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Stress-energy projected components:
Let $\rho =n^u{n}^v{T}_{uv}$ = the stress-energy density (2008 Alcubierre, eqn. 2.4.12)
Let $j_a=-n^u{T}_{uv}{{\gamma }_a}^v$ = the stress-energy spatial current (2008 Alcubierre, eqn. 2.4.12)
Let $S_{ab}=T_{uv}{{\gamma }_a}^u{{\gamma }_b}^v$ = the spatial stress tensor
Let $S=T_{uv}{\gamma }^{uv}={S^a}_a$ = the spatial stress tensor trace
So $T_{uv}{g}^{uv}=T_{uv}({\gamma }^{uv}-n^u{n}^v)=S-\rho$

$\perp j_a={{\gamma }_a}^b{j}_b$
$=-{{\gamma }_a}^b{n}^u{T}_{uv}{{\gamma }^v}_b$
$=-n^u{T}_{uv}{{\gamma }^v}_a$
$=j_a$
so $j_a$ is spatial

$j_a{n}^a=-n^u{T}_{uv}{{\gamma }^v}_a{n}^a=0$ so $j_a$ is spatial

So $T_{uv}={\delta }_u^a{\delta }_v^b{T}_{ab}$
$=({{\gamma }_u}^a-n_u{n}^a)({{\gamma }_v}^b-n_v{n}^b)T_{ab}$
$=({{\gamma }_u}^a{{\gamma }_v}^b-{{\gamma }_u}^a{n}_v{n}^b-n_u{n}^a{{\gamma }_v}^b+n_u{n}^a{n}_v{n}^b)T_{ab}$
$={{\gamma }_u}^a{{\gamma }_v}^b{T}_{ab}-{{\gamma }_u}^a{n}_v{n}^b{T}_{ab}-n_u{n}^a{{\gamma }_v}^b{T}_{ab}+n_u{n}^a{n}_v{n}^b{T}_{ab}$
$=S_{uv}+n_v{j}_u+n_u{j}_v+n_u{n}_v\rho$

So $T_{uv}-\frac{1}{2}{g}_{uv}{T}_{ab}{g}^{ab}$
$=S_{uv}+n_v{j}_u+n_u{j}_v+n_u{n}_v\rho -\frac{1}{2}({\gamma }_{uv}-n_u{n}_v)(S_{ab}+n_b{j}_a+n_a{j}_b+n_a{n}_b\rho )({\gamma }^{ab}-n^a{n}^b)$
$=S_{uv}+n_v{j}_u+n_u{j}_v+n_u{n}_v\rho -\frac{1}{2}({\gamma }_{uv}-n_u{n}_v)(S_{ab}{\gamma }^{ab}-S_{ab}{n}^a{n}^b+n_b{j}_a{\gamma }^{ab}-n_b{j}_a{n}^a{n}^b+n_a{j}_b{\gamma }^{ab}-n_a{j}_b{n}^a{n}^b+n_a{n}_b\rho {\gamma }^{ab}-n_a{n}_b\rho n^a{n}^b)$
$=S_{uv}-\frac{1}{2}{\gamma }_{uv}(S-\rho )+n_v{j}_u+n_u{j}_v+n_u{n}_v\rho +\frac{1}{2}{n}_u{n}_v(S-\rho )$
$=S_{uv}+n_v{j}_u+n_u{j}_v+n_u{n}_v\rho -\frac{1}{2}({\gamma }_{uv}-n_u{n}_v)(S-\rho )$
$=S_{uv}+n_v{j}_u+n_u{j}_v+n_u{n}_v\rho -\frac{1}{2}{g}_{uv}(S-\rho )$

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Einstein Field Equations in projected components:

Einstein Field Equations, contracting both indexes with the normal vector:
$n^u{n}^v{G}_{uv}=8\pi n^u{n}^v{T}_{uv}$
$n^u{n}^v{G}_{uv}=8\pi \rho$
using $2n^u{n}^v{G}_{uv}=R^{\perp }+K^2-K^{ab}{K}_{ab}$:
$R^{\perp }+K^2-K^{ab}{K}_{ab}=16\pi \rho$ (2010 Baumgarte & Shapiro, eqn. 2.125)
$R^{\perp }+K^2-K^{ab}{K}_{ab}-16\pi \rho =0$
Hamiltonian constraint: Let...
$\mathcal{H}=n^u{n}^v{G}_{uv}-8\pi \rho$ (2008 Alcubierre, eqn. 2.5.10)
$\mathcal{H}=\frac{1}{2}(R^{\perp }+K^2-K^{ab}{K}_{ab})-8\pi \rho$ (2008 Alcubierre, eqn. 2.5.9)
Now when the Einstein Field Equations are satisfied, we will have $\mathcal{H}=0$

Einstein Field Equations, contracting one index with normal and projecting the other:
$-n^u{{\gamma }_a}^v{G}_{uv}=-n^u{{\gamma }_a}^{v}8\pi T_{uv}$
using $j_a=-n^u{{\gamma }_a}^v{T}_{uv}$
$-n^u{{\gamma }_a}^v{G}_{uv}=8\pi j_a$
using ${{\gamma }_a}^u{n}^v{G}_{uv}={\mathrm{\nabla }}_a^{\perp }K-{\mathrm{\nabla }}_b^{\perp }{K_a}^b$
${\mathrm{\nabla }}_b^{\perp }{K_a}^b-{\mathrm{\nabla }}_a^{\perp }K=8\pi j_a$
${\mathrm{\nabla }}_b^{\perp }{K_a}^b-{\mathrm{\nabla }}_a^{\perp }K-8\pi j_a=0$
Momentum constraint: Let...
${\mathcal{M}}_a=-n^u{{\gamma }_a}^v{G}_{uv}-8\pi j_a$ (2008 Alcubierre, eqn. 2.5.11)
${\mathcal{M}}_a={\mathrm{\nabla }}_b^{\perp }{K_a}^b-{\mathrm{\nabla }}_a^{\perp }K-8\pi j_a$
When the Einstein Field Equations are satisfied, we will have ${\mathcal{M}}_a=0$

