Rational normal curve as determinantal variety

Question

We work here over complex numbers. Let $\Omega(Z)$

\begin{pmatrix} L_1 & L_2 & ... & L_n \\ M_1 & M_2 & ... & M_n\\ \end{pmatrix}

be a $2 \times n$ matrix of homogeneous linear forms $L_i(Z), M_j(Z) $ in variables $Z_0, Z_1,..., Z_n $ on $\mathbb{P}^n_{\mathbb{C}}$ satisfying the condition that for all $[\lambda, \mu] \in \mathbb{P}^1$ the linear forms $\lambda L_1+ \mu M_1, ..., \lambda L_n+ \mu M_n$ are linearly independent.

In Joe Harris' Algebraic Geometry, Exercise 1.25 (p 14) is asserted that the determinantal variety

$$\Sigma_1(\Omega) = \{[Z_0,..., Z_n] : \operatorname{rank} (\Omega(Z))) =1 \} \subset \mathbb{P}^n ,$$

that's the vanishing set of $2 \times 2$ minors of $\Omega(Z))$ is rational normal curve.

I not understand why $ \Sigma_1(\Omega)$ is a rational normal curve.
Recall that a rational normal curve is defined as image of

$$ f: \mathbb{P}^1 \to \mathbb{P}^n, \lambda \mapsto [A_0(\lambda):...: A_d(\lambda)] $$

where $A_0,..., A_n \in \mathbb{C}[X_0, X_1]$ are linearly independent homogeneous polynomials of degree $n$ on $\mathbb{P}^1$, equivalently $A_0,..., A_n$ form a basis of of the space of $n$-forms on $\mathbb{P}^1$.

The argument which Harris sketched looks incomplete: The argument was simply that for each $\lambda \in \mathbb{P}^1$ the equations

$$ \lambda L_1+ \mu M_1 = ... = \lambda L_n+ \mu M_n = 0 $$

determines the unique point $p_{[\lambda: \mu]}$, so we obtain the map $ f: \mathbb{P}^1 \to \mathbb{P}^n, [\lambda: \mu] \mapsto p_{[\lambda: \mu]}$. But why then $f$ is given by a basis of polynomials of degree $n$ on $\mathbb{P}^1$? I agree that $f$ is given as a column of certain adjugate matrix which has homogeneous polynomials of degree $n$ as entries. But why they are linearly independent?

Answer 1

I think following "elementary" argument should work:

That the homogeneous degree $n$ polys $A_i(\lambda), \ i \in 0,...,n$ constructed above not form a basis is equivalent to that the image of $f$ - the subvariety $\Sigma_1(\Omega) \subset \Bbb P^n$ - is degenerate, ie contained in at least one hyperplane. Let's assume this for sake of trying to deduce a contradiction. We consider the matrix $\Omega(Z)$

\begin{pmatrix} L_1 & L_2 & ... & L_n \\ M_1 & M_2 & ... & M_n\\ \end{pmatrix}

encoding the linear independent equations determining the constructed curve $\Sigma_1(\Omega) $. We observe that elementary row- and column operations giving new matrix $\Omega'$ clearly not change $\Sigma_1(\Omega) $. Let
$H_i(Z)$ with $i=1,..., m <d$ the maximal number of linearly independent forms in $Z_0,..., Z_d$, such that $\Sigma_1(\Omega) \subset V_+(H_i)$. By linearity, $H_i$ generate all forms with this property. Then it's easy to see that $\Omega $ is adjoint to $\Omega'$ given by

\begin{pmatrix} L_1 & ... L_{n-m} & \overline{H_1} &... & \overline{H_m} \\ M_1 & ... M_{n-m} & H_1 &... & H_m\\ \end{pmatrix}

where $\overline{H_j} \in \langle H_1,..., H_m \rangle$, as by construction $\Sigma_1(\Omega) \subset V_+(H_j)$ implies $\Sigma_1(\Omega) \subset V_+(\overline{H_j} )$.

Then it should be doable to check that if $\langle H_1,..., H_m \rangle = \langle \overline{H_1} ,..., \overline{H_m} \rangle$, then there exist a $[\lambda:\mu]$ such that $\lambda H_1 + \mu \overline{H_1} , ..., \lambda H_m + \mu \overline{H_m} $ are not linearly independent, a contradiction!

( eg in case $m=2$ this boils down to find $\lambda, \mu$ and $t \neq 0$ such that equality

$$ t(\lambda H_1 + \mu \overline{H_1})= \lambda H_2 + \mu \overline{H_2} $$

what can be turned in $2 \times 2$ matrix equation after having expressed $ \overline{H_1}, \overline{H_2}$ as linear combinations of $H_1,H_2$ - say $\overline{H_1} =aH_1+bH_2$ and $\overline{H_2}=cH_1+dH_2$ - and then making this replacement leads to a determinant zero condition wrt $t$ in siminar manner one calculates eigenvalues in linear algebra course, and so finally we obtain a nontrivial quadratic equation in $t$.)

Addentum: For arbitrary $m$ it follows from following easy fact:

Let $V \cong \Bbb C^m$ be a complex vector space and $\mathcal{B}_1:=(v_1,..., v_m ), \mathcal{B}_{\infty}:=(w_1,..., w_m )$ two arbitrary ordered sets of complete bases of $V$.
Let for $\lambda:=(\lambda_0:\lambda_1) \in \Bbb P^1(\Bbb C)$ be set

$$\mathcal{B}_{\lambda}:=(\lambda_0 \cdot v_1+ \lambda_1 \cdot w_1,..., \lambda_0 \cdot v_m+ \lambda_1 \cdot w_m) $$

Especially, $\mathcal{B}_{(1:0)}= \mathcal{B}_{1}, \mathcal{B}_{(0:1)}= \mathcal{B}_{\infty}$. Let us denote by $b_k(\lambda)$ the $k$-th member of $\mathcal{B}_{\lambda}$ and and a $m \times m$ matrix $M(\lambda) =(b_1(\lambda) \ ... \ b_n(\lambda))$ matrix with columns $b_k(\lambda)$. Then $\det M(\lambda)$ is clearly polynomial, so has homogeneous zeroes (as over $\Bbb C$), so there exist a $\lambda$ such that $\mathcal{B}_{\lambda}$ has non linear independent members.