Represent $\text{Π}=PG\left(2,q^r\right)$ in $\text{Σ}=PG\left(2r,q\right)$ with $r\ge 2$, $q\ge 2r-1$.

Let $\mathcal{S}$ be a regular spread of $\left(r-1\right)$-spaces of a hyperplane ${\Sigma }^{\prime }=PG\left(2r-1,q\right)$, $\left|\mathcal{S}\right|=q^r+1$. The elements of $\mathcal{S}$ are the points at infinity of $\text{Π}$, the $r$-spaces $S_r$ such that $S_r\cap {\Sigma }^{\prime }\in \mathcal{S}$ are the affine lines of $\text{Π}$, $\mathcal{S}$ is the line at infinity $l_{\infty }$ of $\text{Π}$.

Note that a transversal $r$-space, that is, an $S_r$ with $S_r\cap {\Sigma }^{\prime }\notin \mathcal{S}$ can represent a Baer subplane $\beta$ of $\text{Π}$ only if $r$ is even. This is because $\beta$ would have order $q^{\frac{r}{2}}$, and hence the $\left(r-1\right)$-space $S_r\cap {\Sigma }^{\prime }$ must intersect exactly $q^{\frac{r}{2}}+1$ elements of $\mathcal{S}$.

Fix an $\left(r-1\right)$-space $S_{r-1}^{\infty }$ of $\mathcal{S}$, and choose a curve ${\mathcal{C}}_{\infty }^{r-1}\subset S_{r-1}^{\infty }$ of order $r-1$. Let $S_r^0$ be an $r$-space such that $S_r^0\cap {\Sigma }^{\prime }=S_{r-1}^0\ne S_{r-1}^{\infty }$, and let curve ${\mathcal{C}}_0^r\subset S_r^0$ be a curve satisfying ${\mathcal{C}}_0^r\cap S_{r-1}^0=\varnothing$.

Let $\pi =PG\left(2,q\right)$ be a non-affine subplane of $\text{Π}$ of order $q$ with $P_{\infty }\in l_{\infty }$ as its unique point at infinity, corresponding to the $\left(r-1\right)$-space $S_{r-1}^{\infty }$. Then $\pi$ is represented by the ruled variety $\mathcal{V}=V_2^{2r-1}$, obtained by connecting corresponding points of ${\mathcal{C}}_{\infty }^{r-1}$ and of ${\mathcal{C}}_0^r$ (see Result 3).

The point $P_{\infty }$ is the center of a bundle of $q+1$ lines in $\pi$. The remaining $q^2$ lines of $\pi$ each determine a unique point on the line $l_{\infty }$. Therefore, there exists a subset $\bar{\mathcal{S}}\subset \mathcal{S}$, with $\left|\bar{\mathcal{S}}\right|=q^2$ corresponding to these points.

NOTE 1—For each element $S_{r-1}\in \bar{\mathcal{S}}$, there exists a unique $r$-space $S_r$ that meets the ruled variety $\mathcal{V}$ in a directrix curve ${\mathcal{C}}^r$, which corresponds to a line of $\pi$. This is the only line of $\pi$ whose point at infinity is $S_{r-1}$. Similarly, for $S_{r-1}^0$, there exists a unique $r$-space $S_r^0$ that intersects $\mathcal{V}$ in the directrix curve ${\mathcal{C}}_0^r\subset S_r^0$.

Since ${\mathcal{C}}_{\infty }^{r-1}$ is a rational normal curve, it consists of $q+1$ points, no $r$ of which in a hyperplane $S_{r-2}$. This holds under the assumption $q\ge r-1$, which is satisfied by the hypothesis $q\ge 2r-1$ (cf. Definition 2.2).

Choose a subset consisting of $r-1$ independent points. Let denote the $\left(r-2\right)$-space of $S_{r-1}^{\infty }$ generated by the points of . For $r=2$ the set is a singleton.

Let be the set of the $r-1$ generatrix lines joining the points of and the corresponding $r-1$ points of ${\mathcal{C}}_0^r$. Let ${\mathcal{G}}^{\prime }=\left\{{g^{\prime }}_r,\cdots ,{g^{\prime }}_{q+1}\right\}$ be the set of the remaining $q+2-r$ generatrix lines.

The hyperplane intersects the ruled variety $\mathcal{V}$ in the union of the curve ${\mathcal{C}}_0^r$ and the set of generatrix lines (cf. Result 1, 2)).

Consider the subspace , the direct sum of the two subspaces.

The bundle ${ℬ}_{2r-2}$ of hyperplanes with axes $S_{2r-2}$ contains $q+1$ hyperplanes. Among them are the hyperplane ${\Sigma }^{\prime }$ and the hyperplane .

Each hyperplane $H^{\prime }\in {ℬ}_{2r-2}\backslash {\Sigma }^{\prime }$ intersects all $q+1$ generatrix lines . If $H^{\prime }\ne H$ and contains no generatrix line, then the $r-1$ lines of meet $H^{\prime }$ in the points of . The remaining $q+2-r$ lines in ${\mathcal{G}}^{\prime }$ meet $H^{\prime }$ in affine points. Denote this set of affine points by $\mathcal{Q}=\left\{Q_r,\cdots ,Q_{q+1}\right\}$ such a set.

