it. On $S^1\times S^2$, can it? My answer shows the answer is yes, but I don't think it's obvious. And to show this isn't an idle question, on $S^2$, you can not make such a choice, due to the Hairy Ball theorem. $\endgroup$ May 2 '18 at 0:03
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Let $Z$ be a nontrivial vector field on $S^1$ and $X, Y$ linearly independent vector fields on $S^2$. The three vector fields $(Z, 0), (0, X), (0, Y)$ on $S^1 \times S^2$ are linearly independent. This means: in the tangent space at any $(p, q)\in S^1 \times S^2$, the vectors $(Z_p,0), (0, X_p), (0, Y_p)$ are linearly independent.

It's more difficult to give formulas for three vector fields whose values at each point are linearly independent. But the Cartesian product of any finite number of spheres of various dimensions is parallelizable provided one of the spheres is odd-dimensional.

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    $\begingroup$ Since there are no non-vanishing continuous vector fields on $S^2$, there must be a point where $X$ is the $0$ vector, and hence $(0,X) = (0,0)$ at that point, so you no longer have independent vector fields. $\endgroup$ Jan 20 '15 at 2:59

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