[proofplan] Both sides of the claimed identity are bilinear in $(u, v)$, so it suffices to verify the equation on the standard basis pairs $(e_i, e_j)$ for $1 \le i, j \le 3$. We compute the right-hand side by expanding $u^\flat \wedge v^\flat$ in the basis $\{e^i \wedge e^j\}_{i<j}$, applying the [Hodge Star on Basis Elements](/theorems/2739), and comparing with the standard component formula for $u \times v$. The match on each basis pair, combined with bilinearity, gives the identity on all of $\mathbb{R}^3 \times \mathbb{R}^3$. [/proofplan]

[step:Fix bases and record the action of $\flat$ and $*$ on basis elements]Let $(e_1, e_2, e_3)$ denote the standard orthonormal basis of $\mathbb{R}^3$, and let $(e^1, e^2, e^3)$ denote the [dual basis](/theorems/414) of $(\mathbb{R}^3)^*$, characterised by $e^i(e_j) = \delta^i_j$. Since the basis $(e_1, e_2, e_3)$ is orthonormal with respect to $\langle \cdot, \cdot \rangle$, the musical isomorphism $\flat$ satisfies \begin{align*} e_i^\flat = e^i \qquad (i = 1, 2, 3), \end{align*} because $e_i^\flat(e_j) = \langle e_i, e_j \rangle = \delta_{ij} = e^i(e_j)$. By the [Hodge Star on Basis Elements](/theorems/2739), for the standard orientation $e^1 \wedge e^2 \wedge e^3$ on $\mathbb{R}^3$, \begin{align*} *(e^1 \wedge e^2) &= e^3, & *(e^2 \wedge e^3) &= e^1, & *(e^3 \wedge e^1) &= e^2. \end{align*} By antisymmetry of the wedge product, $e^1 \wedge e^3 = -e^3 \wedge e^1$, so $*(e^1 \wedge e^3) = -e^2$.[/step]

[guided]We need explicit formulas to compute with. The [dual basis](/theorems/414) $(e^1, e^2, e^3)$ is defined by $e^i(e_j) = \delta^i_j$. To identify $e_i^\flat$ with $e^i$ we check they agree on a basis: $e_i^\flat(e_j) = \langle e_i, e_j \rangle = \delta_{ij}$ since $(e_i)$ is orthonormal, and $e^i(e_j) = \delta^i_j$ by definition; hence $e_i^\flat = e^i$. For the Hodge star, the standard orientation on $\mathbb{R}^3$ is the volume form $\omega = e^1 \wedge e^2 \wedge e^3$. The defining property is $\alpha \wedge *\beta = \langle \alpha, \beta \rangle_{\Lambda^k} \omega$. On the orthonormal basis of $\Lambda^2(\mathbb{R}^3)^*$ given by $\{e^1 \wedge e^2, e^2 \wedge e^3, e^3 \wedge e^1\}$, the [Hodge Star on Basis Elements](/theorems/2739) yields the values \begin{align*} *(e^1 \wedge e^2) &= e^3, & *(e^2 \wedge e^3) &= e^1, & *(e^3 \wedge e^1) &= e^2. \end{align*} We will also need $*(e^1 \wedge e^3)$. Using antisymmetry $e^1 \wedge e^3 = -e^3 \wedge e^1$ and linearity of $*$, we get $*(e^1 \wedge e^3) = -e^2$.[/guided]

[step:Expand $u^\flat \wedge v^\flat$ in the basis of $\Lambda^2(\mathbb{R}^3)^*$]Write $u = \sum_{i=1}^3 u_i e_i$ and $v = \sum_{j=1}^3 v_j e_j$ with $u_i, v_j \in \mathbb{R}$. Since $\flat$ is linear and $e_i^\flat = e^i$, \begin{align*} u^\flat = \sum_{i=1}^3 u_i e^i, \qquad v^\flat = \sum_{j=1}^3 v_j e^j. \end{align*} Using bilinearity of the wedge product and the antisymmetry relation $e^i \wedge e^j = -e^j \wedge e^i$ (so in particular $e^i \wedge e^i = 0$), \begin{align*} u^\flat \wedge v^\flat = \sum_{i,j=1}^3 u_i v_j\, e^i \wedge e^j = \sum_{1 \le i < j \le 3} (u_i v_j - u_j v_i)\, e^i \wedge e^j. \end{align*} Writing out the three terms, \begin{align*} u^\flat \wedge v^\flat = (u_1 v_2 - u_2 v_1)\, e^1 \wedge e^2 + (u_1 v_3 - u_3 v_1)\, e^1 \wedge e^3 + (u_2 v_3 - u_3 v_2)\, e^2 \wedge e^3. \end{align*}[/step]

