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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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    $\begingroup$ Does looking at the dimensions give you any information? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 21, 2010 at 1:23
  • $\begingroup$ Best. Title. Ever. You realize that it's unreadable, right? $\endgroup$
    – Harry Gindi
    Commented Jan 21, 2010 at 1:42
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    $\begingroup$ 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? $\endgroup$
    – Ben Webster ♦
    Commented Jan 21, 2010 at 2:33
  • $\begingroup$ 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 $\endgroup$
    – Dyke Acland
    Commented Jan 21, 2010 at 3:30
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    $\begingroup$ 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... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 21, 2010 at 4:51

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$\mathfrak{su}(3)$ 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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  • $\begingroup$ Could there be an embedding of $\mathfrak{su}(2)$ into some four dimensional algebra $B$ such that $\mathfrak{su}(3) \simeq A \otimes B$? $\endgroup$
    – Dyke Acland
    Commented Jan 21, 2010 at 1:32
  • $\begingroup$ See Mariano's answer. $\endgroup$
    – Steve Huntsman
    Commented Jan 21, 2010 at 1:57
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    $\begingroup$ Now I realize that I didn't misread the original version of the question... $\endgroup$
    – Steve Huntsman
    Commented Jan 21, 2010 at 2:03

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