As wikipedia defines well, the Fano plane is a small symmetric block design, specifically a 2-(7,3,1)-design. The points of the design are the points of the plane, and the blocks of the design are the lines of the plane.

an example of Fano plane with 7 number

Now what I'm trying to find is the number of different Fano plane with points labeled $1$ to $7$. By "different" I mean every set of blocks should be disjoint from every other; that no two planes have common blocks.

Put another way, I want to know how large a set of Fano planes can be if the sets of lines are pairwise disjoint.

Every Fano plane has 7 blocks and here I listed all the blocks you could see in the picture:

\begin{align*} \{6, 4, 3\} \\ \{3, 2, 5\} \\ \{5, 1, 6\} \\ \{6, 7, 2\} \\ \{3, 7, 1\} \\ \{5, 7, 4\} \\ \{1, 4, 2\} \\ \end{align*}

Please use a combinatorial approach, not groups and algebra.

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    $\begingroup$ I assume two arrangements are considered equivalent (and count just as one) if the set of blocks in one matches that in the other, right? $\endgroup$ – Mark Fischler May 22 '16 at 13:35
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    $\begingroup$ You wouldn't happen to be familiar with the automorphism/collineation group the the Fano plane, would you? If you know it has order $168$, it's not hard to count non isomorphic Fano planes. But that pushes the difficulty to getting the $168$, if you're not already familiar. $\endgroup$ – pjs36 May 22 '16 at 13:51
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    $\begingroup$ @pjs36 I'm not sure what you mean by "non isomorphic Fano planes", since all Fano planes are isomorphic. $\endgroup$ – Morgan Rodgers May 22 '16 at 14:03
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    $\begingroup$ @HenningMakholm (b) is correct a new Fano is accepted if and only if ALL the sets be different from previous Fano.... thank you for your making clear... $\endgroup$ – Niloofar_jz May 22 '16 at 14:11
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    $\begingroup$ @Niloofar_jz The complexity of the code does not come into play at all, since you are dealing with a fixed problem. It is meaningless to say the time complexity is "exponential" because there is no changing parameter; there are 7! permutations to check, this should be easy on any laptop. $\endgroup$ – Morgan Rodgers May 22 '16 at 14:19

Any set of two pairwise disjoint Fano planes is maximal; three or more Fano Planes must share at least one line. Henning's answer is streets ahead in terms of succinctness (and quite clever to boot), but frankly, I didn't do all of this work not to post an answer :)

But before I launch into a tedious case-by-case proof that there do not exist $3$ or more pairwise-disjoint Fano planes I would like to talk a bit about what I dream might yield a nicer approach for a someone with more insight than me (possibly my future self).

It is fairly well known that any pair of Fano planes share exactly $0, 1$ or $3$ lines (in fact, I found today that each Fano plane is disjoint from exactly $8$ others, shares a single line with exactly $14$ others, and shares $3$ lines with exactly $7$ others).

To quote a passage from Buckard Polsters article, YEA WHY TRY HER RAW WET HAT: A Tour of the Smallest Projective Space,

There is a unique partition of the $30$ labelled Fano planes into $2$ sets $X$ and $Y$ of 15 each such that any $2$ Fano planes in $1$ of the sets have exactly $1$ line in common.

You can take the $15$ single-intersection planes and make a nice picture of something called a generalized quadrangle from them (this one's "the doily", also from his article)

Doily from Polster

whose points are all labeled Fano planes and whose lines are Lines the planes share (there's a big picture in the pdf showing the planes). It's obviously silent on pairs of planes that are disjoint, but I'd like to believe there's some insight to be gained by considering how the two partitions interact with each other. As a sidenote, the article is both very informative and quite fun to read.

The usual definition is that two Fano planes are considered equivalent if they have the same set of lines. Under this definition, there are $30$ non-equivalent Fano planes that we'll arbitrarily separate into one of two categories:

fundamentally different planes

The first do not contain the line $\{1,2,3\}$ and there are $4! = 24$ of these. The second do contain the line $\{1,2,3\}$ and there are $3! = 6$ of these.

Let us now find pairwise disjoint (those sharing no lines) Fano planes. We'll search in the first category of planes, those not containing the line $\{1,2,3\}$. I find it more convenient to use the set $\{1,2,3,x,y,z,w\}$ instead of $\{1,2,3,4,5,6,7\}$ as point labels.

Without loss of generality, we will start with the plane

starting  plane

whose lines are

\begin{array}{ccc} 12x & 13y & 23z\\ 1zw & 2yw & 3xw\\ &xyz& \end{array}

and attempt to find a disjoint plane by permuting $\{x, y, z, w\}$. Such a permutation must have no fixed points: if $x$ did not move, the line $12x$ would be preserved (and similarly for $y$ and $z$), while if $w$ were fixed, the line $xyz$ would be preserved.

