On a conjecture of Kelly on (1, 3)-representation of Sylvester–Gallai designs

Abstract

We give an exact criterion of a conjecture of L. M. Kelly to hold true which is stated as follows. If there is a finite family $\Sigma$ of mutually skew lines in $\mathbb {R}^d,d\ge 4$ such that the 3-flat spanned by every two lines in $\Sigma$, contains at least one more line of $\Sigma$, then we have that all of the lines of $\Sigma$ are contained in a single 3-flat if and only if the arrangement of 3-flats is central. Finally, this article leads to an analogous question for higher dimensional skew affine spaces, where we prove that, for (2, 5)-representations of Sylvester–Gallai designs in $\mathbb {R}^6$, the analogous statement does not hold.

Appendix

1.1 Appendix A: Another proof of the Main Theorem 1.7

In this section, we first give a proof of the main theorem when $d=4$ and later prove it, in general. Now we prove a lemma.

Lemma A.1

Let $\Sigma$ be a finite family of mutually skew lines in $\mathbb {R}^d$ such that the 3-flat spanned by every two lines in $\Sigma$, contains at least one more line of $\Sigma$. Then either all the lines are contained in a single 3-flat or every line is contained in at least three different 3-flats.

Proof

Let L be a line which is contained in exactly two 3-flats $H_1$ and $H_2$. Then we have a partition of the set

$$\begin{aligned} \Sigma \backslash \{L\}=\Sigma _1\cup \Sigma _2 \end{aligned}$$

where the line L and all the lines in $\Sigma _1$ are contained in the 3-flat $H_1$ and the same line L and all the lines in $\Sigma _2$ are contained in another 3-flat $H_2$. Now we have $\mid \Sigma _i\mid \ge 2,i=1,2$. Let $L_i\in \Sigma _i,i=1,2$. Then the 3-flat $H\notin \{H_1,H_2\}$ which is spanned by $L_1$ and $L_2$ must contain a third line from $\Sigma$. So either H contains two lines from the set $\Sigma _1\cup \{L\}$ or two lines from the set $\Sigma _2\cup \{L\}$. This implies either $H=H_1$ or $H=H_2$ which is a contradiction. Hence the lemma follows. $\square$

We state a very important theorem which is required to prove main Theorem 1.7 for $d=4$.

Lemma A.2

Suppose there is a finite family $\Sigma =\{L_1,L_2,\ldots ,L_n\}$ of mutually skew lines in $\mathbb {R}^4$ such that the 3-flat spanned by every two lines in $\Sigma$, contains at least one more line of $\Sigma$. Let $\{H_1,H_2,\ldots ,H_m\}$ be the distinct 3-flats and $m\ge 2$. If the hyperplane arrangement of the 3-flats is central, that is, $O\in \underset{i=1}{\overset{m}{\cap }} H_i$, then there exists a line $L\subset \mathbb {R}^4$ different from $L_i$ and passing through O such that the line $L_i$ and L are coplanar for each $1\le i\le n$.

We first prove the main theorem for $d=4$, using Theorem A.2 and then prove Theorem A.2.

Proof of the Main Theorem for $d=4$

Suppose there are at least two distinct 3-flats. Using Theorem A.2, let O be the origin and

$$\begin{aligned} L=\{t(0,0,0,1)\mid t\in \mathbb {R}\}. \end{aligned}$$

Let $P_i$ be the plane containing origin spanned by the lines L and $L_i$ for $1\le i\le n$. We have by $P_i\ne P_j,1\le i\ne j\le n$ because the lines $L_i,1\le i\le n$ are mutually skew. Now we project $\mathbb {R}^4$ perpendicular to the line L onto $\mathbb {R}^3$. Let

$$\begin{aligned} \pi :\mathbb {R}^4\longrightarrow \mathbb {R}^3 , \end{aligned}$$

$\pi (x,y,z,t)=(x,y,z)$ be the projection. Then we get a finite configuration of n points

$$\begin{aligned} \{\pi (P_i)\mid 1\le i\le n\}\subset \mathbb {P}^2(\mathbb {R}). \end{aligned}$$

The points $\{\pi (P_i)\mid L_i\in \Lambda \}$ are collinear in $\mathbb {P}^2(\mathbb {R})$ where $\Lambda$ is the set of lines in a 3-flat spanned by a pair of mutually skew lines. Hence the configuration forms an SGC in $\mathbb {P}^2(\mathbb {R})$. Now by using the basic Sylvester–Gallai theorem we conclude that all points $\pi (P_i),1\le i\le n$ are collinear in $\mathbb {P}^2(\mathbb {R})$. Hence all the lines $L_1,L_2,\ldots ,L_n$ lie in one three dimensional space containing origin which is a contradiction to the existence of at least two distinct 3-flats. This completes the proof of the main theorem. $\square$

We prove Theorem A.2.

