is tangent bundle of $S^n$ an algebraic variety?

Question

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.

Answer 1

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.

Answer 2

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.