Einstein Field Equations, projecting both indexes.
${{\gamma }_a}^u{{\gamma }_b}^v{G}_{uv}=8\pi {{\gamma }_a}^u{{\gamma }_b}^v{T}_{uv}$
using $S_{ab}=T_{uv}{{\gamma }_a}^u{{\gamma }_b}^v$:
${{\gamma }_a}^u{{\gamma }_b}^v{G}_{uv}=8\pi S_{uv}$
${{\gamma }_a}^u{{\gamma }_b}^v{G}_{uv}-8\pi S_{uv}=0$
Energy constraint: Let...
${\mathcal{E}}_{ab}={{\gamma }_a}^u{{\gamma }_b}^v{G}_{uv}-8\pi S_{uv}$ (2008 Alcubierre, eqn. 2.5.12)
... this one is mentioned in 2008 Alcubierre, eqn. 2.5.12, but no mention of which values for $\perp G_{uv}$ are used ... is it the R definition? is it the K definition?
When the Einstein Field Equations are satisfied, we will have ${\mathcal{E}}_{ab}=0$

starting with (from the "normal projections" worksheet):
${R^{\perp }}_{ab}+K{K}_{ab}-K_{ac}{K^c}_b={{\gamma }_a}^q{{\gamma }_b}^s{R}_{qs}+n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}$
$n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}={R^{\perp }}_{ab}+K{K}_{ab}-K_{ac}{K^c}_b-{{\gamma }_a}^q{{\gamma }_b}^s{R}_{qs}$
Substitute the trace-reversed Einstein field equations, $R_{ab}=8\pi (T_{ab}-\frac{1}{2}{g}_{ab}T)$
$n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}={R^{\perp }}_{ab}+K{K}_{ab}-K_{ac}{K^c}_b-{{\gamma }_a}^u{{\gamma }_b}^{v}8\pi (T_{uv}-\frac{1}{2}{g}_{uv}{T}_{cd}{g}^{cd})$ (2010 Baumgarte & Shapiro, 2.103)
$n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}={R^{\perp }}_{ab}+K{K}_{ab}-K_{ac}{K^c}_b-8\pi (S_{ab}-\frac{1}{2}{\gamma }_{ab}(S-\rho ))$
... and this route is the one that turns into the time-derivative of the extrinsic curvature ...
... but how do we come up with the ${\mathcal{E}}_{ab}$ constraints from only projected, non-time-derivative terms?
We don't. The ${\mathcal{E}}_{ab}$ constraints give rise to the $K_{ab}$ ivp. So using ADM formalism this constraint is always satisfied.

${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=n^u{\partial }_u{K}_{ab}+K_{ub}{\partial }_a{n}^u+K_{au}{\partial }_b{n}^u$
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=n^u{\mathrm{\nabla }}_u{K}_{ab}+K_{ub}{\mathrm{\nabla }}_a{n}^u+K_{au}{\mathrm{\nabla }}_b{n}^u$
using $K_{ab}=-{\mathrm{\nabla }}_a{n}_b-n_a{a}_b$ so ${\mathrm{\nabla }}_a{n}_b=-K_{ab}-n_a{a}_b$
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=n^u{\mathrm{\nabla }}_u(-{\mathrm{\nabla }}_a{n}_b-n_a{a}_b)+{K^u}_b(-K_{au}-n_a{a}_u)+{K^u}_a(-K_{bu}-n_b{a}_u)$
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=-n^u{\mathrm{\nabla }}_u{\mathrm{\nabla }}_a{n}_b-a_b{n}^u{\mathrm{\nabla }}_u{n}_a-n_a{n}^u{\mathrm{\nabla }}_u{a}_b-{K^u}_b{K}_{au}-{K^u}_a{K}_{bu}-{K^u}_b{n}_a{a}_u-{K^u}_a{n}_b{a}_u$ (2010 Baumgarte & Shapiro, eqn. 2.74)
using $a_a=n^u{\mathrm{\nabla }}_u{n}_a$
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=-n^u{\mathrm{\nabla }}_u{\mathrm{\nabla }}_a{n}_b-a_a{a}_b-n_a{n}^u{\mathrm{\nabla }}_u{a}_b-2K_{au}{K^u}_b-{K^u}_b{n}_a{a}_u-{K^u}_a{n}_b{a}_u$
using $n_d{R^d}_{bua}={\mathrm{\nabla }}_u{\mathrm{\nabla }}_a{n}_b-{\mathrm{\nabla }}_a{\mathrm{\nabla }}_u{n}_b$
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=-n_d{n}^u{R^d}_{bua}-n^u{\mathrm{\nabla }}_a{\mathrm{\nabla }}_u{n}_b-a_a{a}_b-n_a{n}^u{\mathrm{\nabla }}_u{a}_b-2K_{au}{K^u}_b-{K^u}_b{n}_a{a}_u-{K^u}_a{n}_b{a}_u$
TODO after a lot of work
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=-\frac{1}{\alpha }{\mathrm{\nabla }}_a^{\perp }{\mathrm{\nabla }}_b^{\perp }\alpha -K_{ac}{K^c}_b+n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}$ (2010 Baumgarte & Shapiro, eqn. 2.82)
using the definition of $n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}$ above ... without stress-energy:
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=-\frac{1}{\alpha }{\mathrm{\nabla }}_a^{\perp }{\mathrm{\nabla }}_b^{\perp }\alpha -K_{ac}{K^c}_b+{R^{\perp }}_{ab}+K{K}_{ab}-K_{ac}{K^c}_b-{{\gamma }_a}^u{{\gamma }_b}^v{R}_{uv}$
... and with stress-energy:
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=-\frac{1}{\alpha }{\mathrm{\nabla }}_a^{\perp }{\mathrm{\nabla }}_b^{\perp }\alpha -K_{ac}{K^c}_b+{R^{\perp }}_{ab}+K{K}_{ab}-K_{ac}{K^c}_b-8\pi (S_{ab}-\frac{1}{2}{\gamma }_{ab}(S-\rho ))$
${\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}=-\frac{1}{\alpha }{\mathrm{\nabla }}_a^{\perp }{\mathrm{\nabla }}_b^{\perp }\alpha +{R^{\perp }}_{ab}+K{K}_{ab}-2K_{ac}{K^c}_b-8\pi (S_{ab}-\frac{1}{2}{\gamma }_{ab}(S-\rho ))$