Lemma 3.1. If $r>2$, then the set ${ℬ}_{2r-2}\backslash \left\{{\Sigma }^{\prime },H\right\}$ contains both hyperplanes that include one generatrix line from and hyperplanes that contain no generatrix line from .

If $r=2$, there exists exactly one hyperplane in ${ℬ}_{2r-2}\backslash {\Sigma }^{\prime }$ that contains a generatrix line.

Proof. From Result 1, it follows that a hyperplane $H^{\prime }\in {ℬ}_{2r-2}\backslash \left\{{\Sigma }^{\prime },H\right\}$ intersects the ruled variety $\mathcal{V}$ either in a rational normal curve ${\mathcal{C}}^{2r-1}$ of degree $2r-1$, or in a curve of order $m<n-1$, which is met by all the generatrix lines and in $n-1-m$ generatrix lines.

Let $r>2$. Assume that $H^{\prime }\cap \mathcal{V}$ contains at least two generatrix lines, . Since all generatrix lines intersect each directrix curve, denote $A_i$ and $A_j$ the points on $g_i$ and $g_j$, respectively, that lie on the directrix curve ${\mathcal{C}}_0^r\subset S_r^0$. Note that $A_i,A_j\notin \mathcal{Q}$. Because $S_{r-1}^0\subset H^{\prime }$ and the line $A_i{A}_j$ meets $S_{r-1}^0$, it follows that $H^{\prime }$ contains the entire $S_0^r$ and $H^{\prime }=H$, which contradicts the assumption.

Hence $H^{\prime }\cap \mathcal{V}$ contains at most one generatrix line.

Now, assume that each of the $q-1$ hyperplanes of ${ℬ}_{2r-2}\backslash \left\{{\Sigma }^{\prime },H\right\}$ contains exactly one generatrix line from . Let $H^{\prime }$ and $H^{″}$ be two distinct hyperplanes from , and let $g_1\subset H^{\prime }$, $g_2\subset H^{″}$ be two generatrix lines such that .

Two distinct hyperplanes in ${ℬ}_{2r-2}\backslash \left\{{\Sigma }^{\prime },H\right\}$ can intersect only in the space $S_{2r-2}$. Therefore such generatrix lines they contain must be distinct. Since $q\ge 2r-1>r$, we have , which implies that there more generatrix lines than elements in , contradicting the definition of . This contradiction completes the argument.

Hence in ${ℬ}_{2r-2}\backslash \left\{{\Sigma }^{\prime },H\right\}$ there are both hyperplanes containing a generatrix of and hyperplanes that contain none.

If $r=2$, , then there is only one generatrix line $g$ through . Hence, there exists a unique hyperplane in ${ℬ}_{2r-2}\backslash {\Sigma }^{\prime }$ that contains $g$.

The following cases should be considered.

Theorem 3.2. i) If $r\ge 2$ and $H^{\prime }\cap \mathcal{V}$ contains no generatrix line, then $H^{\prime }\cap \mathcal{V}$ is a rational normal curve ${\mathcal{C}}^{2r-1}$.

ii₁) If $r>2$ and $H^{\prime }\cap \mathcal{V}$ contains a generatrix line , then $H^{\prime }\cap \mathcal{V}=g\cup {\mathcal{C}}^{2r-2}$.

In both cases, the curve consists of the $r-1$ points of and of $q+2-r$ affine points, each lying on one of the $q+2-r$ generatrix lines of ${\mathcal{G}}^{\prime }$ with no two of them lying on the same directrix.

ii₂) If $r=2$ and $H^{\prime }\cap \mathcal{V}$ contains a generatrix line $g$, then $H^{\prime }\cap \mathcal{V}=g\cup \mathcal{C}$ where $\mathcal{C}$ is a conic directrix.

Proof. i) Assume that $H^{\prime }$ contains no generatrix line from .

The hyperplane $H^{\prime }$ intersects all $q+1$ generatrix lines of ${\mathcal{G}}_{}\cup {\mathcal{G}}^{\prime }$. The $r-1$ lines of are intersected at the points of , while the remaining $q+2-r$ lines of ${\mathcal{G}}^{\prime }$ are intersected at affine points. Denote $\mathcal{Q}=\left\{Q_r,\cdots ,Q_{q+1}\right\}$ the set of these affine points.

Then, from Result 1, 1), it follows that $H\cap \mathcal{V}={\mathcal{C}}^{2r-1}$, where is a rational normal curve. The condition $q\ge 2r-1$, although assumed as hypothesis, can be proved although it was taken as a hypothesis, can be proven (cf. [10] , Theorem 21.1.1, (i)).

Assume that a point $Q\in \mathcal{Q}$ lies on a generatrix line . Then the line $g$, containing two points in $H^{\prime }$, must lie entirely in $H^{\prime }$, which contradicts the assumption. A similar contradiction arises if either a point $Q$ and a point , or two points $Q_j,Q_h\in \mathcal{Q}$, lie on the same generatrix $g$.

Hence, the $q+2-r$ affine points of $\mathcal{Q}$ are distributed one per line over the $q+1-\left(r-1\right)=q+2-r$ generatrix lines $\left\{{g^{\prime }}_r,\cdots ,{g^{\prime }}_{q+1}\right\}$.

