Saturday, June 7, 2025

Entry 139

The general quintic can be reduced to the following one-parameter forms

$$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0\tag1$$

$$x^5-5\alpha x -\alpha = 0\tag2$$

$$x^5+5\sqrt{\alpha}\, x^2 -\sqrt{\alpha} = 0\tag3$$

$$\; x^5+5x+\left(\frac1{\sqrt{\alpha}}-64\sqrt{\alpha}\right)^{1/2} = 0\tag4$$

$$x^5-5\alpha x^3+10\alpha^2x-\alpha^2=0\tag5$$ with the last found by yours truly. They have neat discriminants

$$\begin{align}D_1 &= 5^5\,(1-1728\alpha)^2\,\alpha^8\\ D_2 &= 5^5\,(1-256\alpha)\,\alpha^4\\ D_3 &= 5^5\,(1-108\alpha)\,\alpha^2\\ D_4 &= 5^5\,(1+64\alpha)^2\,\alpha^{-1}\\ D_5 &= 5^5\,(1-36\alpha)(1-32\alpha)\,\alpha^8\end{align}$$ The integers \((1728, 256, 108, 64)\) appear in Ramanujan's theory of elliptic functions to alternative bases and we will connect these quintics to the McKay-Thompson series of class 1A, 2A, 3A, 4A, 6A for the Monster in subsequent entries. 

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