The different products on $k$-alternating forms and their relationship with the exterior product

Question

Let $V$ be a vector space and $k\in \mathbb{N}$. Denote $\Lambda^k V$ the exterior $k
$-power of $V$.

Let $f:\Lambda^k V^*\to (\Lambda^k V)^*$ be the map such that a $k$-covector $\eta_1\wedge \cdots \wedge \eta_k$ is sent to the $k$-alternating form (identified as a dual element of $\Lambda^k V$) $(v_1,\cdots,v_k)\mapsto \sum_{\sigma} \epsilon(\sigma) \eta_1(v_{\sigma(1)}) \cdots \eta_k(v_{\sigma(k)})$. (It seems that there is another convention, with a factor $\frac{1}{k!}$ in that formula.)

There are (see here) two ways to define natural products between $k$-alternating forms. Let us denote $\wedge_1$ the product based on the formula with the $Alt$ operator (in the previous link), and $\wedge_2$ the product based on the formula with a sum over shuffle permutations (in the previous link).

Question : What is true and what is false in the following statements ? Let $\omega \in \Lambda^k V^*$ and $\omega'\in \Lambda^{k'} V^*$.

1) $f(\omega\wedge \omega') = f(\omega)\wedge_1 f(\omega')$

1') Same as 1) but with the $1/k!$ factor in the definition of $f$.

2) $f(\omega\wedge \omega')=f(\omega) \wedge_2 f(\omega')$

2') Same as 2) but with the $1/k!$ factor in the definition of $f$.

Answer 1

Let's be more precise about all the conventions involved. There are two common conventions for the duality between $\Lambda(V^{*})$ and $\Lambda(V)^{*}$:

1. The first which we denote by $f_1 \colon \Lambda(V^{*}) \rightarrow \Lambda(V)^{*}$ is defined on $k$-decomposable elements by $$ (f_1(\eta^1 \wedge \cdots \wedge \eta^k))(v_1 \wedge \dots \wedge v_k) = \det(\eta^i(v_j)) = \sum_{\sigma \in S_k} \varepsilon(\sigma) \eta^1(v_{\sigma(1)}) \cdots \eta^k(v_{\sigma(k)}) $$ and extended linearly.
2. The second which we denote by $f_2 \colon \Lambda(V^{*}) \rightarrow \Lambda(V)^{*}$ is defined on $k$-decomposable elements by $$ (f_1(\eta^1 \wedge \cdots \wedge \eta^k))(v_1 \wedge \dots \wedge v_k) = \frac{1}{k!} \det(\eta^i(v_j)) $$ and extended linearly.

There are also two common conventions for the product of two alternating multilinear forms $\omega \in \Lambda^k(V)^{*}, \mu \in \Lambda^l(V)^{*}$:

1. The first which we denote by $\wedge_1$ is defined by $$ (\omega \wedge_1 \mu)(v_1, \dots, v_{k+l}) = \frac{(k+l)!}{k!l!} \operatorname{Alt}(\omega \otimes \mu)(v_1, \dots, v_{k+l}) = \\ \frac{1}{k! l!} \sum_{\sigma \in \operatorname{S}_{k+l}} \varepsilon(\sigma) \omega(v_{\sigma(1)} \cdots \omega(v_{\sigma(k)}) \mu(v_{\sigma(k+1)}) \cdots \mu(v_{\sigma(k+l)}) = \\ \sum_{\sigma \in \operatorname{Sh}_{k,l}} \varepsilon(\sigma) \omega(v_{\sigma(1)} \cdots \omega(v_{\sigma(k)}) \mu(v_{\sigma(k+1)}) \cdots \mu(v_{\sigma(k+l)}). $$
2. The second which we denote by $\wedge_2$ is defined by $$ (\omega \wedge_2 \mu)(v_1, \dots, v_{k+l}) = \operatorname{Alt}(\omega \otimes \mu)(v_1, \dots, v_{k+l}) = \\ \frac{1}{(k+l)!} \sum_{\sigma \in \operatorname{S}_{k+l}} \varepsilon(\sigma) \omega(v_{\sigma(1)} \cdots \omega(v_{\sigma(k)}) \mu(v_{\sigma(k+1)}) \cdots \mu(v_{\sigma(k+l)}).$$

Clearly we have $\omega \wedge_1 \mu = \frac{(k+l)!}{k! l!} \omega \wedge_2 \mu$. Now, given $\eta \in \Lambda^k(V^{*}), \eta' \in \Lambda^l(V^{*})$, we have:

1. $f_1(\eta \wedge \eta') = f_1(\eta) \wedge_1 f_1(\eta')$.
2. $f_2(\eta \wedge \eta') = f_2(\eta) \wedge_2 f_2(\eta')$.

From here you see that conventions $f_1$ and $\wedge_1$ should be used together to get an algebra isomorphism between $\Lambda(V^{*})$ and $\Lambda(V)^{*}$. Those conventions work over any any field or ring (the defining formulas don't involve any division) and have various other advantages. Alternatively, you can use $f_2$ and $\wedge_2$ together which have the advantage of making the projection $\operatorname{Alt} \colon \operatorname{Mult}^{*}(V) \rightarrow \operatorname{Alt}^{*}(V)$ into an algebra homomorphism. I've never seen someone using $f_1$ and $\wedge_2$ or $f_2$ and $\wedge_1$.