So, my question is the following:

Suppose that we have two Lie algebras $(\mathfrak{g}_1,[\bullet,\bullet]_1)$ and $(\mathfrak{g}_2,[\bullet,\bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(\mathfrak{g}_1\otimes\mathfrak{g}_2,[\bullet,\bullet]_{1\otimes 2}).$$ The underlying vector space $\mathfrak{g}_1\otimes\mathfrak{g}_2$ is constructed using the map $\otimes:\mathfrak{g}_1\times\mathfrak{g}_2\to\mathfrak{g}_1\otimes\mathfrak{g}_2$ and consists of the vectors $\{X_1\otimes X_2|X_1\in\mathfrak{g}_1,X_2\in\mathfrak{g}_2\}$. My question is on how to define the Lie bracket $[\bullet,\bullet]_{1\otimes 2}$ correctly, so that the vector space $\mathfrak{g}_1\otimes\mathfrak{g}_2$ becomes a Lie algebra.

  • $\begingroup$ Have you looked at what happens when $\mathfrak{g}_1$ and $\mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it. $\endgroup$ – md2perpe Aug 12 '18 at 17:13
  • $\begingroup$ As far as I know there is no way to do this. $\endgroup$ – Qiaochu Yuan Aug 12 '18 at 17:13
  • $\begingroup$ @QiaochuYuan So, we just suppose that $(\mathfrak{g}_1\otimes\mathfrak{g}_2,[\bullet,\bullet]_{1\otimes 2})$ is a Lie algebra but we cannot define the Lie bracket? $\endgroup$ – G K Aug 12 '18 at 17:20
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    $\begingroup$ Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general? $\endgroup$ – Tobias Kildetoft Aug 12 '18 at 17:28
  • $\begingroup$ @TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct? $\endgroup$ – G K Aug 12 '18 at 17:30

There is a way to define the Lie bracket on the tensor product as follows. Suppose that $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:\mathfrak{g}_1\times \mathfrak{g}_2\longrightarrow \mathfrak{g}_1$ and $B_2:\mathfrak{g}_1\times \mathfrak{g}_2\longrightarrow \mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by $$ [g_1\otimes g_2, g_1'\otimes g_2']:= B_1(g_1,g_2)\otimes B_2(g_1',g_2')\quad \text{for } g_1,g_1' \in \mathfrak{g}_1 \text{ and } g_2,g_2' \in\mathfrak{g}_2.$$ For example if $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.


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