I want to express coordinate the (co)tangent bundle of sphere. I think

$TS^2$ (or $T^*S^2$) $=\{(x_1,x_2,x_3,v_1, v_2 , v_3 | x_1^2+x_2^2+x_3^2=1, <(x_1,x_2,x_3),(v_1,v_2,v_3)>=0 \}$ and using polar coordinate, express as $=\{(sin\theta cos\phi, sin\theta sin\phi, cos\theta, a cos\theta cos\phi -b sin\theta sin\phi, a cos\theta sin\phi +bsin\theta cos\phi, -asin\theta)| a,b,\theta,\phi \in \mathbb{R}\}.$

Is this expression correct? I know $TS^2$ is not trivial bundle, but as above expression, The bundle seems isomorphic to $S^2 \times \mathbb{R}^2$.

Also, can I possible to express elements $\partial\over{\partial x_i} $ and $\partial\over{\partial v_i} $of $T(TS^2)$ as linear combinations of $ \partial\over{\partial \theta}$, $\partial\over{\partial \phi}$, $\partial\over{\partial a}$, and $\partial\over{\partial b}$?


The expression you give is correct. However, spherical coordinates break down when $\sin(\theta)=0$. Thus the conclusion you should draw is that the tangent bundle of $S^2\backslash\{N,S\}$ is trivial, where $N$ and $S$ are the north/south poles. (This is expected, since the twice-punctured sphere deformation-retracts to $S^1$, and $TS^1$ is trivial.)

I'm not sure how to make sense of your quantities $\partial/\partial x_i$ and $\partial/\partial v_i$. To understand the problem, recall the definition of partial derivative: if you have a coordinate system $y_1,\ldots,y_n$, then $\partial/\partial y_i$ means to differentiate in the $y_i$-coordinate while holding the other coordinates fixed. (Partial derivatives depend on choice of coordinates!) In your case, the coordinates $x_1,x_2,x_3,v_1,v_2,v_3$ are redundant (6 coordinates on a 4-manifold). If I hold 5 coordinates fixed and try to vary the sixth, then I will violate the constraint equations.

Instead, I'm guessing that you'll want to compute the coordinate one-forms $dx_i$, $dv_i$ as linear combinations of $d\theta$, etc. In contrast to partial derivatives, the one-form of a particular coordinate doesn't depend on the other coordinates. Moreover, they are extremely simple to compute with the $d$ operator. For example, your last equation reads $$ v_3=-a\sin(\theta), $$ so $$ dv_3 =-\sin(\theta)\,da-a\cos(\theta)\,d\theta. $$

Finally, if you're interested in parameterizing all of $TS^2$, then polar coordinates are not so suitable. One approach is to use two coordinate charts, e.g. stereographic projection. My preferred method is to use the Lie group $\mathrm{SO}(3)$, which I describe as follows.

A better way to think about $TS^2$

You have given a description of $TS^2$ as a subset of $\mathbb{R}^6$ with two constraint equations. In contrast, the description I'm about to give is a projection from $\mathrm{SO}(3)\times\mathbb{R}^2$ to $TS^2$. The first ingredient is a projection $\mathrm{SO}(3)\to S^2$ which takes an orthogonal matrix $M^i_j$ to its first column $M^i_1$. The clever observation here is that the other two columns $M^i_2$ and $M^i_3$ give an orthonormal basis for the tangent space. Thus I define a map $\mathrm{SO}(3)\times\mathbb{R}^2\to TS^2$ as follows: $$ \mathrm{SO}(3)\times\mathbb{R}\times \mathbb{R}\to \mathbb{R}^3 \times \mathbb{R}^3\\ (M,\alpha,\beta)\mapsto(M_1,\alpha M_2+\beta M_3) $$ It's straightforward to check that the right-hand side takes values only in $TS^2\subset\mathbb{R}^3\times\mathbb{R}^3$. In your terminology, $\vec{x}=M_1$ and $\vec{v}=\alpha M_2+\beta M_3$, so you need to check $|\vec{v}|=1$ and $\vec{v}\cdot\vec{x}=0$.

The only remaining issue to understand is the redundancy of this 5-D parameterization. In other words, when do $(M,\alpha,\beta)$ and $(M',\alpha',\beta')$ map to the same point in $TS^2$? The answer is quite simple. Consider the matrix $$ R=\left( \begin{matrix} 1 & 0 & 0 \\ 0 & \cos{\omega} & \sin{\omega} \\ 0 & -\sin{\omega} & \cos{\omega} \\ \end{matrix}\right). $$ Rewriting the expression for $\vec{v}$, we get $$ \vec{v}=M\left(\begin{matrix}0\\\alpha\\\beta\end{matrix}\right)=\left(MR^T\right)\left(R\left(\begin{matrix}0\\\alpha\\\beta\end{matrix}\right)\right)\equiv M' \left(\begin{matrix}0\\\alpha'\\\beta'\end{matrix}\right), $$ where we have defined $$ M'=MR^T,\\ \alpha'=\cos(\omega)\alpha+\sin(\omega)\beta,\\ \beta'=-\sin(\omega)\alpha+\cos(\omega)\beta. $$ Thus it is clear that for any $\omega$, that $(M',\alpha',\beta')$ and $(M,\alpha,\beta)$ go to the same point. Thus we have an $\mathrm{SO}(2)$ subgroup acting on $\mathrm{SO}(3)\times\mathbb{R}^2$ which preserves our mapping, so the map I described is well-defined on the quotient $$ \frac{\mathrm{SO}(3)\times\mathbb{R}^2}{\mathrm{SO}(2)}\to TS^2. $$ This map is actually a diffeomorphism.

Note 1: Topologically, $\mathrm{SO}(3)=\mathbb{RP}^3$. Moreover, there is a simple parameterization $S^3\to\mathrm{SO}(3)$, where $S^3$ is the group of unit quaternions. Any unit quaternion $q\in S^3$ determines the $\mathrm{SO}(3)$ matrix associated to the linear map $x\mapsto qx\bar{q}$ where $x$ is a general imaginary quaternion.

Note 2: The above description can be summarized in fancy language as "$TS^2$ is the vector bundle associated to the Hopf fibration and the standard representation of $\mathrm{SO}(2)$."


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