The Veronese Embedding

Now we’ve got some basic theory under our belts. Time to get a handle on specific objects. Nothing gives better intuition into algebraic geometry than working out a bunch of examples and seeing how they fit together. We’ll begin with the Veronese Embedding, which will give us a bunch of varieties that are isomorphic to projective space.

We begin by choosing some $n$ and looking at $\mathbb{P}^n$. Then, we have $n+1$ variables. We define the $d$-uple Veronese embedding to be the map taking $(x_0:x_1:\ldots:x_n)$ to (in some order) the point in a bigger projective space with each coordinate given by a monomial of degree $d$. The big projective space will be dimension $N=\binom{n+d}{n}-1$.

Before analyzing this map at all, let’s look at a quick example. If we start with $\mathbb{P}^1$ and look at the 3-uple embedding, we get a map $\mathbb{P}^1\to \mathbb{P}^3$ that takes $(x_0:x_1)\mapsto (x_0^3:x_0^2x_1:x_0x_1^2:x_1^3)$. The image of this map is called the twisted cubic curve in $\mathbb{P}^3$. More generally, the image of Veronese embeddings of $\mathbb{P}^1$ is called a Veronese Curve, or a rational normal curve.

Similarly, let’s look at the 2-uple embedding of $\mathbb{P}^2$ into $\mathbb{P}^5$. It takes $(x_0:x_1:x_2)\mapsto (x_0^2:x_0x_1:x_0x_2:x_1^2:x_1x_2:x_2^2)$. This is called the Veronese surface.

In fact, for convenience, we’ll take the ordering of the monomials to be the lexicographic order. That is, we think of each monomial as its ordered set of powers (so $x_0^2x_1^4x_2^0$ is $(2,4,0)$) and then order them by first looking to see which is bigger in the first coordinate, if they’re equal, move to the next, and so on. We can do this because choosing another order is just moving the Veronese variety around in projective space without changing it at all.

So now we want to actually check that the image is a variety. To check this, we just need to check that it is the zero set of some homogeneous prime ideal. The trick we’ll use to get the ideal is one that comes up regularly in algebraic geometry, and will be central to applications of the Elimination Theorem, which we will talk about later.

As $\mathbb{P}^n$ is the collection of lines in $\mathbb{A}^{n+1}$, we can look at maps of affine varieties for the moment, and then pass back to them if everything turns out homogenous. So we have a map $\mathbb{A}^{n+1}\to \mathbb{A}^{N+1}$. This gives us a map of rings $k[y_0,\ldots,y_N]\to k[x_0,\ldots,x_n]$ given by taking each $y_i$ to a different monomial of degree $d$ in the $x_j$‘s.

Now we quote two essential theorems of commutative algebra. Let $R$ be a ring and $I$ an ideal. Then $I$ is prime if and only if $R/I$ has no zero divisors. That is, there are no pairs of nonzero elements which multiply to zero. The second result we’ll need is that, given a surjective ring homomorphism $R\to S$, and letting $I$ be the ideal of elements sent to zero (called the kernel), then $R/I$ is isomorphic to $S$.

How do these fit together? Well, we can restrict the map to only hitting the image, and so making it surjective, and define $P$ to be the kernel. Then $k[y_0,\ldots,y_N]/P$ is isomorphic to some subring of $k[x_0,\ldots,x_n]$. However, this ring has no zero divisors, and so it’s subrings can’t either. This means that $k[y_0,\ldots,y_N]/P$ is an integral domain, and so $P$ is prime.

To see that it’s homogeneous, and so gives us a projective variety and not just an affine variety, we can take $f\in k[y_0,\ldots,y_N]$. We can write it as the sum of terms, each of which is homogeneous, as $f=\sum f_i$. It’s good enough to just show that $f$ is in the ideal if and only if each homogeneous part is in the ideal, because then it’s certainly generated by homogeneous polynomials. So we have $\phi(f)=\sum \phi(f_i)$, where $\phi$ is the map in question. As $\phi(f_i)$ has degree $di$, no terms in this sum can cancel, so if $\sum \phi(f_i)=0$, then $\phi(f_i)=0$ for all $i$. Thus, $f\in P$ if and only if $f_i\in P$ for all $i$. So the ideal is prime and homogeneous.

