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I want to do something very concrete: write down a smooth scheme of given degree and dimension in projective n-space.

A natural way to go about this is to try and write down a complete intersection, but not all degrees/dimensions can be gotten this way. For instance, I want to write down a smooth, non-degenerate cubic surface in $\mathbb{P}^4$.

What's a systematic way to go about this kind of problem?

Tony
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  • Why do you want intersections only when they are complete? – Colin McLarty Oct 28 '14 at 15:44
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    @ColinMcLarty: I guess all the OP was saying was that it's trivial to read off the degree and dimension of a complete intersection, whereas for non-complete intersections it can be extremely difficult. –  Oct 28 '14 at 16:59
  • Asal expresses my point perfectly. – Tony Oct 28 '14 at 17:26

2 Answers2

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I don't know what a systematic method would be, but in this particular case, the Segre embedding $\mathbb{P}^2 \times \mathbb{P}^1 \to \mathbb{P}^5$ has degree $3$. So, by Bertini's theorem, a generic hyperplane slice of $\mathbb{P}^2 \times \mathbb{P}^1$ will be a degree $3$ surface in $\mathbb{P}^4$.

David E Speyer
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  • I must admit I didn't work out the details, but it seems like a similar idea --- taking hyperplane sections of rational scrolls in their tautological embeddings --- might work in general. –  Oct 28 '14 at 16:16
  • How general is general? I think that gives you a nondegenerate smooth surface of degree $\delta$ in $\mathbb{P}^n$ whenever one exists. I was interpreting the problem as about varieties of any dimension, but I guess the OP will have to say what level of generality was meant. – David E Speyer Oct 28 '14 at 16:23
  • Maybe I should have been clearer --- by "scroll" I didn't just mean surface scroll, but any (embedded copy of a) projective bundle over $\mathbf P^1$. As I say, though, I didn't work it out; there may be an obstruction. –  Oct 28 '14 at 16:30
  • I see. Then I don't know whether it works, but feel free to figure it out! – David E Speyer Oct 28 '14 at 16:51
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    Well, my high-level problem-solving strategy was to leave the original comment, which would then goad you into figuring it out. Sadly, that seems to have failed. Anyway, +1. –  Oct 28 '14 at 16:54
  • Thanks, this is nice for the cubic surface in P^4, and I would definitely be interested in a general construction for any possible (dim, deg) in P^n! – Tony Oct 28 '14 at 17:29
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Let me try to generalise David Speyer's construction a little to show how one can achieve a certain range of degrees and dimensions.

First let me change the notation a little bit: rather than asking for a subvariety $X$ of dimension $b$ and degree $d$ in $\mathbf P^n$, let's ask for a projective vareity $X$ of dimension $b$, codimension $c$, and degree $d$. (Of course the data $(b,n,d)$ and $(b,c,d)$ are equivalent information; I just find it easier to write down the answer in terms of the latter.) Let's call such a variety $X_{b,c,d}$.

Then my claim is the following:

Claim: For any $d \geq 2$, there exist smooth projective varieties $X_{b,c,d}$ in the following cases:

  • $b=d$ and $c=d-1$;
  • $b=d-1$ and $c=d-1$;
  • $b<d-1$ and $b+1 \leq c \leq d-1$.

Remark 1: The maximum codimension of a nondegenerate subvariety of degree $d$ in projective space is $d-1$. This explains the bound on $c$ above.

Remark 2: The claim says we can always get what we want as long as the codimension $c$ is (roughly) at least as big as the dimension $b$. What about when codimension is small compared to dimension? Well, Hartshorne's conjecture predicts that when $b > \frac{2}{3}(b+c)$, in other words when $c<\frac12 b$, any smooth $X_{b,c,d}$ must be a complete intersection. So one should not expect to be able to do this for all $c$. On the other hand, there is plenty of room between $\frac12 b$ and $b+1$; maybe one could do something clever in that range, but I have no idea. It seems much harder.

Proof of Claim: Suppose $b,c,d$ are given. To start with, let $S_d$ be the embedding of $\mathbf P^1 \times \mathbf P^{d-1}$ in $\mathbf P^{2d-1}$ given by the line bundle $O(1,1)$. One can check this has degree $d$, so it is an $X_{d,d-1,d}$ as in the first bullet point. As in David Speyer's answer, one can then take a general linear section of this to get an $X_{d-1,d-1,d}$ as in the second bullet point.

Now assume $b<d-1$ and $b+1 \leq c \leq d-1$. Start with the $X_{d-1,d-1,d}$ as above. Take a section by a general linear subspace of the right dimension to get $X_{b,d-1,d}$. This has the right degree and dimension, but the wrong codimension. To fix this, we project away from a point not on $X$, repeatedly; this lowers the codimension by 1 at each stage. As long as $c>b+1$, the secant variety of $X_{b,c,d}$ is not all of $\mathbf P^{b+c}$, so we can project like this and get a smooth $X_{b,c-1,d}$.

  • I guess this answer somewhat sidesteps the issue of how to write down these varieties. But I claim that there is no obstruction to turning the above description into an explicit (albeit maybe computationally expensive) procedure. –  Oct 29 '14 at 10:54