Arcadian Functor

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Marni D. Sheppeard

Wednesday, November 28, 2007

M Theory Lesson 131

Speaking of the j invariant, recall that

$J(q) = j(q) - 744$

where $744 = 248 \times 3$, so in terms of the modular forms $\theta_0$ and $\theta_1$,

$J(q) = -3 [9 \frac{(\theta_{0}^{4} + 8 \theta_{0} \theta_{1}^{3})^{3}}{(\theta_{1}^{4} - \theta_{0}^{3} \theta_{1})^{3}} + 248]$

M theorists will recognise the number 248 as the dimension of the Lie group E8. Thus polynomials in $J(q)$ are a bit like polynomials in $\Theta$ and $\Pi$ (otherwise known as Eisenstein series $E_4$ and $E_6$). In fact, a classical result (mentioned in this paper) says that the modular discriminant of the denominator is given by

$\Delta = \frac{1}{1728} (\Theta^{3} - \Pi^{2})$

whereas the numerator is simply $\Theta^{3}$. The series $\Theta$ is sometimes written, for $q = e^{2 \pi i \tau}$ and $y = \textrm{Im} (\tau)$,

$\Theta (q) = 1 - \frac{3}{\pi y} - 24 \sum_{1}^{\infty} \frac{r q^{r}}{1 - q^{r}}$

For $y = 1$ (and ignoring the summation, which roughly equals 0.0018779 at $\tau = i$) this equals $1 - \frac{3}{\pi}$, which Louise Riofrio often likes to tell us is equal to 4.507034% (the observed fraction of baryonic matter). Is this another crackpot coincidence? Note that $0.0018779 \times 24$ is also equal to 4.507%, showing that $\Theta (q)$ is close to zero for $\tau = i$.

3 Comments:

Anonymous Tony Smith said...

Sorry for intruding on an M-theory entry,
but it did mention E8 and I just put up on dotMac at

web.mac.com/t0ny5m17h/Site/E8GLTSCl8Cl16.pdf

a pdf file about Garrett's model in terms of E8(8) and how it can be used to calculate stuff.

Tony Smith

PS - I hope I did the link coding correctly.

PPS - I apologize to Louise for using the term "Dark Energy", but when I use it I see it as a more-commonly-recognized term that is synonymous with ""variable c".

November 28, 2007 5:50 PM  
Blogger Kea said...

That's great, Tony. Thanks.

November 29, 2007 7:42 AM  
Anonymous Tony Smith said...

Thanks for allowing my comment with a link to my latest paper to be on your blog.

As of now (8 AM EST or so Thursday morning) I have had no other replies from anyone (Jacques Distler has not removed it from his blog, but he has not replied either, and Bee has not replied to my email, and Garrett has not replied to a comment I posted on his blog.

Anyhow, I put up another longer version at

web.mac.com/t0ny5m17h/Site/GLE8Cl8TSxtnd.pdf

that has more details of calculations, etc.,
and
also a pretty comet picture for fun.

I do not expect to post any further versions (unless somebody shows me mistakes that I think are important enough to warrant posting corrected versions).

Thanks very much for allowing these impositions on your blog's hospitality.

Tony Smith

November 30, 2007 2:47 AM  

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