To carefully check that this gives us the variety we want is a bit trickier and technical, and so I just won’t touch it.

Now on to some properties of Veronese varieties. The Veronese Surface mentioned before has an extremely nice property: if $C\subset v_2(\mathbb{P}^2)$, where $v_d$ is the $d$-uple Veronese embedding, is a curve on the Veronese surface, then there exists a hypersurface $H$ in $\mathbb{P}^5$ such that $H\cap v_2(\mathbb{P}^2)=C$.

This isn’t even hard to prove, in fact. This curve is a curve in the projective plane, and so is defined by a polynomial $f(x_0,x_1,x_2)$ which is homogeneous. The same zero locus is cut out by $f^2$, and we can write this as $f^2(x_0,x_1,x_2)=g(x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2)$. Thus, we get a polynomial $g$ on $\mathbb{P}^5$, whose zero locus contains the curve. We choose the irreducible component of this hypersurface which intersects the Veronese surface, and the intersection is the curve. This property is quite nice.

Really, all we’ve done is embedded one projective space into a bigger one, so isomorphism invariants won’t tell us anything new. However, one thing of value that depends on the embedding is degree, so before we stop, let’s work out the degree of $v_d(\mathbb{P}^n)$.

Now, the Hilbert Function of $\mathbb{P}^n$ is $\binom{n+m}{n}$ as a function of $m$. We’ll use this to work out what happens under the Veronese map. Pick a polynomial of degree $m$ on $\mathbb{P}^N$. We can restrict it to the image of the Veronese map and get a polynomial on $\mathbb{P}^n$. Writing it using the coordinates on $\mathbb{P}^n$, the degree of the polynomial becomes $dm$, where $d$ is the degree of the Veronese map. As such, we have the multiply the variable in the Hilbert Function by $d$, and so we find that $P_{v_d(\mathbb{P}^n)}(m)=\binom{n+dm}{n}$. Expanding this as a polynomial, the lead term is $\frac{d^n}{n!}m^n$, and so the degree of a Veronese variety is $d^n$.

In the future, as a shorthand, we will denote the $d$-uple Veronese Embedding of $\mathbb{P}^n$ sometimes by $v_d(\mathbb{P}^n)$ and sometimes by $V_{d,n}$. 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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12 Responses to The Veronese Embedding

1. ulfarsson says:

A typo in line 1, should say: “under our belts”. In paragraph 4 your missing a \latex. In paragraph 10 you need to add a “\” to your “phi”.

2. Charles says:

Corrected, thanks.

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4. Anonymous says:

I think that there is a error on the dimension $N$ of the target space if the Veronese map. Shouldn’t it be $\binom{n+d}{d}-1$ ?

• upaudel says:

Of course, binomial formula is symmetric, which precisely means two expressions coincide.

5. upaudel says:

I would like to see your post on the “elimination theorem”. Could you give me the link?

• upaudel says:

Found. :)

6. Anonymous says:

I feel a bit confused. If I understand correctly the degree of the Veronese variety v_2(P^2) is 4. On the other hand, if you intersect v_2(P^2) with a line in P^5 you will get 3 points. Where is the problem ? (or maybe there is no problem : but for me if a variety has degree d this mean that the intersection with a plane of complementary dimension gives precisely d points. Maybe I am mistaken).

7. Anonymous says:

What is the map $\phi$ supposed to be?

8. malkoun says:

Hi Charles. I enjoyed reading this post. I was hoping you would write more about the biregularity property of the veronese embedding, which is mentioned in the Wikipedia article of the Veronese embedding, but I wanted to see a careful statement, including carefully stating what “points in general position” means. Do you happen to have a reference for that please?