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A set of $d \times d$ real or complex matrices is commonly called irreducible if those matrices do not jointly preserve a linear subspace with dimension strictly between zero and $d$. A stronger hypothesis which is useful in multiplicative ergodic theory - for example, in Furstenberg's theorem on random matrix products - is that the matrices do not jointly preserve a finite union of nontrivial proper subspaces. This condition is referred to as strong irreducibility. Both of these properties also pass to the semigroup generated by the set. I would like to ask:

Is there a precise characterisation of sets (or semigroups) of matrices which are irreducible but not strongly irreducible?

An irreducible set of matrices which generates a finite semigroup will not be strongly irreducible. (Clearly if we can achieve this situation by multiplying each matrix by a nonzero scalar then the matrix set is also not strongly irreducible.) Another example of irreducibility without strong irreducibility is that in which $\mathbb{R}^d$ or $\mathbb{C}^d$ splits into a direct sum of $k$-dimensional subspaces which are permuted by the different matrices. Do these two mechanisms exhaust the range of possible examples?

Thanks in advance!

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  • $\begingroup$ It seems to me irreducible finite semigroups of matrices in dimension >1 are not strongly irreducible because they will preserve the union of all translates of a line. But they need not permute a direct sum decomposition. $\endgroup$ Feb 10, 2016 at 14:31
  • $\begingroup$ I've edited the post slightly to reflect Benjamin's comment. $\endgroup$
    – Ian Morris
    Feb 10, 2016 at 15:02
  • $\begingroup$ @BenjaminSteinberg, could you expand on your comment a little -- I don't quite understand it. Nice question by the way, Ian. $\endgroup$
    – Nick Gill
    Feb 10, 2016 at 15:26
  • $\begingroup$ Nick: the final paragraph of my question previously read "An irreducible set of matrices which generates a finite group..." $\endgroup$
    – Ian Morris
    Feb 10, 2016 at 15:28
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    $\begingroup$ If we take the tensor product of a set with the first property and a set with the second, do we not just get the second property with larger $k$? $\endgroup$
    – Ian Morris
    Feb 11, 2016 at 0:06

2 Answers 2

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The answer to your second question is no.

Let $E_1$, $E_2$, $E_3$ be pairwise transverse $2$-dimensional subspaces of $\mathbb{R}^4$. Consider the following semigroup: $$ \Sigma := \{M \in \mathrm{Mat}(4,4) ; \; M(E_i)=E_i, i=1,2,3\}. $$

[Edit] The following remark will be useful: Any linear transformation $T_1 : E_1 \to E_1$ can be uniquely extended to an element of $\Sigma$. Indeed, with respect to the splitting $E_1\oplus E_2$, the space $E_3$ is the graph of some isomorphism $L \colon E_1 \to E_2$. Then $\Sigma$ is exactly the set of linear transformations $T\colon \mathbb{R}^4 \to \mathbb{R}^4$ whose block form with respect to the splitting $E_1\oplus E_2$ is: $$ T = \begin{pmatrix} T_1 & 0 \\ 0 & L T_1 L^{-1} \end{pmatrix}. $$

Now, for definiteness, we choose the following three subspaces, given in coordinates $(x_1,x_2,x_3,x_4)$ by: \begin{align*} E_1 &:= \{ x_1 = 0 , \ x_2 = x_4 \}, \\ E_2 &:= \{ x_2 = 0 , \ x_3 = x_4 \}, \\ E_3 &:= \{ x_3 = 0 , \ x_1 = x_4 \}, \end{align*}

Consider the following orthogonal matrix which permutes the $E_i$'s: $$ R := \begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ Note that $R$ preserves the orthogonal splitting $F \oplus G$, where \begin{align*} F &:= \{x_1+x_2+x_3=0, \ x_4=0\},\\ G &:= \{x_1=x_2=x_3\}, \end{align*} being a rotation of $120$ degrees on $F$, and the identity on $G$.

Let $\Gamma := \Sigma \cup \{R\}$. Then:

  1. $\Gamma$ is not strongly irreducible, since $E_1 \cup E_2 \cup E_3$ is preserved.
  2. $\Gamma$ is irreducible. Indeed:

(a) $\Sigma$ preserves no $1$- nor $3$-dimensional subspace. This is easy to check using the "useful remark" above.

(b) the only $2$-dimensional subspaces preserved by $R$ are $F$ and $G$, but these are not preserved by $\Sigma$.

  1. The semigroup generated by $\Gamma$ is infinite (even after identifying rescaled matrices).
  2. $\Gamma$ preserves no union of subspaces forming a splitting of $\mathbb{R}^4$.

Therefore $\Gamma$ is a counterexample to your proposed classification.

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In case your set consists of non-singular (=invertible) matrices, the condition of strong irreducibility is equivalent to the condition that the connected component of identity of the Zariski closure of the group generated by your set acts irreducibly on ${\mathbb C}^d$. In both your cases, this is satisfied.

Suppose $G$ is the Zariski closure and $G^0$ is the connected component of identity; it is a fact that $G/G^0$ is finite. Suppose that $G^0$ does not act irreducibly, and therefore, suppose that $W$ is an irreducible invariant subspace. Then the set of $G$ translates of $W$ is finite (since $G/G^0$ is finite) and hence all of $G$ preserves the union $\cup _{g\in G} g(W) $. In particular, all your matrices preserve this finite union.

Conversely, suppose the group $\Gamma $generated by the matrices preserves a finite union of proper subspaces. Such a union is a Zariski closed subspace of the projective space associated to the vector space. Hence the Zariski closure of $\Gamma$ preserves this finite union, hence so does $G^0$.

If a connected group preserves a finite union of proper subspaces, it is (tedious but) easy to show that it preserves a proper subspace.

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  • $\begingroup$ The question is asked for an arbitrary semigroup of matrices (possibly non-invertible) $\endgroup$
    – YCor
    Mar 2, 2016 at 1:50
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    $\begingroup$ Yes, that is why I added (non-singular) in brackets. It seems to me that this is the "generic" case; in any case, MO questions and answers are not meant to be viewed as "examination papers", with exact answers to the exact question, unless the answer addresses something completely different $\endgroup$ Mar 2, 2016 at 1:54
  • $\begingroup$ No problem with partial answers, but it's not transparent in your formulation. I'd just start with "In case your semigroup is made of invertible matrices, ...". Your current statement seems to hold in general, in terms of the invertible matrices in the semigroup (which can be far from enough, e.g., if the semigroup contains no invertible matrix). $\endgroup$
    – YCor
    Mar 2, 2016 at 2:13
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    $\begingroup$ The OP mentioned Furstenberg's theorem where the elements are non -singular matrices; this should be read in this context. $\endgroup$ Mar 2, 2016 at 3:40
  • $\begingroup$ An answer for non-invertible matrices would be ideal but the invertible case is still very interesting. $\endgroup$
    – Ian Morris
    Mar 2, 2016 at 9:57

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