Proof

Let $L_1,L_2,\ldots ,L_n$ be n mutually skew lines in $\mathbb {R}^4$ for some $n\ge 7$. In fact we can even assume that $n\ge 8$ using Corollary 2.4 because there is at least two distinct 3-flats, all of them containing origin. So using Lemma A.1, we have that each line $L_i,1\le i\le n$ is in at least three different 3-flats. Let us assume that the 3-flats of all these lines are $H_1,H_2,\ldots ,H_m$ which are given as follows, without loss of generality, by gathering as much definite set theoretic knowledge as possible about the configuration:

$$\begin{aligned} \begin{aligned} H_1&\overset{\supset }{\longleftrightarrow } \{L_1,L_2,L_3,\ldots ,L_{a_1}\}=\Sigma _1 \\ H_2&\overset{\supset }{\longleftrightarrow } \{L_1,L_{a_1+1},L_{a_1+2},\ldots ,L_{a_2}\}=\Sigma _2\\ H_3&\overset{\supset }{\longleftrightarrow } \{L_1,L_{a_2+1},L_{a_2+2},\ldots ,L_{a_3}\}=\Sigma _3\\ \vdots&\longleftrightarrow \vdots \\ H_{k=b_1}&\overset{\supset }{\longleftrightarrow } \{L_1,L_{a_{k-1}+1},L_{a_{k-1}+2},\ldots ,L_{a_k=n}\}=\Sigma _{b_1}\\ H_{b_1+1}&\overset{\supset }{\longleftrightarrow } \{L_2,L_{a_1+1},\ldots \}=\Sigma _{b_1+1}\\ H_{b_1+2}&\overset{\supset }{\longleftrightarrow } \{L_2,L_{a_1+2},\ldots \}=\Sigma _{b_1+2}\\ \vdots&\longleftrightarrow \vdots \\ H_{b_2}&\overset{\supset }{\longleftrightarrow } \{L_2,\ldots \}=\Sigma _{b_2}\\ H_{b_2+1}&\overset{\supset }{\longleftrightarrow } \{L_3,L_{a_1+1},\ldots \}=\Sigma _{b_2+1}\\ H_{b_2+2}&\overset{\supset }{\longleftrightarrow } \{L_3,L_{a_1+2},\ldots \}=\Sigma _{b_2+2}\\ \vdots&\longleftrightarrow \vdots \\ H_{b_3}&\overset{\supset }{\longleftrightarrow } \{L_3,\ldots \}=\Sigma _{b_3}\\ \vdots&\longleftrightarrow \vdots \\ H_{b_{a_1-1}}&\overset{\supset }{\longleftrightarrow } \{L_{a_1-1},\ldots \}=\Sigma _{b_{a_1-1}}\\ H_{b_{a_1-1}+1}&\overset{\supset }{\longleftrightarrow } \{L_{a_1},L_{a_1+1},\ldots \}=\Sigma _{b_{a_1-1}+1}\\ H_{b_{a_1-1}+2}&\overset{\supset }{\longleftrightarrow } \{L_{a_1},L_{a_1+2},\ldots \}=\Sigma _{b_{a_1-1}+2} \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \vdots&\longleftrightarrow \vdots \\ H_{b_{a_1}}&\overset{\supset }{\longleftrightarrow } \{L_{a_1},\ldots \}=\Sigma _{b_{a_1}}\\ \vdots&\longleftrightarrow \vdots \\ H_m&\overset{\supset }{\longleftrightarrow } \{L_{*},\ldots \}=\Sigma _{m}. \end{aligned} \end{aligned}$$

