[proofplan] We compare the shuffle expansions of $\alpha \wedge \beta$ and $\beta \wedge \alpha$ when evaluated on an arbitrary tuple $(v_1, \dots, v_{p+q}) \in V^{p+q}$. Composition with the block transposition $\tau \in S_{p+q}$ that sends the first $p$ positions to the second block of length $p$ and the last $q$ positions to the first block of length $q$ furnishes a sign-tracking bijection $\mathrm{Sh}(q,p) \to \mathrm{Sh}(p,q)$. Since this block transposition is a product of exactly $pq$ adjacent transpositions, $\operatorname{sgn}(\tau) = (-1)^{pq}$. Reindexing the shuffle expansion of $\alpha \wedge \beta$ along this bijection transforms it term-by-term into the shuffle expansion of $\beta \wedge \alpha$, multiplied by the global sign $(-1)^{pq}$. [/proofplan] [step:Fix an evaluation tuple and write out both shuffle expansions] Fix vectors $v_1, \dots, v_{p+q} \in V$. By the defining [shuffle formula for the wedge product](/theorems/3558), \begin{align*} (\alpha \wedge \beta)(v_1, \dots, v_{p+q}) &= \sum_{\sigma \in \mathrm{Sh}(p,q)} \operatorname{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}), \\ (\beta \wedge \alpha)(v_1, \dots, v_{p+q}) &= \sum_{\sigma' \in \mathrm{Sh}(q,p)} \operatorname{sgn}(\sigma')\, \beta(v_{\sigma'(1)}, \dots, v_{\sigma'(q)})\, \alpha(v_{\sigma'(q+1)}, \dots, v_{\sigma'(q+p)}). \end{align*} Since both sides of the claimed identity are multilinear and alternating forms in $(v_1, \dots, v_{p+q})$, it suffices to prove the identity after evaluation on every such tuple. [guided] The wedge product is, by definition (via the [Shuffle Identity](/theorems/2499)), built from these explicit shuffle sums. To prove $\alpha \wedge \beta = (-1)^{pq}\beta \wedge \alpha$ as elements of $\Lambda^{p+q}(V^*)$, two alternating multilinear forms on $V$ agree iff they take the same value on every $(p+q)$-tuple of vectors. So we fix an arbitrary tuple $(v_1, \dots, v_{p+q})$ and aim to show that the two displayed sums differ by the factor $(-1)^{pq}$. The strategy is to set up a bijection $\mathrm{Sh}(q,p) \to \mathrm{Sh}(p,q)$ that, term-by-term, identifies the summand for $\sigma'$ on the right with $(-1)^{pq}$ times the summand for the corresponding $\sigma$ on the left. The bijection is given by precomposition with a single permutation — the block transposition swapping the first $q$ slots with the last $p$ slots — whose sign is $(-1)^{pq}$. [/guided] [/step] [step:Define the block transposition $\tau$ and compute its sign] Define $\tau \in S_{p+q}$ by \begin{align*} \tau(i) = \begin{cases} i + q & \text{if } 1 \leq i \leq p, \\ i - p & \text{if } p+1 \leq i \leq p+q. \end{cases} \end{align*} In words, $\tau$ sends the ordered tuple $(1, \dots, p, p+1, \dots, p+q)$ to $(q+1, \dots, q+p, 1, \dots, q)$. We claim \begin{align*} \operatorname{sgn}(\tau) = (-1)^{pq}. \end{align*} To see this, factor $\tau$ as a product of adjacent transpositions. For each $j \in \{1, \dots, p\}$, the element originally in position $j$ must move rightward past each of the $q$ elements originally in positions $p+1, p+2, \dots, p+q$. Moving past one element requires one adjacent transposition, so moving past all $q$ requires $q$ adjacent transpositions. Doing this for $j = p, p-1, \dots, 1$ realises $\tau$ as a product of