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I want to prove that "Given a bundle $E$, for any line bundle $L$ the projectivizations of $E$ and $E$ tensor $L$ are isomorphic i.e $P(E)\cong P(E\otimes L)$". By bundle you can as well assume it to be a smooth vector bundle and their projectivization to be a fibre bundle (in the topological sense) with fibre 1-dim subspaces of the vector space fibres The statement can also be seen on Wikipedia page. The reference they give is for Hartshorne, Algebraic Geometry. But unfortunately, I am not aware of that.

Can anyone suggest a proof involving only fibre bundles or maybe just give a good reference with the desired approach, how can one construct a morphism between these two bundles? Or at least how we can see that morphism geometrically.

Edit 1: I found this interesting argument, I am not getting how the induced map looks like?, is it $[v]\rightarrow[u\otimes v$]? If it is, then how can I define an isomorphism at the total space level will $(x,[v])\rightarrow(x,[u\otimes v$]) work?enter image description here Any suggestions will be of great help.

Thanks and regards in advance

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  • $\begingroup$ Can you say a little more about what you know, where you're trying to use this result (or what lead you to want to learn it), and the level of explanation you're looking for? There are several signs in your post that maybe something is up here. $\endgroup$
    – KReiser
    Commented Apr 8, 2022 at 7:12
  • $\begingroup$ I am aware vector bundles and their construction using functors, I am basically studying projectivizations of bundles in detail and I got to know about this interesting result somewhere but could not find any proof which I could understand. The linked wikipedia page talks about the isomorpshism coming from universal property can you explain that? I am just looking for the intuition behind this result and a bundle theoretic proof if possible. @KReiser $\endgroup$ Commented Apr 8, 2022 at 7:27
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    $\begingroup$ The relevant point here is really what you've written under Sasha's post: "I don't know Algebraic geometry". If you ask a question in the algebraic-geometry tag without adding that information, you're very likely to get an answer which is not going to help you much! I think your question would be much improved by adding a lot more detail about your background - for instance, when you're talking bundles, are you talking bundles in the sense of algebraic topology? What exactly do you mean? $\endgroup$
    – KReiser
    Commented Apr 8, 2022 at 7:40
  • $\begingroup$ I think you are right, although I had mentoined that I am not aware of Algebraic geometry already. I should add few details, thanks @KReiser $\endgroup$ Commented Apr 8, 2022 at 7:45
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    $\begingroup$ The answers to your questions in "Edit 1" are "yes". $\endgroup$
    – Max
    Commented Apr 16, 2022 at 14:00

2 Answers 2

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The projective bundle $\mathbb{P}_X(E)$ represents the functor $$ \Phi_E \colon \mathrm{Schemes} \to \mathrm{Sets}, \qquad S \mapsto \{(f,F,\phi) \mid f \colon S \to X,\ \phi \colon f^*E \to F \}, $$ where $F$ is a line bundle on $S$ and $\phi$ is an epimorphism of vector bundles. The functors $\Phi_E$ and $\Phi_{E \otimes L}$ are isomorphic via $$ (f,F,\phi) \mapsto (f,F \otimes f^*L, \phi \otimes \mathrm{id}_{f^*L}) $$ hence the representing schemes are isomorphic as well.

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  • $\begingroup$ thanks a lot for your answer, but I don't know Algebraic geometry, maybe you can help me in some another way possible say, giving a geometric argument or The linked wikipedia page talks about the isomorpshism coming from universal property can you explain that? or something bundle theoretic @Sasha $\endgroup$ Commented Apr 8, 2022 at 7:24
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$\DeclareMathOperator{\Hom}{Hom}$There's a nice explicit map $P\Hom(L,E)\to P(E)$ that takes a homomorphism to its image. Then use that $E\otimes L \cong \Hom(L^\vee,E)$ where $L^\vee$ is the dual.

But if you work through what this does, it's simply the map $P(E\otimes L)\to P(E)$ that takes $[u\otimes v] \mapsto [u]$ for nonzero $u$ and $v$.

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