Extrinsic curvature initial value formulation:
using ${\mathcal{L}}_{\overrightarrow{n}}=\frac{1}{\alpha }({\mathcal{L}}_{\overrightarrow{t}}-{\mathcal{L}}_{\overrightarrow{\beta }})$
${\partial }_t{K}_{ab}-{\mathcal{L}}_{\overrightarrow{\beta }}{K}_{ab}=\alpha {\mathcal{L}}_{\overrightarrow{n}}{K}_{ab}$
${\partial }_t{K}_{ab}=-{\mathrm{\nabla }}_a^{\perp }{\mathrm{\nabla }}_b^{\perp }\alpha +\alpha ({R^{\perp }}_{ab}+K{K}_{ab}-2K_{ac}{K^c}_b-8\pi (S_{ab}-\frac{1}{2}{\gamma }_{ab}(S-\rho )))+{\mathcal{L}}_{\overrightarrow{\beta }}{K}_{ab}$

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Ricci tensor contracted with normal:
$n^a{n}^b{R}_{ab}$
$=n^a{n}^b{R^c}_{acb}$
$=n^b({\mathrm{\nabla }}_c{\mathrm{\nabla }}_b{n}^c-{\mathrm{\nabla }}_b{\mathrm{\nabla }}_c{n}^c)$
$=n^b{\mathrm{\nabla }}_c{\mathrm{\nabla }}_b{n}^c-n^b{\mathrm{\nabla }}_b{\mathrm{\nabla }}_c{n}^c$
$={\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c)-({\mathrm{\nabla }}_c{n}^b)({\mathrm{\nabla }}_b{n}^c)-{\mathrm{\nabla }}_b(n^b{\mathrm{\nabla }}_c{n}^c)+({\mathrm{\nabla }}_b{n}^b)({\mathrm{\nabla }}_c{n}^c)$
$={\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c)-(-{K_c}^b+n_c{a}^b)(-{K_b}^c+n_b{a}^c)-{\mathrm{\nabla }}_b(n^b{\mathrm{\nabla }}_c{n}^c)+(-{K_b}^b+n_b{a}^b)(-{K_c}^c+n_c{a}^c)$
$={\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c)-({K_c}^b{K_b}^c-n_c{a}^b{K_b}^c-n_b{a}^c{K_c}^b+n_b{n}_c{a}^b{a}^c)-{\mathrm{\nabla }}_b(n^b{\mathrm{\nabla }}_c{n}^c)+K^2$
$={\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c)-{\mathrm{\nabla }}_b(n^b{\mathrm{\nabla }}_c{n}^c)-K_{ab}{K}^{ab}+K^2$
$={\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c)-{\mathrm{\nabla }}_c(n^c{\mathrm{\nabla }}_b{n}^b)-K_{ab}{K}^{ab}+K^2$
$n^a{n}^b{R}_{ab}=K^2-K_{ab}{K}^{ab}+{\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c-n^c{\mathrm{\nabla }}_b{n}^b)$ (2008 Alcubierre, eqn 2.7.3)

start with from the "normal projections" worksheet:
$R=R^{\perp }-K_{ab}{K}^{ab}+K^2-2n^a{n}^b{R}_{ab}$
...substitute contracted Ricci tensor...
$R=R^{\perp }-K_{ab}{K}^{ab}+K^2-2(K^2-K_{ab}{K}^{ab}+{\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c-n^c{\mathrm{\nabla }}_b{n}^b))$
$R=R^{\perp }-K_{ab}{K}^{ab}+K^2-2K^2+2K_{ab}{K}^{ab}-2{\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c-n^c{\mathrm{\nabla }}_b{n}^b)$
$R=R^{\perp }+K_{ab}{K}^{ab}-K^2-2{\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c-n^c{\mathrm{\nabla }}_b{n}^b)$ (2008 Alcubierre, eqn 2.7.4)
Looks like now we disregard the $2{\mathrm{\nabla }}_c(n^b{\mathrm{\nabla }}_b{n}^c-n^c{\mathrm{\nabla }}_b{n}^b)$
because it can be converted to a total integral via divergence theorem.
but this is only the case if we are integrating over a volume, i.e. true for $\int Ld{x}^4$ and therefore $\int Rd{x}^4$
So how can it still be true for $\mathcal{L}=\int \sqrt{-g}Rd{x}^4$? Does $\int {\mathrm{\nabla }}_c\sqrt{-g}d{x}^4$ disappear as well?
$R=R^{\perp }+K_{ab}{K}^{ab}-K^2$

Ok a suspicious thing about the "cancel total derivatives", that would imply:
$n^a{n}^b{R}_{ab}=K^2-K_{ab}{K}^{ab}$
Is only true in the context of a spacetime-volume integral?
If so then how true is all of numerical relativity, where we are doing numerical spatial volume integration / spatial PDEs?

