note-math

Gradient and divergence are ${𝐿}^2$ adjoint to each other.

$\int 𝑑\text{ Vol }𝑔\left(𝑋,\partial 𝑓\right)=\int 𝑑\text{ Vol }𝑔\left({\partial }^{†}𝑋,𝑓\right)$

or

$\int 𝑑\text{ Vol }𝑔\left(𝑋,\text{ grad }𝑓\right)=-\int 𝑑\text{ Vol }𝑔\left(\text{div }𝑋,𝑓\right)$

Generalize to general tensors. For example, suppose "covariant differentiation" adds ${\left(\top 𝑀\right)}^{⊺}$ to the first position, i.e.

$\nabla :\text{ tensor }\to {\left(\top 𝑀\right)}^{⊺}\otimes \text{ tensor}$

Then the adjoint is

${\nabla }^{†}:{\left(\top 𝑀\right)}^{⊺}\otimes \text{ tensor }\to \text{ tensor}$

let ${⟨,⟩}_{{𝐿}^2}=\int 𝑑\text{ Vol }𝑔(,)$

adjoint :=

Which is the Lagrangian for Laplacian? We can arbitrarily add the action mass term ${𝑚}^2|𝑇{|}^2$.

• $|\nabla 𝑇{|}^2𝑑\text{ Vol}$. eq ${\nabla }^{†}\nabla 𝑇$ or ${\nabla }^{†𝜇}{\nabla }_{𝜇}𝑇=0$
• $|{\nabla }^{†}𝑇{|}^2𝑑\text{ Vol}$. eq $\nabla {\nabla }^{†}𝑇=0$ or ${\nabla }_{𝜇}{\nabla }^{†𝜇}𝑇=0$
• $\left(|\nabla 𝑇{|}^2+|{\nabla }^{†}𝑇{|}^2\right)𝑑\text{ Vol}$. eq $\left({\nabla }^{†}\nabla +\nabla {\nabla }^{†}\right)𝑇=0$ or $\left({\nabla }^{†𝜇}{\nabla }_{𝜇}+{\nabla }_{𝜇}{\nabla }^{†𝜇}\right)𝑇=0$

Special case

$$\nabla :\stackrel{𝑘}{⨂}{\left(\top 𝑀\right)}^{⊺}\to \stackrel{𝑘+1}{⨂}{\left(\top 𝑀\right)}^{⊺}$$

adjoint

$${\nabla }^{†}:\stackrel{𝑘+1}{⨂}{\left(\top 𝑀\right)}^{⊺}\to \stackrel{𝑘}{⨂}{\left(\top 𝑀\right)}^{⊺}$$

In coordinates

$$\begin{array}{ccc}\nabla \left(𝑑{𝑥}^{𝑖}\right)& =& -{\mathrm{\Gamma }}_{{𝑖}^{\prime }{𝑖}^{″}}^{𝑖}𝑑{𝑥}^{{𝑖}^{\prime }}\otimes 𝑑{𝑥}^{{𝑖}^{″}}\\ \nabla \left({𝑎}_{𝑖}𝑑{𝑥}^{𝑖}\right)& =& \left({\partial }_{{𝑖}^{\prime }}{𝑎}_{{𝑖}^{″}}-{𝑎}_{𝑖}{\mathrm{\Gamma }}_{{𝑖}^{\prime }{𝑖}^{″}}^{𝑖}\right)𝑑{𝑥}^{{𝑖}^{\prime }}\otimes 𝑑{𝑥}^{{𝑖}^{″}}\end{array}$$

adjoint

$$\begin{array}{ccc}{\nabla }^{†}\left(𝑑{𝑥}^{𝑖}\right)& =& -{𝑔}^{𝑖{𝑖}^{\prime }}{\mathrm{\Gamma }}_{{𝑖}^{\prime }𝑖}^{𝑖}\\ {\nabla }^{†}\left({𝑎}_{𝑖}𝑑{𝑥}^{𝑖}\right)& =& {𝑔}^{𝑖{𝑖}^{\prime }}\left({\partial }_{{𝑖}^{\prime }}{𝑎}_{𝑖}-{\mathrm{\Gamma }}_{{𝑖}^{\prime }𝑖}^{𝑖}\right)\end{array}$$

