Fractional chromatic number of fullerenes

Computations of fractional chromatic numbers this week tell me that for Fullerene Graphs the value is $5/2$. I have computed $100$ of these or more. Is there any theorem that would say this? Any information on formulas for fractional chromatic numbers of families of graphs would be welcome. I am aware that the Kneser graphs $K(a,b)$ have fractional chromatic number $a/b$. For the definition of a Fullerene graph see this MathWorld link.

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If you don't get an answer here, you might try MathOverflow (but be sure to mention at each site that you have posted to the other). – Gerry Myerson Dec 3 '11 at 8:50
Are all your fullerenes on 60 vertices? – Chris Godsil Dec 4 '11 at 16:29
The ones I checked are up to 50 vertices. I am told that perhaps things don't get interesting until they are much larger, so maybe there is nothing to this "conjecture" about 5/2. – stan wagon Dec 5 '11 at 1:07
– J. M. Dec 20 '11 at 6:59

1 Answer

Any fullerene contains pentagons. Each of the 5 vertices on a pentagon touches 2 others. At most 2 vertices on any pentagon can share a colour since adjacent vertices have different colours. Therefore the chromatic number of a pentagon is 5/2.

If some of the vertices of a pentagon are coloured then, if for example a vertex is red and blue, the 2 opposite vertices must be coloured so that one is red and the other is blue. A fullerene can be coloured with 5/2 colours by first colouring a pentagon, then all the adjacent pentagons, then any adjacent to the adjacent pentagons... then any adjacent hexagons, then another pentagon and its adjacent pentagons, then adjacent hexagons etc.

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Checking over my database I found that 3 of the 100-200 fullerenes I examined did not have 5/2...  So this is intriguing. If your proof, Angela, holds water, then there will be something funny about these 3 cases. I double-checked with others that my fractional chi is correct on these 3. Example: – stan wagon Dec 9 '11 at 16:43
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