I'm trying to show that the tangent bundle, $TS^n$ of the n-sphere $S^n$ is diffeomorphic to the set $\sum z_i^2 = 1$ in $\mathbb{C}^{n+1}$.

It's relatively straightforward to see that the tangent bundle of the sphere can be identified with:

$$TS^n = \{ (x_0,...,x_n,y_0,...,y_n) : x_i,y_i \in \mathbb{R}, \sum x_i^2 = 1, \sum x_i y_i = 0 \}$$

Now to show this diffeomorphism I tried the natural thing of writing $z_j = x_j + iy_j$ but now we have $\sum z_j^2 = 1 - \sum y_i^2$ so it only lies in the required subspace if we restrict the tangent spaces of the sphere. I'm wondering how to write down a different map that does this?

I'm also a little concerned about how to show such a map is a diffeomorphism, how could I show that the identification I've made above as the tangent bundle embedded in $\mathbb{R}^{2(n+1)}$ is smooth? It's probably obvious but I'm struggling to see it!

Thanks for any help

  • $\begingroup$ Maybe I'm wrong, but shouldn't it be $\sum x_i^2=1$? $\endgroup$ May 14, 2016 at 12:22
  • $\begingroup$ Wait, why are $x_i$ and $y_i$ in $\mathbb{R}^{n+1}$? $\endgroup$ May 14, 2016 at 12:25
  • $\begingroup$ I think it definitively should be $x_i,y_i\in\mathbb{R}$ and $\sum x_i^2=1$. $\endgroup$ May 14, 2016 at 12:30
  • $\begingroup$ Thanks for your comments, yes there were a couple of typos which I've now corrected! $\endgroup$
    – Wooster
    May 14, 2016 at 12:58
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    $\begingroup$ Yes, you mean the hyperquadric $\sum z_j^2 = 1$ in $\Bbb C^{n+1}$. This is very non-compact and hardly a unit circle or unit sphere. $\endgroup$ May 14, 2016 at 17:00

1 Answer 1


Let $Q\subseteq\mathbf C^{n+1}$ be the affine quadric defined by the equation $\sum z_i^2=1$. The map $$ f\colon TS^n\rightarrow Q $$ defined by $$ z=f(x,y)=x\sqrt{1+||y||^2}+y\sqrt{-1} $$ does the job, where $||y||^2=\sum y_i^2$. Indeed, one has $f(x,y)\in Q$ since $$ \sum_{i=0}^n z_i^2=\sum_{i=0}^n x_i^2(1+||y||^2)-y^2_i+2x_iy_i\sqrt{-1}\sqrt{1+||y||^2}=\\ 1+||y||^2-||y||^2+2\sqrt{-1}\sqrt{1+||y||^2}\sum x_iy_i=1, $$ for $(x,y)\in TS^n$.

The map $f$ is a diffeomorphism since its inverse is $$ g\colon Q\rightarrow TS^n $$ defined by $$ g(z)=\left(\frac{x}{\sqrt{1+||y||^2}}, y\right), $$ where $z=x+y\sqrt{-1}$. One has $g(z)\in TS^n$ since $$ ||x||^2-||y||^2=1 $$ and $$ 2\sqrt{-1}\sum x_iy_i=0 $$ for $z\in Q$.

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    $\begingroup$ Excellent: +1. I have already seen that calculation, but I don't remember where. Would you happen to know in what context these calculations arise? $\endgroup$ May 14, 2016 at 23:59
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    $\begingroup$ @Georges Elencwajg: I don't know about the specific calculation, but the general context is the one of Totaro's good complexifications (see math.ucla.edu/~totaro/papers/public_html/complex.pdf). A good complexification of a smooth manifiold $M$ is a smooth affine real algebraic variety $X$ such that $X(\mathbf R)$ is diffeomorphic to $M$, and the inclusion $X(\mathbf R)\rightarrow X(\mathbf C)$ is a homotopy equivalence. The manifold $S^n$ is particularly simple example of a manifold admitting a good complexification. $\endgroup$ May 15, 2016 at 5:55
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    $\begingroup$ Dear Johannes: thanks a lot for your explanations and reference. I find it quite interesting that this rather elementary question is related to quite advanced research by a brilliant mathematician like Totaro, who seems to have a knack for addressing difficult mathematical problems through rather unsophisticated considerations. $\endgroup$ May 15, 2016 at 6:49
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    $\begingroup$ there is a slight error in the equation (I won't mess with your post). In the last terms it is $\dots 2x_iy_i\sqrt{-(1+|y|^2)}$ and not $2x_iy_i\sqrt{-1}$. That does not affect the result of course (the inner product of $x$ and $y$ is still $0$). Thanks for the beautiful answer $\endgroup$
    – Jose
    Jan 8, 2019 at 6:57
  • $\begingroup$ @Jose You're right. Thank you! $\endgroup$ Jan 8, 2019 at 9:28

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