Suppose $\rho : \Gamma \longrightarrow SL(4, \mathbb{C})$ is a representation. Here $\Gamma = \langle A, B\; |\; A^{2} = B^{3}\rangle$. Suppose $A^{2} = I_{4} = B^{3}$, where $I_{4}$ means 4 $\times$ 4 identity matrix. Matrix $A$ has three possibilities and matrix $B$ has five possibilities. For all other pairs apart from the conjugate matrices below:

\begin{align*} \rho(A) \sim \left(\begin{array}{cccc} 1 & & &\\ & 1 & & \\ & & -1 & \\ & & & -1 \end{array}\right) \mbox{and} \;\; \rho(B) \sim \left(\begin{array}{cccc} \omega & & &\\ & \omega & & \\ & & \omega^{2} & \\ & & & \omega^{2} \end{array}\right), \end{align*} where $\omega + \omega^{2} + 1 = 0$, I can show that the representation is either reducible or irreducible. For the representation whose images are conjugate to the given above matrices, I claim that is reducible, but I don't know how to prove this claim. Could anyone help me?

  • $\begingroup$ If $A$ and $B$ are both diagonal, then the rep is surely reducible. For some other choices you may (will!?) get an irreducible rep. Also, nothing stops you from varying the multiplicities of the eigenvalues of $A$ and $B$. Anyway, the group $\langle A,B\mid A^2=B^3=1\rangle$ is isomorphic to the group $SL_2(\Bbb{Z})/\langle-I_2\rangle$. Already a challenging enough group (before you start with $\Gamma$). $\endgroup$ – Jyrki Lahtonen Jan 3 '18 at 11:52
  • $\begingroup$ Feel free to fix $A$ and change $B$ via conjugation. You will discover that for generic choice of $B$, the representation you get is irreducible. $\endgroup$ – Moishe Kohan Jan 3 '18 at 17:37
  • $\begingroup$ @Jyrki Lahtonen. I tried using the intersection of the two 2-dimensional eigenspaces of $A$ and the two 2-dimensional eigenspaces of $B$ to show reducibility, but to no avail. $\endgroup$ – DYBnor Jan 4 '18 at 1:52
  • $\begingroup$ You do not have "the representation given above": All what you described are representatives of conjugacy classes of the images of $A$ and $B$ in $SL(4,C)$, In order to define a representation, you have to prescribe actual images of the generators. In the present form, your question is not answerable. $\endgroup$ – Moishe Kohan Jan 6 '18 at 14:04
  • $\begingroup$ @MoisheCohen. You are correct. In particular I want the images $\rho(A)$ and $\rho(B)$ to be conjugated to the above matrices respectively, so that $\rho(A), \rho(B)$ satisfy the given relation in the presentation of $\Gamma$. $\endgroup$ – DYBnor Jan 7 '18 at 7:15

This is a correction of my original answer, where I misread the matrix $b$, I somehow read that it has three distinct eigenvalues, not two.

First of all, there is no "the representation $\rho$ with $\rho(A), \rho(B)$ conjugate to above matrices": There is a continuum of such representations, even taken modulo conjugation. One can ask, however, the following questions:

  1. Is it true that some representations $\rho$ with $\rho(A), \rho(B)$ conjugate to above matrices $a, b$ are reducible? This question has positive answer by taking a representation with image in the subgroup of diagonal matrices (just take $\rho(A)$, $\rho(B)$ diagonal).

  2. Is it true that all representations $\rho$ with $\rho(A), \rho(B)$ conjugate to above matrices $a, b$ are reducible?

The second question also has a positive answer, but the proof is more complicated.

Theorem. Every representation $\rho: \Gamma\to SL(4, {\mathbb C})$ which sends $A$ to a matrix conjugate to $a$ and sends $B$ to a matrix conjugate to $b$ is reducible: It has an invariant plane in ${\mathbb C}^4$.

