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## Symmetrization and antisymmetrization of the vector differential in GA.

There was an error in yesterday’s post. This decomposition was correct:
d\Bv
+
\spacegrad \cdot \lr{ d\Bx \wedge \Bv }.

However, identifying these terms with the symmetric and antisymmetric splits of $$\spacegrad \otimes \Bv$$ was wrong.
Brian pointed out that a purely incompressible flow is one for which $$\spacegrad \cdot \Bv = 0$$, yet, in general, an incompressible flow can have a non-zero deformation tensor.

Also, given the nature of the matrix expansion of the antisymmetric tensor, we should have had a curl term in the mix and we do not. The conclusion must be that \ref{eqn:dyadicVsGa:460} is a split into divergence and non-divergence terms, but we really wanted a split into curl and non-curl terms.

## Symmetrization and antisymmetrization of the vector differential in GA: Take II.

Identification of $$\ifrac{1}{2} \lr{ \spacegrad \otimes \Bv + \lr{ \spacegrad \otimes \Bv }^\dagger }$$ with the divergence was incorrect.

Let’s explicitly expand out our symmetric tensor component fully to see what it really yields, without guessing.
\begin{aligned}
d\Bx \cdot
\inv{2}
&=
d\Bx \cdot
\inv{2}
\lr{
\begin{bmatrix}
\partial_i v_j
\end{bmatrix}
+
\begin{bmatrix}
\partial_j v_i
\end{bmatrix}
} \\
&=
dx_i
\inv{2}
\begin{bmatrix}
\partial_i v_j +
\partial_j v_i
\end{bmatrix}
\begin{bmatrix}
\Be_1 \\
\Be_2 \\
\Be_3
\end{bmatrix}.
\end{aligned}

The symmetric matrix that represents this direct product tensor is
\inv{2}
\begin{bmatrix}
\partial_i v_j +
\partial_j v_i
\end{bmatrix}
=
\inv{2}
\begin{bmatrix}
2 \partial_1 v_1 & \partial_1 v_2 + \partial_2 v_1 & \partial_1 v_3 + \partial_3 v_1 \\
\partial_2 v_1 + \partial_1 v_2 & 2 \partial_2 v_2 & \partial_2 v_3 + \partial_3 v_2 \\
\partial_3 v_1 + \partial_1 v_3 & \partial_3 v_2 + \partial_2 v_3 & \partial_3 v_1 + \partial_1 v_3 \\
\end{bmatrix}
.

This certainly isn’t isomorphic to the divergence. Instead, the trace of this matrix is the portion that is isomorphic to the divergence. The rest is something else. Let’s put the tensors into vector form to understand what they really represent.

For the symmetric part we have
\begin{aligned}
d\Bx \cdot
\inv{2}
&=
dx_i
\inv{2}
\begin{bmatrix}
\partial_i v_j +
\partial_j v_i
\end{bmatrix}
\begin{bmatrix}
\Be_1 \\
\Be_2 \\
\Be_3
\end{bmatrix} \\
&=
\inv{2} \lr{
\lr{ d\Bx \cdot \spacegrad } \Bv + \spacegrad \lr{ d\Bx \cdot \Bv }
},
\end{aligned}

and, similarily, for the antisymmetric tensor component, we have
\begin{aligned}
d\Bx \cdot
\inv{2}
&=
dx_i
\inv{2}
\begin{bmatrix}
\partial_i v_j –
\partial_j v_i
\end{bmatrix}
\begin{bmatrix}
\Be_1 \\
\Be_2 \\
\Be_3
\end{bmatrix} \\
&=
\inv{2} \lr{
\lr{ d\Bx \cdot \spacegrad } \Bv – \spacegrad \lr{ d\Bx \cdot \Bv }
} \\
&=
\inv{2}
d\Bx \cdot \lr{ \spacegrad \wedge \Bv}.
\end{aligned}

We find an isomorphism of the antisymmetric term with the curl, but the symmetric term has a divergence component, plus more.

If we want to we can split the symmetric component into it’s divergence and non-divergence terms, we get
\begin{aligned}
d\Bx \cdot \Bd
&=
\inv{2}
\lr{
\lr{ d\Bx \cdot \spacegrad } \Bv + \spacegrad \lr{ d\Bx \cdot \Bv }
} \\
&=
\inv{2}
\lr{
d\Bx \lr{ \spacegrad \cdot \Bv } + \spacegrad \cdot \lr{ d\Bx \wedge \Bv } + \spacegrad \lr{ d\Bx \cdot \Bv }
} \\
&=
\inv{2}
\lr{
} \\
&=
\inv{2}
\lr{
},
\end{aligned}

so for incompressible flow, the GA representation is a single grade one selection

It is a little unfortunate that we cannot factor out the $$d\Bx$$ term. We can do that for the
GA representation of the antisymmetric tensor contribution, which is just
\BOmega
=

Let’s see what the antisymmetric tensor equivalent looks like in the incompressible case, by subtracting a divergence term
\begin{aligned}
d\Bx \cdot \lr{ \spacegrad \wedge \Bv } – d\Bx \lr{ \spacegrad \cdot \Bv }
&=
&=
&=
\end{aligned}

so we have

Both the symmetric and antisymmetric tensors have compressible components.

## Summary.

We found that it was possible to split the vector differential into a divergence and incompressible components, as follows
\begin{aligned}
d\Bv
&= \lr{ d\Bx \cdot \spacegrad } \Bv \\
+
\spacegrad \cdot \lr{ d\Bx \wedge \Bv }.
\end{aligned}

With
\begin{aligned}
d\Bv
&= d\Bx \cdot
\lr{
\inv{2} \lr{ \spacegrad \otimes \Bv + \lr{ \spacegrad \otimes \Bv }^\dagger }
+
\inv{2} \lr{ \spacegrad \otimes \Bv – \lr{ \spacegrad \otimes \Bv }^\dagger }
} \\
&= d\Bx \cdot \lr{ \Bd + \BOmega },
\end{aligned}

we found the following correspondences between the symmetric and antisymmetric tensor product components
\begin{aligned}
d\Bx \cdot \Bd &=
\inv{2} \lr{
\lr{ d\Bx \cdot \spacegrad } \Bv + \spacegrad \lr{ d\Bx \cdot \Bv }
} \\
&=
\inv{2}
\lr{
}
\end{aligned},

and
\begin{aligned}
d\Bx \cdot \BOmega
&=
\inv{2} d\Bx \cdot \lr{ \spacegrad \wedge \Bv } \\
&=
\inv{2} \lr{
}.
\end{aligned}

In the incompressible case where $$\spacegrad \cdot \Bv = 0$$, we have