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Consider the complex space $\mathbb{C}^{n+1}$ with the usual hermitian inner product $\langle \cdot,\cdot\rangle$. We can construct the projective space $\mathbb{C}P^{n}$ by the equivalence relationship $\phi \sim \psi \iff \phi=a\psi$ for some $a\in\mathbb{C}$ and taking the quotient: $$ \mathbb{C}P^{n} = \mathbb{C}^{n+1}/\sim. $$ I want to prove that the tangent space $T_{[z]}\mathbb{C}P^{n}$ at a point $[z] \in \mathbb{C}P^n$ ($[z]$ denotes the equivalency class of $z$) is isomorphic to the space $z^{\perp}$. I've been trying to follow the steps described in the answer to this related question but I haven't had much success. My issue is that, given the canonical projection $$ \pi:\mathbb{C}^{n+1}\backslash \{0\}\to \mathbb{C}P^n\\ z\mapsto[z], $$ how does one compute the explicit expression for the pushforward $\pi_*$? Given that the pushforward is defined by $$ \pi_*:T_z\mathbb{C}^{n+1}\to T_{[z]}\mathbb{C}P^n, $$ once i have the expression for $\pi_*$, how can I show that $T_{[z]}\mathbb{C}P^n \cong z^{\perp}$?

I'm sorry if this is an obvious question or if it doesn't make sense, I've only recently started to study projective spaces and geometry, so any help is appreciated!

Gustavo Schranck Habermann
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  • Of course you mean $a\ne 0$. Of course, any two $n$-dimensional complex vector spaces are isomorphic, so you're actually asking for more. In fact, the isomorphism you're wanting isn't really right. You want $T_{[z]}\Bbb CP^n \cong \text{Hom}(\text{Span}(z),\text{Span}(z)^\perp)$. At any rate, you can find various posts on this, but perhaps my answer here. Note that $\mathscr L$ is the tautological line bundle whose fiber at $[z]$ is $\text{Span}(z)$. – Ted Shifrin Mar 10 '23 at 19:50
  • The tangent space $T_{[z]}\mathbb{C}P^{n}$ at a point $[z] \in \mathbb{C}P^n$ is trivially isomorphic to the space $z^{\perp}$ (simply because both vectors spaces have the same dimension). What you mean is to show that $\pi_*$ maps $z^{\perp}$ isomorphically onto $T_{[z]}\mathbb{C}P^{n}$. You should edit your question. – Paul Frost Mar 17 '23 at 11:42

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