Let $(e_1,e_2,e_3)$ be a basis of $V$ and $(\varepsilon^1,\varepsilon^2, \varepsilon^3)$ be the dual basis. Let $(f_1,f_2)$ be a basis of $W$ and $(\phi^1,\phi^2)$ the dual basis. Let $L: V \rightarrow W$ be the linear map given by
$$
L(e_1) = f_2,\ L(e_2) = f_1,\ L(e_3) = f_1 + f_2.
$$
Define $L^*: W^* \rightarrow V^*$ in terms of the dual bases.
The pullback of an alternating $2$-tensor $\omega \in \Lambda^2W^*$ by a map $L: V \rightarrow W$ is defined to be the alternating $2$-tensor $L^*\omega \in \Lambda$, where for any $v_1, v_2 \in V$,
$$
\langle L^*\omega, v_1\otimes v_2\rangle = \langle \omega, (L(v_1))\otimes(L(v_2))\rangle.
$$
Using the definition of $L$ given by above, find a formula for
$L^*(\phi^1\wedge\phi^2)$
in terms of $\varepsilon^1, \varepsilon^2, \varepsilon^3$.
I try to solve this question:
I know that
$\phi^1\wedge\phi^2=\phi^1\otimes\phi^2-\phi^2\otimes\phi^1$
And I set $L^*(\phi^1\wedge\phi^2)$=($x_1\varepsilon^1+x_2\varepsilon^2+x_3\varepsilon^3$, $y_1\varepsilon^1+y_2\varepsilon^2+y_3\varepsilon^3$).
Then I rewrote this formula
$\langle L^*\omega, v_1\otimes v_2\rangle = \langle \omega, (L(v_1))\otimes(L(v_2))\rangle$
as:
$$(x_1\varepsilon^1+x_2\varepsilon^2+x_3\varepsilon^3)v_1
+(y_1\varepsilon^1+y_2\varepsilon^2+y_3\varepsilon^3)
v_2=\phi^1(L(v_1))\phi^2(L(v_2))-\phi^1(L(v_2))\phi^2(L(v_1))$$
Then set the basis of $V\otimes V$: $(e_1,e_1)$,$(e_2,e_2)$,$(e_3,e_3)$,$(e_1,e_2)$,$(e_1,e_3)$,
$(e_2,e_3)$ as input $(v_1,v_2)$ of former formula to solve the linear equation to get $X$ and $Y$.
But I failed, so I would like to ask where I did wrong.
Best Answer
Take $$\omega := \phi^1\wedge \phi^2 = \phi^1\otimes \phi^2 - \phi^2 \otimes \phi^1,$$ which is an element of $\Lambda^2(T^*W)$. Its components are found by feeding it $f_i$'s. For the pullback, which is an element of $\Lambda^2(T^*V)$, the components are found (or rather defined) by feeding it $e_i$'s as prescribed by the formula: $$\big(L^*\omega\big)_{ij} = L^*\omega(e_i,e_j) := \omega(L(e_i),L(e_j))$$ This differs in appearance from your formula, but the idea is the same (push forward args and apply omega).
Explicitly, we have: $$L(e_1) = f_2,\text{ }\text{ }\text{ }L(e_2) = f_1,\text{ }\text{ }\text{ }L(e_3)=f_1+f_2,$$ Hence for example: $$\color{blue}{(L^*\omega)_{11}} = (\phi^1\otimes\phi^2 - \phi^2\otimes\phi^1)(L(e_1),L(e_1))$$ $$= \phi^1(f_2)\cdot\phi^2(f_2) - \phi^2(f_2)\cdot\phi^1(f_2) = \delta^1_2\cdot \delta^2_2 - \delta^2_2\cdot\delta^1_2 = 0-0 = \color{blue}{0}$$ ($\delta^i_j$ is the Kronecker delta). The other eight components are found similarly.
I would like to add that this is a skew-symmetric tensor, so you need only find $(L^*\omega)_{12}$, $(L^*\omega)_{13}$, and $(L^*\omega)_{23}$ since diagonal elements are all zero and the rest are negatives of the above.