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 $p$.

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    $\begingroup$ As Geoff points out, the 1964 paper by Srinivasan is probably the optimal source here. You might also look at a much later paper which approaches the tensor product situation from a linear algebra viewpoint: ams.org/mathscinet-getitem?mr=2434111 $\endgroup$ Feb 9, 2015 at 13:42

1 Answer 1


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.


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