Segre Embedding And Heights

Product

The Segre embedding allows us to define the product of projective varieties reasonably, and we will discuss it right now. To begin with we consider the product of $\mathbb{P}^m$ an $\mathbb{P}^n$. In this section the ground field is an arbitrary algebraically closed field.

Definition 1. The Segre embedding is defined as follows:

$$\begin{aligned} \iota:\mathbb{P}^m \times \mathbb{P}^n &\to \mathbb{P}^N \\ ([X_0:\cdots:X_m],[Y_0:\cdots:Y_n]) &\mapsto [X_0Y_0:X_0Y_1:\cdots:X_mY_n] \end{aligned}$$

Clearly, $N=(m+1)(n+1)-1=mn+m+n$. The image on the right hand side has $X_iY_j$ ordered lexicographically.

First of all we make sure that this function is well-defined, otherwise our work will be useless.

Proposition 1. The Segre embedding is a well-defined injective map.

Proof. Assume $X_i’=\lambda X_i$ and $Y_j’=\mu Y_j$ for some $\lambda,\mu \ne 0$, then

$$\begin{aligned} \iota([X_0':\dots:X_m'],[Y_0':\dots:Y_n'])&=[X_0'Y_0':\dots:X_m'Y_n'] \\ &=[\lambda\mu X_0Y_0:\dots:\lambda\mu X_mY_n] \\ &=\lambda\mu[X_0Y_0:\dots:X_mY_n] \\ &=[X_0Y_0:\dots:X_mY_n] \\ \end{aligned}$$

Next suppose that $[X_iY_j]=[X_i’Y_j’]$. Without loss of generality we can assume that $X_0 \ne 0$ and $X_0’ \ne 0$ so that we can put them to $1$. Then by looking at first $n+1$ elements we can identify $[Y_0:\dots:Y_n]$ with $[Y_0’:\dots:Y_n’]$. Then using $X_1Y_i=\lambda X_1’Y_i’$ we can immediately identify $X_1$ and $X_1’$. Likewise, other components are identified. $\square$

Next we study the image further using linear algebra

Image of the Segre Embedding And Rank of A Matrix

We can write elements in $\mathbb{P}^N$ as a $(m+1) \times (n+1)$ matrix, which can make things easier:

$$\begin{bmatrix} Z_{00} & \dots & Z_{0n} \\ \vdots & \ddots & \vdots \\ Z_{m0} & \cdots & Z_{mn} \end{bmatrix}.$$

Therefore the image of $\iota$ is given by $Z_{ij}=X_iY_j$. Through an elementary observation, we see the matrix

$$\begin{bmatrix} X_0Y_0 & \cdots & X_0Y_n \\ \vdots & \ddots & \vdots \\ X_mY_0 & \cdots & X_mY_n \end{bmatrix}$$

has rank $1$. The question is, is the converse true? For this reason we study the set

$$Z=Z(\{Z_{ij}Z_{kl}-Z_{kj}Z_{il}:1 \le i,k \le m,1 \le j,l \le n\}).$$

Note $Z_{ij}Z_{kl}-Z_{kj}Z_{il}$ is the determinant of all $2\times2$ submatrices the matrix $[Z_{ij}]$. This $Z$ contains all $[Z_{ij}]$ with rank $1$. To show the converse, we consider the standard affine cover. Let $U_{kl}=Z(Z_{kl})$ and put $V_{kl}=\mathbb{P}^N \setminus U_{kl}$. Then $\{V_{kl}\}$ is the standard affine cover of $\mathbb{P}^N$ as we all know. Then likewise we use the affine open subset $U_k \subset \mathbb{P}^m$ and $U_l’ \subset \mathbb{P}^n$, to obtain

$$V_{kl} \cap Z\to U_k \times U_l', \quad [Z_{ij}] \mapsto ([Z_{0l}:\dots:Z_{ml}],[Z_{k0}:\dots:Z_{kn}])$$

This is indeed the inverse map of $\iota$ on $U_k \times U_l’$. Hence the converse is true.

Therefore, the image of the Segre embedding is a projective variety. As a classic example, the image of $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$ is determined by the polynomial $xy-zw=0$.

Height

In this section we offer a way to understand the Segre embedding in number fields. To begin with, we need some definition.

Height is computed by absolute values on a field so we first normalise all absolute values on $\mathbb{Q}$. Recall that two absolute values $|\cdot|_1$ and $|\cdot|_2$ are equivalent if there exists some $\lambda>0$ such that $|\cdot|_1=|\cdot|_2^\lambda$. The question is, which $\lambda$ should we pick.

