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Let $\mathbb{P}^n$ be the complex projective $n$-space and let $\mathcal{H}$ be the dual bundle of the tautologic line bundle over $ \mathbb{P}^n$.

I have a feeling that the author of a paper I'm reading is calling a section of $\mathcal{H}^{\otimes r}$ a "homogeneous polynomial of degree $r$ in $n+1$ variables". I can (kind of) roughly see why one might conceive of a section of this bundle as such.

My question is: is there an exact correspondence between homogeneous polynomials of degree $r$ in $n+1$ variables and sections of $\mathcal{H}^{\otimes r}$, or is it just convenient and intuitive shorthand, or neither?

JKEG
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  • The global sections of $\mathcal{O}(d)$ on $\Bbb P^n$ are homogeneous polynomials of degree $d$ in $n+1$ variables. See the duplicate for an explanation, especially mercio's answer. – KReiser Aug 08 '20 at 01:13
  • @KReiser It is perfect! Thank you. – JKEG Aug 08 '20 at 01:21
  • @KReiser Quick question: I'm assuming the $P_i$ are given by restricting the section to the open set $U_i$, but if so, why will they be polynomial on $x_0/x_i\dots x_n/x_i$? – JKEG Aug 08 '20 at 02:02
  • @KReiser Wouldn't they just be.. holomorphic? – JKEG Aug 08 '20 at 02:10
  • @KReiser Would that $\mathcal O (m)$ discussed in that question be the same as the $\mathcal H ^{\otimes m}$ of this question? Now they seem different as in each coordinate patch I want to conside all holomorphic functions not only polynomial functions on the coordinates. – JKEG Aug 08 '20 at 02:14
  • The result is the same (though the proof is slightly different) no matter whether you're working in the algebraic or holomorphic category. Your comments hint that you're working in the holomorphic category, but you don't ever actually say so (nor does your post - I saw [algebraic-geometry] and not [complex-geometry] and may have been slightly hasty) - if you are indeed interested in the holomorphic and not algebraic setting, please [edit] your post to make it clear. – KReiser Aug 08 '20 at 02:41
  • @KReiser Local sections don't have to be polynomials in the complex-geometry case, right? – JKEG Aug 08 '20 at 04:51
  • Correct, so you have to make a slight adjustment to make that proof work, but the same idea (the only analytic functions which patch together are the polynomials of the specified degree) is still true and still gives the result. – KReiser Aug 08 '20 at 05:03

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