My question is the following. Which representations of $Sp(2g, \mathbb Z)$ are extendable to representations of $Sp(2g, \mathbb C)$ or $Sp(2g, \mathbb R)$. Is there a general theory and a good beginning reference for these type of questions. Some Representations of $Sp(2g, \mathbb Z)$ dont seem to extend such as coset spaces associated to quotients with respect to congruence subgroups. The question is are these the only ones in some sense. I am a novice in this branch of mathematics so any references would be appreciated. The question arose because I want to understand complex vector bundles admitting a flat Connection on $\mathcal A_g(\mathbb C)$.
The result you are looking for is the Margulis superrigidity theorem. See Chapter 13 of the book "Introduction to Arithmetic Groups" by Dave Witte Morris for more details. 


Perhaps I can take the liberty of amplifying on Andy Putman's answer. We are looking for, say,(finite dimensional complex) representations of $\Gamma = Sp_{2g}({\mathbb Z})$ for $g\geq 2$, so we are in "higher rank". Using superrigidity, it can be proved that all representations of $\Gamma$ are completely reducible. We may therefore, need only describe the irreducible complex representations $\rho $ of $\Gamma$. Any such $\rho$ is a tensor product of the form $\tau \otimes \sigma$ where $\tau$ is an irreducible {\bf algebraic} representation of $Sp_{2g}({\mathbb C})$ (hence given by highest weight theory) and $\sigma$ is an irreducible representation of $Sp_{2g}({\mathbb Z}/m{\mathbb Z})$ (a finite congruence quotient of $\Gamma $). This can be easily deduced from the reference of Prof. Jim Humphreys (BassMilnorSerre paper) 


$g$
here, since the case of a rank one simple Lie group is far more problematic than higher ranks (beginning with the congruence subgroup problem). The literature is formidable, so the bookinprogress suggested by Andy is a good place to start. Though the question in your header looks straightforward, it conceals a lot of heavy mathematics. (By the way, you might add liegroups to your long list of tags.) – Jim Humphreys Dec 6 '12 at 20:50