Back from our "normal projection" worksheet, start with our projected Einstein curvature:
${{\gamma }_a}^u{{\gamma }_b}^v{G}_{uv}={R^{\perp }}_{ab}+K{K}_{ab}-K_{ac}{K^c}_b-\frac{1}{2}{\gamma }_{ab}R-n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}$
Substitute $R=R^{\perp }+K_{ab}{K}^{ab}-K^2$
${{\gamma }_a}^u{{\gamma }_b}^v{G}_{uv}={R^{\perp }}_{ab}+K{K}_{ab}-K_{ac}{K^c}_b-\frac{1}{2}{\gamma }_{ab}(R^{\perp }+K_{ab}{K}^{ab}-K^2)-n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}$
${{\gamma }_a}^u{{\gamma }_b}^v{G}_{uv}={R^{\perp }}_{ab}-\frac{1}{2}{\gamma }_{ab}{R}^{\perp }+\frac{1}{2}{\gamma }_{ab}{K}^2+K{K}_{ab}-\frac{1}{2}{\gamma }_{ab}{K}_{ab}{K}^{ab}-K_{ac}{K^c}_b-n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}$
... how to get rid of that last $n^p{n}^r{{\gamma }_a}^q{{\gamma }_b}^s{R}_{pqrs}$ ?
Maybe you can't. As 2008 Alcubierre after eqn. 2.5.12 says, the ${\mathcal{E}}_{ab}=0$ turn into the ADM equations.

Lagrangian is equal to scalar curvature:
$L=R$

Densitized Lagrangian:
$\mathcal{L}=\sqrt{-g}L$
$=\sqrt{-g}R$
$=\alpha \sqrt{\gamma }R$
...substitute Gauss-Codazzi equations...
$=\alpha \sqrt{\gamma }(R^{\perp }+K_{ab}{K}^{ab}-K^2$
$\mathcal{L}=\alpha \sqrt{\gamma }(R^{\perp }+K_{ij}{K}^{ij}-K^2)$ (2008 Alcubierre, eqn. 2.7.6)

Vacuum Action:
$S=\int \sqrt{-g}Rd{x}^4$
$S=\int \int \alpha \sqrt{\gamma }(R^{\perp }+K_{ij}{K}^{ij}-K^2)d{x}^{3}dt$

${\dot{\gamma }}_{ij}=-2\alpha K_{ij}+2{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)}$
$K_{ij}=-\frac{1}{2\alpha }{\dot{\gamma }}_{ij}+\frac{1}{\alpha }{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)}$
$\frac{\partial }{\partial {\dot{\gamma }}_{ij}}{K}_{kl}=-\frac{1}{2\alpha }{\delta }_k^i{\delta }_l^j$

conjugate momentum:
${\pi }_{ij}=\frac{\partial \mathcal{L}}{\partial {\dot{\gamma }}^{ij}}$
$=\frac{\partial }{\partial {\dot{\gamma }}^{ij}}(\alpha \sqrt{\gamma }(R^{\perp }+K_{kl}{K}^{kl}-K^2))$
$=\frac{\partial }{\partial {\dot{\gamma }}^{ij}}\alpha \sqrt{\gamma }(R^{\perp }+{\gamma }_{km}{\gamma }_{ln}{K}^{kl}{K}^{mn}-K^{kl}{\gamma }_{kl}{K}^{mn}{\gamma }_{mn})$
${\pi }_{ij}=\alpha \sqrt{\gamma }(\frac{\delta }{\delta {\dot{\gamma }}^{ij}}({\gamma }_{km}{\gamma }_{ln}{K}^{kl}{K}^{mn})-\frac{\delta }{\delta {\dot{\gamma }}^{ij}}(K^{kl}{\gamma }_{kl}{K}^{mn}{\gamma }_{mn}))$
${\pi }_{ij}=\alpha \sqrt{\gamma }({\gamma }_{km}{\gamma }_{ln}\frac{\delta }{\delta {\dot{\gamma }}^{ij}}(K^{kl}{K}^{mn})-{\gamma }_{kl}{\gamma }_{mn}\frac{\delta }{\delta {\dot{\gamma }}^{ij}}(K^{kl}{K}^{mn}))$
${\pi }_{ij}=\alpha \sqrt{\gamma }({\gamma }_{km}{\gamma }_{ln}-{\gamma }_{kl}{\gamma }_{mn})\frac{\delta }{\delta {\dot{\gamma }}^{ij}}(K^{kl}{K}^{mn})$
${\pi }_{ij}=\alpha \sqrt{\gamma }({\gamma }_{km}{\gamma }_{ln}-{\gamma }_{kl}{\gamma }_{mn})(K^{kl}\frac{\delta }{\delta {\dot{\gamma }}^{ij}}{K}^{mn}+K^{mn}\frac{\delta }{\delta {\dot{\gamma }}^{ij}}{K}^{kl})$
${\pi }_{ij}=-\alpha \sqrt{\gamma }({\gamma }_{km}{\gamma }_{ln}-{\gamma }_{kl}{\gamma }_{mn})(\frac{1}{2\alpha }{K}^{kl}{\delta }_i^m{\delta }_j^n+\frac{1}{2\alpha }{K}^{mn}{\delta }_i^k{\delta }_j^l)$
${\pi }_{ij}=-\alpha \frac{1}{2\alpha }\sqrt{\gamma }(({\gamma }_{km}{\gamma }_{ln}-{\gamma }_{kl}{\gamma }_{mn})K^{kl}{\delta }_i^m{\delta }_j^n+({\gamma }_{km}{\gamma }_{ln}-{\gamma }_{kl}{\gamma }_{mn})K^{mn}{\delta }_i^k{\delta }_j^l)$
${\pi }_{ij}=-\frac{1}{2}\sqrt{\gamma }(({\gamma }_{km}{\gamma }_{ln}{K}^{kl}{\delta }_i^m{\delta }_j^n-{\gamma }_{kl}{\gamma }_{mn}{K}^{kl}{\delta }_i^m{\delta }_j^n)+({\gamma }_{km}{\gamma }_{ln}{K}^{mn}{\delta }_i^k{\delta }_j^l-{\gamma }_{kl}{\gamma }_{mn}{K}^{mn}{\delta }_i^k{\delta }_j^l))$
${\pi }_{ij}=-\frac{1}{2}\sqrt{\gamma }(K_{ij}-{\gamma }_{ij}K+K_{ij}-{\gamma }_{ij}K)$
${\pi }_{ij}=-\sqrt{\gamma }(K_{ij}-{\gamma }_{ij}K)$
${\pi }^{ij}=-\sqrt{\gamma }(K^{ij}-{\gamma }^{ij}K)$ (2008 Alcubierre, eqn. 2.7.8)
Notice that ${\pi }^{ij}$ is a tensor density with weight 1.

