Let $V$ be a vector space and $V^*$ be the dual of $V.$ For $\omega \in \wedge^k V^* \subseteq V^{* \otimes k} \cong \left (V^{\otimes k} \right )^*$ and $(v_1 \otimes v_2 \otimes \cdots \otimes v_n ) \in V^{\otimes k}$ we denote $\omega (v_1 \otimes v_2 \otimes \cdots \otimes v_k)$ by $\left \langle \omega, v_1 \otimes v_2 \otimes \cdots \otimes v_k \right \rangle.$
Proposition $:$ Let $\omega \in \wedge ^k V^*$ and $\eta \in \wedge^l V^*.$ Then for all $v_1, v_2, \cdots, v_{k+l}$ we have $$\left \langle \omega \wedge \eta, v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l} \right \rangle = \frac {k!\ l!} {(k+l)!} \sum_{\sigma \in S_{k,l}} \varepsilon (\sigma) \left \langle \omega, v_{\sigma (1)} \otimes v_{\sigma (2)} \otimes \cdots \otimes v_{\sigma (k)} \right \rangle \times \left \langle \eta, v_{\sigma (k+1)} \otimes v_{\sigma {(k+2)}} \otimes \cdots \otimes v_{\sigma {(k+l)}} \right \rangle$$ where $S_{k,l} : = \left \{\sigma \in S_{k+l}\ \bigg |\ \sigma (1) \lt \sigma (2) \lt \cdots \lt \sigma (k), \sigma (k+1) \lt \sigma (k+2) \lt \cdots \lt \sigma (k+l) \right \}$ is the set of all $(k,l)$-shuffles and $\varepsilon (\sigma) = \text {sign} (\sigma),$ $\sigma \in S_{k+l}.$
Proof $:$ For each $\sigma \in S_{k+l}$ let $\tilde {\sigma} \in S_{k,l}$ be the element such that $\left \{\sigma (1), \sigma (2), \cdots , \sigma (k) \right \} = \left \{\tilde {\sigma} (1), \tilde {\sigma} (2), \cdots, \tilde {\sigma} (k) \right \}.$ Consequently $\left \{\sigma (k+1), \sigma (k+2), \cdots , \sigma (k+l) \right \} = \left \{\tilde {\sigma} (k+1), \tilde {\sigma} (k+2), \cdots, \tilde {\sigma} (k+l) \right \}.$ Let $\sigma_1' \in S_{k+l}$ be such that $\sigma_1' (\tilde {\sigma} (j)) = \sigma (j),\ 1 \leq j \leq k$ and $\sigma_1' (\tilde {\sigma} (j)) = \tilde {\sigma} (j),\ k+1 \leq j \leq k+l$ and $\sigma_2' \in S_{k+l}$ be such that $\sigma_2' (\tilde {\sigma} (j)) = \tilde {\sigma} (j),\ 1 \leq j \leq k$ and $\sigma_2' (\tilde {\sigma} (j)) = \sigma (j),\ k+1 \leq j \leq k+l.$ Then $\sigma = \sigma_1' \sigma_2' \tilde {\sigma}.$ Hence $\varepsilon (\sigma) = \varepsilon (\sigma_1') \varepsilon (\sigma_2') \varepsilon (\tilde {\sigma}).$
Now $$\begin{align*} \left \langle \omega \wedge \eta, v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l} \right \rangle & = \color {red} {\frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \left \langle \omega \wedge \eta, v_{\sigma (1)} \otimes v_{\sigma(2)} \otimes \cdots \otimes v_{\sigma (k+l)} \right \rangle} \\ & = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \left \langle \omega, v_{\sigma (1)} \otimes v_{\sigma(2)} \otimes \cdots \otimes v_{\sigma(k)} \right \rangle \times \left \langle \eta, v_{\sigma (k+1)} \otimes v_{\sigma(k+2)} \otimes \cdots \otimes v_{\sigma(k+l)} \right \rangle \\ & = \color{red} {\frac {1} {(k+l)!} \sum\limits_{\substack {\tilde {\sigma} \in S_{k,l}} \\ \ {\sigma_1', \sigma_2'}} \varepsilon (\tilde {\sigma})\ \varepsilon (\sigma_1') \left \langle \omega, v_{\sigma_1' \left (\tilde {\sigma} (1) \right )} \otimes v_{\sigma_1' \left (\tilde {\sigma} (2) \right )} \otimes \cdots \otimes v_{\sigma_1' \left (\tilde {\sigma} (k) \right )} \right \rangle \times \varepsilon (\sigma_2') \left \langle \eta, v_{\sigma_2' \left (\tilde {\sigma} (k+1) \right )} \otimes v_{\sigma_2' \left (\tilde {\sigma} (k+2) \right )} \otimes \cdots \otimes v_{\sigma_2' \left ( \tilde {\sigma} (k+l) \right )} \right \rangle} \\ & = \color{red} {\frac {k!\ l!} {(k+l)!} \sum\limits_{\sigma \in S_{k,l}} \varepsilon (\sigma) \left \langle \omega, v_{\sigma (1)} \otimes v_{\sigma(2)} \otimes \cdots \otimes v_{\sigma(k)} \right \rangle \times \left \langle \eta, v_{\sigma (k+1)} \otimes v_{\sigma(k+2)} \otimes \cdots \otimes v_{\sigma(k+l)} \right \rangle} \end{align*}$$
This completes the proof.
