# Total spaces of tangent/cotangent bundles in a course where all varieties are quasi-projective

$\def\PP{\mathbb{P}}$In a course where all varieties are quasi-projective (as in Shafarevich Volume I), I am trying to figure out whether I can justify talking about the total spaces of the tangent and cotangent bundles to a smooth variety.

The tangent bundle to a (quasi)-affine variety is fine. If $X$ is (quasi)-affine, embedded as a subvariety of some vector space $V$, then the tangent bundle of $V$ is isomorphic to $V \oplus V$, and $TX$ is a subvariety of $TV$.

If $X$ embeds in $\PP(W)$, then the tangent bundle to $X$ embeds in the tangent bundle to $T \PP(W)$, so the question is whether there is a sufficiently nice proof that $T \PP(W)$ is quasi-projective. The best I can do is the following: For $\bar{w}$ in $\PP(W)$, the tangent space $T_{\bar{w}} \PP(W)$ is canonically isomorphic to $\mathrm{Hom}(k \bar{w}, W/k \bar{w})$. For $(\bar{w}, \phi) \in T \PP(W)$, choose a lift $\vec{w}$ of $\bar{w}$ to $W$ and a lift $\tilde{\phi} \in \mathrm{Hom}(k \vec{w}, W)$ of $\phi$. Send $(\bar{w}, \phi)$ to $(\vec{w} \otimes \vec{w},\ \vec{w} \wedge \tilde{\phi}(\vec{w}))$ in $\mathbb{P}(\mathrm{Sym}^2 W \oplus \bigwedge^2 W)$. Then the projectivization and the wedge product cancel out our choices, and I believe I can show this is a regular immersion of $T \PP(W)$ into $\mathbb{P}(\mathrm{Sym}^2 W \oplus \bigwedge^2 W)$. This construction is too ugly to present; does anyone know a nicer one?

The cotangent bundle seems to be much worse. It doesn't even make sense unless $\Omega^1$ is reflexive, but let's say $X$ is smooth.

It is true that, if $X$ is affine, then the total space of any vector bundle $E \to X$ is affine. (Use the fact that affineness of a morphism can be checked on any affine cover, see for example Prop 7.3.4 in Vakil, and contrast with Jason Starr's answer here.)

Also, if $X$ is quasi-projective, then the total space of any vector bundle $\pi: E \to X$ is quasi-projective. (Proof sketch: Let $\mathcal{E}^{\vee}$ be the sheaf of sections of the dual bundle. For $L$ sufficiently ample, $\mathcal{E}^{\vee} \otimes L$ is globally generated. I think we can then embed $E$ in $\PP(H^0(X, L)^{\vee} \oplus H^0(X, \mathcal{E}^{\vee} \otimes L)^{\vee})$.)

But these arguments are clearly not for a first term. Is there some more elementary way to do it, or does this just wait until next term?

• Cannot this be somehow derived from the case of the universal principal bundle over the Grassmanian (which is the (Stiefel) manifold of linearly independent tuples)? – მამუკა ჯიბლაძე Nov 6 '14 at 19:23
• Interesting idea! Yes, possibly. I've already built the universal sub-bundle. It looks like, to do what you suggest, I'd have to build the universal quotient bundle as well. – DES-SupportsMonicaAndTransfolk Nov 6 '14 at 21:26
• @user52824 Absolutely! So far I have made this point concerning: The definition of product (why do we have to Segre embed) and the definition of Grassmannian (why can't we just glue $\binom{n}{d}$ affine spaces)? I also plan to make it when talking about normalization. I hadn't realized that total spaces of bundles was another such setting, but now I have another chance to complain about it. – DES-SupportsMonicaAndTransfolk Nov 7 '14 at 14:19
• @მამუკაჯიბლაძე A globally generated vector bundle is always a quotient of a universal quotient bundle; that's why I needed to get to the quotient bundles. In general, I think the best you can say is that, if $E$ is a vector bundle on a quasi-projective variety $X$ then $E \otimes L$ is globally generated for sufficiently ample $L$. This comes up if you try to define Chern classes in algebraic geometry by pulling back from Grassmannians. – DES-SupportsMonicaAndTransfolk Nov 7 '14 at 14:31
• @მამუკაჯიბლაძე I actually do think it is progress of a sort. $T^{\ast} X$ is a quotient of $T^{\ast} \mathbb{P}(W)$ so we have plausibly reduced to (1) Embedding $T^{\ast} \mathbb{P}(W)$ and (2) embedding the universal quotient bundle. Not that this is simple enough to belong in a class, but it's fun. – DES-SupportsMonicaAndTransfolk Nov 7 '14 at 14:39