exactly $pq$ adjacent transpositions. By the [Sign Homomorphism](/theorems/778), each adjacent transposition has sign $-1$, hence $\operatorname{sgn}(\tau) = (-1)^{pq}$. [guided] We want a permutation that converts the position blocks for a $(q,p)$-shuffle into the position blocks for a $(p,q)$-shuffle after precomposition. The map $\tau$ defined above does exactly this: the first $p$ positions are sent to the second block $\{q+1, \dots, q+p\}$ of a $(q,p)$-shuffle, and the last $q$ positions are sent to the first block $\{1, \dots, q\}$. Equivalently, $\tau$ is the unique permutation that re-orders the list $(1, 2, \dots, p+q)$ as $(q+1, q+2, \dots, q+p, 1, 2, \dots, q)$. Why is $\operatorname{sgn}(\tau) = (-1)^{pq}$? Arrange the elements on a line. Each of the $p$ elements in the first block must pass over each of the $q$ elements in the second block. One passage is one adjacent swap, contributing one sign flip. The total number of adjacent swaps is $p \cdot q$, so the accumulated sign is $(-1)^{pq}$. The [Sign Homomorphism](/theorems/778) tells us that $\operatorname{sgn}: S_n \to \{\pm 1\}$ is a group homomorphism with $\operatorname{sgn}((i, i+1)) = -1$ on every adjacent transposition. Writing $\tau$ as a product of $pq$ adjacent transpositions therefore yields $\operatorname{sgn}(\tau) = (-1)^{pq}$, and the count is tied to this explicitly defined block transposition. [/guided] [/step] [step:Define a bijection $\mathrm{Sh}(q,p) \to \mathrm{Sh}(p,q)$ by precomposition with $\tau$] Define the map \begin{align*} \Phi: \mathrm{Sh}(q,p) &\to S_{p+q}, \\ \sigma' &\mapsto \sigma' \circ \tau. \end{align*} We claim that $\Phi$ takes values in $\mathrm{Sh}(p,q)$ and is a bijection onto $\mathrm{Sh}(p,q)$. Let $\sigma' \in \mathrm{Sh}(q,p)$, so $\sigma'(1) < \sigma'(2) < \cdots < \sigma'(q)$ and $\sigma'(q+1) < \cdots < \sigma'(q+p)$. Set $\sigma := \sigma' \circ \tau$. For $1 \leq i \leq p$, $\sigma(i) = \sigma'(\tau(i)) = \sigma'(i + q)$, and since $\sigma'(q+1) < \sigma'(q+2) < \cdots < \sigma'(q+p)$, the values $\sigma(1), \dots, \sigma(p)$ are strictly increasing. For $p+1 \leq i \leq p+q$, $\sigma(i) = \sigma'(\tau(i)) = \sigma'(i - p)$, and since $\sigma'(1) < \sigma'(2) < \cdots < \sigma'(q)$, the values $\sigma(p+1), \dots, \sigma(p+q)$ are strictly increasing. Therefore $\sigma \in \mathrm{Sh}(p,q)$. The map $\Phi$ is injective because right-multiplication by the fixed permutation $\tau$ is injective on $S_{p+q}$. For surjectivity, given any $\sigma \in \mathrm{Sh}(p,q)$, set $\sigma' := \sigma \circ \tau^{-1}$; an identical argument (interchanging the roles of $p$ and $q$ and using $\tau^{-1}$, which sends $(1, \dots, q, q+1, \dots, p+q)$ to $(p+1, \dots, p+q, 1, \dots, p)$) shows $\sigma' \in \mathrm{Sh}(q,p)$, and $\Phi(\sigma') = \sigma$. Hence $\Phi$ is a bijection $\mathrm{Sh}(q,p) \to \mathrm{Sh}(p,q)$. [guided] The idea is simple: a $(p,q)$-shuffle is a permutation that increases on the first $p$ positions and on the last $q$ positions; a $(q,p)$-shuffle increases on the first $q$ positions and the last $p$. To convert between them, we rearrange the positions themselves by the block transposition $\tau$ that interchanges the "first $p$" block with the "last $q$" block. Precomposing $\sigma'$ with $\tau$ swaps