$-{𝑔}^{𝑖{𝑖}^{\prime }}{\mathrm{\Gamma }}_{𝑖{𝑖}^{\prime }}^{𝑗}=\frac{1}{\left|𝑔\right|}{\partial }_{𝑖}\left(\left|𝑔\right|{𝑔}^{𝑖𝑗}\right)$

$\odot ,\wedge$ are the basic building blocks for constructing mixed symmetry.

Using algebraic techniques, define

$$\begin{array}{c}{\nabla }_{\odot }:\stackrel{𝑘}{⨀}{\left(\top 𝑀\right)}^{⊺}\to \stackrel{𝑘+1}{⨀}{\left(\top 𝑀\right)}^{⊺}\\ {\nabla }_{\wedge }:\stackrel{𝑘}{\bigwedge }{\left(\top 𝑀\right)}^{⊺}\to \stackrel{𝑘+1}{\bigwedge }{\left(\top 𝑀\right)}^{⊺}\end{array}$$

Using

$$\text{sym}\left(𝑑{𝑥}^{𝑖}\otimes 𝑑{𝑥}^{{𝑖}^{\prime }}\right)=𝑑{𝑥}^{𝑖}\odot 𝑑{𝑥}^{{𝑖}^{\prime }}$$

Define

$${\nabla }_{\odot }=\text{ sym}\nabla$$

In coordinates

$${\nabla }_{\odot }\left(𝑑{𝑥}^{𝑖}\right)=-{\mathrm{\Gamma }}_{{𝑖}^{\prime }{𝑖}^{″}}^{𝑖}𝑑{𝑥}^{{𝑖}^{\prime }}\odot 𝑑{𝑥}^{{𝑖}^{″}}$$

and

$${\nabla }_{\odot }\left({𝑎}_{𝑖}𝑑{𝑥}^{𝑖}\right)=\left({\partial }_{{𝑖}^{\prime }}{𝑎}_{{𝑖}^{″}}-{𝑎}_{𝑖}{\mathrm{\Gamma }}_{{𝑖}^{\prime }{𝑖}^{″}}^{𝑖}\right)𝑑{𝑥}^{{𝑖}^{\prime }}\odot 𝑑{𝑥}^{{𝑖}^{″}}$$

"product rule"

$$\begin{array}{ccc}{\nabla }_{\odot }\left({𝑎}_{𝑖𝑗}𝑑{𝑥}^{𝑖}\odot 𝑑{𝑥}^{{𝑖}^{\prime }}\right)& =& {\nabla }_{\odot }{𝑎}_{𝑖𝑗}\odot 𝑑{𝑥}^{𝑖}\odot 𝑑{𝑥}^{{𝑖}^{\prime }}\\ & & +{𝑎}_{𝑖𝑗}\left({\nabla }_{\odot }𝑑{𝑥}^{𝑖}\right)\odot 𝑑{𝑥}^{{𝑖}^{\prime }}\\ & & +{𝑎}_{𝑖𝑗}𝑑{𝑥}^{𝑖}\odot \left({\nabla }_{\odot }𝑑{𝑥}^{{𝑖}^{\prime }}\right)\end{array}$$

Then define the adjoint ${\nabla }_{\odot }^{†}$. Also there is

$${\nabla }_{\odot }^{†}=\text{ sym}{\nabla }^{†}$$

"product rule" (Question)

$${\nabla }_{\odot }^{†}\left({𝑎}_{𝑖{𝑖}^{\prime }}𝑑{𝑥}^{𝑖}\odot 𝑑{𝑥}^{{𝑖}^{\prime }}\right)={\nabla }^{†}\left({𝑎}_{𝑖{𝑖}^{\prime }}𝑑{𝑥}^{𝑖}\right)\odot 𝑑{𝑥}^{{𝑖}^{\prime }}+𝑑{𝑥}^{𝑖}\odot {\nabla }^{†}\left({𝑎}_{𝑖{𝑖}^{\prime }}𝑑{𝑥}^{{𝑖}^{\prime }}\right)$$

