# Projective Normality

What is the significance of studying projective normality of a variety ? How does it relate to non-singularity, rationality of a variety ?

• See this MO question. – Brandon Carter Jun 2 '13 at 12:20
• @BrandonCarter Your link is mainly about normality but not projective normality which is stronger than normality and is dependent of the projective embedding. – Yuchen Liu Jun 2 '13 at 14:32

First let me emphasize that projective normality is defined only for varieties $X\subset \mathbb P^n$ embedded in some projective space, not for an abstract variety.
Let us say that such a subvariety is $k$-normal ($k\gt 0$) if the canonical morphism $$\Gamma(\mathbb P^n,\mathcal O_{\mathbb P^n}(k))\to \Gamma(X,\mathcal O_X(k))$$ is surjective.
Equivalently $H^1(\mathbb P^n,\mathcal I_X(k))=0$, where $\mathcal I_X$ is the ideal sheaf defining $X$.
The equivalence follows by taking the appropriate segment of the long exact sequence of cohomology associated to $$0\to \mathcal I_X(k)\to O_{\mathbb P^n}(k)\to \mathcal O_X(k)\to 0$$ and remembering that $H^1(\mathbb P^n, \mathcal O(k))=0$.
Then $X$ is said to be projectively normal if it is $k$-normal for all $k\gt 0$ and if it is normal.

For example any curve $X\subset \mathbb P^2$ of degree $d$ is $k$-normal for all $k\gt 0$ : indeed $$H^1(\mathbb P^2,\mathcal I_X(k))=H^1(\mathbb P^2,\mathcal O_{\mathbb P^2}(-d+k))=0$$
Hence a plane projective curve is projectively normal if and only if it is normal.
More generally any normal complete intersection in $\mathbb P^n$ is projectively normal (the converse is false: see Edit below).
Also, any Segre embedding $\mathbb P^n\times \mathbb P^m\hookrightarrow \mathbb P^N$ is projectively normal.

However a normal projective variety is not necessarily projectively normal: an example is the image $X\subset \mathbb P^3$ of the embedding $\mathbb P^1 \hookrightarrow \mathbb P^3:(u:v)\mapsto(u^4:u^3v:uv^3:v^4)$ .
This degree $4$ curve is not even $1$-normal: the linear map $\Gamma(\mathbb P^3,\mathcal O_{\mathbb P^3}(1))\to \Gamma( X,\mathcal O_X(1))$ is not surjective because the source has dimension $4$ whereas the target has dimension $h^0( X,\mathcal O_X(1)) =h^0(\mathbb P^1,\mathcal O(4))=5$.

Edit: caution!
We have just seen a non projectively normal embedded copy $X$ of $\mathbb P^1$ in $\mathbb P^3$.
But the standard embedding $\mathbb P^1\hookrightarrow \mathbb P^3: (u:v)\mapsto(u^3:u^2v:uv^2:v^3)$ has as image a cubic curve which is projectively normal (even though it is not a complete intersection !): cf.Hartshorne Chapter III, Ex.5.6.(b)(3) p.231, taking into account that this curve has bidegree (1,2) on the quadric $x_0x_3=x_1x_2$ on which it lies.
This confirms that projective normality is not an intrinsic property of a projective variety but depends on the choice of an embedding of it into $\mathbb P^n$.

• I once tried to make an argument that $\Bbb{P}^1\times \Bbb{P}^1$ is not isomorphic to $\Bbb{P}^2$ using that the projective coordinate ring was an invariant under isomorphism......massive fail that time :) – user38268 Jun 3 '13 at 23:42
• I think there is a small mistake in your answer: Then $X$ is said to be projectively normal if it is $k$-normal for all $k>0$ AND it is normal. (cf. Hartshorne Chapter II, Ex. 5.14(d)) If we don't require $X$ to be normal, then any singular plane curve is $k$-normal for any $k>0$, but they are not projectively normal because they are not normal. An equivalent definition of projectively normal is that $\Gamma_* \mathcal{O}_X$ is integrally closed. – Yuchen Liu Jun 11 '13 at 16:31
• Dear @Yuchen: you are right. I have modified my answer in the direction you suggest . Thanks a lot for your attention. – Georges Elencwajg Jun 18 '13 at 19:12
• You just gave an overview of the subject but this actually doesn't give an answer to OP's question. Probably in the very last line you indicated that it is a property of the embedding but finally what information do we get about the variety ? – icmes imrf Oct 15 '18 at 6:18