The ordinary absolute value $|\cdot|_\infty$ has nothing to worry about. But for $p$-adic absolute values $|\cdot|_p$, instead of using other equivalent ones, we have a restriction that $|p|_p=\frac{1}{p}$. All these absolute values will be denoted by

$$M_\mathbb{Q}=\{|\cdot|_p:p\text{ is prime or}=\infty\}.$$

Likewise we define $M_K$, where throughout $K$ will always be a number field. $M_K$ consists of the ordinary absolute value and extensions of $p$-adic ones defined as follows:

$$|x|_v=|N_{K_v/\mathbb{Q}_p}(x)|_p^{1/[K:\mathbb{Q}]},\quad \forall x\in K,v|p.$$

In particular, $M_K$ satisfies the product formula:

$$\prod_{v \in M_K}|x|_v=1 \text{ or } \sum_{v \in M_K}\log|x|_v=0$$

for all $x \in K^\times$. This restriction allows us to work fine with projective spaces, as we will see later.

Definition 2. The (absolute logarithmic) height of $x\in \mathbb{P}^n_{\overline{\mathbb{Q}}}$, with coordinates $(x_1,\dots,x_m) \in K$, is defined by

$$h(x)=\sum_{v \in M_K}\max_j \log |x_j|_v.$$

Actually, the height function can show the “algebraic complication” of $x$, and is well-defined in many senses.

Proposition 2. The height $h(x)$ is independent of the choice of $K$.

Sketch of the proof. Let $L$ be another number field containing $x_0,\dots,x_m$, then we can assume that $K \subset L$, then $L/K$ is a finite separable extension. But in this case,

$$\sum_{w|v}[L_w:K_v]=[L:K],$$

which implies that

$$\sum_{w|v}\log |x|_w=\log |x|_v.$$

Therefore

$$\sum_{w \in M_L}\max_j\log|x_j|_w=\sum_{v \in M_K}\sum_{w|v}\max_j\log|x_j|_w$$

gives what we want. $\square$

Proposition 3. $h(x)$ is well-defined on $\mathbb{P}^n_{\overline{\mathbb{Q}}}$.

Proof. It remains to show that $h(x)$ is independent of the choice of coordinates. For $\lambda \ne 0$, we see

$$\begin{aligned} h(\lambda{x})&=\sum_{v \in M_K}\max_j\log|\lambda x_j|_v \\ &=\sum_{v \in M_K}\left(\log|\lambda|_v+\max_j\log|x_j|_v \right) \\ &=\sum_{v \in M_K}\log|\lambda|_v+\sum_{v \in M_K}\max_j\log|x_j|_v \\ &=\sum_{v \in M_K}\max_j\log|x_j|_v \\ &=h(x). \end{aligned}$$

Note $\sum_{v \in M_K}\log|\lambda|_v=0$ because o the product formula. $\square$

To highlight the ability of height to measure algebraic complication, let’s mention the following theorem of Kronecker.

Theorem 1 (Kronecker). The height of $\zeta\in\overline{\mathbb{Q}}^\times$ is $0$ if and only if $\zeta$ is a root of unity.

One direction is straightforward. To prove the converse, one may need some combinatorics, symmetric functions and Dirichlet’s pigeon-hole principle. See theorem 2.4 of this note for a proof.

Segre Embedding And Height of Polynomials

Now let’s invite the Segre embedding into the party:

$$\begin{aligned} \iota:\mathbb{P}^n_{\overline{\mathbb{Q}}} \times \mathbb{P}^m_{\overline{\mathbb{Q}}} &\to \mathbb{P}^N_{\overline{\mathbb{Q}}} \\ (x,y) &\mapsto x \otimes y \\ &:=(x_iy_j). \end{aligned}$$

Using the fact that $\max_{i,j}|x_iy_j|_v=\max_i|x_i|_v \cdot \max_j|y_j|_v$, we see immediately that

$$h(x \otimes y) = h(x) + h(y).$$

The Segre embedding is immediately used after introducing the height of a polynomial.

Definition 3. For $f(t_1,\dots,t_n) \in K[t_1,\dots,t_n]$, we write

$$f(t_1,\dots,t_n)=\sum_{\mathbf{j}}a_{\mathbf{j}}\mathbf{t^j}.$$

Then the height of $f$ is defined to be

$$h(f)=\sum_{v \in M_K}\log |f|_v$$

where

$$|f|_v=\max_{\mathbf{j}}|a_{\mathbf{j}}|_v.$$

Likewise, it can show the complication of $f$ in some way, but we are not interested in it at this moment. Notice that products of multivariable polynomials in different variables can be understood as tensor products and therefore we have the following fact

Proposition 4. Let $f(t_1,\dots,t_n)$ and $g(s_1,\dots,s_m)$ be polynomials in different sets of variables, then

$$h(fg)=h(f)+h(g).$$

Note: if $f$ and $g$ share the same variable, the story is different. Say we have $f_1,\dots,f_m \in \overline{\mathbb{Q}}[X_1,\dots,X_n]$, and put $f=f_1 \dots f_n$, then

$$\left|h(f)-\sum_{j=1}^{m}h(f_j)\right| \le d\ln 2$$

where $d$ is the sum of the partial degrees of $f$.