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Jul 16, 2021 at 11:52 comment added Sasha @gigi: Yes, indeed, the first realization is a combination of the two Euler sequences. The second part can be deduced as follows. The Euler sequence gives a filtration on $V \otimes \mathcal{O}(1)$ with factors $\mathcal{O}$ and $T$. Taking its symmetric power one gets a filtration on $S^dV \mathcal{O}(d)$ with factors $S^iT$, $0 \le i \le d$. Taking the quotient one gets the second part of the answer.
Jul 16, 2021 at 11:47 comment added Sasha @JasonStarr: Jason, thanks for the correction!
Jul 16, 2021 at 11:46 history edited Sasha CC BY-SA 4.0
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Jul 16, 2021 at 11:07 comment added Jason Starr Typo correction: in the displayed equation for the associated graded pieces, the summands should be the $i$-fold symmetric powers of the tangent bundle, not the $d$-fold symmetric powers. Also, one frequently useful observation that is not so clear from this answer is that the normal bundle is a subbundle of a direct sum of copies of $\mathcal{O}(2d)$. This comes from the standard quadratic generators for the homogeneous ideal of the Veronese variety.
Jul 16, 2021 at 9:03 comment added gigi thank you, now it's clear. I guess that the first realization of $N$ has to do with the standard normal sequence and Euler sequence. Could you give me a reference where I can find the construction of $N$ you suggested in the second part of the answer?
Jul 16, 2021 at 9:01 vote accept gigi
Jul 16, 2021 at 4:38 history answered Sasha CC BY-SA 4.0