MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the Segre embedding $$ \mathbb P^n\times \mathbb P^m \hookrightarrow \mathbb P^N. $$

It seems to me that from the definition it is clear that

$$ H^0(\mathbb P^N, \mathscr O_{\mathbb P^N}(1)) = H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(1))\otimes H^0(\mathbb P^m, \mathscr O_{\mathbb P^m}(1)), $$

but is it obvious that

$$ H^0(\mathbb P^n\times \mathbb P^m, {\mathscr O_{\mathbb P^{N}}(1)}|_{\mathbb P^n\times \mathbb P^m}) = H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(1))\otimes H^0(\mathbb P^m, \mathscr O_{\mathbb P^m}(1))? $$

More generally:

Question: Is the Segre embedding projectively normal?

EDIT Note that strictly speaking the above condition is only linearly normal, I did mean to ask projective normality, which in this case follows from linear normality. (See J.C. Ottem's answer).

share|cite|improve this question
The question is whether the Segre embedding is linearly normal, not projectively normal. Anyway, this follows from the Künneth formula for coherent sheaves. – Angelo Apr 10 '13 at 6:35
Thank you Angelo. I did mean to ask about projective normality, but did not phrase the question very well. The Künneth formula for coherent sheaves indeed solves the issue. In fact, it seems to me that it also proves projective normality the same way, right? Thanks! – user31847 Apr 10 '13 at 19:33
up vote 4 down vote accepted

Yes. The Segre embedding $i:P:=\mathbb P^n \times \mathbb P^m\to \mathbb P^N$ is defined by the sections of the line bundle $O_P(1,1):=pr_1^*O_{\mathbb P^n}(1)\otimes pr_2 O_{\mathbb P^m}(1)$ on $\mathbb P^n \times \mathbb P^m$ and by definition of $i$ we have $i^*O_{\mathbb P^N}(1)=O(1,1)$. In other words, the equality $$H^0(\mathbb P^n \times \mathbb P^m,i^*O_{\mathbb P^N}(k))=H^0(\mathbb P^n \times \mathbb P^m,O(k,k))$$ holds for any $k\in \mathbb Z$, and so $H^0(\mathbb P^n \times \mathbb P^m,i^*O(k))$ decomposes as the tensor product $H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(k))\otimes H^0(\mathbb P^m, \mathscr O_{\mathbb P^m}(k))$ by Kunneth's theorem. Note however, that the definition of projectively normal requires that the map $$H^0(\mathbb P^N,O(k))\to H^0(P,O(k,k))$$is surjective for all $k\ge 0$ not just for $k=1$ (in which case the embedding is called linearly normal). This is true in our case, since global sections of $O(k,k)$ are polynomials in sections of $O(1,1)$. Of course, when $k=2$, the kernel of the above map is generated by the quadrics defined as the $2\times 2$ minors defining the Segre embedding.

share|cite|improve this answer
Thanks JC, this is exactly what I needed. Also, I meant projective normality for the reason you are saying that it follows from linear normality in this case..... – user31847 Apr 10 '13 at 16:59
More along this line: The cone over the Grassmannian for the Pl\"ucker embedding is even a factorial variety. (And quadratic relations among Pl\'ucker co-ordinates holds.) Book by Lakshmibai on 'Standard Monomial Theory' has more generalisations. Popov's generalisation: same is true for orbits of highest weight vectors in an irreducible representation of simply-connected semisimple algebraic group $G$ (over C?). – P Vanchinathan Apr 11 '13 at 0:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.