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Modular Forms, Fundamental Domain Question

  • Thread starter binbagsss
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Homework Statement



My question is below it makes more sense there, after I have gave my interpretation of the definition of the fundamental domain to confirm my understanding

Homework Equations



The fundamental domain is defined as:

1) Every point in H is equivalent to a point in F.

2) Points in the interior of F are SL2(Z)-inequivalent: Only identifications are via • For ##|Re(τ )| = \pm 1/ 2 ## via ## τ → τ ± 1 ##• On the arc ##|τ | = 1 ##via ## τ → −1/τ ## .And is given by :

## -½ \leq u \geq ½ ##

## |t| \leq 1 ##

The Attempt at a Solution



- Criteria doesn’t specify one-to-one, some elements may map once, more than once, as long as there is at least one from H to F, and, each point in F does not necessarily need to be mapped to- does it happen that each is by any chance ?

- Definition of ‘boundary’ with respect to the criteria, i.e. It is clear all of ##u=-1/2## and ##u=1/2## can be mapped to each other via T and the arc for negative and positive ##u## can be mapped to one another via ##S##
-On this, I assume the definition of the 'boundary' relative to the fundamental domain is not that there must exists some point in the exterior that maps to it (or vice versa), since the boundary itself covers that, however, it is not disallowed and so a point not in F may map to the boundary, and then, by the transitive property of an equivalence relationship, it is not only equivalent to that point, but to one more point- its reflection in the y axis (either via T or S).

for eg I have this question:

funddom17.png

And the answer is to just translate negatively five times, yielding (2i+1)/5 clearly leaving a magnitude of ##1## and so on the arc of F. However the question asks ‘what is the equivalent point in F’ (implying only one such point) but via S we are still on the boundary, a reflection in y and so there are actually two equivalent points in F (2i+1)/5 and (2i-1)/5


Many thanks
 

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