# A question related to Conways 99 graph problem

I have observed that the number of triangles $\frac{vk}{6}$ of a strongly regular graph with parameters $(v,k,1,2)$ is given by the coefficient $2(k-1)$ in the molien series of the "4-D extraspecial group $2^{1+2\cdot 2}$": For example the beginning of the molien series is: $1 + t^2 + 5t^4 + 6t^6 + 15t^8 + 19t^{10} + 35t^{12} + 44t^{14} + 69t^{16} + 85t^{18}\\ + 121t^{20} + 146t^{22} + 195t^{24} + 231t^{26} + 295t^{28} + 344t^{30} + 425t^{32} + O(t^{34})$

We know that for $srg(9,4,1,2)$ we have $\frac{9\cdot4}{6}=6$ triangles and the coeffient $6 \cdot t^6$ tells us that at $6 = 2\cdot(4-1)$ we have $6$ triangles.

We know that for $srg(99,14,1,2)$ we have $\frac{99\cdot14}{6}=231$ triangles and the coeffient $231 \cdot t^{26}$ tells us that at $26 = 2\cdot(14-1)$ we have $231$ triangles.

And so on ...

I have checked this for all known parameters for such hypothetical graphs: The parameters of all such graphs are: $(9,4,1,2),(99,14,1,2),(243,22,1,2),(6273,112,1,2),(494019,994,1,2)$ of which only the $n=9$ and $n=243$ are known. The Conway $99$-graph problem asks if there exists such graph with 99 vertices. The number of triangles in such a (hypthetical) graph is: $6,231,891,117096,81842481$

(1) My (naive) question is, if there is a (deeper) connection, and maybe one can find unified way to construct all such graphs using the group above and the invariant homogenous polynomials (if this is not too much to ask)?

(2) The 6 homogenous invariants of degree 6 are:

$\frac14x_0^2x_1^2x_2^2 + \frac14x_0^2x_1^2x_3^2 + \frac14x_0^2x_2^2x_3^2 + \frac14x_1^2x_2^2x_3^2,$

$\frac14x_0^4x_2^2 + \frac14x_0^2x_2^4 + \frac14x_1^4x_3^2 + \frac14x_1^2x_3^4,$

$\frac14x_0^6 + \frac14x_1^6 + \frac14x_2^6 + \frac14x_3^6,$

$\frac14x_0^3x_1x_2x_3 + \frac14x_0x_1^3x_2x_3 + \frac14x_0x_1x_2^3x_3 + \frac14x_0x_1x_2x_3^3,$

$\frac14x_1^4x_2^2 + \frac14x_1^2x_2^4 + \frac14x_0^4x_3^2 + \frac14x_0^2x_3^4,$

$\frac14x_0^4x_1^2 + \frac14x_0^2x_1^4 + \frac14x_2^4x_3^2 + \frac14x_2^2x_3^4$

My question is, if one can find a way to associate to each such a homogenous invariant a unique triangle. I am struggling to find out what the set of vertices $V$ with $|V|=9$ is.

Edit: I found some way for the case $\deg=6$ with a brute force attack in SAGEMATH. But this attack doesn't work for $\deg=26$ since there I have to compute the common zeros of the 231 homogenous polynomials over some finite field. The problem I am facing is that the ideal generated by the homogenous invariants has dimension 2. But for a finite field there should be at most finitely many zeros. Any idea how to solve this in SAGEMATH (Computing the common zeros over some finite field?)

J is the ideal of interest:

  sage: [I.gens() for I in J.minimal_associated_primes()]
[[x1 + x2, x0 + x3], [x2 + x3, x0 + x1], [x1 + x3, x0 + x2]]
sage: J.dimension()
2


https://oeis.org/A030533

https://pure.tue.nl/ws/portalfiles/portal/2157517

https://people.maths.bris.ac.uk/~matyd/GroupNames/1/ES+(2,2).html

• (1) The Molien series in question is the Molien series associated to the unique faithful irreducible (and, it happens, orthogonal, $4$-dimensional) representation of $2^{1+2·2}_+$. Maybe you should make this clearer because it took me some figuring out what was going on here. (2) But really, what you're observing is that the number $vk/6$ of triangles coincides with $k(2k^2-9(-1)^k+13)/24$, and, since $k$ is always even, really, of $k(2k^2+4)/24$, i.e., that $v=\frac{1}{2}k^2+1$: in this form, the coincidence seems less impressive. (contd.) – Gro-Tsen Jun 12 '18 at 9:43
• @Gro-Tsen: Thank you for your comment. How do you come up with $k(2k^2-9(-1)^k+13)/24$? – orgesleka Jun 12 '18 at 9:47
• (3) That being said, it might be interesting to find out why someone thought it useful to add the Molien series in question to the OEIS (which, I suppose, is how you made your observation): you should probably contact the author of the entry in question to ask why they submitted it. (4) The numerology would seem much more impressive you found some interpretation for the corresponding Molien series for $2^{1+2\cdot 2}_-$: the coefficients are: $k(2k^2-15(-1)^k+7)/24$: if you can find some graph-theoretic interpretation for that, I'd be convinced something is going on. – Gro-Tsen Jun 12 '18 at 9:47
• The formula $k(2k^2-9(-1)^k+13)/24$ is in the OEIS entry you quoted. (I merely checked it on the values you gave.) You'll have to ask the person listed in the OEIS as to how they got it. (But Molien series are known to be rational and it's not hard to compute the denominator, so it's probably not difficult to recover such a formula.) – Gro-Tsen Jun 12 '18 at 9:49
• To answer the SageMath subquestion: you can use enum_projective_finite_field. – Ricardo Buring 2 days ago