Extrinsic curvature in terms of conjugate momentum:
${\gamma }_{ij}{\pi }^{ij}={\gamma }_{ij}(-\sqrt{\gamma }(K^{ij}-{\gamma }^{ij}K))$
solve conjugate momentum trace:
$\pi =-\sqrt{\gamma }({\gamma }_{ij}{K}^{ij}-{\gamma }_{ij}{\gamma }^{ij}K)$
$\pi =-\sqrt{\gamma }(K-3K)$
$\pi =2\sqrt{\gamma }K$
$K=\pi /(2\sqrt{\gamma })$
substitute into equation of conjugate momentum in terms of extrinsic curvature:
${\pi }^{ij}=-\sqrt{\gamma }(K^{ij}-{\gamma }^{ij}K)$
${\pi }^{ij}=-\sqrt{\gamma }{K}^{ij}+\sqrt{\gamma }{\gamma }^{ij}\pi /(2\sqrt{\gamma })$
${\pi }^{ij}=-\sqrt{\gamma }{K}^{ij}+\frac{1}{2}{\gamma }^{ij}\pi$
$K^{ij}=-\frac{1}{\sqrt{\gamma }}({\pi }^{ij}-\frac{1}{2}{\gamma }^{ij}\pi )$ (2008 Alcubierre, eqn. 2.7.9)

Extrinsic cuvrature derivative in terms of conjugate momentum:
${\dot{\gamma }}_{ij}=-2\alpha K_{ij}+2{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)}$
substitute $K^{ij}=-\frac{1}{\sqrt{\gamma }}({\pi }^{ij}-\frac{1}{2}{\gamma }^{ij}\pi )$:
${\dot{\gamma }}_{ij}=2\alpha \frac{1}{\sqrt{\gamma }}({\pi }^{ij}-\frac{1}{2}{\gamma }^{ij}\pi )+2{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)}$

useful identity:
$K^2-K_{ij}{K}^{ij}$
$=(\pi /(2\sqrt{\gamma }){)}^2-\frac{1}{\gamma }({\pi }_{ij}-\frac{1}{2}{\gamma }_{ij}\pi )({\pi }^{ij}-\frac{1}{2}{\gamma }^{ij}\pi )$
$=\frac{1}{4\gamma }{\pi }^2-\frac{1}{\gamma }({\pi }_{ij}{\pi }^{ij}-\frac{1}{2}\pi {\gamma }^{ij}{\pi }_{ij}-\frac{1}{2}\pi {\gamma }_{ij}{\pi }^{ij}+\frac{1}{4}\pi {\gamma }_{ij}{\gamma }^{ij}\pi )$
$=\frac{1}{\gamma }(\frac{1}{4}{\pi }^2-{\pi }_{ij}{\pi }^{ij}+\frac{1}{2}{\pi }^2+\frac{1}{2}{\pi }^2-\frac{3}{4}{\pi }^2)$
$=\frac{1}{\gamma }(\frac{1}{2}{\pi }^2-{\pi }_{ij}{\pi }^{ij})$

Hamiltonian density:
$\mathcal{H}={\pi }^{ij}{\dot{\gamma }}_{ij}-\mathcal{L}$
$=-\alpha \sqrt{\gamma }(R^{\perp }+K_{ij}{K}^{ij}-K^2)-\sqrt{\gamma }(K^{ij}-{\gamma }^{ij}K)(-2\alpha K_{ij}+2{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)})$
$=-\alpha \sqrt{\gamma }(R^{\perp }+K_{ij}{K}^{ij}-K^2)+2\alpha \sqrt{\gamma }(K^{ij}-{\gamma }^{ij}K)K_{ij}-2\sqrt{\gamma }(K^{ij}-{\gamma }^{ij}K){\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)}$
$=-\sqrt{\gamma }(\alpha (R^{\perp }+K_{ij}{K}^{ij}-K^2-2(K^{ij}-{\gamma }^{ij}K)K_{ij})+2(K^{ij}-{\gamma }^{ij}K){\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)})$
$=-\sqrt{\gamma }(\alpha (R^{\perp }+K_{ij}{K}^{ij}-K^2-2K^{ij}{K}_{ij}+2K^2)+2(K^{ij}-{\gamma }^{ij}K){\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)})$
$=-\sqrt{\gamma }(\alpha (R^{\perp }-K_{ij}{K}^{ij}+K^2)+2(K^{ij}-{\gamma }^{ij}K){\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)})$
$=-\sqrt{\gamma }(\alpha (R^{\perp }-K_{ij}{K}^{ij}+K^2)-2{\pi }^{ij}\frac{1}{\sqrt{\gamma }}{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)})$
$=-\sqrt{\gamma }(\alpha (R^{\perp }+\frac{1}{\gamma }(\frac{1}{2}{\pi }^2-{\pi }_{ij}{\pi }^{ij}))-2{\pi }^{ij}\frac{1}{\sqrt{\gamma }}{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)})$
$=\alpha (-\sqrt{\gamma }{R}^{\perp }+\frac{1}{\sqrt{\gamma }}({\pi }_{ij}{\pi }^{ij}-\frac{1}{2}{\pi }^2))+2{\pi }^{ij}{\mathrm{\nabla }}_i^{\perp }{\beta }_j$
using integration by parts: $2{\pi }^{ij}{\mathrm{\nabla }}_j^{\perp }{\beta }_i=-2{\beta }_i{\mathrm{\nabla }}_j^{\perp }{\pi }^{ij}$
(TODO prove that integration-by-parts works with the spatial projected covariant derivative)
$=\alpha (-\sqrt{\gamma }{R}^{\perp }+\frac{1}{\sqrt{\gamma }}({\pi }_{ij}{\pi }^{ij}-\frac{1}{2}{\pi }^2))-2{\beta }^i{\gamma }_{ij}{\mathrm{\nabla }}_k^{\perp }{\pi }^{jk}$

