So far I haven't found a source that gives me the unambiguous definition of a Lie algebra being real. So: do we call a Lie algebra real if it has real structure constants OR, when viewed as a vector space, the field over this vector space is $\mathbb{R}$? For example, the algebra consisting of $\{L_1,L_2,L_3\}$ with $[L_i,L_j]=i\epsilon_{ijk}L_k$: is $\text{span}_{\mathbb{R}}\{L_1,L_2,L_3\}$ a real Lie algebra?
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5$\begingroup$ A Lie algebra is real when it is a real vector space, i.e., the scalars are in $\mathbb{R}$. The structure constants are real in particular. $-iL_1,-iL_2,-iL_3$ form a basis of the real Lie algebra $su(2)= so(3)$. $\endgroup$– Valter MorettiCommented Jan 18, 2022 at 15:27
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$\begingroup$ I get that, but i'm confused whether $\text{span}_\mathbb{R}\{L_1,L_2,L_3\}$ is a real Lie algebra? Given your answer, I would say yes because it is a real vector space but i'm in doubt because in the basis $\{L_1,L_2,L_3\}$ the structure constants are purely imaginary. $\endgroup$– 88888888Commented Jan 18, 2022 at 16:57
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4$\begingroup$ The span you consider is a real vector space but it is not a Lie algebra! Since it is not closed under commutation of elements: $[L_i,L_j]$ does not belong to the span. If instead you consider the real span of $-iL_x, -iL_y, -iL_z$ you have both a Lie algebra and a real vector space: a real Lie algebra. $\endgroup$– Valter MorettiCommented Jan 18, 2022 at 17:24
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An algebra is real when its ground field is the reals. In particular, a Lie algebra is an algebra and hence it is real when it's ground field is the reals. In this case, the structure constants are real and the reverse is true.
(This should be distinguished from a real structure: A real structure of a complex vector space is a real vector space whose complexification is isomorphic to that complex vector space. This has the obvious specialisation to Lie algebras).
answered Jan 18, 2022 at 15:51
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$\begingroup$ But in the example I give, the ground field is the reals however the structure constants are purely imaginary so how is that? $\endgroup$– 88888888Commented Jan 18, 2022 at 16:58
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$\begingroup$ @88888888: Its due to a diffetence of convention between mathematicians and physicists, essentially because physicists prefer lie algebra observables to be hermetian rather than anti-hermetian as it turns out in the mathematicians convention. $\endgroup$ Commented Jan 18, 2022 at 17:57