[guided]We compute $u^\flat \wedge v^\flat$ from scratch. Expanding both factors: \begin{align*} u^\flat \wedge v^\flat = \Bigl(\sum_{i=1}^3 u_i e^i\Bigr) \wedge \Bigl(\sum_{j=1}^3 v_j e^j\Bigr) = \sum_{i,j=1}^3 u_i v_j\, e^i \wedge e^j \end{align*} by bilinearity. The wedge product is alternating, so $e^i \wedge e^i = 0$ and the diagonal terms vanish. For off-diagonal terms, pair $(i, j)$ with $(j, i)$: the contribution is $u_i v_j\, e^i \wedge e^j + u_j v_i\, e^j \wedge e^i = (u_i v_j - u_j v_i)\, e^i \wedge e^j$ using $e^j \wedge e^i = -e^i \wedge e^j$. Collecting only $i < j$ gives the displayed expansion.[/guided]

[step:Apply the Hodge star and identify the result with $(u \times v)^\flat$]Applying $*$ to the expansion from the previous step and using the values recorded in Step 1, \begin{align*} *(u^\flat \wedge v^\flat) &= (u_1 v_2 - u_2 v_1)\, *(e^1 \wedge e^2) + (u_1 v_3 - u_3 v_1)\, *(e^1 \wedge e^3) + (u_2 v_3 - u_3 v_2)\, *(e^2 \wedge e^3) \\ &= (u_1 v_2 - u_2 v_1)\, e^3 - (u_1 v_3 - u_3 v_1)\, e^2 + (u_2 v_3 - u_3 v_2)\, e^1 \\ &= (u_2 v_3 - u_3 v_2)\, e^1 + (u_3 v_1 - u_1 v_3)\, e^2 + (u_1 v_2 - u_2 v_1)\, e^3. \end{align*} The classical cross product is defined componentwise by \begin{align*} u \times v = (u_2 v_3 - u_3 v_2)\, e_1 + (u_3 v_1 - u_1 v_3)\, e_2 + (u_1 v_2 - u_2 v_1)\, e_3 \in \mathbb{R}^3. \end{align*} Applying $\flat$ — which is linear and sends $e_i$ to $e^i$ — \begin{align*} (u \times v)^\flat = (u_2 v_3 - u_3 v_2)\, e^1 + (u_3 v_1 - u_1 v_3)\, e^2 + (u_1 v_2 - u_2 v_1)\, e^3. \end{align*} The two expressions for $*(u^\flat \wedge v^\flat)$ and $(u \times v)^\flat$ agree term by term in the basis $(e^1, e^2, e^3)$. Hence \begin{align*} (u \times v)^\flat = *(u^\flat \wedge v^\flat), \end{align*} which holds for all $u, v \in \mathbb{R}^3$. This completes the proof.[/step]

[guided]We now collapse the computation. Applying linearity of $*$ to the three-term expansion of $u^\flat \wedge v^\flat$ and substituting the values from Step 1, \begin{align*} *(u^\flat \wedge v^\flat) &= (u_1 v_2 - u_2 v_1)\, e^3 + (u_1 v_3 - u_3 v_1)(-e^2) + (u_2 v_3 - u_3 v_2)\, e^1. \end{align*} Note carefully the sign on the middle term: $*(e^1 \wedge e^3) = -e^2$, not $+e^2$. This is the place where the cyclic structure $1 \to 2 \to 3 \to 1$ of the cross product gets encoded — when we re-order the basis $2$-vectors into the cyclic frame $\{e^2 \wedge e^3, e^3 \wedge e^1, e^1 \wedge e^2\}$, the coefficient $(u_1 v_3 - u_3 v_1)$ becomes $-(u_3 v_1 - u_1 v_3)$. After rearranging into the standard order $(e^1, e^2, e^3)$, \begin{align*} *(u^\flat \wedge v^\flat) = (u_2 v_3 - u_3 v_2)\, e^1 + (u_3 v_1 - u_1 v_3)\, e^2 + (u_1 v_2 - u_2 v_1)\, e^3. \end{align*} The classical cross product on $\mathbb{R}^3$ has the well-known componentwise formula \begin{align*} u \times v = \bigl(u_2 v_3 - u_3 v_2,\; u_3 v_1 - u_1 v_3,\; u_1 v_2 - u_2 v_1\bigr). \end{align*} Why does $\flat$ then convert this to exactly the right side? Because $\flat$ acts diagonally on coordinates with respect to the orthonormal basis: $u_i$ multiplying $e_i$ on the vector side becomes the same scalar $u_i$ multiplying $e^i$ on the covector side. So $(u \times v)^\flat$ has the same coefficients in $(e^1, e^2, e^3)$ as $u \times v$ has in $(e_1, e_2, e_3)$, namely $(u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1)$. These coefficients match those of $*(u^\flat \wedge v^\flat)$ exactly. Since both sides of the claimed identity are elements of $(\mathbb{R}^3)^*$ with identical expansions in the [dual basis](/theorems/414), they are equal. As $u, v \in \mathbb{R}^3$ were arbitrary, the identity $(u \times v)^\flat = *(u^\flat \wedge v^\flat)$ holds on all of $\mathbb{R}^3 \times \mathbb{R}^3$.[/guided]