Basic setup of 3+1 formalism of General Relativity

[Thanks to latex2wp by Luca Trevisan, this post can be formatted quite smoothly. It is likely that I will also use this tool in future posts.]

Topic for each post in this blog is intended to be only at the textbook level, i.e. not the research level. However, I might choose to present the materials from my own point of view, style, preference, etc. So the presentation does not necessarily follow any particular textbook. I will try to give motivations on why some symbols are defined, some methods are used, etc.

Some posts might be written in Thai (perhaps after I figure out how to apply latex2wp to Thai language), some in English.

Current version v3: added section 9. This is the final version of this post (for now), though there might still be some sign inconsistencies, etc. I might be able to spot these after a careful check or after gaining more insights on the big picture. In that case, I will come back and make corections.

1. Introduction

General relativity or even special relativity teaches us that time and space are on equal footing. This leads to many nice, short, powerful formula. However, when we want to describe important physical quantities and phenomena, it is inevitable to study the dynamics of spacetime. In such study, the role of time and space are separated.

From the point of view of dynamics, the states of the system are changed or evolved in time. In other words, state is a function of time. For example, the motion of an object is described by the position ${(x,y,z)}$ of the object at each instance of time. That is, in order to completely describe the motion of an object, one needs to know three numbers at each instance of time. If the system consists of many, say ${n,}$ objects, the state of the system at each instance of time is described collectively by the position of each object. This means that at each instance of time, one needs to describe three numbers for each object, i.e. ${(x,y,z)_1, (x,y,z)_2, \cdots,(x,y,z)_n.}$ In this case, the state has an extra label which gives a name to each object (i.e. call them unimaginatively as object ${1,}$ object ${2,}$ ${\cdots}$ object ${n}$).

Let us now turn to a system whose state is described by a field, which is a function of spacetime. For simplicity, let us consider a scalar function ${\phi(t,x,y,z).}$ So at an instance, say ${t=t_0,}$ the state is described by ${\phi(t_0,x,y,z).}$ this means that we need to know a number for each ${(x,y,z).}$ So the spatial coordinates ${(x,y,z)}$ label parts of the system.

So we have learnt that when describing dynamics of a field, time coordinate takes a role of an evolution parameter whereas spatial coordinates are essential to describe the system at each instance. This is also the case in the description of the dynamics of spacetime. So we need to break down the nice formula given by general relativity by separating time and space.

More importantly, the intuitive picture where spacetime is a four-dimensional manifold (roughly speaking we need ${4}$ numbers to describe each point in spacetime) is no longer useful when describing the dynamics of general relativity. Instead, one needs to slice the spacetime to hypersurfaces (as a rough intuitive picture, think of a loaf of bread as representing the spacetime. Each slice will represent each hypersurface.) of constant time. This is the basic idea of the setup of 3+1 formalism of general relativity.

This post aims to describe setup and basic calculations of 3+1 formalism of general relativity. I will be using index-free notation as much as possible when writing this post. I found that the advantage of index-free notation over index notation is that it allows us to directly relate each symbol with the corresponding geometrical object instead of, as usually the case of index notation, viewing symbols as representing arrays of numbers.

For background reading on index-free notation, the first chapter of Advanced General Relativity by Winitzki is a good place to start.

Some prerequisites for this post: Set, Mapping, Manifold, Pull-back, Push-forward, Chain rule.

2. Hypersurface of constant time

Let ${M}$ be a 4-dimensional manifold, and let ${\Sigma}$ be a hypersurface at constant time’.

What do we actually mean by constant time? Roughly speaking, if ${M}$ has coordinates ${(x^\mu)=(t,x,y,z),}$ [NB: To be more precise, ${(x^\mu)}$ is only defined on an open subset of ${M.}$ But let us ignore this and various other precise details in this post.] then ${\Sigma}$ corresponds to some constant value of time ${t=const.}$ and has coordinates ${(x,y,z).}$ To be more precise, let us adopt the viewpoint that whenever we mention coordinates, we always mean coordinate map. So

$\displaystyle x^\mu:M\rightarrow\mathbb{R} \ \ \ \ \ (1)$

for ${\mu = 0,1,2,3.}$ Next, the hypersurface ${\Sigma}$ can be thought of as an embedding [NB: I am not being mathematically careful with the word embedding’. I am not sure if this would coincide with the exact mathematical term.] in ${M}$ by using a map

