is tangent bundle of $S^n$ an algebraic variety? I have found somewhere that $T(S^n)$ is an algebraic variety in $\mathbb{C}^{n+1}$. But now I can not recall the explicit form of this variety and the source of this information. It will be helpful if somebody enlightens me on this fact. 
 A: We have
$$S^n = \{ (x_0, \ldots, x_n) \in \mathbb{R}^{n+1} : x_0^2 + \cdots + x_n^2 = 1 \}$$
and therefore
$$T(S^n) = \{ (x_0, \ldots, x_n, v_0, \ldots, v_n) \in \mathbb{R}^{2 n + 2} : x_0^2 + \cdots + x_n^2 = 1, x_0 v_0 + \cdots  + x_n v_n = 0 \}$$
because the (outward) normal vector to the sphere at $(x_0, \ldots, x_n)$ is just $(x_0, \ldots, x_n)$ itself. Putting
$$y_k = x_k \sqrt{1 + v_0^2 + \cdots + v_n^2}$$
we find that we can rewrite the two equations as a single complex equation, namely
$$(y_0 + i v_0)^2 + \cdots + (y_n + i v_n)^2 = 1$$
and thus $T(S^n)$ is diffeomorphic to the complex affine variety
$$\{ (z_0, \ldots, z_n) \in \mathbb{C}^{n+1} : z_0^2 + \cdots + z_n^2 = 1 \}$$
and therefore has the structure of a complex manifold.
A: For $n=1$ this works because the complement of a point in $\mathbb C$ is homeomorphic to $S^1\times\mathbb R$ and the tangent bundle of the circle is trivial. But already for $n=2$ this seems problematic because the complement of a $\mathbb R$ in $\mathbb {C}^2$ is homeomorphic to the trivial bundle $S^2\times \mathbb{R}^2$ whereas the tangent bundle of $S^2$ is nontrivial.
