Tangent space of $\mathbb{P}(V)$

Question

I encountered this problem studying the local period map and I'm wondering how to solve it.

I would like to prove that, given $V$ a complex vector space and $W \subseteq V$ a one-dimensional subspace, the tangent space of $\mathbb{P}(V)$ at $W$ is $$T_W\mathbb{P}(V)=\text{Hom}(W,V/W).$$

Where do I start and what kind of argument is needed?

Answer 1

I'll expand on my comment above, but the comment gives the (local) intuition. Consider the map $\pi\colon V-\{0\}\to\Bbb P(V)$. Then for any $\xi\in W$, $\pi_{*\xi}\colon T_\xi V\to T_{\pi(\xi)}\Bbb P(V) = T_W\Bbb P(V)$. Note that $\pi_{*\xi}(W)=0$, so $\pi_*$ induces a surjective map (hence isomorphism) $T_\xi V/W = V/W\to T_W\Bbb P(V)$.

Why does $\text{Hom}(W,V/W)\cong W^*\otimes V/W$ enter the picture? Let's define $\phi_\xi\colon T_\xi V \to T_W\Bbb P(V)\otimes W$ for any nonzero $\xi\in W$ by $\phi_\xi(v)=\pi_{*\xi}v\otimes \xi$. Then for any $\lambda\ne 0$, we know $\pi\circ \lambda = \pi$, and so $\lambda\pi_{*(\lambda\xi)}=\pi_{*\xi}$. Thus, $$\phi_{\lambda\xi}(v) = \pi_{*(\lambda\xi)}(v)\otimes (\lambda\xi) = \pi_{*(\lambda\xi)}(\lambda v)\otimes\xi = \lambda\pi_{*(\lambda\xi)}(v)\otimes\xi = \pi_{*\xi}(v)\otimes\xi = \phi_\xi(v).$$ This means that $\phi_\xi$ induces a well-defined isomorphism $V/W \cong T_W\Bbb P(V)\otimes W$, and so $T_W\Bbb P(V)\cong W^*\otimes V/W$.

(To return to my intuitive explanation in terms of curves in $V$, note that if $\alpha$ is a curve in $V$ through $\xi$, then $\lambda\alpha$ is a curve through $\lambda\xi$ (for any nonzero $\lambda$), and $(\lambda\alpha)' = \lambda\alpha'$, so just applying $\pi_*$ to the tangent vector will not give us a well-defined tangent vector at $W$ in $\Bbb P(V)$. This is where the twist comes in. Indeed, if $\lambda$ is not constant, but a nowhere-zero function, then we get $(\lambda\alpha)' = \lambda\alpha' + \lambda'\alpha$, and happily the second term drops out mod $W$.)

This argument gives the Euler sequence, as @Nicolas suggested, and also generalizes immediately to Grassmannians and, indeed, to smooth projective varieties (as I mentioned briefly in my earlier post).

Answer 2

The answer by Ted Shifrin is great.

A more "canonical" way is to use the Euler sequence

$$ 0 \to \mathcal O \to V \otimes \mathcal O(1) \to T \to 0 $$

Where $T$ is the tangent bundle. Hence $T$ is $$(V \otimes \mathcal O(1)) / \mathcal O \cong (V / \mathcal O(-1)) \otimes \mathcal O(1) \cong (V / \mathcal O(-1)) \otimes \mathcal O(-1)^* \cong Hom(\mathcal O(-1), V/ \mathcal O(-1))$$

exactly as expected since by definition the fiber of the line bundle $\mathcal O(-1)$ at $[W]$ is $W$. Here $V$ is the trivial vector bundle over $\mathbb P(V)$ with fibre $V$. This (as Ted Shifrin's argument) also works for Grassmannians.