$\displaystyle \phi:\Sigma\rightarrow M. \ \ \ \ \ (2)$

Let ${(X^i), i=1,2,3}$ be coordinate map of ${\Sigma.}$ The embedding ${\phi}$ is chosen in such a way that

$\displaystyle X^i=\phi^*x^i = x^i\circ\phi, \ \ \ \ \ (3)$

where ${\phi^*}$ denotes the pull-back. It is easy to check that indeed ${X^i:\Sigma\rightarrow\mathbb{R}.}$ Actually, this is still not enough to say that ${\Sigma}$ is a hypersurface at a constant time. To be precise, we require that

$\displaystyle \phi^*t:\Sigma\rightarrow\mathbb{R} \ \ \ \ \ (4)$

is a constant map, i.e. for every ${\sigma\in\Sigma}$ we have ${\phi^*t(\sigma) = c,}$ where ${c}$ is a specific number.

3. Induced metric on hypersurface

Having defined the embedding ${\phi,}$ we can use it to get the induced the metric on ${\Sigma}$ from the metric ${g}$ on ${M.}$ In order to do so, let us first examine how coordinate basis of vector on ${\Sigma}$ transform under the push-forward map ${\phi_*.}$ For any function ${f:M\rightarrow\mathbb{R},}$ consider

$\displaystyle \begin{array}{rl} \displaystyle\left(\phi_*\left(\frac{\partial}{\partial X^i}\right)\right)f &=\displaystyle\left(\frac{\partial}{\partial X^i}\right)(f\circ\phi)\\ &\\ &=\displaystyle\left(\frac{\partial}{\partial X^i}\right)(f\circ x^{-1}\circ x\circ \phi)\\ &\\ &=\displaystyle\left(\frac{\partial f}{\partial x^\mu}\right)\frac{\partial}{\partial X^i}(x^\mu\circ\phi)\\ &\\ &=\displaystyle\left(\frac{\partial}{\partial x^i}\right)f. \end{array} \ \ \ \ \ (5)$

This implies that

$\displaystyle \phi_*\left(\frac{\partial}{\partial X^i}\right) =\frac{\partial}{\partial x^i}. \ \ \ \ \ (6)$

We are now ready to compute the induced metric ${\gamma\equiv\phi^*g}$ on ${\Sigma:}$

$\displaystyle \begin{array}{rl} \gamma_{ij} &=(\phi^*g)\left(\dfrac{\partial}{\partial X^i},\dfrac{\partial}{\partial X^j}\right)\\ &\\ &=g\left(\phi_*\dfrac{\partial}{\partial X^i},\phi_*\dfrac{\partial}{\partial X^j}\right)\\ &\\ &=g\left(\dfrac{\partial}{\partial x^i},\dfrac{\partial}{\partial x^j}\right)\\ &\\ &=g_{ij}. \end{array} \ \ \ \ \ (7)$

4. Lapse and shifts

The unit normal to hypersurface ${\Sigma}$ is given by the 1-form

$\displaystyle n = Ndt, \ \ \ \ \ (8)$

where ${N}$ is a function, called lapse function. In order to see that ${n}$ is really a normal to the hypersurface, we can check that ${n}$ annihilates every vector on the hypersurface (roughly speaking, this condition says that dot product between ${n}$ and any vector on the hypersurface is zero). That is

$\displaystyle \begin{array}{rl} n\left(\phi_*\left(\dfrac{\partial}{\partial X^i}\right)\right) &=Ndt\left(\phi^*\dfrac{\partial}{\partial X^i}\right)\\ &\\ &=Ndt\left(\dfrac{\partial}{\partial x^i}\right)\\ &\\ &=0. \end{array} \ \ \ \ \ (9)$

Let us take the signature of spacetime manifold ${M}$ to be ${(-,+,+,+),}$ and suppose that ${g^{00} <0.}$ Then in order for ${n}$ to be a unit 1-form, it has to satisfy ${g^{-1}(n,n) = -1,}$ and hence

$\displaystyle g^{00} = -\frac{1}{N^2}. \ \ \ \ \ (10)$

Next, one can define the projection operator projecting 1-forms to the direction of ${n.}$ It is given by

$\displaystyle P = -g^{-1}(n)\otimes n, \ \ \ \ \ (11)$

where ${g^{-1}}$ maps a 1-form field on ${M}$ to a vector field on ${M.}$ For details on usage of ${g^{-1}}$ and other related notations see for example the first chapter of Advanced General Relativity by Winitzki. It can easily be checked that ${P^2 = P,}$ and ${\textrm{tr} P = 1.}$ The projector which projects 1-forms to the hypersurface is then given by

$\displaystyle \Pi = 1-P, \ \ \ \ \ (12)$

where ${1}$ is the identity map.