${\nabla }_{\wedge }=\text{ alt}\nabla$

But ${\nabla }_{\wedge }\left(𝑑{𝑥}^{𝑖}\right)=-{\mathrm{\Gamma }}_{{𝑖}^{\prime }{𝑖}^{″}}^{𝑖}𝑑{𝑥}^{{𝑖}^{\prime }}\wedge 𝑑{𝑥}^{{𝑖}^{″}}=0$, due to the symmetry of $\mathrm{\Gamma }$.

So ${\nabla }_{\wedge }=𝑑$ is the exterior derivative.

In coordinates $𝑑\left({𝑎}_{𝑖}𝑑{𝑥}^{𝑖}\right)={\partial }_{{𝑖}^{\prime }}{𝑎}_{𝑖}𝑑{𝑥}^{{𝑖}^{\prime }}\wedge 𝑑{𝑥}^{𝑖}$

$𝑑$ also has an adjoint ${𝑑}^{†}$

The Lagrangian for the Laplacian of ${\nabla }_{\odot },𝑑$?

• $|𝑃\left(𝑇\right){|}^2$. eq ${𝑃}^{†}𝑃\left(𝑇\right)=0$
• $|{𝑃}^{†}\left(𝑇\right){|}^2$. eq $𝑃{𝑃}^{†}\left(𝑇\right)=0$
• $|𝑃\left(𝑇\right){|}^2+|{𝑃}^{†}\left(𝑇\right){|}^2$. eq $\left({𝑃}^{†}𝑃+𝑃{𝑃}^{†}\right)\left(𝑇\right)=0$

The variables in the Lagrangian should probably use ${\nabla }_{\odot }𝑇,𝑑𝜔$ instead of $\nabla 𝑇$

The differentiation of the Lagrangian should probably use $\frac{\partial 𝐿}{\partial \left({\nabla }_{\odot }𝑇\right)},\frac{\partial 𝐿}{\partial \left(𝑑𝜔\right)}$ instead of $\frac{\partial 𝐿}{\partial \left(\nabla 𝑇\right)}$

For scalar fields as zeroth-order tensor fields, assume $\partial =𝑑=\nabla$ and ${\partial }^{†}={𝑑}^{†}={\nabla }^{†}=0$. At this time

• $\partial {\partial }^{†}=0$
• ${\partial }^{†}\partial +\partial {\partial }^{†}={\partial }^{†}\partial$

In coordinates

• gradient $\text{grad }𝑓=\left({𝑔}^{𝑖𝑗}{\partial }_{𝑖}𝑓\right){\partial }_{𝑗}$

• divergence $\text{div }𝑋=\frac{1}{\left|𝑔\right|}{\partial }_{𝑖}\left(\left|𝑔\right|{𝑋}^{𝑖}\right)$

• Laplacian $∆𝑓=\text{ div }\text{ grad }𝑓=\frac{1}{\left|𝑔\right|}{\partial }_{𝑖}\left(\left|𝑔\right|{𝑔}^{𝑖𝑗}{\partial }_{𝑗}𝑓\right)$

At the origin of geodesic coordinates, $𝑔=𝜂,\partial 𝑔=0$

• gradient $\text{grad }𝑓={𝜀}_{𝑖}\left({\partial }_{𝑖}𝑓\right){\partial }_{𝑖}$ where ${𝜀}_{𝑖}\in \left\{\pm 1\right\}$ is the $𝑖$-th component of the quadratic form of $𝑝,𝑞$.

• divergence $\text{div }𝑋={\partial }_{𝑖}{𝑋}^{𝑖}$

• harmonician $∆𝑓=\text{ div }\text{ grad }𝑓={𝜀}_{𝑖}{\partial }_{𝑖}^2𝑓$

The same is true in flat metric coordinates.