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• CommentRowNumber1.
• CommentAuthorBartek
• CommentTimeNov 17th 2017
• (edited Nov 18th 2017)

The definition of a Lie algebra object in a linear category appears to use antisymmetry $[x,y]=-[y,x]$ rather than the alternating property $[x,x]=0$. I can see the reason why: in general we don’t have a “cloning” map $X \to X \otimes X$ available in order to formulate the alternating property. But if the category admits cokernels we should be able to phrase the alternating property in terms of the wedge product, by saying that the bracket descends to a map $\mathfrak{g} \wedge \mathfrak{g} \to \mathfrak{g}$.

Usually I see Lie algebras (for example in Humphreys) defined using the alternating property $[x,x]=0$ which then implies antisymmetry via bilinearity. Doesn’t the definition of a Lie algebra object in a linear category fail to reproduce the usual definition of a Lie algebra in the case where the base commutative ring has characteristic 2? If so, is the antisymmetric definition “morally” the correct one?

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeNov 18th 2017

For instance the concept of super Lie algebra would come out wrong if one did it with the alternating property.

I suppose characteristic 2 is already in some tension with the geometric interpretation of Lie algebras?

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeNov 18th 2017

Not 100% sure, but this probably is relevant

There also exist simple Lie algebras over algebraically closed fields of field characteristic 2, 3, and 5 that are not constructed from a simple reduced root system and are not Cartan algebras. http://mathworld.wolfram.com/SimpleLieAlgebra.html

It could reflect the break of the link to Lie groups

• CommentRowNumber4.
• CommentAuthorBartek
• CommentTimeNov 18th 2017
• (edited Nov 18th 2017)

Just curious Urs, why is the alternating definition not the correct one in the setting of super Lie algebras? In the super setting, if we express the alternating property as the bracket $\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ descending to the quotient $\mathfrak{g} \wedge \mathfrak{g} \to \mathfrak{g}$, and the antisymmetry in terms of the braiding, are these two properties distinct for characteristics other than 2? I don’t really know much about super-mathematics to be honest.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeNov 19th 2017

I just meant that for a super Lie algebra it is in general not the case that $[x,x] = 0$. The simplest example is the supersymmetry algebra in dimension 0+1, which has two generators, say $Q$ and $P$, and the only non-trivial relation is $[Q,Q] = P$.

But, you know, I am all in favour of adding more discussion to the entry, of alternative definitions. If you feel like adding discussion of the alternative definition, please do!