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

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 $$f$$.

## 1 Answer

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 $$\wedge_1$$.

• Thank you for this excellent answer. Just to be complete, how do you show $f_1(\eta\wedge \eta')=f_1(\eta)\wedge_1 f_1(\eta')$ ? (I imagine the technique is the same for $\wedge_2$). – LCO Jul 14 at 13:35
• @LCO: Show first by induction that given $\eta^1, \dots, \eta^k \in V^{*}$ and $v_1, \dots, v_k \in V$ you have $(\eta^1 \wedge_1 \cdots \wedge_1 \eta^k)(v_1, \dots, v_k) = \det (\eta^i(v_j))$. The basic ingredient in the proof is expansion of $\det(\eta^i(v_j))$ by the first row/column. This will imply that $f_1 \left( \eta^1 \wedge \cdots \wedge \eta^k \right) = f_1(\eta^1) \wedge_1 \cdots \wedge_1 f_1(\eta^k) = \eta^1 \wedge_1 \cdots \wedge_1 \eta^k$. The general case follows from linearity of $f_1$ and associativity of $\wedge$. – levap Jul 16 at 13:01