# Is there a Whitney theorem type theorem for projective schemes?

We know that any smooth projective curve can be embedded (closed immersion) in $\mathbb{P}^3$. By definition a projective scheme over $k$ admits an embedding into some $\mathbb{P}^n$. Can we create an upper bound for the $n$ required (perhaps by strengthening the hypotheses) necessary to create an embedding of a smooth projective dimension $k$ scheme into $\mathbb{P}^n$ much like Whitney's theorem tells us we can embed an $n$ dimensional manifold in $\mathbb{R}^{2n}$?

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I wondered this myself when I taught a course on algebraic curves last year. I think the answer is yes and that the same general strategy should work: start out in a high-dimensional projective space and show that if $N$ is large enough compared to the dimension of your variety, a general hyperplane projection will give an embedding. Have you tried just adapting the argument given in Hartshorne? Anyway, a true algebraic geometer will chime in soon enough, I'm sure. –  Pete L. Clark Feb 4 '10 at 19:34

Over an algebraically closed field, any projective smooth variety of dimension $n$ can be embedded in $\mathbb P^{2n+1}$. This is elementary and can be found in Shafarevich's Basic Algebraic Geometry, Chapter II, §5.4 . Of course specific varieties might be embedded in projective spaces of lower dimension.

For an abelian variety however we have a very satisfying complete description of the situation:

For $n=1$, we can embed any abelian (=elliptic) curve in $\mathbb P^{2}$.

For $n=2$, some abelian surfaces but not all of them can be embedded in $\mathbb P^{4}$. The others can only be embedded in $\mathbb P^{5}$. This is due to Horrocks-Mumford.

For $n\geq 3$, no abelian variety of dimension $n$ can be embedded in $\mathbb P^{2n}$. They can only be embedded in $\mathbb P^{2n+1}$. (This theorem was proved by Van de Ven.)

Summary and references (added later) $\quad$ Over an algebraically closed field every projective smooth variety of dimension $n$ can be embedded in $\mathbb P^{2n+1}$. The embedding dimension $2n+1$ is sharp in the sense that for every $n$ there is a projective smooth variety of dimension $n$ not embeddable in $\mathbb P^{2n}$.

[For $n=1$ the sharpness is due to the fact that smooth curves do not embed in $\mathbb P^{2}$ unless their genus is of the form $(d-1)(d-2)/2$. For $n\geq 2$ the sharpness is due to the discussion of abelian varieties above]

G. Horrocks and D. Mumford. A rank 2 vector bundle on P4 with 15,000 symmetries. Topology 12 (1973), 63-81

A. Van de Ven. On the embedding of abelian varieties in projective spaces. Ann. Mat. Pura Appl. (4), 103:127–129, 1975.

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I believe that "algebraically closed field" can be replaced by "infinite field", with essentially the same proof. (I presented the result in this form in the case of smooth curves in a class I taught last year.) If anyone has any questions about this, please let me know. –  Pete L. Clark Feb 5 '10 at 18:36
Dear Pete, does your comment also apply to The Horrocks-Mumford construction and to Van de Ven's result? Anyway, I'd love to see your documents (if available) on the subject. –  Georges Elencwajg Feb 5 '10 at 19:30
It seems to be an interesting coincidence that the dimension of the ambient space is $2n+1$, which matches up with the $2n$ or $2n+1$ of the Whitney embedding theorems. –  Kevin H. Lin Jul 16 '10 at 20:48

The previous responses have answered the question well in the case when the ground field $k$ is algebraically closed or at least infinite, but the answer is different when $k$ is finite. For example, a smooth projective curve $X$ over a finite field $k$ need not embed into $\mathbb{P}^3$, because of a set-theoretic obstruction: $X$ might have more $k$-points than $\mathbb{P}^3$ does! Or $X$ might have more closed points of degree $2$ than $\mathbb{P}^3$ does, and so on.

Nghi Nguyen, in his 2005 Berkeley Ph.D. thesis, proved that these infinitely many set-theoretic obstructions give necessary and sufficient conditions for embeddability:

Let $X$ be a smooth projective scheme of dimension $m$ over a finite field $k$, and let $n \ge 2m+1$. There exists a closed immersion $X \to \mathbb{P}^n$ if and only if for every $e \ge 1$, the number of closed points of degree $e$ on $X$ is less than or equal to the number of closed points of degree $e$ on $\mathbb{P}^n$.

For an exposition of this, see Section 8 of Sieve methods for varieties over finite fields and arithmetic schemes.

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Dear Bjorn, your argument that a curve cannot be embedded in three-dimensional projective space because it has more points than the space is the funniest I have ever seen! Why isn't Mathematics always so amusing? –  Georges Elencwajg Feb 6 '10 at 20:05

There is an obvious obstacle: the nonreduced scheme $k[x_1, x_2, \ldots, x_n]/\langle x_1, x_2, \ldots, x_n \rangle^2$ is $0$-dimensional, but can't be embedded in any space of dimension less than $n$. More generally, if there is a point whose Zariski tangent space has dimension $n$, then we need $n$ coordinates to embed the scheme. So, for example, if $A$ is the subring of $k[t]$ generated by the monomials $t^n$, $t^{n+1}$, $t^{n+2}$, ..., then $\mathrm{Spec} \ A$ is a reduced one dimensional scheme which can't be embedded in less than $n$ dimensions.

Replace "dimension" by "maximal dimension of any Zariski tangent space" and I think there should be a result like this.