The notation for example $H_1 \overset{\supset }{\longleftrightarrow } \{L_1,L_2,L_3,\ldots ,L_{a_1}\}=\Sigma _1$ means that $H_1$ is the 3-flat spanned by any two lines in the set $\Sigma _1$ and the set $\Sigma _1$ gives rise to all pairs of lines, each of which, span the 3-flat $H_1$. So, in particular, $H_1$ contains the lines $L_i, 1\le i\le a_1$. Here we observe the following:

$$\begin{aligned} \begin{aligned} L_1&\overset{\subset }{\longleftrightarrow } \{H_1,H_2,\ldots ,H_{b_1}\}=\Delta _1\\ L_2&\overset{\subset }{\longleftrightarrow } \{H_1,H_{b_1+1},\ldots ,H_{b_2}\}=\Delta _2\\ L_3&\overset{\subset }{\longleftrightarrow } \{H_1,H_{b_2+1},\ldots ,H_{b_3}\}=\Delta _3\\ \vdots&\longleftrightarrow \vdots \\ L_{a_1}&\overset{\subset }{\longleftrightarrow } \{H_1,H_{b_{a_1-1}+1},\ldots ,H_{b_{a_1}}\}=\Delta _{a_1}\\ L_{a_1+1}&\overset{\subset }{\longleftrightarrow } \{H_2,H_{b_1+1},H_{b_2+1},\ldots ,H_{b_{a_1-1}+1},\ldots \}=\Delta _{a_1+1}\\ L_{a_1+2}&\overset{\subset }{\longleftrightarrow } \{H_2,H_{b_1+2},H_{b_2+2},\ldots ,H_{b_{a_1-1}+2},\ldots \}=\Delta _{a_1+2}\\ \vdots&\longleftrightarrow \vdots \\ L_n&\overset{\subset }{\longleftrightarrow } \{H_{b_1},\ldots \}=\Delta _n. \end{aligned} \end{aligned}$$

The notation $L_1 \overset{\subset }{\longleftrightarrow } \{H_1,H_2,\ldots ,H_{b_1}\}=\Delta _1$ means that $L_1$ is contained in any 3-flat $H\in \Delta _1$ and the set $\Delta _1$ has all the 3-flats which contain $L_1$. Here we have $b_1=k,a_k=n$ to avoid notation of repeated subscripts. We have

$$\begin{aligned} \mid \Sigma _i\cap \Sigma _j\mid \le 1,1\le i\ne j\le m \end{aligned}$$

and each $H_i$ is the 3-flat spanned by any two lines in $\Sigma _i,1\le i\le m$. We have that all the $H_i,1\le i\le m$ are distinct and hence

$$\begin{aligned} \dim _\mathbb {R}{(H_i\cap H_j)}=2,1\le i\ne j\le m. \end{aligned}$$

We also have that

$$\begin{aligned} 0\in \underset{i=1}{\overset{m}{\cap }}H_i. \end{aligned}$$

We also assume that $L_1,L_2,\ldots ,L_{n-1}$ does not pass through origin and $L_n$ may or may not pass through origin. Hence we have $\dim _\mathbb {R}{\left( \underset{j\in \Delta _i}{\cap } H_j\right) }=2,1\le i\le n-1$ and $\dim _\mathbb {R}{\left( \underset{j\in \Delta _n}{\cap } H_j\right) }\le 2$. $\square$

Proof

Let $v_j$ be a normal vector of the hyperplane $H_j,1\le j\le m$. The set $\{v_1,v_{b_1},v_{b_2}\}$ is linearly independent and $v_j,1\le j\le m$ is linearly dependent on the set $\{v_1,v_{b_1},v_{b_2}\}$.

Proof

For $1\le i\le n-1$, the set $\{v_j\mid H_j\in \Delta _i\}\subset \mathbb {R}^4$ spans a two dimensional space. Now the set $\{v_1,v_{b_1},v_{b_2}\}$ is linearly independent. Suppose not, then $H_1\cap H_{b_1}\cap H_{b_2}=H_1\cap H_{b_1}=H_1\cap H_{b_2}$. This implies the lines $L_1,L_2$ are coplanar which is a contradiction. Now the vector $v_{b_3}$ is linearly dependent on the set $\{v_1,v_{b_1},v_{b_2}\}$. Suppose not then by applying a suitable linear transformation in $GL_4(\mathbb {R})$ we can assume that $v_1=(1,0,0,0),v_{b_1}=(0,1,0,0),v_{b_2}=(0,0,1,0),v_{b_3}=(0,0,0,1)$. Since $n\ge 4$ we have $\dim _\mathbb {R}{\left( \underset{j\in \Delta _i}{\cap } H_j\right) }=2,i=1,2,3$ and hence $v_2=(a,b,0,0),v_{b_1+1}=(c,0,d,0),v_{b_2+1}=(e,0,0,f)$ with $a,b,c,d,e,f\in \mathbb {R}\backslash \{0\}$. Now the rank of the matrix

$$\begin{aligned} \begin{pmatrix} {a}&{}\quad {b}&{}\quad {0}&{}\quad {0}\\ {c}&{}\quad {0}&{}\quad {d}&{}\quad {0}\\ {e}&{}\quad 0 &{}\quad 0 &{}\quad f \end{pmatrix} \end{aligned}$$