Assume that $H^{\prime }$ contains two points $A,B$ on a directrix curve. Since $H^{\prime }\supset S_{2r-2}\supset S_{r-1}^0$ this directrix must be ${\mathcal{C}}_0^r$, because the line $AB$ intersects $S_{r-1}^0$. Therefore $H^{\prime }$ would contain the entire $S_0^r$, implying that $H^{\prime }=H$, which is a contradiction.

ii₁) If $H^{\prime }$ contains one generatrix , then from Result 1, 2), it follows that $2r-1-m=1$ so that $m=2r-2$. Therefore $H^{\prime }\cap \mathcal{V}=g\cup {\mathcal{C}}^{2r-2}$. From Result 3, since ${\mathcal{C}}^{2r-2}$ is irreducible, it is a rational normal curve and hence lies in a subspace ${S^{\prime }}_{2r-2}$ of $H^{\prime }$. Note that ${\mathcal{C}}^{2r-2}$, in addition to $\mathcal{Q}$, contains the entire set , including the point , otherwise it would have only $q$ points. Moreover, it meets all the generatrix lines.

There are no affine point of ${\mathcal{C}}^{2r-2}$ on $g$; otherwise, ${\mathcal{C}}^{2r-2}$ would have $q+2$ points. Therefore, the line $g$ intersects ${\mathcal{C}}^{2r-2}$ in exactly one point and does not lie in $S_{2r-2}$. The proof of the first property of ${\mathcal{C}}^{2r-2}$ is analogous to the proof in i), where contradictions arise from the possibility that $H^{\prime }$ contains more than one generatrix. The proof of the second one is similar.

ii₂) Let $r=2$. Then and $H^{\prime }=S_3$. The line $g\subset H^{\prime }$ is the unique generatrix line, so the residual curve of $H^{\prime }\cap \mathcal{V}$ is a conic $\mathcal{C}$. Let $\alpha$ denote the plane of $\mathcal{C}$; clearly, $\alpha \subset H^{\prime }$.

Assume that the line $l=\alpha \cap {\Sigma }^{\prime }$ contains the point $P$. If $l$ coincides with the line ${\mathcal{C}}_{\infty }^1$, then $\alpha$ would contain $q$ skew generatrix lines of $V_2^3$, which is a contradiction. If $l$ is a line of ${\Sigma }^{\prime }$ that meets $q+1$ lines of the spread, ${\mathcal{C}}_{\infty }^1$ included, then $\alpha$ is a transversal plane and thus represents an affine Baer subplane of $\text{Π}$. Since it would share $q$ points with the non-affine Baer subplane represented by $V_2^3$, the two would have to coincide, again leading to a contradiction. Therefore $l$ is a line of $\mathcal{S}\backslash {\mathcal{C}}_{\infty }^1$, and we conclude that $H^{\prime }\cap V_2^3=g\cup \mathcal{C}$, where $\mathcal{C}$ is a directrix curve, that, as required, intersects all the generatrix lines.

Let $H^{\prime }$ be a hyperplane in the bundle ${ℬ}_{2r-2}\backslash \left\{{\Sigma }^{\prime },H\right\}$ that contains no generatrix line. Denote by $\mathcal{Q}=\left\{H^{\prime }\cap {g^{\prime }}_i=Q_i|i=r,\cdots ,q+1\right\}$ the set of the $q+2-r$ affine points where $H^{\prime }$ intersects the generatrix lines ${g^{\prime }}_i$.

Let ${\mathcal{Q}}^{\prime }=\left\{{Q^{\prime }}_r,\cdots ,{Q^{\prime }}_{q+1}\right\}$ be the subset of points of the projective plane $\pi \subset \text{Π}$, represented in $\text{Σ}$ by the points of $\mathcal{Q}$, to which we add the point $P_{\infty }$, represented by $S_{r-1}^{\infty }$. Denote by $\mathcal{K}={\mathcal{Q}}^{\prime }\cup \left\{P_{\infty }\right\}$ this subset of points of $\pi$.

Proposition 3.3. The set $\mathcal{K}$ is a $\left(q+3-r\right)$-arc of $\pi$. It is maximal, that is, a $\left(q+1\right)$-arc, when $r=2$. In this case, if $q$ is odd, $\mathcal{K}$ is a conic.

Proof. First, note that $\mathcal{K}$ has cardinality $q+1-\left(r-1\right)+1=q+3-r$.

If three affine points of $\mathcal{K}$ were collinear, then the corresponding points in $\mathcal{Q}$ would lie on a directrix curve, contradicting Theorem 3.2. Similarly, if the points ${Q^{\prime }}_i,{Q^{\prime }}_j,P_{\infty }$ were collinear, their line would pass through $P_{\infty }$, implying that the two affine points $Q_i$ and $Q_j$ of ${\mathcal{C}}^{2r-1}$ lie on a generatrix line of $\mathcal{V}$, again contradicting both our assumption and Theorem 3.2.

The arc $\mathcal{K}$ is maximal when $q+3-r=q+1$, that is, when $r=2$, In this case, $\text{Π}=PG\left(2,q^2\right)$ and $\pi$ is a Baer subplane. If $q$ is odd, $\mathcal{K}$ is a conic.