In general, the projection of ${\partial_t}$ is non-vanishing. Intuitively, this means that ${\partial_t}$ does not necessarily align with the normal ${n = Ndt}$ to the hypersurface. The mismatch can be given as a linear combination of (push-forward of) the basis vectors, eq.(6), on the hypersurface and hence one can write

$\displaystyle \Pi\partial_t = N^i\partial_i, \ \ \ \ \ (13)$

where ${N^i}$ are called shift vectors.

Let us compute

$\displaystyle \begin{array}{rl} P\partial_t &=-g^{-1}(n)n(\partial_t)\\ &\\ &=-Ng^{-1}(n)\\ &\\ &=-N^2g^{-1}(dt)\\ &\\ &=-N^2 g^{0\mu}\partial_\mu\\ &\\ &=-N^2 g^{00}\partial_t - N^2 g^{0i}\partial_i\\ &\\ &=\partial_t - N^2 g^{0i}\partial_i, \end{array} \ \ \ \ \ (14)$

where ${\partial_\mu = \partial/\partial x^\mu.}$ Therefore

$\displaystyle \partial_t = P\partial_t + N^2 g^{0i}\partial_i. \ \ \ \ \ (15)$

By comparison with the identity

$\displaystyle \partial_t = P\partial_t + \Pi\partial_t, \ \ \ \ \ (16)$

it can easily be seen that

$\displaystyle g^{0i} = \frac{N^i}{N^2}. \ \ \ \ \ (17)$

5. Decomposition of spacetime metric

In the previous sections, we have shown how to write some components of metric or inverse metric in terms of the information relating to the hypersurface. The results are

$\displaystyle g^{00} = -\frac{1}{N^2},\qquad g^{0i} = \frac{N^i}{N^2},\qquad g_{ij} = \gamma_{ij} \ \ \ \ \ (18)$

In order to obtain the complete decomposition, let us make repeated use of ${g_{\mu\nu}g^{\nu\rho} = \delta_\mu^\rho.}$ This gives, after some algebra [which is left as a simple exercise to the readers],

$\displaystyle g = -N^2 dt^2 + \gamma_{ij}(N^i dt + dx^i)(N^j dt + dx^j), \ \ \ \ \ (19)$

and

$\displaystyle g^{-1} = -\frac{1}{N^2}\left(\partial_t - N^i\partial_i\right)\otimes\left(\partial_t - N^j\partial_j\right) +\gamma^{ij}\partial_i\otimes\partial_j, \ \ \ \ \ (20)$

where ${\gamma^{ij}}$ is the matrix inverse of ${\gamma_{ij}.}$

6. Alternative lapse and shift

There is an interesting idea raised during one of the discussions in Tah Poe School 4. Thanks to Apimook Watcharangkool, Khamphee Karwan, Pitayuth Wongjun, and others for raising the question, encouragement, and discussion. Basically, one which to investigate what happens if one swaps the role of ${\partial_t}$ and ${dt.}$ I will attempt to investigate this issue in this section. I might also come back or writing a new post after I gain more insights into this issue.

The geometrical meaning of ${dt}$ is as the normal to the hypersurface ${\Sigma}$ at constant time, whereas the geometrical meaning of ${\partial_t}$ is as a vector field whose integral curve is parametrised by time coordinate. In the orthodox point of view, the lapse function is defined based on ${dt.}$ So it is natural to talk about the projection ${P}$ along the normal of ${\Sigma}$ (i.e. along ${dt}$) and the projection ${\Pi}$ to the hypersurface ${\Sigma}$. Now if one defines the lapse function based on ${\partial_t,}$ it is only natural to talk about the projection along ${\partial_t.}$ However it is not geometrically clear (at least not to me) what it means by the projection perpendicular to ${\partial_t,}$ i.e. how to describe the hypersurface perpendicular to ${\partial_t.}$

Putting geometrical consideration aside, let us formally work out the decomposition. In this way, one may define alternative lapse function ${W}$ and alternative shift vectors ${W_i}$ so that

