What matrix groups can be embedded in $Sp_4$?

In a joint paper with Yifan Yang we constructed an "exotic" embedding of $SL_2(\mathbb R)$ in $Sp_4(\mathbb R)$ (in fact, of $PSL_2(\mathbb R)$ in $PSp_4(\mathbb R)$), namely, $$\iota\colon\begin{pmatrix} a & b \cr c & d \end{pmatrix} \mapsto\begin{pmatrix} a^2d+2abc & -3a^2c & abd+\frac12b^2c & \frac12b^2d \cr -a^2b & a^3 & -\frac12ab^2 & -\frac16b^3 \cr 4acd+2bc^2 & -6ac^2 & ad^2+2bcd & bd^2 \cr 6c^2d & -6c^3 & 3cd^2 & d^3 \end{pmatrix}.$$ An equivalent form of the embedding was independently discovered by Don Zagier, and we could not find it in the literature.

Although the properties of the embedding (discussed in the preprint above) are nice by themselves, I am interested in an exhaustive list of possibilities to embed other matrix groups and their direct products in $Sp_4(\mathbb R)$ (or $PSp_4(\mathbb R)$). For example, can the direct product of two copies of $SL_2(\mathbb R)$ be embedded?

As I am not a specialist in Lie groups, I would appreciate plainer sources. Thank you for any help in advance!

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Isn't it enough to check whether the group has a 4-dimensional representation V such that \Lambda^2(V) has positive dimension? –  Qiaochu Yuan Jul 5 '10 at 11:42
Also, related, if you haven't seen it: mathoverflow.net/questions/9378/… –  Qiaochu Yuan Jul 5 '10 at 11:45
@Wadim: are you asking whether SL_2(R) x SL_2(R) embeds into Sp_4(R)? Isn't this obvious? SL_2(R) acts on R^2 and preserves an alternating form. So SL_2(R) x SL_2(R) acts on R^4 and preserves an alternating form---just take the direct sum. Am I missing something? –  Kevin Buzzard Jul 5 '10 at 11:49
PS there are several different "explicit" Sp_4's in the literature---different people take different explicit alternating forms. So I think it's good practice to say explicitly what your J is if you're going to write down explicit matrices as in the question. –  Kevin Buzzard Jul 5 '10 at 11:54
Tilouine is a master of automorphic forms on Sp_4 (so, the number theory of Sp_4, not the group theory). I think he likes his J to be anti-diagonal because then I think it makes the standard parabolic subgroups a bit nicer-looking? Of course you can easily move from one form to the other via some conjugation. I don't know if Tilouine would know too much about these exotic SL_2's though. –  Kevin Buzzard Jul 5 '10 at 12:08
You can embed the direct product of two copies of $SL_2(\mathbb{R})$. One embedding sends the entries to the center $2\times2$ block, the other sends the entries to the corners with the $2\times2$ identity matrix in the center block. Here, the $J$ matrix is the standard anti-diagonal matrix