Arcadian Functor

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Marni D. Sheppeard

Sunday, November 18, 2007

M Theory Lesson 126

As Ebeling explains, by taking the seven rows of the Hamming circulant and adding a check bit to each row we can write down the seven vectors

$v_1 = \frac{1}{\sqrt{2}} (0,1,1,0,1,0,0,1)$
$v_2 = \frac{1}{\sqrt{2}} (0,0,1,1,0,1,0,1)$
$v_3 = \frac{1}{\sqrt{2}} (0,0,0,1,1,0,1,1)$
$v_4 = \frac{1}{\sqrt{2}} (1,0,0,0,1,1,0,1)$
$v_5 = \frac{1}{\sqrt{2}} (0,1,0,0,0,1,1,1)$
$v_6 = \frac{1}{\sqrt{2}} (1,0,1,0,0,0,1,1)$
$v_7 = \frac{1}{\sqrt{2}} (1,1,0,1,0,0,0,1)$

in $\mathbb{R}^{8}$. These satisfy the rule for a root lattice, $v^{2} = 2$. With the change of variables $e_1 = v_1$, $e_2 = v_2 - v_1$, $e_3 = v_3 - v_2$, $e_4 = v_4 - v_3$, $e_5 = v_5 - v_4$, $e_6 = v_6 - v_5$ and $e_7 = v_7 - v_6$, and the addition of the vector

$e_8 = \frac{1}{\sqrt{2}} (-1,-1,0,0,1,0,-1,0)$

we have a basis for the $E8$ lattice. Since $u^{2} = 2$ for such vectors, it follows that $u \cdot v \in \{ 0, \pm 1, \pm 2 \}$. Hopefully these numbers are familiar.

5 Comments:

Blogger Philip said...

Marni, how do I get the LaTeX to render in Explorer?

November 18, 2007 1:19 PM  
Blogger kneemo said...

Smells like ternary logic to me. ;)

November 18, 2007 1:25 PM  
Anonymous Anonymous said...

Sorry, but explorer isn't the way to view MathML. Kea.

November 18, 2007 2:23 PM  
Anonymous Anonymous said...

Smells? Or reeks, perhaps?

November 18, 2007 2:25 PM  
Blogger Philip said...

So I discovered... I'm now using mathml fonts and viewing with Firefox and all is well. Thanks.

By the way, now that you've been doctored what are your plans? There have recently been a few positions advertised in Australia and New Zealand. Sydney Uni advertised two positions in pure maths. Otago and Waikato also had positions.

If you're open to working on things other than theoretical physics, there appear to be quite a few postdocs in statistics, applied maths, financial modelling and so on.

November 18, 2007 2:31 PM  

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