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We found geometric algebra representations for the symmetric and antisymmetric components for a gradient-vector direct product. In particular, given
$$\begin{equation}\label{eqn:tensorComponents:20} d\Bv = d\Bx \cdot \lr{ \spacegrad \otimes \Bv } \end{equation}$$
we found
$$\begin{equation}\label{eqn:tensorComponents:40} \begin{aligned} d\Bx \cdot \Bd &= \inv{2} d\Bx \cdot \lr{ \spacegrad \otimes \Bv + \lr{\spacegrad \otimes \Bv }^\dagger } \\ &= \inv{2} \lr{ d\Bx \lr{ \spacegrad \cdot \Bv } + \gpgradeone{ \spacegrad d\Bx \Bv } }, \end{aligned} \end{equation}$$
and
$$\begin{equation}\label{eqn:tensorComponents:60} \begin{aligned} d\Bx \cdot \BOmega &= \inv{2} d\Bx \cdot \lr{ \spacegrad \otimes \Bv – \lr{\spacegrad \otimes \Bv }^\dagger } \\ &= \inv{2} \lr{ d\Bx \lr{ \spacegrad \cdot \Bv } – \gpgradeone{ d\Bx \Bv \spacegrad } }. \end{aligned} \end{equation}$$

Let’s expand each of these in coordinates to verify that these are correct. For the symmetric component, that is
$$\begin{equation}\label{eqn:tensorComponents:80} \begin{aligned} d\Bx \cdot \Bd &= \inv{2} \lr{ dx_i \partial_j v_j \Be_i + \partial_j dx_i v_k \gpgradeone{ \Be_j \Be_i \Be_k } } \\ &= \inv{2} dx_i \lr{ \partial_j v_j \Be_i + \partial_j v_k \lr{ \delta_{ji} \Be_k + \lr{ \Be_j \wedge \Be_i } \cdot \Be_k } } \\ &= \inv{2} dx_i \lr{ \partial_j v_j \Be_i + \partial_j v_k \lr{ \delta_{ji} \Be_k + \delta_{ik} \Be_j – \delta_{jk} \Be_i } } \\ &= \inv{2} dx_i \lr{ \partial_j v_j \Be_i + \partial_i v_k \Be_k + \partial_j v_i \Be_j – \partial_j v_j \Be_i } \\ &= \inv{2} dx_i \lr{ \partial_i v_k \Be_k + \partial_j v_i \Be_j } \\ &= dx_i \inv{2} \lr{ \partial_i v_j + \partial_j v_i } \Be_j. \end{aligned} \end{equation}$$
Sure enough, we that the product contains the matrix element of the symmetric component of $\spacegrad \otimes \Bv$.

Now let’s verify that our GA antisymmetric tensor product representation works out.
$$\begin{equation}\label{eqn:tensorComponents:100} \begin{aligned} d\Bx \cdot \BOmega &= \inv{2} \lr{ dx_i \partial_j v_j \Be_i – dx_i \partial_k v_j \gpgradeone{ \Be_i \Be_j \Be_k } } \\ &= \inv{2} dx_i \lr{ \partial_j v_j \Be_i – \partial_k v_j \lr{ \delta_{ij} \Be_k + \delta_{jk} \Be_i – \delta_{ik} \Be_j } } \\ &= \inv{2} dx_i \lr{ \partial_j v_j \Be_i – \partial_k v_i \Be_k – \partial_k v_k \Be_i + \partial_i v_j \Be_j } \\ &= \inv{2} dx_i \lr{ \partial_i v_j \Be_j – \partial_k v_i \Be_k } \\ &= dx_i \inv{2} \lr{ \partial_i v_j – \partial_j v_i } \Be_j. \end{aligned} \end{equation}$$
As expected, we that this product contains the matrix element of the antisymmetric component of $\spacegrad \otimes \Bv$.

We also found previously that $\BOmega$ is just a curl, namely
$$\begin{equation}\label{eqn:tensorComponents:120} \BOmega = \inv{2} \lr{ \spacegrad \wedge \Bv } = \inv{2} \lr{ \partial_i v_j } \Be_i \wedge \Be_j, \end{equation}$$
which directly encodes the antisymmetric component of $\spacegrad \otimes \Bv$. We can also see that by fully expanding $d\Bx \cdot \BOmega$, which gives
$$\begin{equation}\label{eqn:tensorComponents:140} \begin{aligned} d\Bx \cdot \BOmega &= dx_i \inv{2} \lr{ \partial_j v_k } \Be_i \cdot \lr{ \Be_j \wedge \Be_k } \\ &= dx_i \inv{2} \lr{ \partial_j v_k } \lr{ \delta_{ij} \Be_k – \delta_{ik} \Be_j } \\ &= dx_i \inv{2} \lr{ \lr{ \partial_i v_k } \Be_k – \lr{ \partial_j v_i } \Be_j } \\ &= dx_i \inv{2} \lr{ \partial_i v_j – \partial_j v_i } \Be_j, \end{aligned} \end{equation}$$
as expected.