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I'm trying to do exercise 3.12 of the first chapter in Bourbaki's Lie algebras/groups book, but apparently there is a sign error... somewhere. Let me introduce the objects in question:

First, lets fix a field $k$,$\mathfrak g$ a lie algebra over $k$ and $M$ a $\mathfrak{g}$-module. We define $C^p(\mathfrak{g}, M)$ as the $k$-vector space of alternating $p$-linear mappings of $\mathfrak{g}^p$ into $M$ (that is, $C^p(\mathfrak{g}, M) = \hom_k(\Lambda^p\mathfrak{g}, M)$ ) . This has a canonical $\mathfrak{g}$-module structure induced by the adjoit representation, that is, if I haven't messed this up, if $x, x_i \in g$ and $u\in C^p(\mathfrak{g}, M)$ we have $$ (x.u)(x_1, \dots, x_p) = x \cdot u(x_1, \dots, x_p) - \sum_{i \leq p}u(x_1, \dots, [x,x_i], \dots, x_n). $$

Lets call this action $\theta$, so that $\theta(x)u$ is determined by the expression above. There is also a (degree $-1$) map induced by each $y \in \mathfrak{g}$, namely $i(y): C^{p}(\mathfrak{g}, M) \to C^{p-1}(\mathfrak{g}, M)$ $$ i(y) \cdot u(x_2, \dots, x_p) = u(y, x_2, \dots, x_p). $$

Okay, now Bourbaki claims that $\theta(x)i(y) - i(y)\theta(x) = i[x,y]$, but when I compute I get \begin{align} \theta(x)i(y)u(x_2, \dots, x_p) =& x \cdot u(y, x_2, \dots, x_p) & - u([x,y], x_2, \dots, x_p) &- \sum_{i=2}^p u(y, x_2, \dots, [x,x_i], \dots) \\ i(y)\theta(x)u(x_2, \dots, x_p) =& x \cdot u(y, x_2, \dots, x_p) &&- \sum_{i=2}^p u(y, x_2, \dots, [x,x_i], \dots) \end{align} so that $\theta(x)i(y) - i(y)\theta(x) = -i[x,y]$. What did I do wrong?

Thiago
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  • Hmm ... how sure are you about the $-$ in the definition of $x. u$? If that were a $+$, everything would work out, right? – Torsten Schoeneberg May 05 '20 at 21:25
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    Yeah, that would be great... However I do not think that is correct. The reason for the minus is that the $k$-vector space of alternating function $\mathfrak{g}^p$ to $M$ is naturally seen as a $\mathfrak{g}^o \times \dots \times \mathfrak{g}^o\times \mathfrak{g}$ modules (that is, a right action when acting on the domain). To compensate this, we define the $\mathfrak{g}$ module structure on this space by flipping the signs. I can't be too precise at a comment, but proposition 3 on item 3 of chapter 1 on Bourbaki's book states it quite clearly for multilinear functions. – Thiago May 05 '20 at 23:51
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    I see (literally: I have the volume here). Hmm, your calculation seems right. Have you tried to continue with the rest of the exercise to see if this ever makes a difference? Maybe it really is just a sign error. – Torsten Schoeneberg May 06 '20 at 00:04
  • yeah, maybe it is (however, I still had to be sure before calling something in Bourbaki an error (i've made that mistake before haha) ) – Thiago May 06 '20 at 00:19
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    Agreed. Been there, done that. I still do think I found kind of an error here: https://math.stackexchange.com/q/764696/96384, but yes, they are very rare, and it is still quite possible both you and I overlook something and Bourbaki is right after all. – Torsten Schoeneberg May 06 '20 at 00:26

1 Answers1

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Your last two displayed equations are wrong. Let's do $p=2$: let $u \in C^2$ and $y, x_2 \in \mathfrak{g}$. Then $i(y)(u) \in C^1$ is the map $$ i(y)(u): x_2 \mapsto u(y, x_2) $$ So $$\begin{align*} (x \cdot (i(y)(u)))(x_2) &= x(i(y)(u)(x_2)) - (i(y)(u))([x,x_2])\\ &= x\cdot u(y,x_2) - u(y, [x,x_2]) \end{align*} $$ This is $\theta(x) i(y) u$, note that there's no term with an $[x,y]$ commutator whereas you have such a term in your second-to-last displayed equation.

On the other hand, $(i(y)(x\cdot u))(x_2) = (x\cdot u)(y,x_2)$ which is $$ x\cdot u(y,x_2) - u([x,y],x_2) - u(y,[x,x_2]) $$ This is $i(y)(\theta(x)(u))$ at $x_2$. Subtracting this from what we computed before gives $u([x,y],x_2) = (i([x,y])(u))(x_2)$.

Matthew Towers
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