TITLE: Jordan decomposition of the tensor product of two matrices
QUESTION [6 upvotes]: I asked this question on Math.SE here, but did not get a lot of attention.
I am interested in the problem of determining the Jordan decomposition of the tensor product of two unipotent matrices over a field of positive characteristic $p$. It is enough to consider the case of single Jordan block.
So let $u$ and $v$ be unipotent Jordan blocks, of sizes $n \times n$ and $m \times m$ respectively. 
In characteristic $0$, if $n \geq m$ then $u \otimes v$ has Jordan blocks of sizes $n + m - 1, n + m - 3, \cdots, n - m + 1$. This formula no longer works in positive characteristic. But apparently it is valid if $p \geq m+n$. 
What is known about the Jordan decomposition of the tensor product in general? Is this open? I would like some useful description of the decomposition in terms of $n, m$ and $p$.

REPLY [7 votes]: The situation (over a field of characteristic $p >0$) is well understood, and spelled out in : Srinivasan, Bhama The modular representation ring of a cyclic p-group. Proc. London Math. Soc. (3) 14 1964 677–688.
The key points (for $p >2$) are that $J_{2}(1) \otimes J_{m}(1) = J_{m+1}(1) \oplus J_{m-1}(1)$ as long as $m <p.$ Then associativity of the tensor product can then be exploited to explain the pattern you describe for $m +n < p.$
However, $J_{2}(1) \otimes J_{p}(1) = J_{p}(1) \oplus J_{p}(1).$ Then for $p < m <p^{2}$ the 
earlier pattern re-emerges, and so on. It is all spelled out in complete detail in the paper by Srinivasan. Here, $J_{m}(1)$ denotes a Jordan block of size $m$ with $1$ on the diagonal.