Exercise 1.24 from Joe Harris' Algebraic Geometry: First Course

Question

I have a question about solvability of Exercise 1.24 (p. 14) from Joe Harris' Algebraic Geometry: A First Course and correctness of following 'synthetic' construction which according to the book (or alternatively due to page 79 in this script) should give us a rational normal curve. (Since we want to work as geometric as possible, all constructions and spaces are defined over $\mathbb{C}$)

Harris wrote:

As indicated, we can generalize this to a construction of rational normal curves in any projective space $\mathbb{P}^d$. Specifically, start by choosing $d$ codimension two linear spaces $ \Lambda_i \cong \mathbb{P}^{d-2} \subset \mathbb{P}^d$. The family $\{H_i(\lambda)\}$ of hyperplanes in $\mathbb{P}^d$ containing $\Lambda_i$ is then parameterized by $\lambda \in \mathbb{P}^1$; choose such parameterizations, subject to the condition that for each $\lambda$ the planes $H_1(\lambda), ... , H_d(\lambda)$ are independent, i.e., intersect in a point $p(\lambda)$. It is then the case that the locus of these points $p(\lambda)$ as $\lambda$ varies in $\mathbb{P}^1$ is a rational normal curve.

Exercise 1.24. Verify the last statement

So our constructed curve $C$ is given by

$$ C:= \bigcup_{\lambda \in \mathbb{P}^{1}} H_1(\lambda) \cap ... \cap H_d(\lambda) $$

On page 10 Harris gave the standard definition of a rational normal curve: A rational normal curve is defined as the image of the map $v_d: \mathbb{P}^1 \to \mathbb{P}^d; \lambda \mapsto [A_0(\lambda):...: A_d(\lambda)]$ with an arbitrary basis $A_0, ... , A_d$ of the space of homogeneous polynomials of degree $d$ on $\mathbb{P}^1 $.

Note that this curve is projectively equivalent to the image of the Veronese map $[X_0:X_1] \mapsto [X_0^d: X_0^{d-1}X_1:..., X_1^d]$.

The exercise:
Suppose we start by choosing $d$ codimension two linear spaces $\Lambda_i \cong \mathbb{P}^{d-2} \subset \mathbb{P}^d, i=1,...,d$ (here I doubt that the $\Lambda_i$ can be choosen really arbitrary!) and consider $d$ families of hyperplanes $\{H_i(\lambda) \}_{\lambda \in \mathbb{P}^1}, i=1, ..., d$ where each $ H_i(\lambda)$ contains $\Lambda_i$ parametrized by the line $\mathbb{P}^1$ such that for each $\lambda \in \mathbb{P}^1$ the planes $H_1(\lambda),..., H_d(\lambda)$ are linear independent, i.e. intersect in a point $p(\lambda)$.

That is we have to check that the constructed curve $C$ above can be realized as the image of such map $v_d$, that means that the map

$$ p: \mathbb{P}^1 \to \mathbb{P}^d, \lambda \to p(\lambda)= H_1(\lambda) \cap ... \cap H_d(\lambda) $$

has the form $\lambda \to [A_0(\lambda):...: A_d(\lambda)]$ for appropriate basis $A_0,..., A_d$ of the space of homogeneous polynomials of degree $d$ on $\mathbb{P}^1$.

The way I tried to solve it contains a serious problem, see below. We parametrize every linear space $\Lambda_i$ as vanishing set of two independent linear polynomials $L_i, M_i \in \mathbb{C} \cdot Z_0 + \mathbb{C} \cdot Z_1... + \mathbb{C} \cdot Z_d $. So $\Lambda_i= V(L_i, M_i)$. Then every our pencil $\{H_i(\lambda) \}_{\lambda \in \mathbb{P}^1}$ can be parametrized as vanishing set

$$ \lambda_0 L_i + \lambda_1 M_i =0 $$

with running $\lambda=[\lambda_0: \lambda_1]$. If we reordner the terms after variables $Z_i$ we obtain $d$ equitions

$$ \lambda_0 L_i + \lambda_1 M_i = r_{i,Z_0} (\lambda)Z_0+ r_{i,Z_1} (\lambda)Z_1 + ... + r_{i, Z_d} (\lambda)Z_d =0 $$