Denote by $t$ the tangent line of $\mathcal{K}$ at the point $P_{\infty }$, and let $g_t$ be the corresponding line of the variety $\mathcal{V}$ in $\text{Σ}$.

Corollary 3.4. The line $t$ is represented in $\Sigma$ by a generatrix line $g_t$ such that $g_t\notin H^{\prime }$.

Proof. The tangent line $t$ to $\mathcal{K}$ at the point $P_{\infty }$ is a line through $P_{\infty }$ and, clearly, it lies in the subplane $\pi$. Since all lines through $P_{\infty }$ in $\pi$ are represented in $\text{Σ}$ by the generatrix lines of $\mathcal{V}$, it follows that $g_t$ is a generatrix line of $\mathcal{V}$. By hypothesis, $H^{\prime }$ contains no generatrix lines; therefore $g_t\notin H^{\prime }$.

Let $\mathcal{C}\subset \pi$ be a non degenerate conic containing the unique point at infinity $P_{\infty }$ of $\pi$. Denote by $\left\{{P^{\prime }}_1,\cdots ,{P^{\prime }}_q\right\}$ the $q$ affine points of $\mathcal{C}$, and by $\left\{{g^{\prime }}_1,\cdots ,{g^{\prime }}_q\right\}$ the $q$ affine lines of $\pi$ joining $P_{\infty }$ with the points $\left\{{P^{\prime }}_1,\cdots ,{P^{\prime }}_q\right\}$. Let $t$ be the tangent line to $\mathcal{C}$ at $P_{\infty }$.

Let ${\mathcal{K}}_{\mathcal{C}}$ be the subset of $\mathcal{V}$ corresponding in $\text{Σ}$ to the points of $\mathcal{C}$, $\mathcal{K}=\left\{P_1,\cdots ,P_q\right\}$ is the set of the $q$ affine points of ${\mathcal{K}}_{\mathcal{C}}$ representing $\left\{{P^{\prime }}_1,\cdots ,{P^{\prime }}_q\right\}$, $\mathcal{G}=\left\{g_1,\cdots ,g_q,g_t\right\}$ the set of the $q+1$ generatrix lines of $\mathcal{V}$ corresponding to $\left\{{g^{\prime }}_1,\cdots ,{g^{\prime }}_q,g_t\right\}$ where $g_t$ is the generatrix representing $t$, $T$ the point $g_t\cap {\mathcal{C}}_{\infty }^{r-1}$.

Theorem 3.5. i) The set ${\mathcal{K}}_{\mathcal{C}}$ consists of $q+1$ points of $\mathcal{V}$. Among them, the $q$ affine points of $\mathcal{K}$ each lie on a distinct generatrix in the set $\left\{g_1,\cdots ,g_q\right\}$. The point $T$, which lies on the curve ${\mathcal{C}}_{\infty }^{r-1}$, is the unique point at infinity of ${\mathcal{K}}_{\mathcal{C}}$.

ii) The set ${\mathcal{K}}_{\mathcal{C}}$ forms a $\left(q+1\right)$-cap, with the line $g_t$ being the tangent to ${\mathcal{K}}_{\mathcal{C}}$ at the point $T$.

Proof. i) Obviously no point of $\mathcal{K}$ belongs to the generatrix $g_t$ as no affine point of the conic $\mathcal{C}$ belongs to the tangent $t$.

Assume two points of $\mathcal{K}$ lie on the same generatrix $g\in \left\{g_1,\cdots ,g_q\right\}$. That would imply that the corresponding two points of the conic $\mathcal{C}$ are collinear with $P_{\infty }$, contradicting the fact that $\mathcal{C}$ is non-degenerate. Hence, the $q$ affine points of $\mathcal{K}$ lie each on a distinct generatrix among $g_1,\cdots ,g_q$, all different from the tangent line $g_t$; that is, $P_i\in g_i$ for $i=1,\cdots ,q$.

Assume that ${\mathcal{K}}_{\mathcal{C}}$ contains a point $T^{\prime }\in {\mathcal{C}}_{\infty }^{r-1}$, with $T^{\prime }\ne T$. Then the generatrix containing $T^{\prime }$ must be a line $g_i\in \mathcal{G}$ for some $i=1,\cdots ,q$. Since $P_i\in g_i$, the entire generatrix $g_i=\overline{T^{\prime }{P}_i}$ would lie in the configuration. This would introduce to $\mathcal{K}$ the remaining $q-1$ affine points of the generatrix $g_i$, whose corresponding points in $\pi$ would lie on a line through $P_{\infty }$. That line would then be added to the conic $\mathcal{C}$, contradicting its non-degeneracy. Therefore $T^{\prime }=T$, and $T$ is the unique point at infinity of ${\mathcal{K}}_{\mathcal{C}}$.

ii) First, note that no two points of $\mathcal{K}$ lie on the same generatrix, since no two affine points of the conic $\mathcal{C}$ corresponding to them are collinear with $P_{\infty }$.

