Does $SL_3(R)$ embed in $SL_2(R)$? - MathOverflow

Does $SL_3(R)$ embed in $SL_2(R)$?

2010-11-26T08:18:29Z

Is there any non-trivial ring such that $SL_{3}(R)$ is isomorphic to a subgroup of $SL_{2}(R)$?

$SL_{3}(\mathbb{Z})$ is not am amalgam, and has the wrong number of order $2$ elements to be a subgroup of $SL_{2}(\mathbb{Z})$.

Is there any non-trivial ring where this occurs? When can this definitely not occur? I am trying to understand if there is any sort of group-theoretically apparent notion of dimension here.

Answer by Boris Bukh for Does $SL_3(R)$ embed in $SL_2(R)$?

2010-11-26T11:17:50Z

Define a sequence of groups $G_i$ and associated group rings $R_i=\mathbb{Q}[G_i]$. To start put $G_0=\mathbb{Q}$. Then define $G_{i+1}=SL_3(R_i)$. The group $SL_3(R_i)$ is a subgroup of $SL_2(R_{i+1})$ because $G_{i+1}$ is a subgroup of $SL_2(R_{i+1})$ (as the group of certain diagonal matrices). Similarly, $G_i$ is a subgroup of $G_{i+1}$, and hence $R_i$ is a subring of $R_{i+1}$. Then $R=\bigcup_i R_i$ is the ring you want.

Answer by Bugs Bunny for Does $SL_3(R)$ embed in $SL_2(R)$?

2010-11-26T12:46:41Z

My guess that it should not be possible because $SL_2(R)$ would not have a 3-soluble nonnilpotent subgroup. I am not sure whether it is true that any soluble non-nilpotent subgroup would lie in Borel subgroup but I imagine that this is right...

Answer by Alex Bartel for Does $SL_3(R)$ embed in $SL_2(R)$?

2010-11-26T16:34:31Z

The following doesn't work:

Assume that $2\neq 0$ in $R$ - this is the only assumption I will need. All this is based on ideas of Kevin Buzzard, Tim Dokchitser and Kevin Ventullo.

Since $1\neq-1\in R$, there is an embedding of $S_4$ into $SL_3(R)$: $$ (1,2,3,4)\mapsto \begin{pmatrix}0&1&0\\-1&0&0\\0&0&1 \end{pmatrix},\;\; (1,2)\mapsto \begin{pmatrix}-1&0&0\\0&0&1\\0&1&0 \end{pmatrix}. $$

As Kevin notes, a matrix of order 2 in $SL_2(R)$ is either a scalar (minimal polynomial linear) or the minimal ~~and therefore also the characteristic~~ polynomial is $x^2-1$. Since the involutions in $S_4$ are not central, the latter must be the case for all of them (we are assuming that $R$ is commutative, otherwise we don't know how to make sense of $SL$, so all scalar matrices are central). ~~But the characteristic polynomial of a matrix in $SL_2$ must have constant term $1\neq-1$ - contradiction.~~

If $2=0$ in $R$, then we can embed $SL_3(\mathbb{F}_2)\cong PSL_2(\mathbb{F}_7)$ in $SL_3(R)$. This group is simple, so no element embeds as scalars into $SL_2(R)$. Thus, the minimal polynomials of all the elements of $SL_3(\mathbb{F}_2)\subseteq SL_2(R)$ are quadratic. Can anyone see how to finish this?

Answer by Qiaochu Yuan for Does $SL_3(R)$ embed in $SL_2(R)$?

2010-11-26T20:32:11Z

(Edit: the arguments below build on observations by several other people in this thread.)

Case: $2 = 0$ in $R$. As has been observed elsewhere, in this case $G = \text{SL}_3(\mathbb{F}_2)$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $4$ and trace $r$. Then $g^4 = 1$ and $g^2 = rg + 1$, hence $g^4 = r^2 g^2 + 1$ and $r^2 g^2 = 0$. This gives $r^4 g^4 = 0$, hence $r^4 = 0$. But $r^2 g^2 = r^3 g + r^2 = 0$, hence $r^4 g = r^3 = 0$, hence $r^3 g = r^2 = 0$, and this is both necessary and sufficient.

The elements of $R$ which square to zero form a nilpotent ideal $I$. It follows that if $G$ embeds into $\text{SL}_2(R)$, the image of any elements of order $4$ in $\text{SL}_2(R/I)$ must, by the above computation, have order dividing $2$. In particular the image of $G$ is not isomorphic to $G$, so it must be trivial since $G$ is simple. Hence the image of $G$ in $\text{SL}_2(R)$ consists only of matrices congruent to the identity $\bmod I$. But any such matrix has trace squaring to zero, hence order dividing $4$, which contradicts the existence of elements of order $7$ in $G$. So no such embedding exists.