Can anybody please explain the equalities in red? I am having hard time understanding these equalities. Thanks for your time.
EDIT $:$ What I know is that $$\omega \wedge \eta = \text {Alt}_{k+l} (\omega \otimes \eta) = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \rho (\sigma) (\omega \otimes \eta)$$ where $\rho(\sigma) : V^{*\otimes {(k+l)}} \longrightarrow V^{*\otimes {(k+l)}}$ is a linear isomorphism given by $$\rho (\sigma) (v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l}) : = v_{\sigma^{-1} (1)} \otimes v_{\sigma^{-1} (2)} \otimes \cdots \otimes v_{\sigma^{-1} (k+l)}.$$ So if $\omega = \omega_1 \otimes \omega_2 \otimes \cdots \otimes \omega_k$ and $\eta = \omega_{k+1} \otimes \omega_{k+2} \otimes \cdots \otimes \omega_{k+l}$ then it follows that $$\begin{align*} \left \langle \omega \wedge \eta, v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l} \right \rangle & = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \prod\limits_{j=1}^{k+l} \omega_{\sigma^{-1} (j)} (v_j) \\ & = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \prod\limits_{j=1}^{k+l} \omega_{j} (v_{\sigma (j)}) \\ & = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \left \langle \omega \otimes \eta, v_{\sigma (1)} \otimes v_{\sigma (2)} \otimes \cdots \otimes v_{\sigma (k+l)} \right \rangle \end{align*}$$ which is not precisely the first equality. Also I have no idea how to get the third and the fourth equalities.
Let $\omega := \omega_1 \otimes \omega_2 \otimes \cdots \otimes \omega_k \in \wedge^k V^*$ and $\eta := \omega_{k+1} \otimes \omega_{k+2} \otimes \cdots \otimes \omega_{k+l} \in \wedge^l V^*.$ Since $\text {Alt}$ is an idempotent linear operator it follows that $\text {Alt}_k (\omega) = \omega$ and $\text {Alt}_l (\eta) = \eta.$ This shows that $\text {Alt}_k (\omega) = \left (\text {Alt}_k \circ \otimes_k \right ) (\omega_1, \omega_2, \cdots, \omega_k)$ and $\text {Alt}_l (\eta) = \left (\text {Alt}_l \circ \otimes_l \right ) (\omega_{k+1}, \omega_{k+2}, \cdots, \omega_{k+l}).$ Before going to prove the equalities we note that the third equality can easily be obtained from the second equality just by decomposing $\sigma$ as $\sigma_1' \sigma_2' \tilde {\sigma},$ where $\tilde {\sigma} \in S_{k,l},$ $\sigma_1'$ is a permutation of $\{\tilde {\sigma} (1), \tilde {\sigma} (2), \cdots, \tilde {\sigma} (k) \}$ and $\sigma_2'$ is a permutation on $\{\tilde {\sigma} (k+1), \tilde {\sigma} (k+2), \cdots, \tilde {\sigma} (k+l) \}.$ So for a given $\tilde {\sigma},$ the number of such $\sigma_1'$ and $\sigma_2'$ are respectively $k!$ and $l!.$ Another thing to note here is that the multi-linear map $\left ( \text {Alt} \circ \otimes \right )$ is alternating linear. So to obtain the fourth equality from the third one let us first analyze $F_{\sigma_1}: = \left \langle \omega, v_{\sigma_1' \left (\tilde {\sigma} (1) \right )} \otimes v_{\sigma_1' \left (\tilde {\sigma} (2) \right )} \otimes \cdots \otimes v_{\sigma_1' \left (\tilde {\sigma} (k) \right )} \right \rangle.$