Assume that $P_i,P_j,P_h\in \mathcal{K}$ are collinear, and let $l$ be the line passing through them. Let ${P^{\prime }}_i,{P^{\prime }}_j,{P^{\prime }}_h$ be the corresponding points on the conic $\mathcal{C}$. The line $l$ is not a generatrix line, and therefore it determines, via its intersection point $l\cap {\Sigma }^{\prime }$ an element ${S^{\prime }}_{r-1}\in \mathcal{S}$, and subsequently a space ${S^{\prime }}_r$ such that ${S^{\prime }}_r\cap {\Sigma }^{\prime }={S^{\prime }}_{r-1}$. This space ${S^{\prime }}_r$ contains a directrix curve ${{\mathcal{C}}^{\prime }}^r$.

The line in $\pi$ through $P_i$ and $P_j$ is represented in $\text{Σ}$ by the curve ${{\mathcal{C}}^{\prime }}^r$, that is the same line in $\pi$ through $P_i,P_h$ and $P_j,P_h$. Hence, the three points ${P^{\prime }}_i,{P^{\prime }}_j,{P^{\prime }}_h\in \mathcal{C}$ must lie on a common line in $\pi$, a contradiction to the fact that $\mathcal{C}$ is a non-degenerate conic.

The line $g_t\in \mathcal{V}$, corresponding of the tangent $t$ to the conic $\mathcal{C}$ at the point $P_{\infty }$, has only the point $T$ in common with ${\mathcal{K}}_{\mathcal{C}}$. Therefore, $g_t$ is the tangent line to ${\mathcal{K}}_{\mathcal{C}}$ at $T$.

To complete the characterization of ${\mathcal{K}}_{\mathcal{C}}\subset \mathcal{V}$, it remains to understand the nature of ${\mathcal{K}}_{\mathcal{C}}$ in the case where it is contained in the intersection of $\mathcal{V}$ with a hyperplane.

Let $S^{\prime }=S_{2r-1}$ be a hyperplane such that ${\mathcal{K}}_{\mathcal{C}}\subset S^{\prime }$. Necessarily $S^{\prime }\ne {\Sigma }^{\prime }$.

The subspace ${S^{\prime }}_{2r-2}=S^{\prime }\cap {\Sigma }^{\prime }$ may either intersect each element of the spread $\mathcal{S}$ in subspaces of the same dimension, or contain one element of $\mathcal{S}$ entirely.

Note that the number of points in ${S^{\prime }}_{2r-2}$ is $q^{2r-2}+q^{2r-3}+\cdots +q+1$.

Since this number is not divisible by $q^r+1=\left|\mathcal{S}\right|$, the first case, where ${S^{\prime }}_{2r-2}$ intersects each element of the spread $\mathcal{S}$ in subspaces of equal dimension, cannot occur. Hence ${S^{\prime }}_{2r-2}$ must contain one element of the spread.

1) Assume that ${S^{\prime }}_{2r-2}\supset S_{r-1}^{\infty }$, so that $S^{\prime }\supset S_{r-1}^{\infty }$.

Then $S^{\prime }$ contains the curve ${\mathcal{C}}_{\infty }^{r-1}$ as well as the $q$ generatrix lines of $\mathcal{G}$ (excluding $g_t$) that connect the $q$ affine points of ${\mathcal{K}}_{\mathcal{C}}$ to the corresponding $q$ points of ${\mathcal{C}}_{\infty }^{r-1}\backslash \left\{T\right\}$. Let $g$ denote the set of the affine points on the line $g_t$. Then $S^{\prime }\cap \mathcal{V}=\mathcal{V}\backslash \left\{g\right\}$, which consists of the $q$ generatrix lines together with ${\mathcal{C}}_{\infty }^{r-1}$, that is, a reducible curve of order $q+r+1>2r-1$, contradicting Result 1.

2) Assume ${S^{\prime }}_{2r-2}$ contains an element of ${S^{\prime }}_{r-1}\in \mathcal{S}\backslash \left\{S_{r-1}^{\infty }\right\}$, so that $S^{\prime }\supset {S^{\prime }}_{r-1}$.

Since $S^{\prime }\ne {\Sigma }^{\prime }$, it follows that $S^{\prime }\cap S_{r-1}^{\infty }$ is a subspace ${\bar{S}}_{r-2}$, which intersects the curve ${\mathcal{C}}_{\infty }^{r-1}$ in $r-1$ points, including the point $T$ (cf. Definition 2.2).

Define ${S^{\prime }}_{2r-2}={S^{\prime }}_{r-1}+{\bar{S}}_{r-2}$.

There are $q+1$ hyperplanes containing ${S^{\prime }}_{2r-2}$, one of which one is ${\Sigma }^{\prime }$ (which contains $S_{r-1}^{\infty }$). Another hyperplane contains an $r$-dimensional subspace ${S^{\prime }}_r$ such that ${S^{\prime }}_r\cap {\Sigma }^{\prime }={S^{\prime }}_{r-1}$. This subspace ${S^{\prime }}_r$ represents a line $d$ of $\text{Π}$, and its intersection with $\mathcal{V}$ is a directrix curve ${\mathcal{C}}_d^r$, which corresponds to the unique line of $\pi$ through the point at infinity represented by .

