Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider a 3-dimensional projective space $X$.

Let $m$ be the smallest number so that there are $m$ pairs of lines $ \ell_1,\ell'_1$, $ \ell_2,\ell_2'$, ... , $\ell_m, \ell'_m$ in $X$:

a) For every $ i=1,2,\dots,m$, $\ell_i \cap \ell'_i = \emptyset$.

b) For every $i,j \le m$, $i \ne j$, $\ell_i \cap \ell'_j \ne \emptyset$.

(If there is no upper bound on $m$ we let $m=\infty$.)


a) Is it always the case that $m=6$?

b) Is it (at least) true that either $m=6$ or $m=\infty$?

c) What is the answer for the projective space over the Quaternions?


1) Since $X$ has dimension greater than 2 it must be Desarguian and therefore there is a division ring $D$ the points and lines in $X$ corresponds to the 1-dimensional and 2-dimensional subspaces over $D^4$.

2) We can always construct 6 pairs by considering a standard basis for $D^4$, the six 2-dimensional spaces $V_1,\dots V_6$ spanned by pairs of basis elements and letting $U_i$ correspond to the complement pair of basis elements to the pair used for $V_i$. Therefore, $m \ge 6$.

3) If $X$ is Papussian, namely if $D$ is a field $F$, then there is an argument based on the exterior algebra that $m \le 6$. The intersection property implies that the vectors in $\bigwedge^2(F^4)$ which correspond to the 2-dimensional vectors represented by the $m$ lines are linearly independent. This is a special case of a theorem by Lovasz. For more details see this blog post.

A question of independent interest that might be related is:

Question d) Is there some analog of the exterior algebra over division rings?

4) Benjy Weiss brought to my attention a paper by S.A. Amitsur Rational identities and applications to algebra and geometry. In the paper it is shown that, for Desarguian geometries, intersection theorems are equivalent to rational identities in the coordinate ring and that any nontrivial intersection theorem, together with the order axioms, implies Pappus' theorem. This may be relevant for showing that always, or at least in some cases, if you cannot find $m$ pairs with $m>6$ then $ m=\infty$. However I cannot tell if Amitsur's theorem covers the case at hand.

share|improve this question
The following MO question is of some relevance mathoverflow.net/questions/65421/… –  Gil Kalai Apr 28 '14 at 13:17
Nice question! An easy bound for the quaternions is $\binom{16}{4}$: Identify $Q$ with $\mathbb{R}^4$ and you are asking for a collection of $8$-planes in $16$-space with certain pairwise intersections, and you can deduce that the corresponding points in $\bigwedge^4 \mathbb{R}^{16}$ are linearly independent. I doubt this is tight, though. –  David Speyer Apr 28 '14 at 14:12
Sorry, that should read $\binom{16}{8}$, not $\binom{16}{4}$ (and $\bigwedge^8 \mathbb{R}^{16}$.) –  David Speyer Apr 28 '14 at 14:29
Thanks, David. Yes this is correct. But we can identify the quaternions with 2 dimensional v.s. over the complex so we get 8 choose 4. and then we can save a bit by intersecting with a generic hyperplane and get 7 choose 3. I wonder if Amitsur's theorem is relevant. –  Gil Kalai Apr 28 '14 at 15:01
In the case of real projective space, if we add the additional assumption that the lines within one of the two subsets of six are skew, then $m\le 6$ follows immediately from the fact that four lines have in general zero or two quadrisecants; see e.g. en.wikipedia.org/wiki/Schl%C3%A4fli_double_six. Perhaps one way of attacking this problem would be to understand which other projective spaces have the same quadrisecant property. –  David Eppstein Apr 28 '14 at 21:19

2 Answers 2

up vote 3 down vote accepted

For a general division algebra, there is no upper bound.

I'll write vectors in $D^4$ as row vectors, with scalar multiplication acting on the left. I'll write $2$-planes in $D^4$ as row spans of $2 \times 4$ matrices.

Lemma The $2$-planes $$\begin{pmatrix} 1 & a & 0 & 0 \\ 0 & 0 & 1 & a \end{pmatrix} \ \mbox{and} \ \begin{pmatrix} 1 & 0 & b & 0 \\ 0 & 1 & 0 & b \end{pmatrix}$$ have nontrivial intersection if and only if $a$ and $b$ commute.

