Given a complex vector bundle $V\rightarrow M$, we can form a fibre bundle $\mathbb{P} V\rightarrow M$, where the fiber over each point is the corresponding projective space. In particular consider the space $ \mathbb{P}(T \mathbb{P}^2)$, the projectivized tangent space of $\mathbb{P}^2$. Can this complex manifold be embedded (holomorphically) in some projective space $\mathbb{P}^N$?

More generally if we can embed $M$ in a projective space, does it imply we can embed $\mathbb{P}(V)$ in a projective space?

Everything is over complex numbers and the projective spaces are all complex projective spaces.

  • $\begingroup$ Since every complex algebraic variety can be embedded in a projective space, unless I'm missing something here the answer to your question is trivially yes. $\endgroup$ Commented Jan 25, 2013 at 17:13
  • $\begingroup$ Is it immediately obvious that the projectivized tangent space of CP^2 is a complex algebraic variety? $\endgroup$
    – Ritwik
    Commented Jan 25, 2013 at 17:32
  • 1
    $\begingroup$ Chuck: there are lots of nonprojective varieties. See this question: mathoverflow.net/questions/3624/nonprojective-surface $\endgroup$
    – user5117
    Commented Jan 25, 2013 at 17:45
  • $\begingroup$ Sorry -- I should have been more clear. I'm using the fact that if E is an algebraic vector bundle on a variety then indeed the projectivization of E is also a variety (this is contained in the $\textbf{Proj}$ construction, cf Section II.7 of Hartshorne). Since the tangent space to $ \mathbb P^n $ is an algebraic bundle the result follows. (As for the projectiveness, one doesn't require a variety to be projective for it to embed in projective space -- eg, every affine variety will embed in projective space too. If we are talking about a closed embedding, of course the answer is different.) $\endgroup$ Commented Jan 25, 2013 at 17:57
  • $\begingroup$ Chuck: right, the answer to the question (in much greater generality) is all contained in Hartshorne II. About my comment: I should have said "lots of proper nonprojective varieties". Of course, that's irrelevant to the question at hand, but I thought it was worth mentioning for casual readers. $\endgroup$
    – user5117
    Commented Jan 25, 2013 at 18:31

1 Answer 1


Yes to both questions. In particular, $\mathbb P(T\mathbb P^2)$ is isomorphic to a hyperplane section of the Segre embedding $\mathbb P^2\times\mathbb P^2\hookrightarrow \mathbb P^8$. Indeed, $\mathbb P(T\mathbb P^2)$ is isomorphic to the variety of complete flags in the vector space $\mathbb C^3$. Suppose now that in the embedding of $\mathbb P^2\times\mathbb P^2$ in $\mathbb P^8$, the first copy of $\mathbb P^2$ is projectivization of $\mathbb C^3$ and the second is projectivization of $(\mathbb C^3)^*$. Then the linear functional on $\mathbb C^3\otimes (\mathbb C^3)^*$ that defines the hyperplane section in question is the "evaluation" bilinear form on $\mathbb C^3\times(\mathbb C^3)^*$.


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