Since $S^{\prime }$ contains the $r-1$ generatrix lines, say $g_1,\cdots ,g_{r-1}$, through the points of ${\bar{S}}_{r-2}$, as well as the corresponding points of the directrix curve ${\mathcal{C}}_d^r$, it follows that

$S^{\prime }\cap \mathcal{V}={\mathcal{C}}_d^r\cup \left\{g_1,\cdots ,g_{r-1}\right\},$

in accordance with Result 1.

Let $A,B\in \mathcal{K}$ be any two affine points of ${\mathcal{K}}_{\mathcal{C}}$, and let $A_{\pi },B_{\pi }$ be the corresponding points in $\pi$ represented by $A$ and $B$, respectively; note that $A_{\pi },B_{\pi }\in \mathcal{C}$. By hypothesis $A$ and $B$ lie in the hyperplane $S^{\prime }$, since $S^{\prime }\supset {\mathcal{K}}_{\mathcal{C}}$. Then the point at infinity $R_{\infty }$ of the line $r=AB$ lies either in ${\bar{S}}_{r-2}$ or in ${S^{\prime }}_{r-1}$; that is, $R_{\infty }\in S_{r-1}^{\infty }\cup {S^{\prime }}_{r-1}$.

The line $r_{\pi }=A_{\pi }{B}_{\pi }$ is a secant of the conic $\mathcal{C}$, and therefore it cannot pass through $P_{\infty }$. This implies that the point at infinity $R_{\infty }$ of the line $r=AB$ cannot lie in $S_{r-1}^{\infty }$.

If, on the other hand, $R_{\infty }\in {S^{\prime }}_{r-1}$, then $A,B\in {\mathcal{C}}_d^r$. Since this would hold for any pair of affine points of $A,B\in \mathcal{K}$, it would follow that $\mathcal{K}={\mathcal{C}}_d^r$. However, in $\pi$, the points of ${\mathcal{C}}_d^r$ are represented by the line $d$, and thus $\mathcal{C}$ would have to coincide with the line $d$, contradicting the assumption that $\mathcal{C}$ is a non-degenerate conic.

Therefore, ${\mathcal{K}}_{\mathcal{C}}$ cannot be contained in such a hyperplane $S^{\prime }$ representing a line of $\pi$.

Based on cases of 1) and 2), we now make the following choices. Fix a subspapce ${S^{\prime }}_{r-1}\in \bar{\mathcal{S}}=\mathcal{S}\backslash S_{r-1}^{\infty }$, and choose a subspace ${\bar{S}}_{r-2}\subset S_{r-1}^{\infty }$ such that ${\bar{S}}_{r-2}$ intersects the curve ${\mathcal{C}}_{\infty }^{r-1}$ in exactly $r-1$ points. Define

${S^{\prime }}_{2r-2}={S^{\prime }}_{r-1}+{\bar{S}}_{r-2}.$

In the bundle ${ℬ}_{2r-2}$ of hyperplanes with axes ${S^{\prime }}_{2r-2}$, there are $q+1$ hyperplanes: one is $\text{Σ}\text{'}$, and another is $H={S^{\prime }}_r+{\bar{S}}_{r-2}$, where ${S^{\prime }}_r$ represents a line $d$ of $\pi$. Thus,

$\left|{ℬ}_{2r-2}\backslash \left\{{\Sigma }^{\prime },H\right\}\right|=q-1.$

Choose a hyperplane $S^{\prime }\in {ℬ}_{2r-2}\backslash \left\{{\Sigma }^{\prime },H\right\}$.

NOTE 2—From NOTE 1, it follows that for each choice of ${S^{\prime }}_{r-1}\in \bar{\mathcal{S}}$, there are $q^2$ possibilities, and for each of these, there are $q-1$ hyperplanes like $S^{\prime }$. Moreover, one must count the possible choices of $r-1$ independent points on ${\mathcal{C}}_{\infty }^{r-1}$, each determining a distinct subspace ${\bar{S}}_{r-2}$, and hence giving rise to different hyperplanes.

Assume that $S^{\prime }$ contains ${\mathcal{K}}_{\mathcal{C}}$. In this case, the point $T=g_t\cap {\mathcal{C}}_{\infty }^{r-1}$ is one of the $r-1$ points of the intersection ${\bar{S}}_{r-2}\cap {\mathcal{C}}_{\infty }^{r-1}$, and the line $g_t$ cannot contain any affine points. Denote by $\left\{g_1,\cdots ,g_{r-2},g_t\right\}$ the $r-1$ generatrix lines passing through the points of ${\bar{S}}_{r-2}\cap {\mathcal{C}}_{\infty }^{r-1}$.

Theorem 3.6. i) $r>2$. Then $S^{\prime }\cap \mathcal{V}$ consists of $r-2$ generatrix lines and a residual curve ${\mathcal{C}}^{r+1}$ representing ${\mathcal{K}}_{\mathcal{C}}$. The curve ${\mathcal{C}}^{r+1}$ meets all the generatrix lines; it is a rational normal curve lying in a subspace $S_{r+1}\subset S^{\prime }$, with ${\mathcal{C}}^{r+1}\cap {\bar{S}}_{r-2}=\left\{T\right\}$.

ii) $r=2$. Then ${\mathcal{K}}_{\mathcal{C}}=S^{\prime }\cap \mathcal{V}$. If $q$ is odd, ${\mathcal{K}}_{\mathcal{C}}$ is a twisted cubic containing in $S^{\prime }$.

