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This question comes from Georgi, Lie Alegbras in Particle Physics. Consider the algebra generated by $\sigma_a\otimes1$ and $\sigma_a\otimes \eta_1$ where $\sigma_a$ and $\eta_1$ are Pauli matrices (so $a=1,2,3$). He claims this is "semisimple, but not simple". To me, that means we should look for an invariant subalgebra (a two-sided ideal). The multiplication table is pretty easy to figure out:

$[\sigma_a,\sigma_b]=i\epsilon_{abc}\sigma_c,$

$[\sigma_a,\sigma_b\otimes\eta_1]=i\epsilon_{abc}\sigma_c\otimes\eta_1$

$[\sigma_a\otimes\eta_1,\sigma_b\otimes\eta_1]=i\epsilon_{abc}\sigma_c\otimes1$

I'm dropping off the identity in all the places where it looks like it should be. So the only subalgebra is the $\mathfrak{su}(2)$ generated by $\sigma_a\otimes 1$, and that is not invariant from the second line above. So this looks like a simple algebra to me. Is there a typo somewhere I do not see?

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Alternatively, you could try to see if it is a direct sum of two simple lie algebras. –  Heidar Nov 23 '11 at 23:12
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By the way, Lie algebra ideals are always two-sided because of the antisymmetry of the bracket. Thus one usually says just "ideals". –  José Figueroa-O'Farrill Nov 24 '11 at 0:07
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2 Answers

up vote 11 down vote accepted

In this relatively simple example, one can observe that the subalgebras $\{\sigma_a \otimes \frac{1\mp\eta_1}{2}\}$ are the two commuting copies of $su(2)$.

For more complicated situations, one actually has an algorithm to veify the simplicity of a Lie algebra. This is because (the root systems of) simple Lie algebras are classified by Cartan, thus one just needs to verify if the root system of the given Lie algebra coincides with one of the types. Actually, an algebra which is semisimple but not simple will necessarily have orthogonal simple roots.

Taking the given example for illustration. One can choose $\sigma_3\otimes 1$ and $\sigma_3\otimes \eta_1$ as the generators of the Cartan's subalgebra. The corresponding root generators can be found as: $\sigma_{\pm} \otimes \frac{1\mp\eta_1}{2}$. The positive roots can be chosen as: $\alpha = [1,1 ]$, and $\beta = [1,-1 ]$. Now we notice that on one hand they are linearly independent thus they are simple roots. On the other hand we have $\alpha.\beta=0$ in contradiction to the property of a simple root system requiring that for any distinct simple roots $\alpha.\beta<0$.

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Unless I am mistaken, your algebra is $\mathfrak{so}(4)=\mathfrak{su}(2)\oplus\mathfrak{su}(2)$. The generators are $\sigma_a\otimes 1 \pm \sigma_a\otimes\eta_1$.

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I agree. The first commutator verifies that $\sigma_a$ are Pauli matrices. $\eta_1$ isn't a Pauli matrix, it is ${\rm diag}(+1,-1)$, as implicitly seen from the last two equations. So one deals with $4\times 4$ block diagonal matrices with a combination of Pauli matrices in each $2\times 2$ block, and that's clearly $SU(2) \times SU(2)$, with your right generalization. Just to answer the OP's question, it is a product so it is not simple. –  Luboš Motl Nov 24 '11 at 7:31
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@Luboš Motl: any Pauli matrix meets two last equations and after all ${\rm diag}(+1,-1)$ is also Pauli matrix $\sigma_z$. –  Alex 'qubeat' Nov 24 '11 at 10:50
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It is rather $so(4,{\mathbb C}) \approx sl(2,{\mathbb C}) \oplus sl(2,{\mathbb C})$, because OP defines it as a complex algebra. –  Alex 'qubeat' Nov 24 '11 at 12:28
    
@AlexV, I believe the usual convention in physics is to have structure constants of the generators of a real Lie algebra be purely imaginary, so that the real span of $i$ times generators is the Lie algebra itself. E.g. the generators of $\mathfrak{su}(2)$ are taken to be $\sigma_a$ and not $i\sigma_a$. In the OP's example, $\mathfrak{so}(4)=span_\mathbf{R}(i\sigma_a\otimes 1,i\sigma_a\otimes\eta_1)$. –  Pavel Safronov Nov 28 '11 at 15:42
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In such a case OP could write $[i\sigma_a,i\sigma_b]=\cdots$. It is not a convention, it is rather a convenient trick to work with Lie algebra of unitary group, because the algebra is anti-Hermitian matrices. If to use the convention without warning we could not distinguish $sl(2,C)$ and $su(2)$ –  Alex 'qubeat' Nov 29 '11 at 10:49
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