# Exercise 1.24 from Joe Harris' Algebraic Geometry: First Course

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.

• I'm pretty sure the $\Lambda_i$ cannot be chosen arbitrarily. Even being pairwise distinct is not enough: Consider the case $d = 3$ and the lines $\Lambda_i$ meet at a single point, i.e. $\Lambda_1 \cap \Lambda_2 \cap \Lambda_3 = \{q\} \subset \mathbb P^3$. Then $q$ will also be contained in each intersection $H_1(\lambda) \cap H_2(\lambda) \cap H_3(\lambda)$. Maybe assume first $\Lambda_i \cap \Lambda_j = \emptyset$ for all $i \neq j$? That should be the most general case. Jul 5, 2021 at 6:29
• Yes, that's a good counterexample. Your proposed suggestions to require additionally $\Lambda_i \cap \Lambda_j = \emptyset$ for $i \neq j$ seems to be specific feature of dimension $3$, since for $4 \le n$ any two linear subspaces $\Lambda_i,\Lambda_j \subset \mathbb{P}^n$ have a nonempfy intersection by dimension formula. Nevertheless one could try to generalize your approach in sense which seems to extend quite natural your requirements on $\Lambda_i$ in dimension $3$: Jul 6, 2021 at 0:18
• Just a suggestion, maybe one can ask if for arbitrary dimension $n$ there exist a function $r: \mathbb{N} \to \mathbb{N}$ with the property that any set $\Lambda_1,..., \Lambda_n \subset \mathbb{P}^n$ of codimension $2$ linear subspaces and hyperplane families $\{H_i(\lambda) \}$ satisfying the requirement from the constuction that for each $\lambda$ the $H_i(\lambda)$ intersect in unique point, the curve $$\bigcup_{\lambda \in \mathbb{P}^{1}} H_1(\lambda) \cap ... \cap H_d(\lambda)$$ is rational normal curve iff Jul 6, 2021 at 0:20
• for all $i_1, ..., i_m$ with $i_k \neq i_l$ for $k \neq l$ and $r(n) \le m$ the intersection $$\Lambda_{i_1} \cap ... \Lambda_{i_m}$$ is required to be empty, but I don't know if it is really a good way to try to generalize it, that's just an idea how I would try to extend your suggestion to arbitrary $\mathbb{P}^n$. Jul 6, 2021 at 0:20
• Please avoid over-editing your post. It creates unnecessary noise in the main site. Thank you,
– Pedro
Jul 11, 2021 at 11:17

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):

$$\overline{R}(\lambda):= \ (r_0(\lambda) \ \ r_1(\lambda) \ ... \ r_{d}(\lambda)) = \ \ \ \begin{pmatrix} r_{1,Z_0} (\lambda) & r_{1,Z_1} (\lambda) & ... & r_{1,Z_d} (\lambda) \\ r_{2,Z_0} (\lambda) & r_{2,Z_1} (\lambda) & ... & r_{2,Z_d} (\lambda)\\ \\ ... \\ \\ ... \\ r_{d,Z_0} (\lambda) & r_{d,Z_1} (\lambda) & ... & r_{d,Z_d} (\lambda)\\ \end{pmatrix}$$

with columns $$r_{d}(\lambda)$$. Assume that the defined variety is degenerated, i.e. is contained in a hyperplane, wlog (otherwise make base change by homogenity) in $$V_+(Z_d) \subset \mathbb{P}^d$$. 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.

On the other hand, there exist a $$\lambda_0$$ such that $$\det \overline{R}_0(\lambda_0)=0$$. In order to keep the rank of $$\overline{R}(\lambda_0)$$ to be still exactly $$d$$, $$\det \overline{R}_0(\lambda_0)=0$$ implies the the existence of a nontrivial linear equation between $$r_1(\lambda_0),r_2(\lambda_0)..., r_{d-1}(\lambda_0)$$.
But this implies that $$\det \overline{R}_i(\lambda_0)=0$$ for all $$i$$, and therefore the rank of $$\overline{R}(\lambda_0)$$ is smaller than $$d$$, a contradiction.