In no case does $g_t$, tangent to ${\mathcal{K}}_{\mathcal{C}}$ at the point at infinity, belong to $S^{\prime }$.

Proof. i) $r>2$. The intersection $S^{\prime }\cap \mathcal{V}$ must either be a curve of order $2r-1$ consisting of a total of $q+1$ points, or consist of $2r-1-m$ generatrix lines and of a residual curve of order $m<2r-1$ (cf. Result 1). Since $S^{\prime }$ contains ${\mathcal{K}}_{\mathcal{C}}$, which includes $q$ affine points, it follows that $S^{\prime }\cap \mathcal{V}$ must consist of the $r-2=2r-1-m$ generatrix lines $\left\{g_1,\cdots ,g_{r-2}\right\}$, and of a residual curve of order $m$ satisfying $2r-1-m=r-2$, hence $m=r+1<2r-1$. Such a curve ${\mathcal{C}}^{r+1}$ must intersect all the generatrix lines (cf. Result 1).

More precisely, the lines in $\left\{g_1,\cdots ,g_{r-2}\right\}$ intersect ${\mathcal{C}}^{r+1}$ at the corresponding points of ${\bar{S}}_{r-2}$. The generatrix $g_t$, which represents the tangent $t$ to the conic $\mathcal{C}$ at its infinite point, does not belong to $S^{\prime }$ (by hypothesis). Since $g_t$ contains no affine points of ${\mathcal{K}}_{\mathcal{C}}$, it intersects ${\mathcal{C}}^{r+1}$ only at the point $T$.

Therefore ${\mathcal{K}}_{\mathcal{C}}$ is represented by ${\mathcal{C}}^{r+1}$, a rational normal curve of $\mathcal{V}$, lying in a subspace $S_{r+1}$ of $S^{\prime }$ (cf. Result 1, 3)). Indeed, since $dim\left(S_{r+1}\cap {\bar{S}}_{r-2}\right)=0$, which is equivalent to ${\mathcal{C}}^{r+1}\cap {\mathcal{C}}_{\infty }^{r-1}=\left\{T\right\}$, the remaining $q$ affine points are distributed one on each of the remaining $q$ generatrix lines.

ii) For $r=2$, case i) can still be applied since $2r-1=r+1$. Specifically, note that $\mathcal{V}=V_2^3$ is embedded in $PG\left(4,q\right)$, where the hyperplane $S^{\prime }$ is a 3-dimensional subspace. The curve at infinity, ${\mathcal{C}}_{\infty }^1={\mathcal{C}}_{\infty }^{2-1}$, is a line $l_{\infty }$. The hyperplane $S^{\prime }$ meets $l_{\infty }$ in exactly one point, $T$, and intersects each generatrix line in exactly one of the $q$ affine point of ${\mathcal{K}}_{\mathcal{C}}$.

From Theorem 3.5, it follows that no three points of ${\mathcal{K}}_{\mathcal{C}}$ are collinear. Therefore, in this case and for $q$ odd, ${\mathcal{K}}_{\mathcal{C}}$ is a twisted cubic curve of $S^{\prime }$, that is, a normal rational curve of $S^{\prime }$ (cf. [10] , Theorem 21.2.3, [5] , Theorem 3.1).

In neither case i) nor ii), by construction, does $g_t$ belong to $S^{\prime }$.

This result shows that for $r>2$ (that is, for $2r-1>r+1$), no hyperplane strictly defines a substructure of the variety $\mathcal{V}$ capable of representing a cap of $q+1$ points, unless one considers a subspace of dimension $r+1$ within a hyperplane $S^{\prime }$, which can be constructed as follows.

First, remind that the curve $C_{\infty }^{r-1}$ of $S_{r-1}^{\infty }$ consists of $q+1$ points, no $r$ of which lie in a hyperplane of $S_{r-1}^{\infty }$, with $q\ge r-1$ (since by hypothesis $q\ge 2r-1$; cf. Definition 2.2).

Choose a subset of $r-1$ points from $C_{\infty }^{r-1}$. Let be the subspace generated by . Choose a subspace $S_{r-1}\in \mathcal{S}\backslash S_{r-1}^{\infty }$. Denote by $\mathcal{G}=\left\{g_1,\cdots ,g_{r-2},g_t\right\}$ the set of the generatrix lines through the points of with .

Theorem 3.7. There exists a hyperplane $H^{\prime }$ containing a subspace $S_{r+1}$, in which lies a rational normal curve $C^{r+1}$. This curve $C^{r+1}$ is a cap of $\mathcal{V}$, consisting of $q+1$ points, exactly one of which is at infinity. The curve ${\mathcal{C}}^{r+1}$ corresponds in the plane $\pi$, to a conic passing through the point at infinity .

