# Veronese and Segre

In Harris's book there is an exercise that the image of the diagonal $\Delta \subset \mathbb{P}^n\times\mathbb{P}^n$ under Segre map is the Veronese variety $\nu_2(\mathbb{P}^n)$. I want to understand this in general. Is this correct for any variety? Precisely let $X\subseteq\mathbb{P}^n$ be a closed projective variety. Is the following correct? The $r$th Veronese of $X$ is the diagonal in the Segre i.e. $\nu_r(\mathbb{P}^n)=\Delta\cap (X\times\ldots\times X)$, Where $\Delta$ is the diagonal of $\mathbb{P}^n\times\ldots\mathbb{P}^n, \; r$-times and $X\times\ldots\times X$ is the $r$-fold (Segre) of $X$. If yes, how one should proceed to prove it. Only hints are enough.

• Compare the linear systems defining the embeddings. Nov 2 '16 at 6:51
• So the answer to this is Yes! ? Nov 2 '16 at 7:07

## 2 Answers

As Sasha said.

The Segre map of $\times_{i=1}^r \mathbb P^n$ is defined by the functions $\prod_{i=1}^rx_{ij(i)}$ where $j$ runs through all $\{1,\dots,r\}\to\{0,\dots,n\}$ functions.

The Veronese is the same if you make all $x_{ij}=x_{i'j}$ for all $i,i'\in \{1,\dots,r\}$.

One way to think of the Segre map $$\mathbb{P}^{n_1} \times \dotsm \times \mathbb{P}^{n_s} \to \mathbb{P}^N$$ is as the map corresponding to multiplication of linear forms. By this I mean, for each $$i=1,\dotsc,s$$, give $$\mathbb{P}^{n_i}$$ a basis of variables $$x_{i,0},\dotsc,x_{i,n_i}$$. The elements of $$\mathbb{P}^{n_i}$$ are linear forms in those variables, up to a scalar factor. The elements of $$\mathbb{P}^N$$ are polynomials in all the $$x_{i,j}$$'s, of total degree $$s$$; with the additional restriction that they are multihomogeneous of multidegree $$(1,1,\dotsc,1)$$. The Segre map is multiplication.

So for example an element in the domain might be $$((2x_{1,0} + 3x_{1,1}) , (x_{2,0} - x_{2,1} + x_{2,2}) , (5x_{3,0}))$$. And it would get mapped to the product of those factors, the polynomial $$(2x_{1,0} + 3x_{1,1})(x_{2,0} - x_{2,1} + x_{2,2})(5x_{3,0})$$.

Now if $$n_1 = n_2 = \dotsb = n_s$$, and we use the same basis for every $$\mathbb{P}^{n_i}$$, then the diagonal consists of multiplications with all factors the same, in other words, taking the $$s$$'th power of a linear form. And that's the Veronese map.

If there's any exercise remaining here, it's to see why these multiplication descriptions of the Segre and Veronese maps are the same as other definitions that might be given.