Proof Clearly, if $ab=ba$, then $(1,a,b,ab)$ is in the row span of both matrices.

Let $p(1,a,0,0) + q (0,0,1,a) = r(1,0,b,0) + s(0,1,0,b)$. Looking at the first three coordinates, $p=r$, $s=pa$ and $q=rb$. Then the equality of the fourth coordinates gives $pba = pab$. If there is a solution with $p \neq 0$, then $ab=ba$. $\square$

So we are done if we can find a division algebra $D$ with elements $(a_1, \ldots, a_N)$ and $(b_1, \ldots, b_N)$ so that $a_i$ and $b_j$ commute for $i \neq j$, but not for $i = j$.

Lemma For any positive integer $N$, there are noncommutative $\mathbb{Q}$-central division algebras $D_1$, $D_2$, ..., $D_N$ such that $D_1 \otimes D_2 \otimes \cdots \otimes D_N$ is a division algebra, where the tensor products are over $\mathbb{Q}$.

Given this lemma, we take $D = D_1 \otimes D_2 \otimes \cdots \otimes D_N$ and take $a_i$ and $b_i$ to be noncommuting element in $D_i$.

Proof of lemma

We use the description of the Brauer group of $\mathbb{Q}$: There is a short exact sequence $$0 \to Br(\mathbb{Q}) \to \bigoplus_p \mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}/2 \to \mathbb{Q}/\mathbb{Z} \to 0$$ where the big direct sum runs over all primes. Choose $2N$ distinct primes $p_1$, $q_1$,..., $p_N$, $q_N$ and $N$ pairwise relatively prime positive integers $c_1$, $c_2$, ..., $c_N$. Let $D_i$ be the division algebra corresponding to $(0,0,\ldots, 0, 1/c_i, -1/c_i, 0, 0, \ldots)$, where the two nonzero entries are in positions $p_i$ and $q_i$.

According to the answers to this question: $D_i$ is a $\mathbb{Q}$-central division algebra of dimension $c_i^2$ and $D:=D_1 \otimes \cdots D_n$ is isomorphic to $M_r(\Delta)$ where $\Delta$ is $\mathbb{Q}$-central division algebra of dimension $LCM(c_1, \ldots, c_N)^2$. Since the $c_i$ are pairwise relatively prime, $LCM(c_1, \ldots, c_N)^2 = (c_1 \ldots, c_n)^2 = \prod \dim D_i = \dim D$, so $r=1$ and $D$ is a division algebra. $\square$

I can show that this is close to optimal in the following sense, although the notion of "close" is not good enough to resolve the quaternion case.

  • I have a case by case bash to show that, if there are $7$ pairs of lines as stated, then all the $\ell_i$ are pairwise skew, as are all the $\ell'_i$.

  • If $(\ell_1, \ell_2, \ell_3)$ and $(\ell'_4, \ell'_5, \ell'_6)$ are pairwise skew and meet in the required manner, then I can show that we may apply a $GL_4(D)$ symmetry to assume that $$\ell_1 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \ \ell_2 = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix} \ \ell_3 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ $$\ell'_4=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} \ \ell'_5=\begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{pmatrix} \ \ell'_6=\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

Then, for $i \geq 7$, we get that $\ell_i$ and $\ell'_i$ are of the forms $$\begin{pmatrix} 1 & a & 0 & 0 \\ 0 & 0 & 1 & a \end{pmatrix} \ \mbox{and} \ \begin{pmatrix} 1 & 0 & b & 0 \\ 0 & 1 & 0 & b \end{pmatrix}$$ respectively.

So, if $m$ pairs of lines are achievable, we can find $m-6$ non-commuting pairs as above. In particular, this gives the bound $m \leq 8$ for the quaternions. However, I haven't been able to figure out whether $m=7$ or $8$ is actually achievable for the quaternions.