Case: $2$ is nilpotent in $R$. It is still true in this case that $SL_3({\mathbb F}_2)$ embeds into $\text{SL}_3(R)$, because the two complex 3-dimensional irreducible representations of $SL_3({\mathbb F}_2)=\text{PSL}_2({\mathbb F}_7)$ are realisable over ${\mathbb Z}_2$ (they need $(1\pm\sqrt{-7})/2$ which are in ${\mathbb Z}_2$), and $R$ is a ${\mathbb Z}_2$-algebra. However, as explained in the follow-up question, $SL_3({\mathbb F}_2)$ is not a subgroup of $\text{SL}_2(R)$ for any $R$.

Case: $2$ is not a zero divisor in $R$. As has been observed elsewhere, in this case $G = S_4$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $2$ and trace $r$. Then $g^2 = 1$ and $g^2 - rg + 1 = 0$, hence $rg = 2$. Squaring gives $r^2 = 4$, so $r$ is also not a zero divisor. It follows that $g$ must be a scalar multiple of the identity, hence central. But $S_4$ contains elements of order $2$ which are not central. (As does $S_3$. Hence this argument also shows, as in Tim Dokchitser's answer, that $S_3$ does not embed into $\text{SL}_2(R)$ in this case.)

In particular the above arguments cover the case that $2$ is invertible, as well as the case that $R$ is an integral domain. The remaining case is that $2 \neq 0$, and it is a zero divisor but is not nilpotent.

One may hope that the reduction to Artinian rings as in the follow-up question and the fact that we know the answer both when $2$ is invertible and for ${\mathbb Z}_2$-algebras can actually finish this off.

Answer by Tim Dokchitser for Does $SL_3(R)$ embed in $SL_2(R)$?

2010-11-26T21:30:01Z

Here is a brute force proof that $SL_3(R)$ does not embed into $SL_2(R)$ for any commutative ring $R$ with 1 where 2 is invertible, inspired by Kevin's universality remarks and Alex's observation that $S_3$ in my comment does not land in $SL_2(R)$.

The claim is that the symmetric group $S_3$ cannot be embedded in $SL_2(R)$ for any $R$: suppose it can be, take 2 general matrices $M=\begin{pmatrix}a&b\cr c&d\end{pmatrix}$ and $T=\begin{pmatrix}e&f\cr g&h\end{pmatrix}$ and consider the relations $M^3=id=T^2$, $M^2T=TM$ and $det M=det T=1$. These are 4+4+4+1+1=14 polynomial relations in 8 variables $a,...,h$, and a Groebner basis computation shows that the ideal they generate is $\langle h^2-1,g,f,h-e,d-1,c,b,a-1\rangle$; in Mathematica this is

$\gt$ M = {{a, b}, {c, d}}; T = {{e, f}, {g, h}}; id = {{1, 0}, {0, 1}};

$\gt$ GroebnerBasis[{M.M.M - id, T.T - id, M.M.T - T.M, Det[T] - 1, Det[M]-1 }]

{-1+h^2,g,f,-e+h,-1+d,c,b,-1+a}

In other words, the relations imply that $a=d=1$ and $b=c=0$ for any $R$, so $M=1$, contradicting the assumption that $S_3\to SL(2,R)$ is injective.

On the other hand, using $$ M=\begin{pmatrix}0&0&1\cr 1&0&0\cr 0&1&0\end{pmatrix},\qquad T=\begin{pmatrix}0&-1&0\cr -1&0&0\cr 0&0&-1\end{pmatrix}, $$ we can embed $S_3$ into $SL(3,R)$ for any ring $R$.

P.S. Hopefully, there is a better proof that $S_3$ does not embed into $SL(2,R)$!

Edit: The Groebner basis computation works over $Z[1/2]$, so this only works for rings $R$ with 2 invertible. Kevin & John: thank you for pointing this out!

Answer by Peter Samuelson for Does $SL_3(R)$ embed in $SL_2(R)$?

2010-11-27T00:08:33Z

EDIT: I misquoted the book, and the answers here show it isn't fixable.

Let $A,B \in SL_2(R)$. This book by Brumfiel and Hilden has the following two facts in it:

$Tr(A)Tr(B) = Tr(AB) + Tr(AB^{-1})$.
~~The pair $A,B$ is uniquely determined, up to simultaneous conjugation, by the elements $Tr(A)$, $Tr(B)$, and $Tr(AB)$.~~

Taking $A^2 = 1$ and $A = B$, the first fact implies that $Tr(A) = \pm 2$, and then the second fact implies that $A$ is conjugate (and hence equal) to $\pm Id$. Since there is an embedding $S_4 \hookrightarrow SL_3(R)$ (which is described in the other answers), we can finish by noting that the images of the transpositions are not central in $SL_3(R)$.

~~I would guess that there's a more elementary way to prove that involutions in $SL_2(R)$ are central, but I don't know one at the moment.~~

Answer by Frogger for Does $SL_3(R)$ embed in $SL_2(R)$?

2010-12-04T15:06:39Z

$\newcommand{\SL}{\mathrm{SL}}$ $\newcommand{\m}{\mathfrak{m}}$ $\newcommand{\F}{\mathbf{F}}$ $\newcommand{\Z}{\mathbf{Z}}$

Assume that $\SL_3(R)$ is a subgroup of $\SL_2(R)$. We wish to obtain a contradiction.