We can naturaly encode the coefficents of $i$-th family as $i$-th row of following matrix $R \in Mat_{d+1}(\mathbb{C}[\lambda])$; the $(n+1)$-row we fill with zeroes:

$$ \begin{pmatrix} r_{1,Z_0} (\lambda) & r_{1,Z_1} (\lambda) & ... & r_{1,Z_d} (\lambda) \\ r_{2,Z_0} (\lambda) & r_{2,Z_1} (\lambda) & ... & r_{2,Z_d} (\lambda)\\ \\ ... \\ \\ ... \\ r_{d,Z_0} (\lambda) & r_{d,Z_1} (\lambda) & ... & r_{d,Z_d} (\lambda)\\ 0 & 0 & ... & 0 \\ \end{pmatrix} $$

Since for every $\lambda$ the $H_1(\lambda), ..., H_d(\lambda)$ are independent, this matrix $A$ has rank $d$.

What we do next looks promissing at first glance; we consider the adjugate matrix $A$ of $R$, that's a unique matrix $A \in Mat_{d+1}(\mathbb{C}[\lambda])$ with

$$ AR=RA= det(R) \cdot I_{n+1} = 0 $$

By definition of adjugate matrix it's $d+1$-th column $A_{\cdot, n+1}$ provides exactly the solution we are looking for: for every $\lambda$ it is up to multiplication by a scalar $c \neq 0$ the solution of linear equations encoded by matrix $R$ and every entry of this column is by construction a homogeneous polynomial of degree $d$ in $\lambda_0, \lambda_1$, let set

$$ (A_0(\lambda), ... , A_d(\lambda)):= A_{\cdot, n+1}^T $$

and pass to it's homogenization.

Problem (a really serious one): Why are the $A_i $ linearly independent, equivalently why they form a basis of the space of homogeneous deg $d$ polynomials in $\lambda_0, \lambda_1$? (see also the comment by lhl73 in this closely related question dealing with nondegeneracy of curves of this type separately)

Indeed, that's equivalent to the property that the image of $\lambda \mapsto [A_0(\lambda):...: A_d(\lambda)]$ is not contained in a proper hyperplane $H \cong \mathbb{P}^{d-1} \subset \mathbb{P}^{n-1}$.

And if we looking again at the construction, Harris nowhere imposed additional assumptions how the $\Lambda_i$ are related to each other; it doesn't rule out eg the bad case $\Lambda_1= \Lambda_2$, since one can find still two families $\{H_i(\lambda)\}, i=1,2$ of hyperplanes containing $\Lambda_1 (=\Lambda_2)$ with $H_1(\lambda) \neq H_2(\lambda)$ for every $\lambda$, so the imposed condition is not violated. But eg if we choose $\Lambda_1= \Lambda_2$ by construction the complete curve $C$ would be comtained in $\Lambda_1$, therefore the $A_i(\lambda)$ constructed as above will cannot be linearly independent and the curve will not be rational normal curve as defined by Harris on page 10.

Question: Does anybody have experience with this Exercise 1.24 and knows how to solve it correctly, or if it's indeed true that the quoted construction not always gives a rational normal curve in sense of Harris book? (there is also nowhere Errata of this book available)

Probably Harris also has forgotten to impose an additional assumption on the spaces $\Lambda_i$ in the construction above or I'm just too stupid to solve the exercise & understand the construction. If that's so, can the construction be slightly modified in a most general way when one would obtain always a rational normal curve? E.g. does the construction give us always rational normal curve if we additionally require that all $\Lambda_i$ should be distinct?

In any case I would be very thankful if anybody who has experience with this Exercise and construction would share how it can be solved correctly.

