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I was searching for a response on the internet but I was not able to find out an explicit answer.

It is known that if $\mathbb{P}^n \subset \mathbb{P}^N$ is embedded linearly then the normal bundle $N_{\mathbb{P}^n/\mathbb{P}^N}\cong \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus (N-n)}$. This can be proved for example via Koszul complex.

My question now is the following: if we embed $\mathbb{P}^n$ into $\mathbb{P}^N$ with higher degree, for example with the Veronese embedding $$v_d:\mathbb{P}^n \rightarrow \mathbb{P}^N:=\mathbb{P}(H^0(\mathcal{O}_{\mathbb{P}^n}(d)))$$ what is it the normal bundle $N_{v_d(\mathbb{P}^n)/ \mathbb{P}^N}$? It is possible that could be $\mathcal{O}_{v_d(\mathbb{P}^n)}(d)^{\oplus(N-n)}$?

I was trying some Koszul approach like in the linear case but for $d>1$ I'm not able anymore to control the free resolution of the Veronese varieties.

Thanks in advance.

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    $\begingroup$ Your guess is incorrect for a $2$-uple Veronese embedding of $\mathbb{P}^1$ in $\mathbb{P}^2$. $\endgroup$
    – Jason Starr
    Commented Jul 15, 2021 at 22:07
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    $\begingroup$ @JasonStarr yes sorry I meant $\mathcal{O}_{v_d(\mathbb{P}^n)}(d)$, I have fixed. Thank you. $\endgroup$
    – gigi
    Commented Jul 15, 2021 at 22:15
  • $\begingroup$ Now the guess is incorrect for the $3$-uple Veronese embedding of $\mathbb{P}^1$ in $\mathbb{P}^3$. $\endgroup$
    – Jason Starr
    Commented Jul 16, 2021 at 0:30
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    $\begingroup$ The normal bundle of the Veronese embedding $\mathbb{P}^2\hookrightarrow \mathbb{P}^5$ is not a direct sum of line bundles. $\endgroup$
    – abx
    Commented Jul 16, 2021 at 4:06

1 Answer 1

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The normal bundle $N$ of the Veronese embedding $\mathbb{P}(V) \to \mathbb{P}(S^dV)$ can be described by the exact sequence $$ 0 \to V \otimes \mathcal{O}(1) \to S^dV \otimes \mathcal{O}(d) \to N \to 0, $$ where the first arrow is the unique nonzero $\mathrm{GL}(V)$-equivariant morphism.

Alternatively, one can describe the normal bundle as an iterated extension of symmetric powers of the tangent bundle $T$ of $\mathbb{P}(V)$ --- there is a filtration on $N$ with associated graded of the form $$ \mathrm{gr}_\bullet(N) = \bigoplus_{i=2}^d S^i T. $$

Finally, let me give a couple of explicit examples. If $\dim(V) = 2$ one has $$ N_{\mathbb{P}(V)/\mathbb{P}(S^dV)} \cong S^{d-2}V \otimes \mathcal{O}(d+2), $$ and if $d = 2$ one has $$ N_{\mathbb{P}(V)/\mathbb{P}(S^dV)} \cong S^2T. $$

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  • $\begingroup$ 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? $\endgroup$
    – gigi
    Commented Jul 16, 2021 at 9:03
  • $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented Jul 16, 2021 at 11:07
  • $\begingroup$ @JasonStarr: Jason, thanks for the correction! $\endgroup$
    – Sasha
    Commented Jul 16, 2021 at 11:47
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    $\begingroup$ @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. $\endgroup$
    – Sasha
    Commented Jul 16, 2021 at 11:52

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