$\displaystyle g_{00} = -\dfrac{1}{W^2},\qquad g_{0i} = \dfrac{W_i}{W^2}. \ \ \ \ \ (21)$

Combined with ${g_{ij} = \gamma_{ij},}$ we have now decomposed the spacetime metric. In order to get the decomposition of the metric inverse, let us make a repeated use of ${g_{\mu\nu}g^{\nu\rho} = \delta_\mu^\rho.}$ This gives

$\displaystyle g^{00} = -\dfrac{W^2}{1+\dfrac{W_j W^j}{W^2}},\qquad g^{0i} = \dfrac{W^i}{1+\dfrac{W_j W^j}{W^2}},\qquad g^{ij} = \gamma^{ij} - \dfrac{\dfrac{W^i W^j}{W^2}}{1+\dfrac{W_k W^k}{W^2}}. \ \ \ \ \ (22)$

We see that the decomposition looks more complicated than its counterpart in eq.(20). So it is quite likely that subsequent calculations which make use of the decomposition will be more complicated. Further investigation will need to be carried out to see whether the decomposition based on the alternative definition of lapse and shifts would lead to any issue.

7. Frame fields

7.1. Frame fields as unit vector fields: a basic example

Undergraduate physics students in Thailand should have all seen usage of frame fields. In fact, they encountered frame fields even before coordinate basis. Frame fields that are introduced to them in the form of unit vectors [It looks to me that sometimes physicists do not distinguish tensors from tensor fields. So to be more precise, what I meant by unit vectors’ is in fact unit vector fields’.] corresponding to each coordinate.

As an illustration, consider polar coordinates ${(r,\theta)}$ in ${\mathbb{R}^2.}$ They are related to Cartesian coordinates as

$\displaystyle x=r\cos\theta,\qquad y=r\sin\theta. \ \ \ \ \ (23)$

[Note that in the above equations, we can also view ${x,y,r,\theta}$ all as functions on ${\mathbb{R}^2.}$ That is they all are maps (I am not being careful with the domain) ${\mathbb{R}^2\rightarrow\mathbb{R}.}$ The RHS of each of the above equation are understood as based on function multiplications: for ${f,g}$ being any map ${\mathbb{R}^2\rightarrow\mathbb{R}}$, and any ${p\in\mathbb{R}^2,}$ function multiplication is given by ${(fg)(p) = f(p)g(p).}$ ]

In the Cartesian coordinates, unit vectors along ${x-}$ and ${y-}$ coordinates are respectively called as ${\hat{i}, \hat{j}.}$ For polar coordinates unit vectors along ${r-}$ and ${\theta-}$ coordinates are respectively called as ${\hat{r}, \hat{\theta}.}$ It can easily be seen simply by drawing that the two sets of unit vectors are related by

$\displaystyle \hat{r} =\cos\theta\hat{i}+\sin\theta\hat{j},\qquad \hat{\theta} =-\sin\theta\hat{i}+\cos\theta\hat{j}. \ \ \ \ \ (24)$

[One may view ${\hat{r},\hat{\theta},\hat{i},\hat{j}}$ as being vector fields, and then makes use of the fact that vector fields form a module over scalar function to realise that the product between a scalar field and a vector field, e.g. ${\cos\theta\hat{i},}$ returns a vector field.]

The figure below [generated from GNU Octave] illustrates the frame field made of ${\hat r}$ and ${\hat \theta.}$

So we see from the example that a frame field assigns a frame (two unit vectors ${\hat r, \hat\theta}$ which are orthogonal to each other) at each point on ${\mathbb R^2}$ (to be more precise, on ${\mathbb R^2/ \{0\}}$). This can easily be extended to other cases, and especially to general relativity, in which a frame field assigns four unit vectors at each point on spacetime.