which positions are the "increasing on $p$" block and which are the "increasing on $q$" block. Concretely, for $\sigma = \sigma' \circ \tau$: - Positions $i \in \{1, \dots, p\}$ in $\sigma$ are filled by $\sigma'$ evaluated at positions $\{q+1, \dots, q+p\}$ — i.e., $\sigma'$'s "second (length-$p$) increasing block" — so $\sigma$ is increasing on $\{1, \dots, p\}$. ✓ - Positions $i \in \{p+1, \dots, p+q\}$ in $\sigma$ are filled by $\sigma'$ evaluated at positions $\{1, \dots, q\}$ — i.e., $\sigma'$'s "first (length-$q$) increasing block" — so $\sigma$ is increasing on $\{p+1, \dots, p+q\}$. ✓ So $\Phi$ lands in $\mathrm{Sh}(p,q)$. Bijectivity is immediate because right-multiplication by an invertible element of a group is a bijection on the group: $\Phi^{-1}: \sigma \mapsto \sigma \circ \tau^{-1}$, and by symmetry $\tau^{-1}$ is itself a block transposition (with roles of $p, q$ swapped), so $\Phi^{-1}$ takes $\mathrm{Sh}(p,q)$ into $\mathrm{Sh}(q,p)$. This bijection is the geometric heart of the proof: it identifies the indexing set of the $\alpha \wedge \beta$ shuffle sum with the indexing set of the $\beta \wedge \alpha$ shuffle sum, in a way that exactly tracks how the $p$ arguments of $\alpha$ and the $q$ arguments of $\beta$ change roles. [/guided] [/step] [step:Reindex the shuffle expansion of $\alpha \wedge \beta$ via $\Phi$] Using the bijection $\Phi: \mathrm{Sh}(q,p) \to \mathrm{Sh}(p,q)$, $\sigma' \mapsto \sigma = \sigma' \circ \tau$, we reindex the shuffle sum for $\alpha \wedge \beta$: \begin{align*} (\alpha \wedge \beta)(v_1, \dots, v_{p+q}) &= \sum_{\sigma \in \mathrm{Sh}(p,q)} \operatorname{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}) \\ &= \sum_{\sigma' \in \mathrm{Sh}(q,p)} \operatorname{sgn}(\sigma' \circ \tau)\, \alpha\bigl(v_{(\sigma' \circ \tau)(1)}, \dots, v_{(\sigma' \circ \tau)(p)}\bigr)\, \beta\bigl(v_{(\sigma' \circ \tau)(p+1)}, \dots, v_{(\sigma' \circ \tau)(p+q)}\bigr). \end{align*} By the [Sign Homomorphism](/theorems/778), $\operatorname{sgn}(\sigma' \circ \tau) = \operatorname{sgn}(\sigma')\operatorname{sgn}(\tau) = (-1)^{pq} \operatorname{sgn}(\sigma')$. Substituting the values of $\tau$ computed in the previous step, \begin{align*} ((\sigma' \circ \tau)(1), \dots, (\sigma' \circ \tau)(p)) &= (\sigma'(q+1), \dots, \sigma'(q+p)), \\ ((\sigma' \circ \tau)(p+1), \dots, (\sigma' \circ \tau)(p+q)) &= (\sigma'(1), \dots, \sigma'(q)). \end{align*} Pulling the global $(-1)^{pq}$ outside the sum: \begin{align*} (\alpha \wedge \beta)(v_1, \dots, v_{p+q}) = (-1)^{pq} \sum_{\sigma' \in \mathrm{Sh}(q,p)} \operatorname{sgn}(\sigma')\, \alpha(v_{\sigma'(q+1)}, \dots, v_{\sigma'(q+p)})\, \beta(v_{\sigma'(1)}, \dots, v_{\sigma'(q)}). \end{align*} [guided] We are converting the index of summation. Every $\sigma \in \mathrm{Sh}(p,q)$ is uniquely of the form $\sigma' \circ \tau$ for some $\sigma' \in \mathrm{Sh}(q,p)$, by the bijection $\Phi$ established above. So the sum over $\sigma \in \mathrm{Sh}(p,q)$ equals the sum over $\sigma' \in \mathrm{Sh}(q,p)$ of the same summand with $\sigma = \sigma' \circ \tau$ substituted everywhere. Three substitutions happen at once: 1. **Sign**: The [Sign Homomorphism](/theorems/778) is the group homomorphism $\operatorname{sgn}: S_{p+q} \to \{\pm 1\}$, so $\operatorname{sgn}(\sigma' \circ \tau) = \operatorname{sgn}(\sigma') \cdot \operatorname{sgn}(\tau) = (-1)^{pq} \operatorname{sgn}(\sigma')$. This is where the desired sign factor $(-1)^{pq}$ first appears. 