3
$\begingroup$

For two associative unital algebras $A$ and $B$, defined over $\mathbb{K} = \mathbb{R}, \mathbb{C}$, is it possible to have an irreducible representation of $A \otimes_{\mathbb{K}}B$ which is not of the form $V \otimes W$, where $V$ is a representation of $A$ and $W$ is a representation of $B$?

$\endgroup$
5
  • 1
    $\begingroup$ Yes, very easily. For example, $(V \otimes W) \oplus (V \otimes W)$ does not in general factorize as $V' \otimes W'$. Did you want some irreducibility condition? $\endgroup$ Jan 15 '20 at 14:39
  • $\begingroup$ Yes, I want irreducibility. I have now written this. $\endgroup$ Jan 15 '20 at 14:42
  • 3
    $\begingroup$ K=R, A=B=C is a standard example where C/R is any nontrivial field extension. $\endgroup$ Jan 15 '20 at 14:51
  • 1
    $\begingroup$ @PeterMcNamara Why writing answers in the comment section? (I ask this on MO since 10 years ...) $\endgroup$ Jan 16 '20 at 12:56
  • 2
    $\begingroup$ @MartinBrandenburg, when I do something like that, it's because I suspect that the question might change. For example, probably the question should have been not just about irreducible representations (as @‌MarkWildon suggested) but about absolutely irreducible representations (as indicated in @‌Mare's answer). $\endgroup$
    – LSpice
    Jan 16 '20 at 18:13
4
$\begingroup$

In case your two algebras $A,B$ are finite dimensional and the field is algebraically closed (or more generally the two algebras are split over the field), then all simple modules over $A \otimes_K B$ are indeed of the form $V \otimes_K W$ for a simple $A$-module $V$ and a simple $B$-module $W$.

This is not true when the algebras are not split: Let $K= \mathbb{R}$ and $A=B=\mathbb{C}= \mathbb{R}[x]/(x^2+1)$.

Then $A \otimes_K B= \mathbb{C}[x]/(x^2+1)=\mathbb{C}[x]/(x+i) \times \mathbb{C}[x]/(x-i)=\mathbb{C} \times \mathbb{C}$.

Thus $A \otimes_K B$ has a simple modules of $K$-dimension two, while all non-zero $A \otimes_K B$-modules of the form $V \otimes_K W$ have $K$-dimension at least 4.

$\endgroup$
4
  • $\begingroup$ What does it mean for an algebra to be split? Isomorphic to a direct power $K^{\oplus n}$? $\endgroup$
    – LSpice
    Jan 16 '20 at 18:14
  • 1
    $\begingroup$ Also, note that @PeterMcNamara gave a variant of your counterexample in the comments. $\endgroup$
    – LSpice
    Jan 16 '20 at 18:15
  • 1
    $\begingroup$ @LSpice It means that the algebra modulo its jacobson radical is a products of matrix rings over the field $K$. What you say would be split and basic. Being split is also often called elementary. $\endgroup$
    – Mare
    Jan 16 '20 at 18:28
  • $\begingroup$ What happens in the infinite dimensional case? $\endgroup$ Jan 19 '20 at 15:28

This site is temporarily in read only mode and not accepting new answers.

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