Math 2.0

Theorem

The Hoffman–Singleton theorem

Statement

As noted above, the multiplicities of the eigenvalues are given by

M±=12[(v−1)±2k+(v−1)(λ−μ)(λ−μ)2+4(k−μ)]M_{\pm} = \frac{1}{2}\left[(v - 1) \pm \frac{2k + (v - 1)(\lambda - \mu)}{\sqrt{(\lambda - \mu)^2 + 4(k - \mu)}}\right]

which must be integers.

In 1960, Alan Hoffman and Robert Singleton examined those expressions when applied on Moore graphs, which are strongly regular graphs that have λ = 0 and μ = 1. Such graphs are free of triangles (otherwise λ would exceed zero) and quadrilaterals (otherwise μ would exceed 1), hence they have a girth (smallest cycle length) of 5. Substituting the values of λ and μ in the equation (v−k−1)μ=k(k−λ−1)(v - k - 1)\mu = k(k - \lambda - 1), it can be seen that v=k2+1v = k^2 + 1, and the eigenvalue multiplicities reduce to

M±=12[k2±2k−k24k−3]M_{\pm} = \frac{1}{2}\left[k^2 \pm \frac{2k - k^2}{\sqrt{4k - 3}}\right]

For the multiplicities to be integers, the quantity 2k−k24k−3\frac{2k - k^2}{\sqrt{4k - 3}} must be rational, therefore either the numerator 2k−k22k - k^2 is zero or the denominator 4k−3\sqrt{4k - 3} is an integer.

If the numerator 2k−k22k - k^2 is zero, the possibilities are:

  • k = 0 and v = 1 yields a trivial graph with one vertex and no edges, and
  • k = 2 and v = 5 yields the 5-vertex cycle graph C5C_5, usually drawn as a regular pentagon.

If the denominator 4k−3\sqrt{4k - 3} is an integer t, then 4k−34k - 3 is a perfect square t2t^2, so k=t2+34k = \frac{t^2 + 3}{4}. Substituting:

M±=12[(t2+34)2±t2+32−(t2+34)2t],\begin{aligned} M_{\pm} &= \frac{1}{2} \left[\left(\frac{t^2 + 3}{4}\right)^2 \pm \frac{\frac{t^2 + 3}{2} - \left(\frac{t^2 + 3}{4}\right)^2}{t}\right], \end{aligned}

so

32M±=(t2+3)2±8(t2+3)−(t2+3)2t=t4+6t2+9±−t4+2t2+15t=t4+6t2+9±(−t3+2t+15t).\begin{aligned} 32 M_{\pm} &= (t^2 + 3)^2 \pm \frac{8(t^2 + 3) - (t^2 + 3)^2}{t} \\ &= t^4 + 6t^2 + 9 \pm \frac{- t^4 + 2t^2 + 15}{t} \\ &= t^4 + 6t^2 + 9 \pm \left(-t^3 + 2t + \frac{15}{t}\right). \end{aligned}

Since both sides are integers, 15t\frac{15}{t} must be an integer, therefore t is a factor of 15, namely t∈{±1,±3,±5,±15}t \in \{\pm 1, \pm 3, \pm 5, \pm 15\}, therefore k∈{1,3,7,57}k \in \{1, 3, 7, 57\}. In turn:

  • k = 1 and v = 2 yields a trivial graph of two vertices joined by an edge,
  • k = 3 and v = 10 yields the Petersen graph,
  • k = 7 and v = 50 yields the Hoffman–Singleton graph, discovered by Hoffman and Singleton in the course of this analysis, and
  • k = 57 and v = 3250 famously predicts a graph that has neither been discovered since 1960, nor has its existence been disproven.

The Hoffman-Singleton theorem states that there are no girth-5 Moore graphs except the ones listed above.

Further reading

  1. Wikipedia, page 1399856, revision 1363981900, Strongly regular graph, section Algebraic properties of strongly regular graphs > The Hoffman–Singleton theorem. adopted
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