7.2. Frame field for 3+1 formalism

For a given manifold, one has a freedom to choose a frame field that one likes. For our purpose, we would like to choose a frame field which is suitable for 3+1 formalism. Recall that we already have a unit vector field associated to the unit one-form field ${n.}$ Let us call this unit vector field as ${u\equiv-g^{-1}n.}$ The fact that it is a unit vector field is reflected by

$\displaystyle \begin{array}{rl} g(u,u) &=g(g^{-1}n,g^{-1}n)\\ &=g^{-1}(n,n)\\ &=-1. \end{array} \ \ \ \ \ (25)$

So we need three more unit vector fields which are orthogonal to each other and are orthogonal to ${u.}$ Let us call them as ${e_I; I=1,2,3.}$

Suppose that we have managed to pick the desired frame field. In this case, the metric inverse will take the form

$\displaystyle g^{-1} = -u\otimes u + \delta^{IJ}e_I\otimes e_J, \ \ \ \ \ (26)$

i.e. ${g^{-1}}$ is diagonalised in the basis ${(u,e_I)}$ [NB: I still do not understand why frame fields diagonalise metric inverse. After all, the definition only requires that vectors in the frame field are orthonormal. Furthermore, if metric is diagonalised, then we should be able to talk about eigenvalue equation. But what is the corresponding eigenvalue equation?]. By comparing with eq.(20), we learn that

$\displaystyle u = \dfrac{1}{N}\partial_t - \dfrac{N^i}{N}\partial_i,\qquad e_I = V_I^i\partial_i, \ \ \ \ \ (27)$

where ${V_I^i}$ are chosen such that ${\gamma^{ij} = \delta^{IJ}V_I^i V_J^j.}$ In fact, the choice’ of ${u}$ is consistent with ${n=Ndt}$ as can easily be seen from the simple manipulation

$\displaystyle \begin{array}{rl} u &=-g^{-1}n\\ &=-Ng^{-1}dt\\ &=-N(g^{00}\partial_t + g^{0i}\partial_i)\\ &\\ &=-N\left(-\dfrac{1}{N^2}\partial_t + \dfrac{N^i}{N^2}\partial_i\right)\\ &\\ &=\dfrac{1}{N}\partial_t - \dfrac{N^i}{N}\partial_i. \end{array} \ \ \ \ \ (28)$

Having chosen a suitable frame field, let us also choose the corresponding coframe field to be ${(n,\sigma^I)}$ such that

$\displaystyle n(u) = 1,\qquad n(e_J) = 0,\qquad \sigma^I(u) = 0,\qquad \sigma^I(e_J) = \delta^I_J. \ \ \ \ \ (29)$

Note that ${n}$ is already chosen correctly by construction. As for ${\sigma^I,}$ it is given in the form

$\displaystyle \sigma^I = \Lambda^I_i\left(N^i dt + dx^i\right), \ \ \ \ \ (30)$

where ${\Lambda^I}$ satisfies ${ V^i_I\Lambda^J_i = \delta^J_I, }$ and ${ V^i_I\Lambda^I_j = \delta^i_j. }$ The derivation of eq.(30) is left as a simple exercise [Hint: write ${\sigma^I = \sigma_\mu^I dx^\mu.}$ Then use eq.(29) and ${dx^\mu(\partial_\nu) = \delta^\mu_\nu}$]. Next, it can easily be seen that the metric from eq.(19) can be written using coframe field as

$\displaystyle g = -n\otimes n + \delta_{IJ}\sigma^I\otimes\sigma^J. \ \ \ \ \ (31)$

It is useful to give more comments on properties of ${\Lambda^I}$ and ${V_I.}$ The indices ${I,J,\cdots}$ are raised/lowered by ${\delta^{IJ}}$ and ${\delta_{IJ}.}$ So ${\delta_{JI}\Lambda^I\equiv\Lambda_J, \delta^{JI}V_I\equiv V^J.}$ Note that the types of tensors do not change. However, indices ${i,j,\cdots}$ are raised/lowered by ${\gamma^{ij}}$ and ${\gamma_{ij}.}$ This operation swaps the the role of ${\Lambda}$ and ${V.}$ That is ${\gamma_{ji}V_I^i =\Lambda_{jI}, \gamma^{ji}\Lambda^I_i =V^{jI}.}$

8. Einstein-Hilbert action in 3+1 formalism

8.1. Rewriting 4d Einstein-Hilbert action

In general relativity, Einstein-Hilbert action is given, up to an overall constant, by

$\displaystyle S = \int d^4 x\sqrt{-g}R, \ \ \ \ \ (32)$

where ${R}$ is Ricci scalar. We want to put the action in the 3+1 formalism. This can be done by separating vector basis ${(\partial_\mu)}$ as ${(\partial_t,\partial_i),}$ and one-form basis ${(dx^\mu)}$ as ${(dt,dx^i).}$ Although it is not necessary, let us make use of frame field in order to obtain the result.

