Wedge and tensor algebra are very new concepts to me and I want to understand how to prove the following property of the wedge product:
$$\omega\wedge(\eta_{1}+\eta_{2})=\omega\wedge\eta_{1}+\omega\wedge\eta_{2}$$
where $$\omega,\eta_{1},\eta_{2}\in\Lambda^{1}(V):=V^{\star}\hspace{5mm}(\diamondsuit)$$
By definition of the wedge product we have
$$\omega\wedge\eta_{1}(v_{1},v_{2})=\frac{(1+1)!}{1!1!}Alt(\omega\otimes\eta_{1})(v_{1},v_{2})=2!\frac{1}{2!}\displaystyle\sum_{\sigma\in S_{2}}\text{sgn}\;\sigma\cdot\omega(v_{\sigma(1)})\cdot\eta_{1}(v_{\sigma(2)})=$$
$$=\omega(v_{1})\eta_{1}(v_{2})-\omega(v_{2})\eta_{1}(v_{1})$$
Then the rhs of $(\diamondsuit)$ should be something like this
\begin{align}
\omega\wedge\eta_{1}(v_{1},v_{2})+\omega\wedge\eta_{2}(v_{1},v_{2})= &\omega(v_{1})\eta_{1}(v_{2})-\omega(v_{2})\eta_{1}(v_{1})+\omega(v_{1})\eta_{2}(v_{2})-\omega(v_{2})\eta_{2}(v_{1})\\
=&\omega(v_{1})[\eta_{1}(v_{2})+\eta_{2}(v_{1})]-\omega(v_{2})[\eta_{1}(v_{1})+\eta_{2}(v_{1})]\\
=&\omega\wedge(\eta_{1}+\eta_{2}))(v_{1},v_{2})&
\end{align}
On the other hand
\begin{align}
\omega\wedge(\eta_{1}+\eta_{2}))(v_{1},v_{2})&=\frac{(1+1)!}{1!1!}Alt(\omega\otimes(\eta_{1}+\eta_{2}))(v_{1},v_{2})\\
&=\displaystyle\sum_{\sigma\in S_{2}}\omega(v_{\sigma(1)})\cdot (\eta_{1}+\eta_{2})(v_{\sigma(2)})\\
&=\omega(v_{1})(\eta_{1}+\eta_{2})(v_{2})-\omega(v_{2})(\eta_{1}+\eta_{2})(v_{1})\\
&=\omega(v_{1})\eta_{1}(v_{2})+\omega(v_{1})\eta_{2}(v_{2})-\omega(v_{2})\eta_{1}(v_{1})-\omega(v_{2})\eta_{2}(v_{1})
\end{align}
Is this approach correct? Also should this work for tensors of higher degrees?
Thank you
Best Answer
It's correct except for three typos. First, in the first line of your last equations, it should be $\text{Alt}(\omega\otimes (\eta_1+\eta_2))$. Second, in two places you've messed up on distributing a negative: line 2 of the second computation, last line of the last computation.