Somewhere in the middle of this we had:
$\mathcal{H}=-\sqrt{\gamma }(\alpha (R^{\perp }-K_{ij}{K}^{ij}+K^2)-2{\pi }^{ij}\frac{1}{\sqrt{\gamma }}{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)})$
...TODO should look like $\mathcal{H}=-2\sqrt{\gamma }(\alpha \mathcal{H}+{\beta }_i{\mathcal{M}}^i)$
...TODO that means defining earlier ... $\mathcal{H}=R^{\perp }-K_{ij}{K}^{ij}+K^2$ ... but next we define it as the densitized curvature ... which is it?

Constraints:
Let $\mathcal{H}=-\sqrt{\gamma }{R}^{\perp }+\frac{1}{\sqrt{\gamma }}({\pi }_{ij}{\pi }^{ij}-\frac{1}{2}{\pi }^2)$
So $\mathcal{H}=-\sqrt{\gamma }(R^{\perp }+K^2-{K^i}_j{K^j}_i)$
So $\mathcal{H}=-\sqrt{\gamma }(R^{\perp }+{ϵ}_{ijm}{ϵ}^{klm}{K^i}_k{K^j}_l)$

Let ${\mathcal{M}}^i=2{\mathrm{\nabla }}_j^{\perp }{\pi }^{ij}$
$\mathcal{H}=\alpha \mathcal{H}-{\beta }_i{\mathcal{M}}^i$
(I'm using Alcubierre's symbols from the numerical relativity community. In the LQG world this usually looks like $H=CN+C^i{N}_i$.)
Assume $\mathcal{H}=0,{\mathcal{M}}^i=0$.
(TODO Alcubierre's book, eqn 2.7.11, defines Hamiltonian as non-densitized and puts the density into this equation. Maybe I should too? Where did I get the densitized version from anyways?)
Let $\mathbf{\text{H}}=\int \mathcal{H}d{x}^3$
$S=\int \mathbf{\text{H}}dt$

Derivatives of momentum constraint:
${\beta }_i{\mathcal{M}}^i=-2{\beta }^k{\gamma }_{ki}{\mathrm{\nabla }}_j^{\perp }{\pi }^{ij}$
$=-2{\beta }_i{\mathrm{\nabla }}_j^{\perp }{\pi }^{ij}$
using integration-by-parts of spatial covariant derivative (TODO verify you can do this)
$=2{\pi }^{ij}{\mathrm{\nabla }}_i^{\perp }{\beta }_j$
$={\pi }^{ij}\cdot 2{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)}$
using ${\mathcal{L}}_{\overrightarrow{\beta }}{\gamma }_{ij}=2{\mathrm{\nabla }}_{(i}^{\perp }{\beta }_{j)}$
$={\pi }^{ij}{\mathcal{L}}_{\overrightarrow{\beta }}{\gamma }_{ij}$
using integration by parts of Lie derivative (TODO verify you can do this)
$=-{\gamma }_{ij}{\mathcal{L}}_{\overrightarrow{\beta }}{\pi }^{ij}$
so $\frac{\partial }{\partial {\gamma }_{ij}}({\beta }_k{\mathcal{M}}^k)=-{\mathcal{L}}_{\overrightarrow{\beta }}{\pi }^{ij}$
and $\frac{\partial }{\partial {\pi }^{ij}}({\beta }_k{\mathcal{M}}^k)={\mathcal{L}}_{\overrightarrow{\beta }}{\gamma }_{ij}$

Let $\{f,g\}$ represent the Poisson bracket of $f$ and $g$, defined as:
$\{f,g\}=\frac{\partial f}{\partial {\gamma }_{ij}}\frac{\partial g}{\partial {\pi }^{ij}}-\frac{\partial f}{\partial {\pi }^{ij}}\frac{\partial g}{\partial {\gamma }_{ij}}$

There is also the option to define an extended phase space: $\{{\gamma }_{ij},{\beta }_i,\alpha \}$ and $\{{\pi }^{ij},p^i,p\}$. (think of some letters for those).
The Poisson bracket can then be defined as:
$\{f,g\}=\frac{\partial f}{\partial {\gamma }_{ij}}\frac{\partial g}{\partial {\pi }^{ij}}-\frac{\partial f}{\partial {\pi }^{ij}}\frac{\partial g}{\partial {\gamma }_{ij}}+\frac{\partial f}{\partial {\beta }_i}\frac{\partial g}{\partial p^i}-\frac{\partial f}{\partial p^i}\frac{\partial g}{\partial {\beta }_i}+\frac{\partial f}{\partial \alpha }\frac{\partial g}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial g}{\partial \alpha }$.
Then choose the phase space hypersurface/subset such that $p=0,p^i=0$. I guess you could pick any values for $p$ and $p^i$ so long as they are constant, such that $\frac{\partial }{\partial p}=0$ and $\frac{\partial }{\partial p^i}=0$. This causes your Poisson bracket value to reduce to the original definition of $\{f,g\}=\frac{\partial f}{\partial {\gamma }_{ij}}\frac{\partial g}{\partial {\pi }^{ij}}-\frac{\partial f}{\partial {\pi }^{ij}}\frac{\partial g}{\partial {\gamma }_{ij}}$. (This doesn't sit right. If a coordinate variable remains constaint, this does not imply that the derivative of another variable wrt that coordinate variable will be constant.)

