It is well known that $T\mathbb{S}^{n-1}$ is diffeomorphic to $M= f^{-1}(1)$ where 
$f:\mathbb{C}^n\rightarrow \mathbb{C}$ is $f(z):=\sum\_{i=1}^{n} z\_{i}^{2}$.
Two questions:


1) Is $M$ a symplectic submanifold of $\mathbb{C}^n\sim \mathbb{R}^{2n}$ (with the standard symplectic structure)?


If the answer is affirmative, we can consider two symplectic structures for $T\mathbb{S}^{n-1}$. The first is the original structure of the tangent or cotangent bundle, the second one is the pull back structure of $M$.


2) Are these structures equivalent? 



symplectic structure of tangent bundle of $\mathbb{S}^{n-1}$
Here is a formula for an explicit symplectomorphism $F$ from $T^\*S^{n-1}$ to the affine quadric $\{\sum z\_{j}^{2}=1\}$ in $\mathbb{C}^n$: 


$$ 
F(p,q) = \left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{1/2} q - i\left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{-1/2}p 
$$


Here I view $T^\*S^{n-1}$ as consisting of pairs $(p,q)\in \mathbb{R}^n\times\mathbb{R}^n$ with $p\cdot q=0$ and $|q|=1$. To check that $F$ is a symplectomorphism one can just check that $F$ pulls back the one-form $\alpha=-\sum y\_jdx\_j$ on $\mathbb{C}^n$ (which is a primitive for the standard symplectic form) to the canonical one-form on $T^\*S^{n-1}$. The coefficients involving $\sqrt{1+4|p|^2}$ are designed to cause the map to have image in the quadric. 


To indicate how I came up with this formula, the reasoning was that a suitably natural symplectomorphism ought to be equivariant with respect to the natural Hamiltonian $O(n)$-actions on both spaces. The easiest way to do this seems to be to take $F(p,q)=f(|p|)q-ig(|p|)p$ for some real functions $f$ and $g$. For the image to be in the quadric one needs $$f(|p|)^2- |p|^2g(|p|)^2=1.$$ Meanwhile if this is to be a symplectomorphism it should pull back the moment map for the $O(n)$-action on $\mathbb{C}^n$ to the moment map for the $O(n)$-action on $T^\*S^{n-1}$. The norms of these moment maps are, respectively $(x+iy)\mapsto |x||y|$ and $(p,q)\mapsto |p|$, giving an equation $|f||g||p|=|p|$, i.e., $$|g|=\frac{1}{|f|}.$$ Solving these for $f$ and $g$ yields the formula at the top, which can then be directly confirmed to have the required properties.


