2
$\begingroup$

I am interested in tuples of real or complex $d\times d$ matrices which are irreducible in the sense that their constituent matrices do not admit a common invariant linear subspace whose dimension is strictly between $0$ and $d$. In particular, I am interested in how the property of irreducibility is affected by some tensor-algebraic procedures. My question is:

If the $N$-tuples $(A_1,\ldots,A_N)$ and $(B_1,\ldots,B_N)$ are irreducible, when is the $N$-tuple $(A_1\otimes B_1,A_2 \otimes B_2,\ldots, A_N \otimes B_N)$ irreducible?

If we are working in the real context then it is trivially true that the tuple of pairwise tensor products is not necessarily irreducible: for example, if $N=1$, $A_1e_1=e_2$, $A_1e_2=-e_1$ then the $1$-tuple $(A_1)$ is irreducible, but $A_1 \otimes A_1$ preserves the span of $\{e_1\otimes e_2,e_2\otimes e_1\}$. Indeed, the $1$-tuple $(A_1\otimes A_1)$ is obviously reducible because every $4\times 4$ real matrix has a nontrivial invariant subspace.

So far I have not found an example in the complex case where the original tuples are irreducible but the product tuple is not. So: when does irreducibility pass to the pairwise tensor product if we are working over the complex field?

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
3
$\begingroup$

It seems that generically the tensor product you describe is not irreducible. Let $\pi$ be a $d$-dimensional irreducible representation of $G = $ SU($n$) (or any compact connected simple Lie group), with $d > 1$. Then $\pi \otimes \pi$ is a reducible representation (this is presumably well known, but also follows from a difficult theorem of Kumar-xxx (can't remember the other guy)). By $\pi \otimes \pi$, I mean the representation of $G$ given by $g \mapsto \pi(g) \otimes \pi (g)$.

The condition that a set of matrices have no common nontrivial invariant subspace means the double centralizer is the full matrix algebra, and if the set of matrices (or the algebra they generate) is stable with respect to transpose conjugate, then this is the same as the matrices generating the matrix algebra (I am working over the complexes here).

Select a finite set of elements of $G$ such that $\{ \pi (u_i)\}$ generates the matrix algebra, $M_d C$ (there are lots of choices). Then $\{\pi (u_i) \otimes \pi (u_i)\}$ has nontrivial common centralizer (hence the set is not irreducible in your sense), since $\pi \otimes \pi$ is reducible.

More generally, we could let $(G,\pi)$ be any compact group (including finite, thus incorporating the example of Professor Robinson) with irreducible $\pi$ such that $\pi \otimes \pi$ is reducible, which is practically all of them (except the $1$-dimensional ones); or even use a second irreducible $\phi$ in place of the second $\pi$; all that is needed is that $\pi \otimes \phi$ be reducible, and this is almost invariably (I first wrote almost invariantly) the case.

Improve this answer
$\endgroup$
5
  • $\begingroup$ When you say "$\pi \otimes \pi$ is reducible", you mean for the diagonal subgroup $\{(g,g) : g \in G \}$. It is irreducible as a representation of $G \times G$ if $\pi$ is irreducible for $G$. $\endgroup$
    – Geoff Robinson
    Commented Oct 23, 2015 at 23:22
  • $\begingroup$ Yes; $\pi \otimes \pi$ as a representation of $G$. With $A_i = u_i = B_i$, this is consistent with the definition in the question. Oh, I see---I interpreted $\{A_1 \otimes B_1, \dots , A_N \otimes B_N\}$ to mean $\{A_i \otimes B_i\}$; but it could have meant $\{ A_i \otimes B_j\}$, which makes it much more likely to be irreducible. The question should be clarified. $\endgroup$
    – David Handelman
    Commented Oct 23, 2015 at 23:29
  • $\begingroup$ I think you interpreted correctly what the question (as written) said, and there was nothing wrong with your reasoning, but nevertheless one should say which group you mean in the statement about $\pi \otimes \pi$. As I said in an earlier (now deleted, since I wrote my answer) comment, if you allow all $A_{i} \otimes B_{j}$, then that set will be irreducible if the $A_{i}$ are all invertible and the $B_{j}$ are all invertible ( and the $A_{i}$'s and $B_{j}$'s separately act irreducibly). $\endgroup$
    – Geoff Robinson
    Commented Oct 23, 2015 at 23:40
  • $\begingroup$ OK; I will have revised the answer. $\endgroup$
    – David Handelman
    Commented Oct 23, 2015 at 23:45
  • $\begingroup$ I was aware of the ambiguity but somehow didn't manage to avoid it in the original question: I am interested in the `diagonal' case $A_i \otimes B_i$ and not in the unrestricted case $A_i \otimes B_j$. $\endgroup$
    – Ian Morris
    Commented Oct 25, 2015 at 9:50
2
$\begingroup$

In general, even in the complex case, the tensor product of two irreducible $2$-tuples need not be irreducible. Consider the case $A_{1} = B_{1} = \left(\begin{array}{clcr} i & 0\\0&-i \end{array}\right)$ and $A_{2} = B_{2} = \left(\begin{array}{clcr} 0 & 1\\-1&0 \end{array}\right)$. Then $(A_{1},A_{2})$ and $(B_{1},B_{2})$ are irreducible in your sense, but $(A_{1} \otimes B_{1},A_{2} \otimes B_{2})$ is not irreducible. The is is because $\langle A_{1} \otimes B_{1}, A_{2} \otimes B_{2} \rangle$ is a $2$-generator nilpotent group of class at most $2$, so has order at most $8$ ( actually, it is a Klein $4$-group), and has proper non-zero invariant subspaces.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .