Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There's an algebraic geometry theorem (I.4.9 in Hartshorne) that says: any variety of dimension r (over an algebraically closed field) is birationally equivalent to a hypersurface in projective space of dimension r+1. The proof is quite algebraic, and I'd like to see some interesting geometric examples.

share|improve this question
add comment

3 Answers

The proof is actually extremley geometric, if you just wave your hand hard enough. Take an r-dimension variety V in P^{r+d}. Pick a generic point Q in P^{r+d} (where d > 1), and project P^{r+d} from this point to P^{r+d-1}. A generic line through Q does not hit V at all, and a generic line that hits V, hits it at exactly one point (I waved my hands right here), so the projection is birational from V to it's image.

share|improve this answer
add comment

Perhaps the first example was in a paper of Corrado Segre, where he analyzed a quartic surface in P^3 with a double conic, as a projection of a complete intersection of two quadrics in P^4. I say first, because in that paper he states that some people may not believe in the existence of 4 dimensional space. but he does not think that is relevant and will use it anyway. Segre's surface, a special example of a del Pezzo surface, is discussed in Semple and Roth, p. 141., where the reference to his "epoch making" paper is given as math. ann. 24, (1884), 313.

Another nice example is the cubic surface in P^3, realized by projecting the veronese embedding of P^2 in P^9 by plane cubics. This time the projections are done successivelky from points on the surface, thus "blowing up" 6 points in the process.

A simpler example is the projection of a rational space cubic from P^3 to a nodal cubic curve in P^2. This illustrates the fact that general projections tend to pick up singularities. Fulton's and Hansen's connectedness theorem has as a consequence that a smooth surface projected from P^5 into P^3 from points off the surface picks up not only a double curve but "pinch" points as well (http://www.math.umn.edu/~roberts/ima_tutorial.html#projections).

A detailed geometric proof that any smooth space curve can be projected birationally into the plane with only nodes as singularities is in Mumford, Alg. Geom. I, Complex projective varieties, pp.132 ff.

share|improve this answer
add comment

Let $X\subset\mathbb{P}^N$ be a variety of dimension $r$ and assume $N\geq r+2$. Let $H$ be a general linear subspace of dimension $N-r-2$. A general linear space of dimension $N-r-1$ spanned by $H$ and a point $x\in X$ intersects $X$ exactly in $x$ bacause $N-r-1+r = N-1 < N$.Fix a general linear subspace $\Lambda$ of dimension $r+1$. Therefore the projection from $H$ $$\pi_H:X\rightarrow \Lambda\cong\mathbb{P}^{r+1},\: x\mapsto \left\langle x,H\right\rangle\cap\Lambda$$ is birational and $X$ is birational to the hypersurface $X^{'} = \overline{\pi_H(X)}\subset\mathbb{P}^{r+1}$. Note that since $H$ does not intersect $X$ we have $deg(X^{'}) = deg(X)$.

For instance projecting the Veronese surface $V\subset\mathbb{P}^5$ from a general line we get a surface $S\subset\mathbb{P}^3$ of degree $4$ and singular along three double lines. The surface $S$ is known as Steiner roman surface.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.