0
$\begingroup$

Let $V$ be a d-dimensional $R$-vector space and $\omega:V^k\to R$ a k-linear form.

$Alt(\omega)$ ist defined by:

$Alt(\omega)(v_1,..,v_k)= 1/k!* \sum_\sigma sign(\sigma)\omega(v_{\sigma(1)},..,v_{\sigma(k)})$

Show that:

i) $Alt(\omega)$ is alternating

ii) if $\omega$ is symmetric, then $Alt(\omega)=0$

iii) For $\nu\in \wedge^kV^*$ is $Alt(\nu) =\nu$ and so $Alt(Alt\omega)= Alt(\omega)$ for every k-Linear form.

I am new to this topic, so I am grateful for any hint. What I know so far is:

i) To show: $Alt(\omega)\in\wedge^kV^* $

A k-linear form is alternating if

$i,j \in$ {1,...,k},$i \not= j$ with $v_i=v_j$ follows $\omega(v_1,..,v_k)=0$

$\endgroup$
0
$\begingroup$

Hint: Suppose that $v_i = v_j$. Let $\tau$ denote the transposition which switches the $i$th and $j$th elements. $v_i = v_j$ means that $$ \omega(v_{1},..,v_{k}) = \omega(v_{\tau(1)},..,v_{\tau(k)}) $$ and for any $\sigma$, we have $$ \omega(v_{\sigma(1)},..,v_{\sigma(k)}) = \omega(v_{\sigma (\tau (1))},..,v_{\sigma(\tau(k))}) $$ However, note that $sgn(\sigma \circ \tau) = -sgn(\sigma)$.

$\endgroup$

Your Answer

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