Poisson bracket of momentum constraint and arbitrary function:
$\{f(\pi ,\gamma ),{\beta }_k{\mathcal{M}}^k\}$
$=\frac{\partial f}{\partial {\gamma }_{ij}}\frac{\partial }{\partial {\pi }^{ij}}({\beta }_k{\mathcal{M}}^k)-\frac{\partial f}{\partial {\pi }^{ij}}\frac{\partial }{\partial {\gamma }_{ij}}({\beta }_k{\mathcal{M}}^k)$
$=\frac{\partial f}{\partial {\gamma }_{ij}}{\mathcal{L}}_{\overrightarrow{\beta }}{\gamma }_{ij}+\frac{\partial f}{\partial {\pi }^{ij}}{\mathcal{L}}_{\overrightarrow{\beta }}{\pi }^{ij}$
$={\mathcal{L}}_{\overrightarrow{\beta }}f$

Derivatives of Hamiltonian constraint:
$\frac{\partial }{\partial {\gamma }_{ij}}\mathcal{H}$
$=\frac{\partial }{\partial {\gamma }_{ij}}(-\sqrt{\gamma }{R}^{\perp }+\frac{1}{\sqrt{\gamma }}({\gamma }_{km}{\gamma }_{ln}{\pi }^{kl}{\pi }^{mn}-\frac{1}{2}{\gamma }_{kl}{\gamma }_{mn}{\pi }^{kl}{\pi }^{mn}))$
$=-(\frac{1}{2}\frac{1}{\sqrt{\gamma }}{R}^{\perp }\frac{\partial }{\partial {\gamma }_{ij}}(\gamma )+\sqrt{\gamma }\frac{\partial }{\partial {\gamma }_{ij}}(R^{\perp }))-\frac{1}{2}\frac{1}{{\gamma }^{3/2}}\frac{\partial }{\partial {\gamma }_{ij}}(\gamma )({\gamma }_{km}{\gamma }_{ln}{\pi }^{kl}{\pi }^{mn}-\frac{1}{2}{\gamma }_{kl}{\gamma }_{mn}{\pi }^{kl}{\pi }^{mn})+\frac{2}{\sqrt{\gamma }}({{\pi }^i}_k{\pi }^{kj}-\frac{1}{2}\pi {\pi }^{ij})$
$=-\frac{1}{2}\frac{1}{\sqrt{\gamma }}{R}^{\perp }\gamma {\gamma }^{ij}-\sqrt{\gamma }\frac{\partial }{\partial {\gamma }_{ij}}(R^{\perp })-\frac{1}{2}\frac{1}{{\gamma }^{3/2}}\gamma {\gamma }^{ij}({\pi }_{kl}{\pi }^{kl}-\frac{1}{2}{\pi }^2)+\frac{2}{\sqrt{\gamma }}({{\pi }^i}_k{\pi }^{kj}-\frac{1}{2}\pi {\pi }^{ij})$
$=-\frac{1}{2}\sqrt{\gamma }{R}^{\perp }{\gamma }^{ij}-\sqrt{\gamma }\frac{\partial }{\partial {\gamma }_{ij}}(R^{\perp })-\frac{1}{2}\frac{1}{\sqrt{\gamma }}{\gamma }^{ij}({\pi }_{kl}{\pi }^{kl}-\frac{1}{2}{\pi }^2)+\frac{2}{\sqrt{\gamma }}({{\pi }^i}_k{\pi }^{kj}-\frac{1}{2}\pi {\pi }^{ij})$

$\frac{\partial }{\partial {\pi }^{ij}}\mathcal{H}$
$=\frac{\partial }{\partial {\pi }^{ij}}(-\sqrt{\gamma }{R}^{\perp }+\frac{1}{\sqrt{\gamma }}({\gamma }_{km}{\gamma }_{ln}{\pi }^{kl}{\pi }^{mn}-\frac{1}{2}{\gamma }_{kl}{\gamma }_{mn}{\pi }^{kl}{\pi }^{mn}))$
$=\frac{1}{\sqrt{\gamma }}({\gamma }_{im}{\gamma }_{jn}{\pi }^{mn}+{\gamma }_{ki}{\gamma }_{lj}{\pi }^{kl}-\frac{1}{2}({\gamma }_{ij}{\gamma }_{mn}{\pi }^{mn}+{\gamma }_{kl}{\gamma }_{ij}{\pi }^{kl}))$
$=\frac{2}{\sqrt{\gamma }}({\pi }_{ij}-\frac{1}{2}{\gamma }_{ij}\pi )$

Derivatives of the Hamiltonian:
$\frac{\partial }{\partial {\gamma }_{ij}}\mathcal{H}=\frac{\partial }{\partial {\gamma }_{ij}}(\alpha \mathcal{H}+{\beta }_i{\mathcal{M}}^i)$
$=\alpha \frac{\partial }{\partial {\gamma }_{ij}}\mathcal{H}$
$=\alpha (-\frac{1}{2}\sqrt{\gamma }{R}^{\perp }{\gamma }^{ij}-\sqrt{\gamma }\frac{\partial }{\partial {\gamma }_{ij}}(R^{\perp })-\frac{1}{2}\frac{1}{\sqrt{\gamma }}{\gamma }^{ij}({\pi }_{kl}{\pi }^{kl}-\frac{1}{2}{\pi }^2)+\frac{2}{\sqrt{\gamma }}({{\pi }^i}_k{\pi }^{kj}-\frac{1}{2}\pi {\pi }^{ij}))$

$\frac{\partial }{\partial {\pi }^{ij}}\mathcal{H}=\frac{\partial }{\partial {\pi }^{ij}}(\alpha \mathcal{H}+{\beta }_i{\mathcal{M}}^i)$
$=\alpha \frac{\partial }{\partial {\pi }^{ij}}\mathcal{H}+{\beta }_i\frac{\partial }{\partial {\pi }^{ij}}{\mathcal{M}}^i$
...