Answer 1

[THIS and approach in first UPDATE seem to contain gaps I not know how to fix, see UPDATE #2 part for argument which - if I'm not overlooking something - goes flawlessly through] I think here is a full solution why the polynomials $A_0(\lambda),..., A_d(\lambda)$ must be linearly independent and therefore build basis of $n$-forms on $\mathbb{P}^1$ (and that solves the "serious problem")

Recall that for every fix $\lambda=[\lambda_0: \lambda_1]$ the point $p_{[\lambda_0: \lambda_1]} \in \mathbb{P}^d$ is the unique solution of independent linear system

$$ \lambda_0 L_1 + \lambda_1 M_1 = ... = \lambda_0 L_d + \lambda_1 M_d= 0 $$

We reorder the terms after variables $Z_0, Z_1, ..., Z_d$ and obtain $d$ equations $$ \lambda_0 L_i + \lambda_1 M_i = r_{i,Z_0} (\lambda)Z_0+ r_{i,Z_1} (\lambda)Z_1 + ... + r_{i, Z_d} (\lambda)Z_d =0 $$

We naturaly encode the coefficents of $i$-th family as $i$-th row of following matrix $R \in Mat_{d+1}(\mathbb{C}[\lambda])$; the $(n+1)$-row we fill with zeroes:

$$ \begin{pmatrix} r_{1,Z_0} (\lambda) & r_{1,Z_1} (\lambda) & ... & r_{1,Z_d} (\lambda) \\ r_{2,Z_0} (\lambda) & r_{2,Z_1} (\lambda) & ... & r_{2,Z_d} (\lambda)\\ \\ ... \\ \\ ... \\ r_{d,Z_0} (\lambda) & r_{d,Z_1} (\lambda) & ... & r_{d,Z_d} (\lambda)\\ 0 & 0 & ... & 0 \\ \end{pmatrix} $$

There exist the adjugate matrix $A$ of $R$, that's a unique matrix $A \in Mat_{d+1}(\mathbb{C}[\lambda])$ with

$$ AR=RA= det(R) \cdot I_{n+1} = 0 $$

By definition of adjugate matrix it's $d+1$-th column $A_{\cdot, n+1}$ provides exactly the solution we are looking for: for every $\lambda$ it is up to multiplication by a scalar $c \neq 0$ the solution of linear equations encoded by matrix $R$ and every entry of this column is by construction a homogeneous polynomial of degree $d$ in $\lambda_0, \lambda_1$, let set

$$ (A_0(\lambda), ... , A_d(\lambda)):= A_{\cdot, n+1}^T $$

and pass to it's homogenization.

That's what we already know. Now the interesting part is why are $A_d(\lambda)$ $\mathbb{C}$-linearly independent. Assume the $A_0(\lambda), ... , A_d(\lambda)$ are lineraly dependent. Then there exist a $G \in GL_{n+1}(\mathbb{C})$ with

$$(G \cdot A_{\cdot, n+1})^T=(\widetilde{A_0}, ...,\widetilde{A_{d-1}}, 0)$$

where $G$ is the conposition of finitely many elementary row manipulation of $A$.

Because $\operatorname{adj}(AB)= \operatorname{adj}(B)\operatorname{adj}(A)$ we conclude that $\widetilde{A}:= \frac{1}{det(G)}GA$ is the adjugate matrix of $\widetilde{R}:=RG^{-1}$.

Since $\widetilde{R}$ differs from $R$ by finitely many column manipulations, we have $\widetilde{R}(\lambda)=$

$$ \begin{pmatrix} \widetilde{r}_{1,1} (\lambda) & \widetilde{r}_{1,2}(\lambda) & ... & \widetilde{r}_{1,d+1}(\lambda) \\ \\ ... \\ \\ ... \\ \widetilde{r}_{d,1} (\lambda) & \widetilde{r}_{d,2} (\lambda) & ... & \widetilde{r}_{1,d+1} (\lambda) \\ 0 & 0 & ... & 0 \\ \end{pmatrix} $$

and by definition of adjugate matrix we obtain $\operatorname{det}(\widetilde{R}_{d+1, d+1})=0$ for the $(d+1, d+1)$-minor of $\widetilde{R}$.

Therefore there exist a $Z:=[z_0: z_1:...: z_{d-1}:0] \in \mathbb{P}^d$ with $\widetilde{R}(\lambda) \cdot Z=0$ for all $\lambda \in \mathbb{P}^1$ simultaneously. Then our map $p$ is constant, a contradiction. Therefore the $A_i$ are linearly independent.

