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I know that, in ${\Bbb R}^2$, in general, $5$ points determine a unique (non-degenerate) conic. I was wondering about the higher-dimensional analogue of this. Is it true, for example, that, in general, $9$ points determine a unique (non-degenerate) quadric?

The last statement in parenthesis seems a little wonky to me, because it seems like you can choose $9$ points on a cylinder, without any $4$ lying on the same plane. My question is how/what is the appropriate generalization of $5$ points determining a conic, and how does one exclude degenerate cases (that is $n$ points satisfying some independence relation in ${\Bbb R}^m$ determine a unique quadric, and the quadric must be non-degenerate).

Rodrigo de Azevedo
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guest1
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    Those numbers (for $n$ dimensions) come from the number of at most quadratic monomials in $n$ variables. When $n=2$ you have three quadratic monomials, $x^2,xy,y^2,$, two linear ones, $x,y$ and the constant. A total of six terms. Because a non-zero constant multiple of an equation won't change the curve, we need five points in 'general position' to nail it down to a single curve (a non-trivial solution to a linear homogeneous system). Here generic means that the five vectors gotten by evaluating those monomials at the points $P_1,\ldots,P_5$ should all be linearly independent. – Jyrki Lahtonen Apr 02 '13 at 17:57
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    (cont'd) With $n$ coordinates the number of quadratic monomials is $n(n+1)/2$, so the minimum number of points we need is $n+n(n+1)/2=n(n+3)/2$. I dare not guess, whether the linear dependence of those vectors can be checked more simply - I would keep the quadratic terms in the vector. I don't know how to (easily) formulate non-degeneracy. In case the quadric hypersurface splits to a union of two linear ones, there will be singular points on the surface, i.e. the Jacobian will not have full rank there. May be an algorithmic condition can be built out of that? – Jyrki Lahtonen Apr 02 '13 at 18:02

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Yes, what you say is true, as long as you are clear about the meaning of "degenerate".

Forgetting for the time being about the problem of not having enough points, the basic point is that there are two equivalent ways to think of degenerate conics in the plane:

1) those which have a singular point (perhaps a "point at infinity"); or

2) those which are defined by a reducible form.

Now, when we pass to three dimensions, both of these notions still make sense, but they are no longer the same! In your example, a cylinder is degenerate in sense 1) above (in fact, in the world of projective varieties, it's the same as a cone) but it is not degenerate in the sense of 2).

So the correct statement will be that 9 points in $\mathbf{R}^3$ determine a quadric, and if those points are in linearly general position, then the quadric is not reducible, i.e. degenerate in sense 2) above.

If you want to make sure they don't lie on a quadric which is degenerate in sense 1), you have to be a little bit more picky; unfortunately at the moment I can't think of a nice geometric characterisation of these sets of points at the moment.

Finally, note that all these things generalise to higher dimensions: the general statement is that ${n+2 \choose 2} -1$ points in general position will determine a unique quadric.

Hope that helps!

  • Yes this is very helpful, and I did confuse the two types of degeneracies. – guest1 Apr 02 '13 at 22:10
  • Glad it was useful! It's difficult not to be confused by this stuff --- it seems as though the terminology used to talk about real quadrics isn't quite consistent with usage in "mainstream" algebraic geometry. –  Apr 02 '13 at 22:14