Proof. Set $r>2$. Consider the set $\mathcal{G}\backslash \left\{g_t\right\}$ consisting of the $r-2$ generatrices $\left\{g_1,\cdots ,g_{r-2}\right\}$, and denote by $S_{r-1}^{\text{*}}$ the subspace they generate. Let $H^{\prime }=S_{r-1}^{\text{*}}+S_{r-1}$ be the hyperplane defined as the span of $S_{r-1}^{\text{*}}$ and a chosen element $S_{r-1}\in \mathcal{S}\backslash S_{r-1}^{\infty }$. Since $H^{\prime }$ contains $2r-1-m=r-2$ generatrix lines, it follows by Result 1 that $H^{\prime }\cap \mathcal{V}$ includes a residual curve ${\mathcal{C}}^{r+1}$ of order $m=r+1$, which meets all the generatrix lines.

Since $H^{\prime }\cap S_{r-1}^{\infty }$ is a subspace of dimension $r-2$, it follows that $H^{\prime }$ contains the entire subspace , and in particular, the point $T$. Therefore, $H^{\prime }$ is a hyperplane of the bundle ${ℬ}_{2r-2}$ of hyperplanes.

The curve ${\mathcal{C}}^{r+1}$ is a rational normal curve, so it lies in a subspace $S_{r+1}\subset H^{\prime }$ that intersects ${S^{\prime }}_{r-2}$ in a single point. If ${\mathcal{C}}^{r+1}\subset S_{r+1}$ met each generatrix in an affine point, it would be a directrix. However, the maximum possible order of a directrix is $r<r+1$, which leads to a contradiction. Therefore, ${\mathcal{C}}^{r+1}$ must intersect one of the generatrices in $\mathcal{G}$ at the point $P=S_{r+1}\cap {S^{\prime }}_{r-2}$ which lies in .

Assume $P\ne T$. Then ${\mathcal{C}}^{r+1}$ meets $g_t$ in an affine point, implying that $g_t$ must lie in $H^{\prime }$. This leads to a contradiction, as including $g_t$ would increase the dimension to $S_{r-1}^{\text{*}}$. Therefore $P=T$.

The remaining $q$ generatrices are each intersected by ${\mathcal{C}}^{r+1}$ at exactly one affine point, so that ${\mathcal{C}}^{r+1}$ contains $q$ affine points and one point at infinity. This configuration does not contradict the presence of the lines $\left\{g_1,\cdots ,g_{r-2}\right\}$ in $H^{\prime }$, as they are already contained within in. Since ${\mathcal{C}}^{r+1}$ is a normal rational curve with $q+1$ points, it clearly forms a $\left(q+1\right)$-cap.

Moreover, $H^{\prime }\ne {\Sigma }^{\prime }$ since ${\Sigma }^{\prime }$ contains no generatrices, and $H^{\prime }\ne H$, as the hyperplane $H$ contains $r-1$ generatrices and the unique directrix curve of order $r$.

If $r=2$, then and $\mathcal{G}=\left\{g_t\right\}$. Although the previous procedure can still be applied, to clarify this case, let $H^{\prime }=S_3$ be one of the $q-1$ hyperplanes around the plane $S_{r-1}+T=S_1+T$ distinct from both ${\Sigma }^{\prime }$ and $H$, so that $H^{\prime }$ contains no conic directrix.

If $g_t\in H^{\prime }$, then $2r-1-m=3-m=1$, so that the residual curve $C$ would have order $m=2r-2=2$. That is, $C$ would be a conic meeting all the generatrix lines (cf. Result 1). Therefore $C$ would be a directrix and $H^{\prime }=H$, which is a contradiction.

Hence $H^{\prime }$ does not contain $g_t$; it meets the line $C_{\infty }^{r-1}=C^1$ in single point, namely, $T$. Consequently, $H^{\prime }\cap \mathcal{V}$ is an irreducible rational normal curve of order $2r-1=3$ (cf. [5] , Lemma 2.1) having $T$ as its unique point at infinity.

By construction, in both cases, it is immediate to verify that ${\mathcal{C}}^{r+1}$ represents a conic of $\pi$ passing through the point $P_{\infty }$.

Theorem 3.8. There exists a partition of the affine points of the variety $\mathcal{V}=V_2^{2r-1}$ consisting of $q$ rational normal curves of order $r+1$, together one generatrix line.

Proof. In the non-affine subplane $\pi$, let $ℱ$ be the bundle of hyperosculating conics at the point $P_{\infty }$, all sharing the common tangent line $t$ through $P_{\infty }$. It is straightforward to verify that $\left|ℱ\right|=q$. Consequently, the affine points of $ℱ\cup t$ provide a partition of the affine points of $\pi$.

Denote by ${ℒ}_{ℱ}$ the set of $q$ curves of order $r+1$ in the variety $\mathcal{V}$ corresponding to the conics of the bundle $ℱ$. Let $g_t$ be the generatrix line representing the tangent line $t$ at $P_{\infty }$ (cf. Theorem 3.5).

Since two distinct conics in $ℱ$ meet only at the point $P_{\infty }$, the corresponding curves in ${ℒ}_{ℱ}$ have no affine points in common. The total number of affine points of $\mathcal{V}$ are $q^2+q$, each curve in ${ℒ}_{ℱ}$ contains exactly $q$ affine points, so the union of all these curves accounts for $q\cdot q=q^2$. Adding the $q$ affine points of the generatrix line $g_t$, we cover all $q^2+q$ affine points of $\mathcal{V}$.