is three. Hence the space $H_2\cap H_{b_1+1}\cap H_{b_2+1}$ is exactly one dimensional and contains the line $L_{a_1+1}$. But $n>a_1+1$ and the line $L_{a_1+1}$ does not pass through origin which is a contradiction. So $v_{b_3}$ is linearly dependent on the set $\{v_1,v_{b_1},v_{b_2}\}$. Similarly by applying the same argument, we conclude that $v_{b_j}$ is linearly dependent on the set $\{v_1,v_{b_1},v_{b_2}\}$ for $3\le j\le a_1$ since $H_2\cap H_{b_1+1}\cap H_{b_{j-1}+1}$ contains the line $L_{a_1+1}$. Now $v_i$ is linearly dependent on $\{v_1,v_{b_j}\}$ for $b_{j-1}+1\le i\le b_j-1,3\le j\le a_1$ and hence we obtain that $v_i$ is linearly dependent on the set $\{v_1,v_{b_1},v_{b_2}\}$ for $1\le i\le b_{a_1}$. Now we observe the following. For any $2\le i\le n$ there exists a unique $j\in \{1,2,\ldots ,b_1\}$ such that $L_i\subset H_j$. Similarly for $1\le i\le n,i\ne 2$ there exists a unique $j\in \{1,b_1+1,b_1+2,\ldots , b_2\}$ such that $L_i\subset H_j$. Now we observe that the set $\Delta _n\subset \underset{i=1}{\overset{n-1}{\cup }} \Delta _i$. So by applying the dual argument we get that, for $a_1+1\le i\le n-1$, the set $\Delta _i$ contains two hyperplanes one from the set $\Delta _1$ and one from the set $\Delta _2$ different from $H_1$. The normal vector of any hyperplane in $\Delta _i$ is linearly dependent on the normal vectors of those two hyperplanes in $\Delta _i$ coming from $\Delta _1$ and $\Delta _2$. So for $b_{a_1}+1\le i\le m, v_i$ is linearly dependent on the set $\{v_1,v_{b_1},v_{b_2}\}$. This can be concluded irrespective of the a priori fact that $\dim _\mathbb {R}{\left( \underset{j\in \Delta _n}{\cap } H_j\right) }=0$ or 1 or 2. This completes the proof of the claim. $\square$

Without loss of generality by applying a linear transformation in $GL_4(\mathbb {R})$, let $v_1=(1,0,0,0),v_{b_1}=(0,1,0,0),v_{b_2}=(0,0,1,0)$. Then we get using the claim that, the line

$$\begin{aligned} \{t(0,0,0,1)\mid t\in \mathbb {R}\}=L\subset \underset{i=1}{\overset{m}{\cap }} H_i. \end{aligned}$$

Now for $1\le i\le n$, the line $L_i$ and the line L are contained in a two dimensional plane $H_r\cap H_s$ for any $r\ne s$ such that $H_r, H_s\in \Delta _i$. So in that plane the line $L_i$ and L are coplanar. Now $L\ne L_i,1\le i\le n-1$ because L passes through origin and $L_i$ does not. We also have that $L\ne L_n$. This completes the proof of Theorem A.2.$\square$

Now we state the analogous statement of Theorem A.2 in higher dimensions.

Lemma A.3

Let $d\ge 4$ be an integer. Suppose there is a finite family $\Sigma =\{L_1,L_2,\ldots ,$ $L_n\}$ of mutually skew lines in $\mathbb {R}^d$ such that the 3-flat spanned by every two lines in $\Sigma$, contains at least one more line of $\Sigma$. Let $\{H_1,H_2,\ldots ,H_m\}$ be the distinct 3-flats and $m\ge 2$. If the arrangement of 3-flats is central, that is, $O\in \underset{i=1}{\overset{m}{\cap }} H_i$, then there exists a line $L\subset \mathbb {R}^d$ different from $L_i$ and passing through O such that the line $L_i$ and L are coplanar for each $1\le i\le n$.

We first prove the main theorem using Theorem A.3.

Proof of the Main Theorem

This proof is similar to the proof of the main theorem for the value $d=4$ given in the previous section. $\square$

We prove Theorem A.3.

