Cannot decide a certain multilinear map is alternating

Question

Suppose $\mathfrak{g}$ is a Lie algebra and $f$ is a multilinear map from $\mathfrak{g}^k$ to $\mathbb{R}$ and let $\operatorname{Alt}(f)$ be the alternization of $f$. Then consider the map $h: \mathfrak{g}\times\cdots\times \mathfrak{g} \to \mathbb{R}$ defined by $$h(x_1, \ldots, x_{k+1}) =\operatorname{Alt}(f) ([x_1, x_2], x_3, \ldots, x_{k+1}) $$ where [, ] denotes the Lie bracket on $\mathfrak{g}$. Is $h$ alternating?

Answer 1

Let's compute the alternization of $h$ : $$ \begin{array}{rcl} \mathrm{Alt}(h)(x_1,\ldots,x_{k+1}) &=& \displaystyle \frac{1}{(k+1)!} \sum_{s\in\mathfrak{S}_{k+1}} \mathrm{sgn}(s) \,h(x_{s(1)},\ldots,x_{s(k+1)}) \\ &=& \displaystyle \frac{1}{(k+1)!} \sum_{s\in\mathfrak{S}_{k+1}} \mathrm{sgn}(s) \,\mathrm{Alt}(f)([x_{s(1)},x_{s(2)}],x_{s(3)},\ldots,x_{s(k+1)}) \\ \end{array} $$ where $\mathfrak{S}_n$ denotes the symmetric group acting on $n$ objects. Note that this group is generated by $n-1$ transpositions of the form $(i,i+1)$. In consequence, a permutation $s\in\mathfrak{S}_{k+1}$ can be decomposed as a transposition (or not) of the first two indices inside the Lie bracket and a permutation acting on $k$ indices (from $2$ to $k+1$), i.e. $s = \tau \circ \sigma$, where $\tau \in \mathfrak{S}_2 < \mathfrak{S}_{k+1}$ and $\sigma \in \mathfrak{S}_k < \mathfrak{S}_{k+1}$, with $\sigma(1) = 1$; there are $k+1$ such cases, because the said transposition is given by $\tau = (1,m)$, where $m=1,2,\ldots,k+1$.

Now, since the Lie bracket and $\mathrm{Alt}(f)$ are alternating maps, one has $$ \begin{array}{rcl} [x_{s(1)},x_{s(2)}] &=& \displaystyle [x_{\tau(\sigma(1))},x_{\tau(\sigma(2))}] \\ &=& \displaystyle \mathrm{sgn}(\tau) [x_{\sigma(1)},x_{\sigma(2)}] \\ &=& \displaystyle \mathrm{sgn}(\tau) [x_1,x_{\sigma(2)}], \end{array} $$ and $$ \begin{array}{rcl} \mathrm{Alt}(f)([x_{s(1)},x_{s(2)}],x_{s(3)},\ldots,x_{s(k+1)}) &=& \displaystyle \mathrm{Alt}(f)(\mathrm{sgn}(\tau)[x_1,x_{\sigma(2)}],x_{\sigma(3)},\ldots,x_{\sigma(k+1)}) \\ &=& \displaystyle \mathrm{sgn}(\tau) \mathrm{Alt}(f)([x_1,x_{\sigma(2)}],x_{\sigma(3)},\ldots,x_{\sigma(k+1)}) \\ &=& \displaystyle \mathrm{sgn}(\tau) \mathrm{sgn}(\sigma) \mathrm{Alt}(f)([x_1,x_2],x_3,\ldots,x_{k+1}) \\ &=& \displaystyle \mathrm{sgn}(s) \mathrm{Alt}(f)([x_1,x_2],x_3,\ldots,x_{k+1}) \\ &=& \displaystyle \mathrm{sgn}(s) h(x_1\ldots,x_{k+1}) \end{array} $$ since $\mathrm{sgn}(\tau)\mathrm{sgn}(\sigma) = \mathrm{sgn}(\tau\circ\sigma)$, because the map $\mathrm{sgn}$ is a homomorphism. Given that $\mathrm{sgn}(s)^2 = 1 \;\forall s\in\mathfrak{S}_{k+1}$ and recollecting all the terms, we get in the end $$ \mathrm{Alt}(h)(x_1,\ldots,x_{k+1}) = \frac{1}{(k+1)!} \cdot (k+1)! \,h(x_1\ldots,x_{k+1}) = h(x_1\ldots,x_{k+1}), $$ hence $h = \mathrm{Alt}(h)$ and thus $h$ is an alternating map.

Answer 2

You can start with an alternating map $f$ already. Then $\operatorname{Alt}(f) = f$. Now, the $\operatorname{Alt}$ is dropped, and $f$ is alternate. Moreover, it does not really matter the target of $f$, as long as it's a vector space. So make it also $\frak{g}$. Also, consider $k=2$. Now, what is a good multilinear alternating map $\frak{g} \times \frak{g} \to \frak{g}$ ? The Lie bracket works. Now, is the the map

$$(x,y,z) \mapsto [[x,y],z]$$

alternating?

Note: we are not asking about $\operatorname{Alt}(h)$ which will automatically make it alternating.