Let $Sp_2g$ be the symplectic group defined over $\mathbb{Q}$. Consider $SL_2(\mathbb{Z})$ as a subgroup of $Sp_{2g}(\mathbb{Z})$ (the embedding that I have in mind is $A\to \begin{pmatrix} A& 0\\ 0& I_{2g-2} \end{pmatrix}$, but you can take any algebraic group embedding which yields a positive answer to the following question). Is it true that there eixst finitely many $g_i\in Sp_{2g}(\mathbb{Q})$ such that the $\mathbb{Q}$-Zariski closure of $\cup_i g_i SL_2(\mathbb{Z})g_i^{-1}$ is equal to $Sp_{2g}$? i.e. that this union is Zariski-dense in $Sp_{2g}$? If yes, what conditions and restrictions should these elements satisfy? How should the $g_i$ be chosen? I would appreciate any result, comments or reference in this direction.
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$\begingroup$ In order for this question to be well-defined, you might need to say how $SL_{2}(\mathbb{Z})$ is considered as a subgroup of $Sp_{2g}(\mathbb{Z})$. $\endgroup$– Jeremy RouseCommented Jan 30, 2017 at 22:55
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$\begingroup$ @ Jeremy Rouse, I added explanation to the question. But any embedding which gives a positive answer is welcome! $\endgroup$– Darius MathCommented Jan 30, 2017 at 23:50
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4 Answers
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This seems impossible to me, at least with the kind of embedding considered or anything remotely close to it. I'll think classically, in terms of $\mathbb{C}$ points. Your embedding $SL_2(\mathbb{Z})$ sits inside $SL_2(\mathbb{C}) \subset Sp_{2g}(\mathbb{C})$. Note that $SL_2(\mathbb{C})$ is three dimensional. So the Zariski closure of $\bigcup g_i SL_2(\mathbb{Z}) g_i^{-1}$ is contained in the closed subscheme $\bigcup g_i SL_2(\mathbb{C}) g_i^{-1}$. But the latter is a union of finitely many $3$-dimensional varieties, so it can't fill up the $2g^2+g>3$ dimensional variety $Sp_{2g}(\mathbb{C})$.
This argument applies whenever the map $SL_2(\mathbb{Z}) \to Sp_{2g}(\mathbb{Z})$ extends to a map $SL_2(\mathbb{C}) \to Sp_{2g}(\mathbb{C})$. If you just ask for a map of abstract groups, ignoring algebraic geometry, I don't know what to expect.
answered Jan 31, 2017 at 0:44
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1$\begingroup$ @Misha You should post the details! $\endgroup$ Commented Jan 31, 2017 at 2:03
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3$\begingroup$ Indeed, the desired $\mathbb{Q}$ closure is $Sp_{2g}$ if and only if the image of $SL_2(\mathbb{Z})$ is itself Zariski dense (since $Sp_{2g}$ is an irreducible variety). $\endgroup$ Commented Jan 31, 2017 at 2:15
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2$\begingroup$ If the embedding is $\mathbb{Q}$ algebraic, then the answer to the question is never, unless $g=1$. This is the answer of David. There is the possibility that the $\mathbb{Q}$ closure is different from the $\mathbb{C}$-closure; in that case, what you are asking is that under an algebraic (but not defined over $\mathbb{Q}$-) representation, the image of $SL_2(\mathbb{Z})$ should be $\mathbb{Q}$-Zariski dense in $Sp_{2g}$; there do exist such embeddings; take a standard embedding over $\mathbb{Q}$ and conjugate it by a "transcendental" element. $\endgroup$ Commented Jan 31, 2017 at 12:10
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1$\begingroup$ @Venkataraman On reflection, I don't see this either. Certainly, the image of one element of $SL_2(\mathbb{Z})$ isn't $\mathbb{Q}$-Zariski dense after conjugation by a transcendental (for example, its trace will still be an integer), so it isn't obvious to me that the Zariski conjugate of the whole$SL_2(\mathbb{Z})$ becomes $\mathbb{Q}$-Zariski dense. $\endgroup$ Commented Jan 31, 2017 at 16:58