Clearly $$F_{\sigma_1} = \left ( \left (\left (\text {Alt}_k \circ \otimes_k \right ) \circ \mu \left ({\tilde {\sigma}}^{-1} \sigma_1' \tilde {\sigma}, \cdot \right ) \right ) (\omega_1,\omega_2, \cdots, \omega_k) \right ) \left (v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k)} \right ) = \varepsilon \left ({\tilde {\sigma}}^{-1} \sigma_1' \tilde {\sigma} \right ) \left ( \left (\text {Alt}_k \circ \otimes_k \right ) (\omega_1,\omega_2, \cdots, \omega_k ) \right ) \left (v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k)} \right )$$ where for a given $\tau \in S_k$ we have $\mu (\tau,\cdot) (v_1,v_2, \cdots, v_k) = \left (v_{\tau^{-1} (1)}, v_{\tau^{-1} (2)}, \cdots, v_{\tau^{-1} (k)} \right ),$ $(v_1,v_2, \cdots, v_k) \in V^{\times k}.$ Hence for any $\tilde {\sigma} \in S_{k,l}$ and for any permutation $\sigma_1'$ of the set $\{\tilde {\sigma} (1), \tilde {\sigma} (2), \cdots, \tilde {\sigma} (k) \}$ we have $$\varepsilon \left (\sigma_1' \right ) F_{\sigma_1} = \left ( \left ( \text {Alt}_k \circ \otimes_k \right ) (\omega_1, \omega_2, \cdots, \omega_k ) \right ) \left (v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k)} \right ).$$ Similarly if we define $G_{\sigma_2} := \left \langle \eta, v_{\sigma_2' \left (\tilde {\sigma} (1) \right )} \otimes v_{\sigma_2' \left (\tilde {\sigma} (2) \right )} \otimes \cdots \otimes v_{\sigma_2' \left (\tilde {\sigma} (k) \right )} \right \rangle$ then for any $\tilde {\sigma} \in S_{k,l}$ and for any permutation $\sigma_2'$ of the set $\{\tilde {\sigma} (k+1), \tilde {\sigma} (k+2), \cdots, \tilde {\sigma} (k+l) \}$ we have $$\varepsilon \left (\sigma_2' \right ) G_{\sigma_2} = \left ( \left (\text {Alt}_l \circ \otimes_l \right ) (\omega_{k+1}, \omega_{k+2}, \cdots, \omega_{k+l} ) \right ) \left (v_{\tilde {\sigma} (k+1)} \otimes v_{\tilde {\sigma} (k+2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right ).$$ Having obtained all those things the third equality simplifies to $$\begin{align*} \frac {k!\ l!} {(k+l)!} \sum\limits_{\tilde {\sigma} \in S_{k,l}} \varepsilon \left (\tilde {\sigma} \right ) \left ( \left (\text {Alt}_k \circ \otimes_k \right ) (\omega_1, \omega_2, \cdots, \omega_k ) \right ) \left (v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (l)} \right ) \times \left ( \left (\text {Alt}_l \circ \otimes_l \right ) (\omega_{k+1}, \omega_{k+2}, \cdots, \omega_{k+l} ) \right ) \left (v_{\tilde {\sigma} (k+1)} \otimes v_{\tilde {\sigma} (k+2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right ) \end{align*}$$ which further simplifies to $$\frac {k!\ l!} {(k+l)!} \sum\limits_{\tilde {\sigma} \in S_{k,l}} \varepsilon \left (\tilde {\sigma} \right ) \left \langle \omega, v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k)} \right \rangle \times \left \langle \eta, v_{\tilde {\sigma} (k+1)} \otimes v_{\tilde {\sigma} (k+2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right \rangle$$ which again simplifies to $$\frac {k!\ l!} {(k+l)!} \sum\limits_{\tilde {\sigma} \in S_{k,l}} \varepsilon \left (\tilde {\sigma} \right ) \left \langle \omega \otimes \eta, v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right \rangle.$$ Hence $$\left \langle \omega \wedge \eta, v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l} \right \rangle = \frac {k!\ l!} {(k+l)!} \sum\limits_{\tilde {\sigma} \in S_{k,l}} \varepsilon \left (\tilde {\sigma} \right ) \left \langle \omega \otimes \eta, v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right \rangle.$$