2. **Arguments of $\alpha$**: For $i \in \{1, \dots, p\}$, $(\sigma' \circ \tau)(i) = \sigma'(\tau(i)) = \sigma'(i + q)$. So $\alpha$ now sees the values $\sigma'$ takes on its **second** (length-$p$) block. 3. **Arguments of $\beta$**: For $i \in \{p+1, \dots, p+q\}$, $(\sigma' \circ \tau)(i) = \sigma'(\tau(i)) = \sigma'(i - p)$. As $i$ ranges over $\{p+1, \dots, p+q\}$, $i - p$ ranges over $\{1, \dots, q\}$. So $\beta$ now sees the values $\sigma'$ takes on its **first** (length-$q$) block. Finally, $(-1)^{pq}$ does not depend on $\sigma'$, so it factors out of the sum. [/guided] [/step] [step:Identify the reindexed sum with $(-1)^{pq}(\beta \wedge \alpha)$] Since the values of $\alpha$ and $\beta$ are [real numbers](/page/Real%20Numbers), the multiplication $\alpha(\cdot)\beta(\cdot) = \beta(\cdot)\alpha(\cdot)$ is commutative pointwise. Rewriting: \begin{align*} (\alpha \wedge \beta)(v_1, \dots, v_{p+q}) &= (-1)^{pq} \sum_{\sigma' \in \mathrm{Sh}(q,p)} \operatorname{sgn}(\sigma')\, \beta(v_{\sigma'(1)}, \dots, v_{\sigma'(q)})\, \alpha(v_{\sigma'(q+1)}, \dots, v_{\sigma'(q+p)}) \\ &= (-1)^{pq}\, (\beta \wedge \alpha)(v_1, \dots, v_{p+q}), \end{align*} where the final equality is the defining shuffle formula for $\beta \wedge \alpha$ as stated in the [Shuffle Identity](/theorems/2499). Since $(v_1, \dots, v_{p+q}) \in V^{p+q}$ was arbitrary and both $\alpha \wedge \beta$ and $(-1)^{pq}\beta \wedge \alpha$ are alternating multilinear forms on $V$, we conclude \begin{align*} \alpha \wedge \beta = (-1)^{pq}\, \beta \wedge \alpha \end{align*} as elements of $\Lambda^{p+q}(V^*)$, completing the proof. [guided] The last step is the identification of the reindexed sum with the shuffle expansion of $\beta \wedge \alpha$. After the reindexing in the previous step, the sum reads, for each $\sigma' \in \mathrm{Sh}(q,p)$: \begin{align*} \operatorname{sgn}(\sigma')\, \alpha(v_{\sigma'(q+1)}, \dots, v_{\sigma'(q+p)})\, \beta(v_{\sigma'(1)}, \dots, v_{\sigma'(q)}). \end{align*} Compare with the shuffle expansion of $\beta \wedge \alpha$ from Step 1, indexed by $\sigma' \in \mathrm{Sh}(q,p)$: \begin{align*} \operatorname{sgn}(\sigma')\, \beta(v_{\sigma'(1)}, \dots, v_{\sigma'(q)})\, \alpha(v_{\sigma'(q+1)}, \dots, v_{\sigma'(q+p)}). \end{align*} The two summands differ only by the order in which we write the scalar product $\alpha(\dots)\beta(\dots) = \beta(\dots)\alpha(\dots)$, which is valid because both are real (or scalar-field) numbers and ordinary multiplication is commutative — *not* because the wedge product is commutative (it is not, except up to the sign we are deriving). After this rewriting, the reindexed sum matches the shuffle expansion of $\beta \wedge \alpha$ term-by-term. Finally, two alternating multilinear forms on $V$ are equal as elements of $\Lambda^{p+q}(V^*)$ iff they agree on every $(p+q)$-tuple of vectors. We showed agreement after multiplying $\beta \wedge \alpha$ by $(-1)^{pq}$, and our tuple was arbitrary, so the form-level identity $\alpha \wedge \beta = (-1)^{pq}\beta \wedge \alpha$ holds. [/guided] [/step]