Is there a Whitney theorem type theorem for projective schemes? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:09:48Z http://mathoverflow.net/feeds/question/14177 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14177/is-there-a-whitney-theorem-type-theorem-for-projective-schemes Is there a Whitney theorem type theorem for projective schemes? Ryan Eberhart 2010-02-04T19:25:08Z 2010-07-16T15:19:48Z <p>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 $\mathbb{R}^{2n}$?</p> http://mathoverflow.net/questions/14177/is-there-a-whitney-theorem-type-theorem-for-projective-schemes/14180#14180 Answer by David Speyer for Is there a Whitney theorem type theorem for projective schemes? David Speyer 2010-02-04T19:42:55Z 2010-02-04T20:02:07Z <p>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.</p> <p>Replace "dimension" by "maximal dimension of any Zariski tangent space" and I think there should be a result like this.</p> <p><hr></p> <p>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. </p> http://mathoverflow.net/questions/14177/is-there-a-whitney-theorem-type-theorem-for-projective-schemes/14183#14183 Answer by Georges Elencwajg for Is there a Whitney theorem type theorem for projective schemes? Georges Elencwajg 2010-02-04T20:11:03Z 2010-02-06T19:41:13Z <p>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.</p> <p>For an abelian variety however we have a very satisfying complete description of the situation:</p> <p>For $n=1$, we can embed any abelian (=elliptic) curve in $\mathbb P^{2}$.</p> <p>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.</p> <p>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.)</p> <p><strong>Summary and references</strong> (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 $\mathbb P^{2n}$. </p> <p>[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]</p> <p>G. Horrocks and D. Mumford. A rank 2 vector bundle on P4 with 15,000 symmetries. Topology 12 (1973), 63-81</p> <p>A. Van de Ven. On the embedding of abelian varieties in projective spaces. Ann. Mat. Pura Appl. (4), 103:127–129, 1975.</p> http://mathoverflow.net/questions/14177/is-there-a-whitney-theorem-type-theorem-for-projective-schemes/14399#14399 Answer by Pete L. Clark for Is there a Whitney theorem type theorem for projective schemes? Pete L. Clark 2010-02-06T16:30:39Z 2010-02-06T16:30:39Z <p>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.</p> <p>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):</p> <p><a href="http://www.math.uga.edu/~pete/8320notes6.pdf" rel="nofollow">http://www.math.uga.edu/~pete/8320notes6.pdf</a></p> <p>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.</p> <p>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. </p> <p>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.]</p> <p>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. </p> <p>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.]</p> http://mathoverflow.net/questions/14177/is-there-a-whitney-theorem-type-theorem-for-projective-schemes/14405#14405 Answer by Bjorn Poonen for Is there a Whitney theorem type theorem for projective schemes? Bjorn Poonen 2010-02-06T17:37:46Z 2010-02-06T17:37:46Z <p>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 <em>finite</em>. 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.</p> <p>Nghi Nguyen, in his 2005 Berkeley Ph.D. thesis, proved that these infinitely many set-theoretic obstructions give <em>necessary and sufficient</em> conditions for embeddability:</p> <blockquote> <p>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 $\mathbb{P}^n$.</p> </blockquote> <p>For an exposition of this, see Section 8 of <a href="http://www-math.mit.edu/~poonen/papers/sieve.pdf" rel="nofollow">Sieve methods for varieties over finite fields and arithmetic schemes</a>.</p> http://mathoverflow.net/questions/14177/is-there-a-whitney-theorem-type-theorem-for-projective-schemes/32187#32187 Answer by Qing Liu for Is there a Whitney theorem type theorem for projective schemes? Qing Liu 2010-07-16T15:19:48Z 2010-07-16T15:19:48Z <p>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 <b>7</b> (1979), no. 8, 775--790. </p> <p>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 $\dim\emptyset = -\infty$. </p> <p><b> Theorem</b> (Kleiman-Altman) <i> 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 $k$.</i> </p> <p>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. </p> <p>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 $X$. </p>