2
$\begingroup$

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 $f$.

$\endgroup$
1
$\begingroup$

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 $\wedge_1$.

$\endgroup$
  • $\begingroup$ 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 $\wedge_2$). $\endgroup$ – LCO Jul 14 at 13:35
  • $\begingroup$ @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 $\wedge$. $\endgroup$ – levap Jul 16 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.