Last remark: How changes $p$ due to the left action of $G^{-1}$ of $R$. Well, this action changes the columns of $R$ and therefore changes just the coordinates $Z_0, Z_1,..., Z_d$; that means it maps the image under $p$ to a projectively equivalent curve. Since every projectively equivalent curve of a rational normal curve is rnc as well by definition, this action of column of not changes the content of the statement and we are done.

#UPDATE: Following solution should be "bug free" in order to finally close this issue:

We consider the matrix $\overline{R} \in Mat_{d,d+1}(\mathbb{C}[\lambda])$ (...as before, but without the artificial $d+1$-th zero row from above):

$$ \overline{R}(\lambda):= \ (r_0(\lambda) \ \ r_1(\lambda) \ ... \ r_{d}(\lambda)) = \ \ \ \begin{pmatrix} r_{0,1} (\lambda) & r_{1,1} (\lambda) & ... & r_{d,1} (\lambda) \\ r_{0,2} (\lambda) & r_{1,2} (\lambda) & ... & r_{d, 2} (\lambda)\\ \\ ... \\ \\ ... \\ r_{0, d} (\lambda) & r_{1, d} (\lambda) & ... & r_{d,d} (\lambda)\\ \end{pmatrix} $$

with columns $r_{d}(\lambda)$. Assume that the defined variety is degenerated, i.e. is contained in a hyperplane, wlog in $V_+(Z_d) \subset \mathbb{P}^d$ (otherwise make base change by homogenity). Then for every $\lambda$ the vectors $r_0(\lambda), r_1(\lambda),..., r_{d-1}(\lambda)$ would be linearly dependent, equivalently the determinant of $\overline{R}_d(\lambda)$ is zero for all $\lambda $ simultaneously, or, equivalently for every $\lambda$ there exist a projective $(d+1)$-vector $Z(\lambda):=[Z_0(\lambda): Z_1(\lambda):...:Z_{d-1}(\lambda):0]$ with

$$\sum_{i=0}^{d-1} r_i(\lambda) \cdot Z_i(\lambda)=0 $$

By construction, $Z(\lambda) =H(\lambda)$ and $Z(\lambda)$ unique, as otherwise $\overline{R}(\lambda)$ would have rank smaller then $d$ (= at least two inequiv lin equations between the $r_i(\lambda)$.)

Observe, that $Z_j(\lambda) \neq 0$ is an open condition on $\lambda \in \Bbb P^1$, so also the set $U \subset \Bbb P^1 $ of $\lambda$'s such that $Z_j(\lambda) \neq 0$ simultaneuosly for all $j=1,..., d-1$ is open.

Then we can find a $\lambda_0 \in U$ such that $\text{det} \overline{R}_j(\lambda_0)=0$ for some $j \in 0,..., d-1$ where $\overline{R}_j(\lambda)$ is obtained from $\overline{R}(\lambda)$ by discarding the vector $r_j(\lambda)$ and we then would obtain the contradiction to the rank $d$ assumption for $\overline{R}(\lambda)$.

UPDATE #2: The argument from previous update still flaws: A priori the reasoning in prev paragraph just implies that in eq $\sum_{i=0}^{d-1} r_i(\lambda) \cdot Z_i(\lambda)=0 $ for $\lambda=\lambda_0$ the coeff's $a_0(\lambda_0)$ and $a_d(\lambda_0)$ are zero, but that gives a priori not really gives contradiction i can see so far...

I think following argument should be flawless:

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 $C$. We observe that

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

and that elementary row- and column operations giving new matrix $\Omega'$ not change $C$. Let
$H_i(Z)$ with $i=1,..., m <d$ the maximal number of linearly independent forms in $Z_0,..., Z_d$, such that $C \subset V_+(H_i)$. 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_{d-m} & \overline{H_1} &... & \overline{H_m} \\ M_1 & ... M_{d-m} & H_1 &... & H_m\\ \end{pmatrix}

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

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 this 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 $t$.)

Proof for arbitrary $m$ follows from following:

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.

NB 1: By the way, if this argument is correct, it would not only work for rational normal curve, but for any rational normal scroll.

NB #2: I would appreciate to see if it's possible to give a shorter proof. :)