# Does there exist an $A$ and $\mathfrak{su}(2) \subset B$ such that $\mathfrak{su}(3) \simeq A \otimes B$

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?

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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

$\mathfrak{su}(3)$ 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).

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

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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.

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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