Constructing subgroups by intersection

asked 2015-02-27 19:03:03 +0200

Jimeree gravatar image

I'd like to construct a subgroup of $Sp\left(4,\mathbb{Z}\right)$ of the form:

$$G_0\left(N\right) = M\left(N\right) \cap {Sp}\left(4,\mathbb{Z}\right)$$

where $M\left(N\right)$ is a $4\times4$ matrix over the integer ring with elements that are multiples of the integer $N$. I think I know how to construct such an $M\left(N\right)$ for a given $N$, but how does one then construct such a subgroup $G_0\left(N\right)$? Thanks!

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With your definition M(N) is not a group as it does not contain the identity... Do you meant the principal congruence subgroup?

vdelecroix gravatar imagevdelecroix ( 2015-02-27 23:40:52 +0200 )edit

Sorry, I should have put curly brackets around $M\left(N\right)$, so it's just a matrix. Specifically, I want to construct:

$$G_0\left(N\right) = { \left( \begin{array}{cccc} \mathbb{Z} & \mathbb{Z} & \mathbb{Z} & N\mathbb{Z} \ N\mathbb{Z} & \mathbb{Z} & N\mathbb{Z} & N^2 \mathbb{Z} \ \mathbb{Z} & \mathbb{Z} & \mathbb{Z} & N \mathbb{Z} \ \mathbb{Z} & \mathbb{Z} & \mathbb{Z} & \mathbb{Z} \end{array} \right) } \cap {Sp}\left(4,\mathbb{Z}\right)$$

We can define congruence subgroups of the modular group in this way, but I want to do the same thing for subgroups of $Sp\left(4,\mathbb{Z}\right)$. Thanks for your help!

Jimeree gravatar imageJimeree ( 2015-02-28 17:25:57 +0200 )edit

The answer really depends on what kind of computations you want to achieve. Could you precise it in your question? Building such a group in Sage will require some non-trivial amount of work and the only non-trivial operations you might get will come from the software GAP (which is shipped with Sage and used a lot for everything related to group theory). You should have a look at it.

vdelecroix gravatar imagevdelecroix ( 2015-02-28 23:50:45 +0200 )edit

Okay, thanks for your response! What I really want to look at are the generators for such subgroups..

Jimeree gravatar imageJimeree ( 2015-03-02 12:38:57 +0200 )edit