# isogeny and congruence subgroup

Let $G_1$ and $G_1$ be two semisimple algebraic groups defined over $\mathbb{Q}$, suppose we have a surjective homomorphism $f: G_1\to G_2$, with finite kernel contained in the center of $G_1$.

By congruence subgroup of $G_i$, for $i=1,2$, we means $K\cap G_i(\mathbb{Q})$, with $K$ compact open subgroup of $G_i(\mathbb{A}_f)$.

Then consider the images of congruence subgroups of $G_1$ under the map $f$, is it cofinal with the congruence of $G_2$, i.e. every congruence subgroup of $G_2$, it conatines image of some congruence of $G_1$ and every image of congruence subgroup of $G_1$ containes congruence subgroup of $G_2$ ?

If the above is not true, then is it true that almost all (except finite many) congruence subgroups of $G_2$ are contained in the image of some congruence subgroup of $G_1$?

For example $G_1=SL_2(\mathbb{Q})\times SL_2(\mathbb{Q})\times SL_2(\mathbb{Q})$, and $G_1$ has a natural map into $SP_8(\mathbb{Q})$ by tensor product action. Let $G_2$ be the image . Let $C_2$ (resp. $C_1$) be a connected shimura curve using $G_2$ (resp.$G_1$) and congruence subgroup $\Gamma_2$ (resp.$\Gamma_1$) of $G_2$ (resp. $G_1$). Then I want to know is it true that except finite many congruence subgroup $\Gamma_2$ , one can always find some $\Gamma_1$, such that there is an etale map from $C_2$ to $C_1$.

• @Tom: Note that congruence subgroups of $G_2$ can't be finite, in your last paragraph. Is your definition of "congruence subgroup" standard? – Jim Humphreys Nov 30 '13 at 1:07

To deal with isogenies it is useful to start with some of the basic work done on the Congruence Subgroup Problem in the 1960s and later. This problem can be formulated for any connected linear algebraic group defined over a number field such as $\mathbb{Q}$. Much of the work focuses on the difficult case of an (almost) simple group $G$, where it was known long ago for $G=\mathrm{SL}_2$ that not every arithmetic subgroup of $G(\mathbb{Q})$ (commensurate to $G(\mathbb{Z})$) is a congruence subgroup (a subgroup containing a principal congruence subgroup, necessarily of finite index in $G(\mathbb{Z})$). But eventually the split, simply connected groups of rank at least 2 (such as $\mathrm{SL}_n$ with $n \geq 3$) were found to satisfy the congruence subgroup property unless the field is totally imaginary (Bass-Milnor-Serre, Matsumoto). Further work by Prasad-Raghunathan has dealt with non-split groups.

But when $G$ fails to be simply connected the outcome is much less favorable. This is discussed carefully by Serre in the Bourbaki seminar (exp. 330) here. See 1.2(c); later work clarified the status of the strong approximation property invoked here.

In particular, the isogeny from the simply connected covering group $\widehat{G}$ onto $G$ does not behave at all well on congruence subgroups. If I understand correctly what is being asked here, this often provides a strongly negative answer.

• I am sorry for making mistake. I edit hte queston. I am especially concern the example I added. – Lan Nov 30 '13 at 12:35
• To clarify the end of Jim's answer, for a number field $F$ and $n\ge 2$ there are many finite-index non-congruence subgroups $\Gamma_2\subset {\rm{PGL}}_n(O_F)$, yet the cokernel ${\rm{H}}^1(O_F,\mu_n)$ of $f:{\rm{SL}}_n(O_F) \rightarrow {\rm{PGL}}_n(O_F)$ is finite, so replacing $\Gamma_2$ with a finite-index subgroup ensures $\Gamma_2=f(\Gamma_1)$ for $\Gamma_1=f^{-1}(\Gamma_2)$. But $\Gamma_1$ has finite index in ${\rm{SL}}_n(O_F)$, so it is a congruence subgroup if $n \ge 3$ and $F$ has a real place. Then $\Gamma_1$ is congruence with image containing no congruence subgroup. – user76758 Dec 1 '13 at 7:02

The image in $\text{PSL}_2(\mathbb{Q})$ of a congruence subgroup of $\text{SL}_2(\mathbb{Q})$ need not be a congruence subgroup (hence doesn't contain one).

Incidentally, for any reductive group $G$ over $\mathbb{Q}$ the notion of a congruence subgroup of $G(\mathbb{Q})$ is well defined. One can either choose a $\mathbb{Z}$-structure and show that your notion is independent of the choice, or use adeles.

@user76758 Yes, that's a good definition.

@Tom The isogeny $f$ gives a homomorphism on the completions of the groups of rational points for the topologies defined by the congruence subgroups. If $G_1$ is simply connected, the kernel of this map is $N(A_f)$ where $N$ is the (finite) kernel of $f$ and $A_f$ is ring of finite adeles. Since $N(A_f)$ is infinite, this suggests there are a lot more congruence subgroups in $G_1$ than $G_2$. Another way of looking at it, if you only form Shimura curves wrt to the congruence subgroups of $PSL_2$ (rather than $SL_2$), then you are missing a lot of curves.

• But is it true that almost all(except finite many) congruence subgroups of $G_2$ contain image of some congruence of $G_1$? In your example, is it ture almost all congruence subgroups of $PGL_{2}(\mathbb{Q})$ containe image of some congruence subgroup of $SL_2(\mathbb{Q})$? – Lan Nov 29 '13 at 21:53
• @anon: is your implicit definition of "congruence subgroup" $G(\mathbf{Q}) \cap K$ for a compact open subgroup $K$ of $G(\mathbf{A}^{\infty})$? – user76758 Nov 29 '13 at 23:30
• @Tom: Every congruence subgroup of $G_2(\mathbf{Q})$ contains the image of a congruence subgroup of $G_1(\mathbf{Q})$ since its preimage in $G_1(\mathbf{Q})$ is such a subgroup. Indeed, this reduces to checking that $G_1(\mathbf{A}^{\infty}) \rightarrow G_2(\mathbf{A}^{\infty})$ is topologically proper (so preimage of a compact open subgroup is a compact open subgroup), which holds because $G_1 \rightarrow G_2$ is a finite morphism. – user76758 Nov 29 '13 at 23:33
• Can you give an explicit example of a congruence subgroup of $SL_2(\mathbb{Q})$ that does not project to a congruence subgroup of $PSL_2(\mathbb{Q})$? It seems to me that you should even be able to get an example in $SL_2(\mathbb{Z})$ by intersection. In particular, if $K$ is a compact-open subgroup of $SL_2(\mathbb{A}^\infty)$, then $f(K)$ is compact in $PSL_2(\mathbb{A}^\infty)$, so $f(K\cap SL_2(\mathbb{Q}))= f(K)\cap PSL_2(\mathbb{Q})$ is the intersection of a compact subgroup with $PSL_2(\mathbb{Q})$. So I presume you have in mind an example where $f(K)$ is not open? – Ian Agol Nov 30 '13 at 3:41
• I am sorry, I make mistake in formulating the question, I edited it. – Lan Nov 30 '13 at 12:23