We start from collecting the frame field as ${(e_{\hat{I}}) \equiv (u,e_I),}$ and collecting the coframe field as ${(\sigma^{\hat{I}}) \equiv (n,\sigma^I),}$ where ${\hat{I},\hat{J},\cdots = 0,1,2,3.}$ These indices are raised/lowered by ${\eta^{\hat{I}\hat{J}},\eta_{\hat{I}\hat{J}}.}$ Spin connections ${\omega^{\hat{I}}{}_{\hat{J}}}$ are one-forms such that

$\displaystyle \nabla_{X}e_{\hat{J}} =-\omega_{\hat{J}}{}^{\hat{K}}(X)e_{\hat{K}}, \ \ \ \ \ (33)$

for any vector field ${X.}$ The spin connections satisfy

$\displaystyle \omega^{\hat{I}}{}_{\hat{J}} = -\omega_{\hat{J}}{}^{\hat{I}}. \ \ \ \ \ (34)$

Next, we quote some useful equations.

Consider

$\displaystyle \begin{array}{rl} g({\nabla}_{e_{\hat{I}}}e_{{\hat{J}}},e_{\hat{K}}) &= {\nabla}_{e_{\hat{I}}}(g(e_{\hat{J}},e_{\hat{K}})) - ({\nabla}_{e_{\hat{I}}}g)(e_{\hat{J}},e_{\hat{K}})-g(e_{\hat{J}},{\nabla}_{e_{\hat{I}}}e_{\hat{K}})\\ &= {\nabla}_{e_{\hat{I}}}({\eta}_{{\hat{J}}{\hat{K}}}) - 0-g(e_{\hat{J}},{\nabla}_{e_{\hat{I}}}e_{\hat{K}})\\ &= 0 - 0-g(e_{\hat{J}},{\nabla}_{e_{\hat{I}}}e_{\hat{K}})\\ &= -g(e_{\hat{J}},[e_{\hat{I}},e_{\hat{K}}]+{\nabla}_{e_{\hat{K}}}e_{\hat{I}})\\ &= -g(e_{\hat{J}},[e_{\hat{I}},e_{\hat{K}}])-g(e_{\hat{J}},{\nabla}_{e_{\hat{K}}}e_{\hat{I}}), \end{array} \ \ \ \ \ (41)$

where we used Leibniz property of covariant derivative, orthonormality of frame field and torsionless property. Using eq.(33) and eq.(37) gives

$\displaystyle -\omega_{{\hat{J}}{\hat{K}}{\hat{I}}} =-C_{{\hat{J}}{\hat{I}}{\hat{K}}}+\omega_{{\hat{I}}{\hat{J}}{\hat{K}}}. \ \ \ \ \ (42)$

By cyclic permutation, we obtain

$\displaystyle -\omega_{{\hat{K}}{\hat{I}}{\hat{J}}} =-C_{{\hat{K}}{\hat{J}}{\hat{I}}}+\omega_{{\hat{J}}{\hat{K}}{\hat{I}}}, \ \ \ \ \ (43)$

$\displaystyle -\omega_{{\hat{I}}{\hat{J}}{\hat{K}}} =-C_{{\hat{I}}{\hat{K}}{\hat{J}}}+\omega_{{\hat{K}}{\hat{I}}{\hat{J}}}. \ \ \ \ \ (44)$

By following the manipulation (42)+(43)(44), one obtains

$\displaystyle \omega_{{\hat{I}}{\hat{J}}{\hat{K}}} ={\dfrac{1}{2}}\left(C_{{\hat{I}}{\hat{K}}{\hat{J}}}-C_{{\hat{J}}{\hat{K}}{\hat{I}}}-C_{{\hat{K}}{\hat{J}}{\hat{I}}}\right). \ \ \ \ \ (45)$