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2$\begingroup$ the latter is not really a question about $SL_2(Z)$, since it boils down to understanding whether the conjugate of the standard copy $H$ of $SL_2(Q)$ (as given in the question) by a transcendental element (in some sense) is $Q$-Zariski dense. Anyway, since the set of $Q$-algebraic subgroups is countable and since the set of $g$ conjugating $H$ into a given algebraic proper subgroup is a proper subvariety, whenever we take $g$ outside this countable union we have $gHg^{-1}$ $Q$-Zariski dense. $\endgroup$– YCorCommented Jan 31, 2017 at 19:38
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Even for any algebraic group embedding of $SL_2$ into $Sp_{2g}$, with an arbitrary amount of conjugation by arbitrary elements, the answer is no. Consider the matrix $\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}$ in $SL_2$. Its image under the algebraic group embedding is some $2g \times 2g$ matrix whose eigenvalues are powers of $\lambda$. The characteristic polynomial of this matrix is some $2g$-tuple of polynomials in $\lambda$, all defined over $\mathbb Z$, whose image is a $1$-dimensional subvariety, defined over $\mathbb Q$, of the space of polynomials.
Because elements conjugate to this one are dense in $SL_2(\mathbb C)$, the image of every element of $SL_2(\mathbb C)$ under the characteristic polynomial map is contained in this $1$-dimensional $\mathbb Q$-subvariety. The same is true for any number (even an infinite number) of conjugate copies of $SL_2$. Because the characteristic polynomials of elements of $Sp_{2g}$ form a $g$-dimensional variety, there is no way this union can be $\mathbb Q$-Zariski dense.
answered Feb 4, 2017 at 15:43
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As requested, here is a construction (a sketch only) of faithful representations $PSL(2, {\mathbb Z})$ to $PSp(n, {\mathbb Z})$ with Zariski dense images. I will add some details when and if I have time.
I will work modulo the center(s), taking care of the latter takes extra effort. First note that $\Gamma=PSp(n, {\mathbb Z})$ contains involutions $\sigma$ which act as Cartan involutions on the associated symmetric space of $Sp(n, {\mathbb R})$. Let $\tau$ denote an order 3 element of $PSp(n, {\mathbb Z})$ coming from the block-diagonal embedding of $SL(2, {\mathbb Z})$ as in your question. Now, choose a generic sufficiently long (say, in the word metric, but you can also use a matrix norm) element $\gamma\in \Gamma$ and consider the representation $PSL(2, {\mathbb Z})\cong Z_3 \star Z_2\to \Gamma$ sending the generator of $Z_3$ to $\tau$ and the generator of $Z_2$ to the conjugate $\gamma \sigma \gamma^{-1}$. The claim is that such representations do the job. A proof is (mostly) a certain ping-pong argument.
Of course, I recognize that this is not what OP had in mind, but it is a response to a remark in David's answer.
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$\begingroup$ what do you mean by $\mathbb{Z}_{2}\star \mathbb{Z}_{3}$? Sorry, since I am not an Experte in the field! $\endgroup$ Commented Jan 31, 2017 at 19:38
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2$\begingroup$ $\star$ (more usually $\ast$) denotes the free product of groups, see en.wikipedia.org/wiki/Free_product. Here $\mathbb{Z}_n$ is a (somewhat controversial) notation for the cyclic group on $n$ elements (common for some group theoretists, but in conflict with $n$-adic numbers). $\endgroup$– YCorCommented Jan 31, 2017 at 19:39
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$\begingroup$ @YCor, thank you very much, yes of course I know what $Z_2$ means, I meant only the star notation. $\endgroup$ Commented Jan 31, 2017 at 19:51
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A complement to David's answer. Zariski dense (faithful) representations of the $2, 4, 5$ triangle group are constructed by Long and Thistlethwaite's Zariski dense surface subgroups in $SL(4, \mathbb{Z}).$ This is not exactly easy (the paper is nicely written, though).
answered Jan 31, 2017 at 4:01