Proof. First of all, suppose that $M\in SL(4,{\mathbb C})$ is a diagonalizable matrix with eigenvalues $\lambda, \lambda$, $\mu, \mu$ (listed with their multiplicity), such that $\lambda\ne \mu$. Let $E_\lambda, E_\mu$ denote the eigenspaces of $M$ corresponding to $\lambda$ and $\mu$ respectively. Then the set of $M$-invariant planes in ${\mathbb C}^4$ is a subvariety consisting of three components. Two of them are singletons, $\{E_\lambda\}, \{E_\mu\}$, and the remaining component $C_M$ is a 2-dimensional smooth subvariety isomorphic to $({\mathbb C} P^1)^2$. Elements of $C_M$ are the planes $V\subset {\mathbb C}^4$ intersecting both $E_\lambda, E_\mu$ along lines. The subvariety $C_M$ sits naturally in the Grassmannian $Gr_2({\mathbb C}^4)=G(4,2)$ (2-planes in ${\mathbb C}^4$). Its Poincare dual is a cohomology class $[C_M]^*\in H^4(G(4,2))$.

I will need some basic knowledge of the cohomology ring $H^*(G(4,2))$ conveniently discussed in here. You can find much more on cohomology groups of Grassmannians elsewhere as well. (Just google "cohomology ring of Grassmannian".)

The Grassmannian $G(4,2)$ is a smooth connected 4-dimensional complex projective variety (all dimensions are complex in what follows). The $H^2(G(4,2))$ is generated by the Schubert class $a_2$ which is Poincare dual to the complex 3-dimensional cycle $S_P$ consisting of 2-planes which have nonzero intersection with a fixed 2-plane $P\subset {\mathbb C}^4$. (Actually, for the following proof you do not need to know that the dual of $[S_P]$ generates $H^2(G(4,2))$, it suffices to know that the class $[S_P]$ is nonzero (as the fundamental class of any irreducible projective subvariety in a complex projective variety). Then the fact that $H^2(G(4,2))\cong {\mathbb Z}$ will imply that the Poincare dual of $[S_P]$ is a positive multiple of the generator $a_2$.)

The main thing to know is that $$ a_2^4\ne 0\in H^8(G(4,2)). $$ (See the link above.) In particular, if $P_1,...,P_4$ are four planes in ${\mathbb C}^4$ then $$ S_{P_1}\cap S_{P_2}\cap S_{P_3}\cap S_{P_4}\ne \emptyset. $$

Suppose now that $M_1, M_2$ are two diagonzalizable matrices in $SL(4,C)$ each of which has exactly two distinct eigenvalues $\lambda_1, \mu_1$ and $\lambda_2, \mu_2$, of multiplicity 2 each.

Lemma. The linear transformations represented by $M_1, M_2$ preserve a 2-dimensional subspace $Q$ in ${\mathbb C}^4$.

Proof. Observe that $$ C_{M_k}= S_{E_{\lambda_k}}\cap S_{E_{\mu_k}}, k=1, 2. $$ Hence, as we observed above, $$ C_{M_1}\cap C_{M_2}\ne \emptyset. $$ Therefore, there exists a plane $Q\subset {\mathbb C}^4$ which belongs to both $C_{M_1}, C_{M_2}$, hence, invariant under both $M_1, M_2$. qed

Specializing to the case $\lambda_1=1, \mu_1=-1$, $\lambda_2=\omega, \mu_2=\omega^2$, where $\omega\ne 1$ is a cube root of $1$, we obtain the assertion of the Theorem. qed

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  • $\begingroup$ Could you clarify why for each matrix M conjugate to b the subvariety F_{k,M} has complex dimension ≤1 while dimF_{k,a} ≤1 if k≠2 and dimF_{2,a} =2? $\endgroup$ – DYBnor Mar 1 '18 at 11:18
  • $\begingroup$ @DYBnor: It is case-by-case analysis (with about 8 different cases to consider). I am not going to do all these cases, which cases do you find most difficult? $\endgroup$ – Moishe Kohan Mar 1 '18 at 13:19
  • $\begingroup$ Thanks for the reply. I have no idea about the cases you are talking about. Could you do any two cases? For instance, dimF_{2,a} =2 and dimF_{1,M} $\leq$ 1. Thank you $\endgroup$ – DYBnor Mar 2 '18 at 0:32
  • $\begingroup$ @DYBnor: OK, I will do these two cases in a couple of days when I have a bit more time. $\endgroup$ – Moishe Kohan Mar 2 '18 at 0:37
  • $\begingroup$ @DYBnor: See the edit. $\endgroup$ – Moishe Kohan Mar 3 '18 at 19:42

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