MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm in the middle of trying to prove something at the moment and am looking for a decomposition of the Lie algebra $\mathfrak{su}(3)$ into a tensor product of some algebra $A$, and another $B$ containing $\mathfrak{su}(2)$, or some such result. Does anyone know of anything?

share|cite|improve this question
Does looking at the dimensions give you any information? – Mariano Suárez-Alvarez Jan 21 '10 at 1:23
Best. Title. Ever. You realize that it's unreadable, right? – Harry Gindi Jan 21 '10 at 1:42
This is one of those questions where I wish the poster had said how they came across the question, since I simply can't imagine how it happened. What question about $\mathfrak{su}(3)$ could possibly be hard enough to justify all of this? – Ben Webster Jan 21 '10 at 2:33
I was looking for an embedding of the coordinate ring of $SU(2)$ into the coordinate ring of $SU(3)$, and thought I find one using a dual map – Dyke Acland Jan 21 '10 at 3:30
FIIW, my comment above refers to an old, rather different version of the question. @Dyke: it tends to be not a good idea to change questions like that... – Mariano Suárez-Alvarez Jan 21 '10 at 4:51
up vote 7 down vote accepted

$\mathfrak{su}(3)$ cannot be decomposed as a tensor product, since there are no nonabelian simple Lie algebras of dimension 4, 2 or 1 (thus, any 8-d Lie algebra which is a tensor product has a proper ideal, which $\mathfrak{su}(3)$ doesn't).

share|cite|improve this answer

$\mathfrak{su}(2)$ is contained in $\mathfrak{su}(3)$, so you can take $A=k$ and $B=\mathfrak{su}(3)$ to answer the question in your title.

share|cite|improve this answer

There is no such nontrivial algebra, because $\dim \mathfrak{su}(2) = 3$ and $\dim \mathfrak{su}(3) = 8$.

As Mariano pointed out (and I missed), considering $\mathfrak{su}(2) \subset \mathfrak{su}(3)$ works trivially.

share|cite|improve this answer
Could there be an embedding of $\mathfrak{su}(2)$ into some four dimensional algebra $B$ such that $\mathfrak{su}(3) \simeq A \otimes B$? – Dyke Acland Jan 21 '10 at 1:32
See Mariano's answer. – Steve Huntsman Jan 21 '10 at 1:57
Now I realize that I didn't misread the original version of the question... – Steve Huntsman Jan 21 '10 at 2:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.