Proof

Just similar to the proof of Theorem A.2, here also we have a similar set theoretic knowledge of the lines and their 3-flats. We assume as usual that O is the origin and $L_i,1\le i\le n-1$ do not pass through origin and $L_n$ may or may not pass through origin. Here we have that all the 3-flats $H_i,1\le i\le m$ are distinct and $\dim _\mathbb {R}{(H_i\cap H_j)} = 2$ for all $i\ne j$ such that $H_i,H_j\in \Delta _p$ for any $1\le p\le n-1$ and in general we have $\dim _{\mathbb {R}}({H_i\cap H_j}) \le 2$ unlike the case when $d=4$. It follows that $\dim _\mathbb {R}{\left( \underset{j\in \Delta _i}{\cap } H_j\right) } = 2,1\le i\le n-1$ and $\dim _\mathbb {R}{\left( \underset{j\in \Delta _n}{\cap } H_j\right) } \le 2$. Now we prove the following claim. $\square$

Proof of the Main Theorem

Let $V_i=H_i^{\perp }\subset \mathbb {R}^d,1\le i\le m$. Then

• (1) $\dim _\mathbb {R}{(V_1+V_{b_1}+V_{b_2})}=l-1$.

• (2) $V_i \subsetneq V_1+V_{b_1}+V_{b_2}$ for $1\le i\le m$.

Proof of the Main Theorem

We prove (1) first. The space $(V_1+V_{b_1}+V_{b_2})=(H_1\cap H_{b_1}\cap H_{b_2})^{\perp }$. We have

$$\begin{aligned} \begin{aligned} \dim _\mathbb {R}{(H_1\cap H_{b_1}\cap H_{b_2})}&=\dim _\mathbb {R}{(H_1\cap H_{b_1})}+\dim _\mathbb {R}{(H_1\cap H_{b_2})}\\&-\mathrm {dim}_{\mathbb {R}}((H_1\cap H_{b_1})+(H_1\cap H_{b_2})). \end{aligned} \end{aligned}$$

Now for $i=1,2, H_1\cap H_{b_i}$ is the plane containing $L_i$ and the origin O. Hence the space $(H_1\cap H_{b_1})+(H_1\cap H_{b_2})$ is exactly the 3-flat $H_1$ spanned by the lines $L_1$ and $L_2$ containing O which is therefore three dimensional. So we have $\dim _\mathbb {R}{(H_1 \cap H_{b_1}\cap H_{b_2})}=1$ and hence $\dim _\mathbb {R}{(V_1+V_{b_1}+V_{b_2})}=l-1$.

Now we prove (2). First we observe that for $1\le p \le n-1, i\ne j,r\ne s,i,j,r,s\in \Delta _p$ the $V_i+V_j=V_r+V_s,\dim _\mathbb {R}{(V_i+V_j)}=l-2$. This is because $H_i\cap H_j=H_r\cap H_s=\underset{t\in \Delta _p}{\cap } H_t$ and its dimension is 2. Now we have the following sequence of subspace inclusions:

$$\begin{aligned} \begin{aligned} V_{b_3}&\subsetneq V_1+V_{b_2+1} \subset V_1+V_2+V_{b_1+1}~(\text {because }V_{b_2+1}\subsetneq V_2+V_{b_1+1})\\&=V_1+V_2+V_{b_2}~(\text {because }V_1+V_{b_1+1}= V_1+V_{b_2})\\&= V_1+V_{b_1}+V_{b_2}~(\text {because }V_1+V_2= V_1+V_{b_1}). \end{aligned} \end{aligned}$$

Similarly, we get $V_{b_j}\subsetneq V_1+V_{b_1}+V_{b_2}$ for $3\le j\le a_1$. This implies that $V_i\subsetneq V_1+V_{b_1}+V_{b_2}$ for $1\le i\le b_{a_1}$. We have $\Delta _n\subset \underset{i=1}{\overset{n-1}{\cup }}\Delta _i$ and for $a_1+1\le i\le n-1$, the set $\Delta _i$ contains two 3-flats one from the set $\Delta _1$ and one from the set $\Delta _2$ different from $H_1$. Hence $V_i\subsetneq V_1+V_{b_1}+V_{b_2},1\le i\le m$. This proves the claim. $\square$

Coming to the proof of Theorem A.3, let $L=(V_1+V_{b_1}+V_{b_2})^{\perp }\subseteq \underset{i=1}{\overset{m}{\cap }} H_i$. For $1\le i\le n$, the line L and $L_i$ are coplanar in the plane $H_r\cap H_s$ for $r\ne s,r,s\in \Delta _i$. Clearly $L\ne L_i,1\le i\le n-1$ and by coplanarity of L with say $L_1$, we must have $L\ne L_n$. This proves Theorem A.3.