Re: Lattice Gauge Theory

Hi Gerard,

I posted an earlier response to this where I pointed out that your
example is not valid because your 4-paths consist of 2-paths for which
@^2 != 0 and that is not acceptible. However, there is an even worse
problem with your example that Urs pointed out to me via email and I
missed the first time.

Gerard Westendorp <[EMAIL PROTECTED]> wrote:
> Urs Schreiber wrote:
> [..]
> > Dimension on discrete complexes is naturally defined as the maximum p for
> > which the complex contains p-paths that are in the kernel of @^2. Or dually,
> > the maximum p for which we can find discrete p-forms on the complex.
> 
> But isn't it possible to define 4-paths with 
> @^2=0 on a 2d lattice? For example, if you take a square grid, with
> vertex coordinartes (i,j), tou could make 4-paths A and B:
> 
> A = (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2)
> B = (0,0) -> (1,0) -> (2,0) -> (2,1) -> (2,2)
> 
> And then compose A-B, which would have @^2=0. 

Sorry, actually A-B does not satisfy @^2 = 0.

Remember that @^2 = t^2 - s^2. So we can compute the boundary of each
piece: A,B.

t^2 A = (0,2) -> (1,2) -> (2,2)

s^2 A = (0,0) -> (0,1) -> (0,2)

t^2 B = (2,0) -> (2,1) -> (2,2)

s^2 B = (0,0) -> (1,0) -> (2,0)

I hope you can see that @^2(A-B) = (t^2 - s^2)(A-B) != 0.

Therefore, your example of a 4-path A-B on a 2d lattice does not
satisfy @^2 = 0.

It is great that you are challenging us :)

based on your example, I can give another example that I imagine you
might dream of after getting this response.

Ack! But I gotta run! Maybe I'll wait and see if you can think of
other examples that might cause us some trouble in the meantime :)

Best regards,
Eric