Thus our permutation is either a $4$-cycle ($a \mapsto b \mapsto c \mapsto d \mapsto a$), or a product of disjoint $2$-cycles, as these are the only permutations of $4$ letters with no fixed points.

The list of products of disjoint $2$-cycles is rather small

\begin{align*}x \leftrightarrow y &\text{ and } z \leftrightarrow w \\ x \leftrightarrow z &\text{ and } y \leftrightarrow w \\ x \leftrightarrow w &\text{ and } y \leftrightarrow z \end{align*}

and for each, one of the lines containing $w$ is preserved (e.g., for $x \leftrightarrow y \text{ and } z \leftrightarrow w$, the line $1zw$). Thus we must use some $4$-cycle to permute the letters, and without loss of generality, we may choose $x \mapsto y \mapsto z \mapsto w \mapsto x$, yielding the plane

enter image description here

with lines \begin{array}{ccc} 12w & 13x & 23y\\ 1yz & 2xz & 3zw\\ &xyw& \end{array}

which is easily seen to be disjoint from our starting plane.

Now we attempt to add a third plane, one that still avoids the line $\{1,2,3\}$. We have enough lines to avoid that it's best to keep them in mind, and attempt to place $x, y, z, w$ around the restrictions:


In the picture, the green letters are the only possible values for a certain spot (we can't have $x$ in the line with $1$ and $2$, since we've already used the line $12x$, and so on, for the straight lines). The fact that the two circles have been lines $xyz$ and $xyw$ means that we can't use both $x$ and $y$ in the circular line. Thus, each point has two possibilities. By making a single choice, the rest of the points are completely determined.

Alas, neither choice has a happy ending:

no good

So to recap, we've only managed to find a pitiful pair of disjoint Fano planes from the $24$ that avoid the line $\{1,2,3\}$. But maybe we can find a plane containing $\{1,2,3\}$ that's disjoint from ours (spoiler alert: We cannot).

Using the template above (but $x, y, z, w$ instead of $4, 5, 6, 7$) for a plane containing $\{1,2,3\}$, we write possibilities for the three to-be-labeled points in green. We also have the same Forbidden lines from our two disjoint planes.

no good 2: the ungoodening

Using each of the Forbidden lines with two letters (e.g., $2$ can't go at the top because the line $2yw$ has already been used), we are able to narrow down the possibilities.

But, to our shared dismay, $2$ is needed in two places, and we cannot find another Fano plane, disjoint from our two. As we have been working in full generality, no set of two disjoint Fano planes can be expanded to a set of three or more pairwise disjoint Fano planes.

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    $\begingroup$ I think the quote from Polster actually settles it immediately: Once you have picked one plane from the $X$ set and one plane from the $Y$ set, every remaining plane is in either $X$ or $Y$, so it necessarily shares a line with at least one of the planes already selected. $\endgroup$ – Henning Makholm May 24 '16 at 7:14
  • $\begingroup$ @Henning I think you're right. I initially wasn't quite sure (and still am not) whether a pair from either partition would share just $1$ line (or instead if all the $0$ and $3$ sharing was happening within a single partition). But I think you're right, which does indeed give a maximum of two disjoint as an immediate corollary! $\endgroup$ – pjs36 May 24 '16 at 15:00

It is not possible to make more than two Fano planes on the same seven points that don't share any line.

If one of them is the standard numbering (where three points are collinear iff their bitwise XOR is 0):

   1 2
5   3   6

then the complete set of planes that are compatible with this (up to automorphisms of each plane) is

   3           3           3           3
  1 2         1 2         1 2         1 2
   5           4           7           6
6  4  7     7  5  6     4  6  5     5  7  4

   3           3           3           3
  1 2         1 2         1 2         1 2
   6           7           4           5
7  4  5     6  5  4     5  6  7     4  7  6

and every pair of these have a line in common.

The 8 compatible planes are enumerated by first selecting what the third point in the line containing 1,2 is (giving one of the four columns), then selecting what the third point in the line containing 3 and the previously chosen point (two choices possible since the corresponding line in the base plane must be avoided), and finally only one placement of the two remaining points is possible.

There's a symmetry to the arrangement: The compatible planes arise from the standard plane by applying a permuation with cycle type $1+1+2+3$ where the one of the transposed points is collinear with the two fixed points. Each compatible plane arises in $21$ ways by this construction, corresponding to $21$ choices of the two fixed points.

The above representations of the $8$ compatible planes correspond to composing the permutations $(3\,4)(5\,6\,7)$ and $(3\,4)(7\,6\,5)$ with each of the four automorphisms of the standard plane that stabilize $1$, $2$, and $3$.

  • $\begingroup$ With thanks to @pjs36 for pointing out several errors in my previous answer. $\endgroup$ – Henning Makholm May 23 '16 at 22:51
  • $\begingroup$ This is a very nice approach; definitely quite refined and a nice improvement! (And you flatter me: It was but a single error!) $\endgroup$ – pjs36 May 24 '16 at 0:52

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