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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.

Thanks!

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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

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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 $\omega_X\otimes L$).

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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

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