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I am a newcomer for the course of Lie algebra. A subalgebra of a Lie algebra is subspace with the same Lie product.

Can we have a subspace of a Lie algebra which is not a subalgebra ?

Priya Pandey
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    A plane in $\mathbb{R}^3$ (the Lie bracket is the cross product) should do the job. – Kandinskij Sep 11 '22 at 05:51
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    "A subalgebra of a Lie algebra is a subalgebra having the induced product" is not the correct definition. It's not a definition at all, actually. – Torsten Schoeneberg Sep 11 '22 at 06:53
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    Since no-one has said it explicitly, the correct definition of a Lie subalgebra is a subspace that is closed under the Lie bracket. That is $[x,y] \in V$ for each $x,y \in V$. – Callum Sep 11 '22 at 10:57

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Take the Lie algebra $\mathfrak{so}_3(\Bbb R)$ with basis $\{x,y,z\}$ and Lie brackets $$ |x,y]=z,\; [z,x]=y,\; [y,z]=x. $$ Now consider the subspace $V$ spanned by $x$ and $y$ of dimension $2$. It is not a Lie subalgebra, since $[x,y]\not\in V$.

In fact, $\mathfrak{so}_3(\Bbb R)$ has no $2$-dimensional subalgebra at all. This is often an argument to show that it cannot be isomorphic to $\mathfrak{sl}_2(\Bbb R)$, which certainly has $2$-dimensional subalgebras.

Dietrich Burde
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