Considering the complex sphere $$S_\mathbb{C}^{n-1} = \left\{ z \in \mathbb{C}^n | \sum_{i=1}^n |z_i|^2 =1 \right\}$$ and its associated tangent bundle $T S_\mathbb{C}^{n-1}$, can one find a section?
I think I understand that in the real case, $\mathbb{S}_{n-1}$ in $\mathbb{R}^n$, the tangent bundle is trivial for $n=1,2,4,8$.
Is that right? Can one give a similar result for the complex sphere?
Thanks,
If $z_i = x_i + iy_i$, then $|z_i|^2 = x_i^2 + y_i^2$, so
\begin{align*} S^{n-1}_{\mathbb{C}} &= \left\{(z_1, \dots, z_n) \in \mathbb{C}^n \mid \sum_{i=1}^n|z_i|^2 = 1\right\}\\ &= \left\{(x_1, \dots, x_n, y_1, \dots, y_n) \in \mathbb{R}^{2n} \mid \sum_{i=1}^nx_i^2 + y_i^2 = 1\right\}\\ &= S^{2n-1}. \end{align*}
So $TS^{n-1}_{\mathbb{C}} = TS^{2n-1}$ which is trivial if and only if $n = 1, 2, 4$. However, for any value of $n$, $TS^{n-1}_{\mathbb{C}}$ admits at least one nowhere-zero section.