Given a basis $a_i$ for the lie algebra $\mathfrak{s}\mathfrak{u}(d)$, does the set of elements $a_i \otimes a_j$ form a basis for $\mathfrak{s}\mathfrak{u}(d^2)$?
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What are the dimensions of these two algebras ? – Sylvain L. Jun 24 '15 at 16:03
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Ah, I see now that I should say that I wish to include the identity matrix in my basis set so that the set of elements in $a_i$ is $d^2$ not $d^2-1$. How should I modify the phrasing of my question? – Joel Klassen Jun 24 '15 at 16:09
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Also, I noticed another mistake – Joel Klassen Jun 24 '15 at 16:14
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(small note: you can save some typesetting pain by writing \mathfrak{su} vs. \mathfrak{s}\mathfrak{u}) – Simon S Jun 24 '15 at 16:21
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$\mathfrak{u}(d)$ get the dimension right. But then, the notion of a tensor product of Lie algebras, not sure it exists. It works though for usual algebras. So, $M_m\otimes M_n \simeq M_{mn}$ for algebras of matrices over a field. – orangeskid Jun 24 '15 at 16:41