The poster clarifies below that he means smooth varieties. In this case, the answer is yes. If $X$ is a smooth projective variety of dimension $d$ over an infinite field then it can be embedded in dimension $2d+1$. The idea of the proof is to embed in $\mathbb{P}^{N-1}$ and consider the Grassmannian of projections $\mathbb{P}^{N-1} \to \mathbb{P}^{2d+1}$. This has dimension $(2d+2)(N-2d-2)$; one shows that the conditions that the projection is not defined on $X$, identifies two points of $X$, or is not injective somewhere on the Zariski tangent space of $X$ all have lower dimension.

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@DS: OK. I was thinking about the case of a smooth variety (Ryan said "smooth projective curve"; I didn't notice that he didn't repeat "smooth" after that). –  Pete L. Clark Feb 4 '10 at 19:45
Sorry--I did mean smooth. It was in the curve case but as Pete said and you noticed I forgot to explicitly include it in the hypothesis of my question. –  Ryan Eberhart Feb 4 '10 at 19:55
David:regarding the embedding dimension of singular varieties, the local embedding dimensions are not enough. Take a chain of n+1 projective lines, then the tangent spaces have dimension at most 2, but any ample divisor D on this curve has degree at least n, and L(D) has dimension at least n+2 (up to $\epsilon$, I didn't check carefully), so the curve can not be embedded in $\mathbb P^{n+1}$. Maybe you mean irreducible varieties. –  Qing Liu Feb 4 '10 at 22:06
I didn't make any specific claim about the embedding dimension of singular varieties. I said "I think there should be a result like this" and then explained the smooth case. That said, I don't buy your example. There is no reason I have to embed by a complete linear system. It looks to me like your example embeds in P^3: just take a generic n+1 points in P^3 and join them up by lines. –  David Speyer Feb 4 '10 at 22:16
@Liu: We try to avoid the "s word" here on MO, even applied to oneself. :) –  Pete L. Clark Feb 6 '10 at 16:04

This is a reply to a question posed in Georges' answer. It started as a comment, but I was worried about space limitations. Upon reflection, it's better to have it as an answer, because -- unlike the comment I posted to Georges' answer -- I will be automatically notified of any responses to it.

Since Georges asked for it, the [rough!] lecture notes where I discuss the facts that any smooth curve over an infinite field can be embedded in $\mathbb{P}^3$ and "immersed" in $\mathbb{P}^2$ with only ordinary double point singularities are available here (see Section 5):

http://www.math.uga.edu/~pete/8320notes6.pdf

As you will see, I am merely repeating the argument in Hartshorne -- omitting the trickier details of the immersion result -- and explaining why the ground field need not be algebraically closed but does need to be infinite.

Concerning Horrocks-Mumford and Van de Ven: I was not familiar with these results until Georges' post. But all the non-embeddability statements carry over immediately: if you have an embedding into $\mathbb{P}^n$ over the ground field, then the base change to the algebraic closure is still an embedding, of course.

This leaves the question of the positive part of the Horrocks-Mumford result. In strongest form, the question is: is it true that for any field $k$, there is an abelian surface over $k$ that can be embedded in $\mathbb{P}^4$? [I can certainly do it with $\mathbb{P}^2 \times \mathbb{P}^2$ -- take a product of two elliptic curves -- and it is conceivable to me that one might be able to get from this an embedding into $\mathbb{P}^4$ by composing with a well-chosen birational isomorphism, but I haven't even tried to decide whether this would work.]

I would have to see the proof of H-M to see whether it can be adapted to answer this question. Can you post a link to the paper? Or, if you need to know ASAP, ask Bjorn Poonen -- he eats questions like this for breakfast.

Finally, let me remark that over a non-algebraically closed field, a principal homogeneous space under an abelian variety may have higher embedding dimension than the (Albanese) abelian variety itself. The easy example of this is that if a smooth curve of genus one can be emedded in $\mathbb{P}^2$, then for geometric reasons it must embed as a cubic and therefore has a rational point of degree at most $3$. [Actually, it is possible that this is the only example. By the same theorems Georges quoted above, the only other possibility is a phs which does not embed in $\mathbb{P}^4$ while its Albanese abelian surface does.]

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Thanks for your interesting reply and your lecture notes, Pete. I hope your students appreciate their luck! I have added the references to the two articles I mentioned in my post. I should have done that immediately, of course (as you are too polite to mention!) –  Georges Elencwajg Feb 6 '10 at 19:54

To contradict what I said on embedding of singular varieties, here is a theorem of Kleiman and Altman "Bertini theorems for hypersurface sections containing a subscheme". Comm. Algebra 7 (1979), no. 8, 775--790.

Let $X$ be an algebraic variety over a field $k$. For any $x\in X$, define the local embedding dimension $e(x)$ of $X$ at $x$ by $e(x)=\dim (\Omega_{X/k}^1\otimes k(x))$ (so if $k$ is perfect, then $e(x)$ is just the dimension of the Zariski tangent space at $x$). It is easy to see that for any integer $e$, the set $X_e$ of $x\in X$ such that $e(x)=e$ is contructible. By convention $\dim\emptyset = -\infty$.

Theorem (Kleiman-Altman) Suppose $k$ is infinite and $X$ is quasi-projective (resp. projective) over $k$. Let $r$ be the maximum of $\dim (X_e) +e$ for all $e\ge 0$. Then $X$ can be embedded in a smooth quasi-projective (resp. projective) variety $Z$ of dimension $r$ over $k$.

For example, a reduced projective curve over an infinite perfect field can be embedded in a smooth projective surface if and only if the tangent space at every point has dimension at most 2.

In general, combining with the result on embedding of smooth projective varieties, one gets an embedding of $X$ in a projective space of dimension bounded by $\dim X$ and local embedding dimensions of $X$.

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