nForum - Search Results Feed (Tag: lie-algebra) 2021-09-16T17:54:20-04:00 https://nforum.ncatlab.org/ Lussumo Vanilla & Feed Publisher The definition of a Lie algebra object https://nforum.ncatlab.org/discussion/8160/ 2017-11-17T20:06:09-05:00 2017-11-19T11:11:39-05:00 Bartek https://nforum.ncatlab.org/account/1354/ The definition of a Lie algebra object in a linear category appears to use antisymmetry [x,y]=&minus;[y,x][x,y]=-[y,x] rather than the alternating property [x,x]=0[x,x]=0. I can see the reason ...

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?

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Lie algebra cohomology via smooth sets https://nforum.ncatlab.org/discussion/7499/ 2016-11-13T22:56:55-05:00 2016-11-16T00:21:22-05:00 amg https://nforum.ncatlab.org/account/827/ The cohomology of a Lie algebra gg is by definition that of its Chevalley–Eilenberg complex, CE *(g)CE^*(g). It’s fairly straightforward to see that this cdga can also be obtained as the ...

The cohomology of a Lie algebra $g$ is by definition that of its Chevalley–Eilenberg complex, $CE^*(g)$. It’s fairly straightforward to see that this cdga can also be obtained as the $G$-invariant differential forms on $G$, for any Lie group $G$ integrating $g$:

$CE^*(g) \cong \Omega^*(G)^G$

(where the $G$-action on forms is given by pullback along the multiplication map).

It occurred to me that this should be expressible in the language of smooth sets. First of all we have $\Omega^k(G) = hom_{SmthSet}(G,\Omega^k)$, and the $G$-action is now given by precomposition. This means that instead of taking $G$-invariants of the hom-set we could just map out of $G/G = pt$. But this clearly doesn’t recover $CE^k(g)$.

For a moment I thought that maybe the issue is that I’m forgetting the cohesiveness of the group $G$ here: so really, I should be looking at an internal hom $[G,\Omega^k]$, and this carries a $G$-action which is witnessed by having it as the fiber of some object (its (homotopy) quotient) over $\mathbf{B}G$. After all, in the previous paragraph I was asking a smooth group to act on a discrete set, which should be a no-no.

But this would come from looking at internal hom in ${\mathbf H}_{/{\mathbf B}G}$ of $G/G$ into $\Omega^k \times {\mathbf B}G$ (which classifies the trivial action of $G$ on $\Omega^k$). And so once again we see $G/G$ appearing, which even speaking cohesively is just the terminal object. So this doesn’t fix the problem either.

Does anyone see what I’m doing wrong here?

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