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Heegner Numbers: Why 744?

  1. Jun 22, 2012 #1
    Just curious if anyone had thoughts to offer on where the 744 comes from. I know it's the second term in the q expansion of the j-invariant, but since I am not much familiar with the intricacies of the maths for either q or j, I'm kind of flummoxed past that.

    TIA for any thoughts,
    AC
     
  2. jcsd
  3. Jun 22, 2012 #2

    [tex]744=2^3\cdot 3 \cdot 31=12^2\cdot 31[/tex] Now, the number [itex]\,12\,[/itex] has a special role in all this mess of elliptic curves. For example, [itex]\,12^3=1728\,[/itex] divides the value of [itex]\,j(\tau)\,[/itex] , with [itex]\,\tau=\,[/itex] being a complex number with positive imaginary part, connected to a elliptic curve [itex]\,E=\mathbb Z \oplus \tau\mathbb Z[/itex].

    This is very interesting and beautiful stuff, but way too messy and advanced to develop it here.

    DonAntonio
     
  4. Jun 22, 2012 #3
    Thank you for the reply Don Antonio.

    Funny you should mention the number 12, because it is the divisors of 12 (modulo 12) that led to my interest in the Heegner numbers (not unrelated to the observations of Jeremy Ebert on another thread you are familiar with.) Take as our definition of "prime" the primes at the beginning of the 20th Century (thus 1 is indexed as p_0) and now insert 0, 1, 2, 3, 4, 6 as prime indices: 1, 2, 3, 5, 7, 13. Iterate again as prime indices: 2, 3, 5, 11, 17, 41, which are Euler's Lucky Primes. These relate quite simply to the last 6 Heegner numbers by the rule 4p - 1.

    If, on the other hand, you take the divisors of 12 sans modulo, you get Euler's Convenient Primes after the first iteration. These are primes not of form a^1b + b^1c + a^1c. I include the 1's as exponents to suggest the possibility of a link to Klein's Quartic Curve since, if you switch the 1's to 3's, that's exactly what you get.

    - AC

    P.S. Seems to be a little typo above. 744 = 12 * 2 * 31 = 24*31 rather than 12^2 * 31
     
    Last edited: Jun 22, 2012
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