Let me use the following notation for orthonormal basis $\{\sigma_0,\sigma_1,\dots\}$ and this one for a general curvilinear basis $\{\mathbf{e}_0,\mathbf{e}_1,\dots\}$. The basis are constrained by the commutation relations: $\sigma_\mu \cdot \sigma_\nu=\delta_\mu^\nu$ and $\mathbf{e}_\mu\cdot \mathbf{e}_\nu = g_{\mu\nu}$, respectively, thus they are geometric algebras. We note that one can write the curvilinear basis as a linear combination of the orthonormal basis. Here is an example of a multivector of the clifford algebra of 2 dimensions:
$$ \mathbf{u}=a+x\sigma_1+y\sigma_2+b\sigma_1\wedge \sigma_1 $$
Intuitively, we could expect that the typical curvilinear multivector for this basis is obtained via simply one-to-one replacement of the basis, such that:
$$ \mathbf{v}=a+x\mathbf{e}_1+y\mathbf{e}_2+b\mathbf{e}_1\wedge \mathbf{e}_2 $$
However, I have noticed a number of undesirables that occur as a result of this basis replacement, which makes me suspect there is a much better bending process.
For instance, the complex number $z=a+b\sigma_1 \wedge \sigma_2$ rewriten in terms of curvilinear basis as $z'=a+b\mathbf{e}_1\wedge \mathbf{e}_2$ is specially problematic. The complex norm goes from the familiar $a^2+b^2$ to :
$$ \begin{align} (z')^*z'&=(a+b\mathbf{e}_1\wedge \mathbf{e}_2)(a-b\mathbf{e}_1\wedge \mathbf{e}_2)\\ &=a^2-b^2\mathbf{e}_1\wedge \mathbf{e}_2\wedge \mathbf{e}_1\wedge \mathbf{e}_2\\ &=a^2-b^2\mathbf{e}_1\wedge \mathbf{e}_2\wedge \mathbf{e}_1\wedge \mathbf{e}_2\\ &=a^2+b^2 (\sqrt{|\det g|})^2 \end{align} $$
The multiplier $(|\det g|)$ only acts on $b^2$ and not on $a^2$. To me, it makes little sense that a curvilinear basis bends the imaginary part of a complex number just fine, but is unable to touch the real for some reason. If we instead make the mapping from $z \to z'$ as follows:
$$ z \to \sqrt{|\det g|}z $$
then we get:
$$ \begin{align} z'&= \sqrt{|\det g|}a + \sqrt{|\det g|}b \sigma_1\wedge \sigma_2\\ &=\sqrt{|\det g|}a + b \mathbf{e}_1\wedge \mathbf{e}_2 \end{align} $$
and the adjusted norm becomes $||z'||^2= (a^2+b^2) (\sqrt{|\det g|})^2$, and the metric acts on the whole complex number.
- So a possible requirement to our bending procedure is to multiply the multivector by $\sqrt{|\det g|}$.
This kind of problem occurs at multiple levels within a multivector. For instance, consider the following vector of $Cl_4$:
$$ \mathbf{u}=X_1 \sigma_1 + X_2\sigma_2+V_1 \sigma_2\sigma_3\sigma_4 + V_2 \sigma_1\sigma_3\sigma_4 \tag{1} $$
which, because of the relation $\sigma_1=I\sigma_2\sigma_3\sigma_4$ and $\sigma_2=I\sigma_1\sigma_3\sigma_4$ I can rewrite as:
$$ \mathbf{u}=(X_1 + I V_1) \sigma_1 + (X_2+ I V_2)\sigma_2 \tag{2} $$
Now, if I replace the basis of (1) to curvilinear coordinates, I get:
$$ \begin{align} \mathbf{u}&=X_1 \mathbf{e}_1 + X_2\mathbf{e}_2+V_1 \mathbf{e}_2\mathbf{e}_3\mathbf{e}_4 + V_2 \mathbf{e}_1\mathbf{e}_3\mathbf{e}_4\\ &=(X_1 + IV_1 \sqrt{|\det g[\mathbf{e}_2,\mathbf{e}_3,\mathbf{e}_4]|} ) \mathbf{e}_1 + (X_2+ I V_2 \sqrt{|\det g[\mathbf{e}_1,\mathbf{e}_3,\mathbf{e}_4]|})\mathbf{e}_2 \end{align} $$
Here again, the imaginary part of the complex number $X_1+IV_2$ bends as a result of the replacement, but not it's real part. If I instead start with (2) then make the replacement, I get:
$$ \mathbf{u}=(X_1 + I V_1) \mathbf{e}_1 + (X_2+ I V_2)\mathbf{e}_2 $$
then neither parts bend.
One solution that comes to mind is as follows:
- If we introduce a convoluted tensor product (not quite certain how to achieve this in the details, but may have to also include the requirement #1 identified above), we can perhaps arrange for multiplication of $X_1$ and $V_1$ by $\sqrt{|\det g[\mathbf{e}_2,\mathbf{e}_3,\mathbf{e}_4]|}$ and of $X_2$ and $V_2$ by $\sqrt{|\det g[\mathbf{e}_1,\mathbf{e}_3,\mathbf{e}_4]|}$.
Maybe I am missing something obvious; my general question is how to 'correctly' move from an orthogonal basis to a curvilinear basis for a multivector?
edit:
I have not mentioned it, but the notion of 'invariance with respect to a change in coordinates' may also be important. The multivector with curvilinear coordinates is not invariant with respect to such a change in the metric, upon a one-to-one replacement with the orthogonal basis to curvilinear basis.
What is the general multivector in a curvilinear geometric algebra such that its norm is invariant with respect to a change of coordinate basis?