MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a general theory of embeddings of the (total variety of) the tangent bundle on a (nonsingular) projective variety into projective space? I suppose what I really mean is (and to be more precise about the projective space), if $\mathbb{P}$ is the projective completion of $T_X\rightarrow X$, then what can be said about the relative dimension of "infinity": $D=\mathbb{P}\backslash T_X$?

I apologize if this question is vague. Any thoughts or references will be greatly appreciated.


share|cite|improve this question
Is the term "projective completion" well-defined here? If you take different embeddings of TX in projective space, you are likely to get different results. – Charles Staats Apr 24 '10 at 16:39
up vote 5 down vote accepted

The simplest way to get a "projective completion" is to consider the projectivization on $X$ of $T_X \oplus L$ for some line bundle $L$ on $X$. In this case the complement will be the projectivization of $T_X$ and will have codimension 1. Sometimes you can contract this completion to get smaller complement, e.g. if $X$ is a curve (choose $L$ in such a way that $\omega_X\otimes L$ is very ample, then $P(T_X \oplus L) = P(T_X\otimes L^{-1} \oplus O)$ which is a blowup of the projective cone over $X$ in the embedding given by $\omega_X\otimes L$).

share|cite|improve this answer
Thank you for your answer, Sasha. Can you please clear up one point for me? If X is n-dimensional, then the total manifold of T_X\oplus L is (2n+1)-dimensional, and so projectivizing the fibres of this bundle will give us a bundle with projective fibres whose total space is 2n-dimensional? Why then is T_X of codimension 1 in this compactification? I'm sure I'm just making a silly mistake in how I am thinking of this. Thanks. – user5395 Apr 25 '10 at 18:19
Certainly, $T_X$ is of codimension 0, it is $P(T_X)$ which is of codimension 1. Let me repeat this once again, $P(T_X \oplus L)$ contains both the total space of $T_X$ and the projectivization of $T_X$ and is the union of these two sets. The first is an open subset and the second is a closed subset. This is just a relative version of the fact that $P(V \oplus k) = P^n$ contains $V = A^n$ as an open subset and $P(V) = P^{n-1}$ as its closed complement. – Sasha Apr 26 '10 at 15:13
I see now. Thanks! – user5395 Apr 27 '10 at 17:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.