Lie bracket of momentum constraint and momentum constraint:
(TODO was this $\mathcal{H}$ the Hamiltonian constraint scalar, coupled with $\alpha$, or was it $\mathcal{H}$ the Hamiltonian?).
$\{{\beta }_k{\mathcal{M}}^k,b_k{\mathcal{M}}^k\}$
$=\frac{\partial }{\partial {\gamma }_{ij}}({\beta }_k{\mathcal{M}}^k)\frac{\partial }{\partial {\pi }^{ij}}(b_l{\mathcal{M}}^l)-\frac{\partial }{\partial {\pi }^{ij}}({\beta }_k{\mathcal{M}}^k)\frac{\partial }{\partial {\gamma }_{ij}}(b_l{\mathcal{M}}^l)$
$=-{\mathcal{L}}_{\overrightarrow{\beta }}{\pi }^{ij}\frac{\partial }{\partial {\pi }^{ij}}(b_l{\mathcal{M}}^l)-{\mathcal{L}}_{\overrightarrow{\beta }}{\gamma }_{ij}\frac{\partial }{\partial {\gamma }^{ij}}(b_l{\mathcal{M}}^l)$
$=-{\mathcal{L}}_{\overrightarrow{\beta }}(b_k{\mathcal{M}}^k)$
$=[b_k{\mathcal{M}}^k,\overrightarrow{\beta }]$
TODO show this is equal to $=[\overrightarrow{b},\overrightarrow{\beta }{]}_k{\mathcal{M}}^k$
(i.e. show that ${\mathcal{L}}_{\overrightarrow{\beta }}{M}^k=-2{\mathcal{L}}_{\overrightarrow{\beta }}{\mathrm{\nabla }}_j^{\perp }{\pi }^{jk}=0$)

Lie bracket of momentum constraint and Hamiltonian constraint:
(TODO was this $\mathcal{H}$ the Hamiltonian constraint scalar, coupled with $\alpha$, or was it $\mathcal{H}$ the Hamiltonian?).
$\{{\beta }_k{\mathcal{M}}^k,\alpha \mathcal{H}\}$
$=\frac{\partial }{\partial {\gamma }_{ij}}({\beta }_k{\mathcal{M}}^k)\frac{\partial }{\partial {\pi }^{ij}}(\alpha \mathcal{H})-\frac{\partial }{\partial {\pi }^{ij}}({\beta }_k{\mathcal{M}}^k)\frac{\partial }{\partial {\gamma }_{ij}}(\alpha \mathcal{H})$
$=-{\mathcal{L}}_{\overrightarrow{\beta }}{\pi }^{ij}\frac{\partial }{\partial {\pi }^{ij}}(\alpha \mathcal{H})-{\mathcal{L}}_{\overrightarrow{\beta }}{\gamma }_{ij}\frac{\partial }{\partial {\gamma }^{ij}}(\alpha \mathcal{H})$
$=-{\mathcal{L}}_{\overrightarrow{\beta }}(\alpha \mathcal{H})$
TODO show this is equal to $-\mathcal{H}{\mathcal{L}}_{\overrightarrow{\beta }}\alpha$
i.e. show ${\mathcal{L}}_{\overrightarrow{\beta }}\mathcal{H}={\mathcal{L}}_{\overrightarrow{\beta }}(-\sqrt{\gamma }{R}^{\perp }+\frac{1}{\sqrt{\gamma }}({\pi }_{ij}{\pi }^{ij}-\frac{1}{2}{\pi }^2))=0$

Lie bracket of Hamiltonian constraint and Hamiltonian constraint:
(TODO was this $\mathcal{H}$ the Hamiltonian constraint scalar, coupled with $\alpha$, or was it $\mathcal{H}$ the Hamiltonian?).
$\{\alpha \mathcal{H},a\mathcal{H}\}$
$=\frac{\partial }{\partial {\gamma }_{ij}}(\alpha \mathcal{H})\frac{\partial }{\partial {\pi }^{ij}}(a\mathcal{H})-\frac{\partial }{\partial {\gamma }_{ij}}(a\mathcal{H})\frac{\partial }{\partial {\pi }^{ij}}(\alpha \mathcal{H})$
$=\frac{\partial }{\partial {\gamma }_{ij}}(\alpha \mathcal{H})\frac{\partial }{\partial {\pi }^{ij}}(a\mathcal{H})-\frac{\partial }{\partial {\gamma }_{ij}}(a\mathcal{H})\frac{\partial }{\partial {\pi }^{ij}}(\alpha \mathcal{H})$
TODO show this is equal to $(\alpha {\partial }^{k}a-a{\partial }^k\alpha ){\mathcal{M}}_k$

TODO:
$\dot{f}=?\frac{d}{dt}f=\{f,\mathcal{H}\}$, and how does this relate to the Lie derivative, and to the total derivative definition $\frac{d}{dt}=\frac{\partial }{\partial t}-{\mathcal{L}}_{\overrightarrow{\beta }}$ and the Lie deriative statements above?
implied by $\mathcal{H}={\mathcal{M}}^i=0$:
$\dot{\mathcal{H}}=\{\mathcal{H},\mathcal{H}\}=0$
${\dot{\mathcal{M}}}^i=\{{\mathcal{M}}^i,\mathcal{H}\}=0$



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