The Segre Embedding

Today, we talk about the Segre Embedding. This will let us say a bit more about the products of varieties that were mentioned before.

We start out with the variety $\mathbb{P}^n\times\mathbb{P}^m$ , which we can define as before with the universal property. We define a map $\sigma:\mathbb{P}^n\times\mathbb{P}^m\to \mathbb{P}^{(m+1)(n+1)-1}$ by $((x_0:\ldots:x_n),(y_0:\ldots:y_m))\mapsto (\ldots:x_iy_j:\dots)$ . We denote the image of this map by $\Sigma_{n,m}$ .

We’ll use $z_{ij}$ to denote the homogeneous coordinates on $\mathbb{P}^{(m+1)(n+1)-1}$ . With these coordinates, $\Sigma_{n,m}$ is given as the zero set of the polynomials $z_{ab}z_{cd}-z_{cb}z_{ad}$ as $a,b,c,d$ vary. In fact, $\sigma$ is an isomorphism of $\mathbb{P}^n\times\mathbb{P}^m$ with $\Sigma_{n,m}$ .

The Segre embedding gives us a way of constructing products of varieties explicitly. Let $V$ and $W$ be quasi-projective varieties (so they could be affine, projective, open subsets of either, whatever). Then, they are contained in some projective spaces $V\subset\mathbb{P}^n$ and $W\subset\mathbb{P}^m$ . So then we can restrict the Segre embedding to $V\times W$ , and we obtain a quasi-projective variety in $\mathbb{P}^{(m+1)(n+1)-1}$ . This is isomorphic to $V\times W$ , and so we can take this to actually be $V\times W$ . This demonstrates that products of quasi-projective varieties always exist and are quasi-projective.

Let’s look more closely at the simplest Segre variety. That is, $n=m=1$ . This is an embedding $\mathbb{P}^1\times\mathbb{P}^1\to \mathbb{P}^3$ . The image is given by the equation $xw-yz=0$ , where $(x:y)$ were coordinates on the first factor and $(z:w)$ were from the second.

This shows that the image actually contains two families of lines. If we fix a point in either space, we get a linear map from $\mathbb{P}^1\to\mathbb{P}^3$ , and so we get two collections of lines. Different lines in each collection are disjoint, and lines in different families intersect at a single point.

Now we want to look at the degree of $\Sigma_{n,m}$ . If we take a homogeneous polynomial of degree $\ell$ on $\mathbb{P}^{(m+1)(n+1)-1}$ , it will restrict, on $\Sigma_{n,m}$ , to a polynomial which is homogeneous in each collection of variables separately and of degree $\ell$ in each. So the Hilbert Polynomial is the product $\binom{m+\ell}{m}\binom{n+\ell}{n}$ . This expands to $\frac{1}{m!n!}\ell^{m+n}+\ldots$ . The degree is the lead coefficient multiplied by the degree factorial, and so is $\frac{(m+n)!}{m!n!}=\binom{m+n}{n}$ .

The Segre embedding also interacts nicely with the Veronese embedding. We can obtain all sorts of new varieties isomorphic to $\mathbb{P}^n\times\mathbb{P}^m$ by combining the Veronese embedding and Segre embedding in various ways.

A nice fact about the two is that the diagonal $\Delta\subset \mathbb{P}^n\times\mathbb{P}^n$ has image under the Segre embedding in $\mathbb{P}^{n^2+2n}$ that is equal to $V_{2,n}$ if you choose the right subspace of $\mathbb{P}^{n^2+2n}$ .

That’s it for now, but we’ll return to Segre varieties later, if for no other reason because they tend to form nice special cases of other constructions.

About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.

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9 Responses to The Segre Embedding

ulfarsson says:

February 3, 2008 at 4:26 am

In paragraph 3 you should have a “\Sigma” rather than a “\sigma”.

Charles says:

February 4, 2008 at 2:10 am

That’s not a mistake, the statement is that the map $\sigma$ is an isomorphism $\mathbb{P}^n\times\mathbb{P}^m\to \Sigma_{n,m}$ .

twin7819 says:

December 18, 2009 at 6:00 pm

The ideal generated by z_{ab}z_{cd}-z_{cb}z_{ad} is prime, right? How should I prove that?

Charles Siegel says:

December 19, 2009 at 4:31 pm

The easiest way (to me) is to prove that these polynomials cut out precisely the image of the Segre map, which is a bijection and algebraic, and then use the fact that $\mathbb{P}^n\times\mathbb{P}^m$ is irreducible, so its image must be, and thus, the ideal must be prime. (Well, strictly, this says that the map has prime radical by the Nullstellensatz, but then you use that the generators are quadratic to show that it’s already a radical ideal)

twin7819 says:

January 3, 2010 at 3:10 am

Thank for the reply. How does the fact that generators being quadratic lead to the ideal being radical?
Anyway, I found a paper which proves determinantal ideal over integral domain is prime, and the proof is far from trivial!

Charles Siegel says:

January 4, 2010 at 4:24 pm

Ahh, actually, it’s those specific polynomials, I was just saying things that weren’t entirely accurate. I know it is prime (as that paper proves) but this case is easier. However, I’m blanking on it at the moment, and if it comes back to me, I’ll post something here. You might have some luck asking on Math Overflow (so long as you craft the question well enough that it is clearly not homework)

- twin7819 says:
  
  January 4, 2010 at 4:39 pm
  
  The paper’s proof was elementary enough (though not easy) for me to follow and digest, so I’m happy now. Thanks for your comment, though!
  
Saeed says:

May 25, 2010 at 1:41 pm

If we take the transversal intersection of $\mathbb{P}^n \times \mathbb{P}^m$ and $\mathbb{P}^{mn+m+n-2}$ , we have the generators of the complex cobordism group, which are called Milnor’s manifolds. Do they bound some real manifold?

Charles Siegel says:

May 25, 2010 at 1:58 pm

I actually know nothing about this…I think you need to check the Stiefel-Whitney classes to determine this. Also, are you certain that it’s manifolds? Because this would be a single connected manifold in projective space. Also, have you tried asking this on Math Overflow? Someone there might be able to help (of course, to write a good question there, include some background and motivational material)

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