share|improve this answer
A less number theoretic example of a division algebra with $2N$ elements obeying $a_i b_j = a_j b_i$ for $i \neq j$ is the fraction field of the quantum torus. The quantum torus is $R:=\mathbb{Q}(q)\langle x_1, \ldots, x_{2n}\rangle$ with relations $x_i x_j = x_j x_i$ for $i \neq 2n+1-j$ and $x_i x_{2n+1-i} = q x_{2n+1-i} x_i$. $R$ is an Ore domain (see the appendix of arxiv.org/abs/math/0404446 ), so it has a skew field of fractions. –  David Speyer Apr 29 '14 at 1:52
Amazing!! Many thanks, David. Very nice result. –  Gil Kalai Apr 29 '14 at 9:00
I am quite a bit confused, probably because OP is asking for smallest $m$, but ... wouldn't $a_1 = i$, $b_1 = j$, $a_2 = j$, $b_2 = i$ give you what you want? –  Vít Tuček May 1 '14 at 17:59
@VítTuček That lets me define $(\ell_1, \ell_2, \ell_3, \ell_7, \ell_8)$ and $(\ell'_4, \ell'_5, \ell'_6, \ell'_7, \ell'_8)$ meeting in the required manner. But I don't know whether or not I can fill in $(\ell_4, \ell_5,\ell_6)$ and $(\ell'_1, \ell'_2, \ell'_3)$. –  David Speyer May 1 '14 at 18:13
Right. Sorry for my silly question. I should really practice counting to ten more often. ;) –  Vít Tuček May 1 '14 at 18:24

Here is the case-by-case bash promised in my other answer. I am showing that, if there exist $\ell_i$ and $\ell'_i$ as stated with $m \geq 7$, then all of the $\ell_i$ are pairwise skew, as are all of the $\ell'_i$.

Claim 1 There do not exist $(\ell_i, \ell_j, \ell_k)$ which are all three contained in a plane, and all three pass through a common point.

Proof If there were, then $\ell'_k$ would meet $\ell_i$ and $\ell_j$ but not $\ell_k$; but in the scenario above, all lines which meet $\ell_i$ and $\ell_j$ also meet $\ell_k$. $\square$

Claim 2 There are not three lines in a common plane.

Proof Suppose that $\ell_1$, $\ell_2$ and $\ell_3$ were in $H$. By Claim 1, they do not pass through a common point. So all lines which meet $(\ell_1, \ell_2, \ell_3)$ lie in $H$. In particular, $\ell'_4$, $\ell'_5$, $\ell'_6$ and $\ell'_7$ lie in $H$, and no three of them pass through a common point. Then $\ell_4$ meets $(\ell'_5, \ell'_6, \ell'_7)$, so it lies in $H$. But then $\ell_4$ and $\ell'_4$ meet, a contradiction. $\square$

Claim 3 There are not three lines passing through a common point.

Proof This is the polar dual to Claim 2. $\square$

We now prove the main result. Suppose, for the sake of contradiction, that $\ell_1$ and $\ell_2$ are not skew; set $p = \ell_1 \cap \ell_2$ and $H = \mathrm{Span}(\ell_1, \ell_2)$. By Claims 2 and 3, $\ell_3$ does not pass through $p$ nor lie in $H$; set $q = \ell_3 \cap H$ and $J = \mathrm{Span}(p, \ell_3)$. Then any plane meeting $(\ell_1, \ell_2, \ell_3)$ must either

(1) pass through $p$ and lie in $J$ or

(2) pass through $q$ and lie in $H$.

So each of $\ell'_4$, $\ell'_5$, $\ell'_6$ and $\ell'_7$ are of type (1) or (2). If three of them are of the same type, then we violate Claim 1, so there are two of each; say $\ell'_4$ and $\ell'_5$ pass through $p$ and lie in $J$ while $\ell'_6$ and $\ell'_7$ pass through $q$ and lie in $H$.

Note that $\ell'_6 \cap \ell'_7 = q$, $\mathrm{Span}(\ell'_6, \ell'_7) = H$, $\ell'_5 \cap H = p$ and $\mathrm{Span}(q, \ell'_5) = J$. In short, $(\ell'_7, \ell'_6, \ell'_5, q,p,H,J)$ has the same properties as $(\ell_1, \ell_2, \ell_3, p, q, H, J)$. So, repeating the logic with the new inputs, $\ell_4$ must either pass through $p$ and lie in $H$, or pass through $q$ and lie in $J$. In the former case, $(\ell_1, \ell_2, \ell_4)$ violate Claim 1; in the latter case $\ell_4$ and $\ell'_4$ meet.

We have reached a contradiction, and we see that the $\ell_i$ and $\ell'_i$ are pairwise skew.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.