# Defining the Lie bracket on a tensor product Lie algebra

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.

• 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. – md2perpe Aug 12 '18 at 17:13
• As far as I know there is no way to do this. – Qiaochu Yuan Aug 12 '18 at 17:13
• @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? – G K Aug 12 '18 at 17:20
• 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? – Tobias Kildetoft Aug 12 '18 at 17:28
• @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? – 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.