DOC

doc/discretization.md

@@ -20,14 +20,8 @@ and $h$ are the lattice spacings. See the figure below:
 
 ![Lattice discretization](img/discretization.svg "FIG 1. Lattice discretization.")
 
-We choose $`Q(j) = Q(\vec{r}_j)`$ to be values of $`Q`$ at the lattice
-sites. We define the Wilson link matrices
-$`U_{ij}
-  =U_{ji}^{-1}
-  =U(\vec{r}_{i},\vec{r}_{j})
-  =
-  \mathrm{Pexp}[i\int_{L(\vec{r}_i,\vec{r}_j)}d\vec{r}'\,\cdot\vec{\mathcal{A}}(\vec{r}')]
-`$
+We choose $`Q(j) = Q(\vec{r}_j)`$ to be values of $`Q`$ at the lattice sites. We define the Wilson link matrices
+$`U_{ij} =U_{ji}^{-1} =U(\vec{r}_{i},\vec{r}_{j}) = \mathrm{Pexp}[i\int_{L(\vec{r}_i,\vec{r}_j)}d\vec{r}'\,\cdot\vec{\mathcal{A}}(\vec{r}')]`$
 where $`L(\vec{r}_i,\vec{r}_j)`$ is straight line from $`\vec{r}_j`$ to
 $`\vec{r}_i`$ and $`\mathrm{Pexp}`$ is the path-ordered integral.
 The neighbor cells of $`j`$ are $`i\in{}\mathrm{neigh}(j)`$
@@ -66,7 +60,7 @@ D
 ```
 where $`d`$ is the space dimension.
 
-Expanding in small $h$ we have
+Expanding in small $`h`$ we have
 ```math
 S_2
 \simeq
@@ -130,7 +124,7 @@ which then allows expressing $`F_{ij}`$ in terms of the link matrices.
 
 ## Hall term
 
-When discretizing the $Q\hat{\nabla}_iQ\hat{\nabla}_jQ$ part, we can
+When discretizing the $`Q\hat{\nabla}_iQ\hat{\nabla}_jQ`$ part, we can
 consider a discrete derivative
 ```math
   D_{ij}
@@ -160,13 +154,13 @@ have
   \mathrm{tr}
   (U_{lk} U_{kj} - U_{li} U_{ij}) D_{ji} Q(i) D_{il}
 ```
-So the structure is $\frac{-i}{h^4}$ $\times$ (gauge factors for
-counterclockwise loop $-$ return back clockwise) $\times$ $DQD$ for
-three sites in counterclockwise order.  Since only $Q(i)$, $Q(j)$,
-$Q(l)$ appear here, one can also think of this as a plaquette corner
-with $Q(i)$ being the corner.
+So the structure is $`\frac{-i}{h^4}`$ $`\times`$ (gauge factors for
+counterclockwise loop $`-`$ return back clockwise) $`\times`$ $`DQD`$ for
+three sites in counterclockwise order.  Since only $`Q(i)`$, $`Q(j)`$,
+$`Q(l)`$ appear here, one can also think of this as a plaquette corner
+with $`Q(i)`$ being the corner.
 
-To do the sum over the $\mu$, $\nu$ indices, one can average over
+To do the sum over the $`\mu`$, $`\nu`$ indices, one can average over
 corners of the plaquettes as follows, which gives the final discretization
 of the Hall term:
 ```math
@@ -197,7 +191,7 @@ of the Hall term:
   \Bigr)
   \,,
 ```
-where $\mathrm{plaqc}(i;jkl)$ are the plaquette corners. The last line
+where $`\mathrm{plaqc}(i;jkl)`$ are the plaquette corners. The last line
 follows with direct calculation, and considering the cyclic
 permutations.
 
@@ -224,7 +218,7 @@ We can also calculate the matrix current $j\mapsto{}i$ between neighboring cells
   ]\rvert_{\xi=0}
 ```
 %
-where $T^a$ is an appropriate matrix generator, and the transformation
+where $`T^a`$ is an appropriate matrix generator, and the transformation
 is inserted only to one of the links.
 
 The local gauge invariance now implies that there is an exact discrete
@@ -248,11 +242,13 @@ However, the definition of the matrix current is \emph{asymmetric}, so
 that $`J_{ij}\ne{}-J_{ji}`$ as they differ by a quantity of order
 $`\mathcal{O}(h^2)`$. The asymmetry arises when $`T^a`$ does not commute
 with $`U_{ij}`$, so it is an issue only for the SU(2) component. The
-problem is to some degree just in the interpretation of what $J_{ij}$
-means: it is the incoming current measured at site $`i`$, whereas
+problem is to some degree just in the interpretation of what $`J_{ij}`$
+means: 
+
+It is the incoming current measured at site $`i`$, whereas
 $`J_{ji}`$ is measured as site $`j`$.  Since the parallel transport of
 spin between these two locations can imply rotation due to the gauge
-field, there's no reason why we should have $`J_{ij} = -J_{ji}`$, except
+field, there's no reason why we should have $`J_{ij}=-J_{ji}`$, except
 in the continuum limit where the two points become close to each
 other.
 
@@ -261,10 +257,10 @@ other.
 
 The above is sufficient for numerical implementation: it is not
 necessary to find out the saddle point equation symbolically. It can
-be deduced by parametrizing first so that $Q^2=1$ and then using
+be deduced by parametrizing first so that $`Q^2=1`$ and then using
 automatic differentiation (AD). Only a routine evaluating the value of
 the action is then needed, in terms of the AD dual variables.
 
 When looking for the saddle point in equilibrium, the problem becomes
-even simpler; $Q$ has then additional restrictions, and we are looking
-for the minimum of the free energy $S$.
+even simpler; $`Q`$ has then additional restrictions, and we are looking
+for the minimum of the free energy $`S`$.