Here is the strategy. Suppose that $R$ contains a subring of the form $A \oplus B$ where $2A = 0$. Then $SL_3(\F_2)$ is a subgroup of $\SL_3(A)$, which is a subgroup of $\SL_3(A \oplus B)$, which is a subgroup of $\SL_3(R)$. Hence, under our assumption on $R$, $\SL_3(\F_2)$ is a subgroup of $\SL_2(R)$, and this is ruled out by Silence Dogood's answer. $R$ trivially admits such a decomposition when $2 = 0$. Hence we may assume that $2 \ne 0$, and thus that $S_4$ is a subgroup of $\SL_3(R)$, and hence of $\SL_2(R)$.

If $S \subset R$ contains a subring of the form $A \oplus B$ with $2A = 0$, then so does $R$. Thus, WLOG, assume that $R$ is generated by the entries of $g-1$ where $g \in S_4 \subset \SL_2(R)$. Let $K \subset S_4$ denote the Klein $4$-subgroup. Then any map $S_4 \rightarrow G$ is injective if and only if the restriction $K \rightarrow G$ is non-zero (Obvious). $K$ is the only non-trivial normal subgroup of $S_4$ which is a $p$-group (Obvious). By construction, $R$ is Noetherian. If $x \in R$ is any element, and $\m$ is a maximal ideal containing the annihilator of $x$, then $x$ is non-zero in the localization $R_{\m}$. Hence there exists an $\m$ such that $K \rightarrow \SL_2(R_{\m})$ is non-zero, so $S_4 \rightarrow \SL_2(R_{\m})$ is injective. (Choose $x$ to be a non-zero matrix entry of $g-1$ for $g \in K$.) Let $A = R_{\m}$, and let $k = A/\m$. Consider the projection map $S_4 \rightarrow \SL_2(k)$, and let $H$ denote the kernel. Let $g$ be an element of $H$ which is not the identity (if such an element exists). By the Krull intersection theorem (as in SD's answer), there exists a minimal integer $n$ such that $$g - 1 \equiv 0 \mod \m^n, \qquad (g - 1) \not\equiv 0 \mod \m^{n+1}.$$ If $i$ is co-prime to the characteristic of $k$, then it is a unit in $A$, and $$g^i - 1 = (1 + (g-1))^i - 1 \equiv i (g-1) \mod \m^{n+1} \not\equiv 0 \mod \m^{n+1}.$$ It follows that the order of $g$ is some power of the characteristic (or is trivial if $\mathrm{char}(k) = 0$), and hence $H$ is a $p$-group. Hence either $S_4$ injects into $\SL_2(k)$, or $k$ has characteristic $2$ and $H = K$. The former does not occur. We shall prove that $2 = 0$ in $A$. The image of $S_4$ in $\SL_2(k)$ is $S_3$. $S_4$ contains an element $M$ of order $2$ which maps to an element of order $2$ in $S_3$ (for example, any $2$-cycle). The matrix $M$ has order two, and hence satisfies the polynomial $M^2 - 1 = 0$. Yet $M$ also has determinant one, and thus also satisfies the polynomial $M^2 - \mathrm{trace}(M) M + 1 = 0$, by Cayley--Hamilton. It follows that $\mathrm{trace}(M) M = 2 \ne 0$ (by assumption). Yet $M$ has at least one entry that is a unit, and thus $(\mathrm{trace}(M)) = (2)$ in $A$, and it follows easily that the image of $M$ is a scalar matrix in $\SL_2(k)$. Since $k$ has characteristic $2$, this implies that the image of $M$ in $\SL_2(k)$ is trivial ($v^2 = 1$ implies that $v = 1$), a contradiction. Hence $2 = 0$ in $A$.

We have now shown that $2 = 0$ in $A = R_{\m}$. Suppose we can show in addition that $A$ has finite length, that is $A/\m^k = A$ for some $k$. Assume this is so. Let $x_1, \ldots, x_n$ be generators of $\m^k \subset R$. By definition, $x_i$ maps to zero in the localization map $R \rightarrow R_{\m} = A$. Thus there exists an element $y_i \notin \m$ such that $y_i x_i = 0$. Let $y = y_1 \times \ldots \times y_n$. Since $y_i \notin \m$, the product $y \notin \m$. It follows that $$y + \m^k = R,$$ as the ideal on the LHS is not contained in any maximal ideal. On the other hand, $y$ annihilates $\m^k$ by construction. Thus, by the Chinese remainder theorem, $$R = R/y \m^k = R/y \oplus R/\m^k = R/y \oplus A.$$

Since $2 = 0$ in $A$, this shows that $R$ has the required decomposition. Thus we will be done if we can show that $A$ has finite length. Equivalently, we are done if we can show that the non-unit elements of $A$ are nilpotent.

It seems according to Tim that this won't work, since $S_4$ injects into $\SL_2(\F[[x]])$ via the map $$(12) \mapsto \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)$$ and $$(1234) \mapsto \left( \begin{matrix} 1+x+x^2 & 1+x^2 \\ x^2 & 1+x+x^2 \end{matrix} \right)$$

Hence this answer, for the time being, is a complete fail.