Let us now compute Ricci scalar. For this, consider

$\displaystyle \begin{array}{rl} d\omega_{{\hat{I}}{\hat{J}}}(e^{{\hat{J}}},e^{{\hat{I}}}) &=e^{{\hat{J}}}({\omega}_{{\hat{I}}{\hat{J}}}(e^{{\hat{I}}})) - e^{{\hat{I}}}({\omega}_{{\hat{I}}{\hat{J}}}(e^{{\hat{J}}})) - {\omega}_{{\hat{I}}{\hat{J}}}([e^{{\hat{J}}},e^{\hat{I}}])\\ &=e^{{\hat{J}}}({\omega}_{{\hat{I}}{\hat{J}}}{}^{{\hat{I}}}) - e^{{\hat{I}}}({\omega}_{{\hat{I}}{\hat{J}}}{}^{{\hat{J}}}) - {\omega}_{{\hat{I}}{\hat{J}}}(C^{{\hat{K}}{\hat{J}}{\hat{I}}}e_{\hat{K}})\\ &=-2e_{\hat{I}}({\omega}^{\hat{I}}{}_{\hat{J}}{}^{\hat{J}}) - {\omega}_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{K}}{\hat{J}}{\hat{I}}}\\ &=2e_{\hat{I}}(C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}}) - C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{J}}{\hat{K}}{\hat{I}}} + {\dfrac{1}{2}} C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{I}}{\hat{J}}{\hat{K}}}, \end{array} \ \ \ \ \ (46)$

and

$\displaystyle \begin{array}{rl} ({\omega}_{{\hat{I}}{\hat{K}}}{\wedge}{\omega}^{\hat{K}}{}_{\hat{J}})(e^{\hat{J}},e^{\hat{I}}) &={\omega}_{{\hat{I}}{\hat{K}}}{}^{\hat{J}}{\omega}^{\hat{K}}{}_{\hat{J}}{}^{\hat{I}}-{\omega}^{\hat{K}}{}_{\hat{J}}{}^{\hat{J}}{\omega}_{{\hat{I}}{\hat{K}}}{}^{\hat{I}}\\ &={\omega}_{{\hat{I}}{\hat{J}}{\hat{K}}}{\omega}^{{\hat{J}}{\hat{K}}{\hat{I}}} + C_{\hat{J}}{}^{{\hat{J}}{\hat{K}}}C_{\hat{I}}{}^{\hat{I}}{}_{\hat{K}}\\ &={\dfrac{1}{2}}{\omega}_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{K}}{\hat{J}}{\hat{I}}} + C_{\hat{J}}{}^{{\hat{J}}{\hat{K}}}C_{\hat{I}}{}^{\hat{I}}{}_{\hat{K}}\\ &={\dfrac{1}{2}} C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{J}}{\hat{K}}{\hat{I}}} - \dfrac{1}{4}C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{I}}{\hat{J}}{\hat{K}}} + C_{\hat{J}}{}^{{\hat{J}}{\hat{K}}}C_{\hat{I}}{}^{\hat{I}}{}_{{\hat{K}}}. \end{array} \ \ \ \ \ (47)$

Therefore,

$\displaystyle R =2e_{\hat{I}}(C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}}) + \dfrac{1}{4} C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{I}}{\hat{J}}{\hat{K}}} -{\dfrac{1}{2}} C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{J}}{\hat{K}}{\hat{I}}} + C_{\hat{J}}{}^{{\hat{J}}{\hat{K}}}C_{\hat{I}}{}^{\hat{I}}{}_{{\hat{K}}}, \ \ \ \ \ (48)$

and hence, Einstein-Hilbert action becomes

$\displaystyle S=\int d^4x{\cal L}, \ \ \ \ \ (49)$

where

$\displaystyle {\cal L} =2\sqrt{-g}e_{\hat{I}}(C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}}) + \dfrac{1}{4} \sqrt{-g} C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{I}}{\hat{J}}{\hat{K}}} -{\dfrac{1}{2}} \sqrt{-g}C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{J}}{\hat{K}}{\hat{I}}} + \sqrt{-g}C_{\hat{J}}{}^{{\hat{J}}{\hat{K}}}C_{\hat{I}}{}^{\hat{I}}{}_{{\hat{K}}}. \ \ \ \ \ (50)$

Consider the first term on RHS:

$\displaystyle \begin{array}{rl} 2\sqrt{-g}e_{\hat{I}}(C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}}) &=2\sqrt{-g}e_{\hat{I}}^{\mu}{\partial}_{\mu}(C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}})\\ &=tot.- 2{\partial}_{\mu}(\sqrt{-g}e_{\hat{I}}^{\mu}) C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}}. \end{array} \ \ \ \ \ (51)$

Consider

$\displaystyle \begin{array}{rl} C^{\hat{K}}{}_{{\hat{K}}{\hat{J}}} &=\sigma^{\hat{K}}([e_{\hat{K}},e_{\hat{J}}])\\ &=\sigma^{\hat{K}}_{\mu} 2e_{[{\hat{K}}|}^{\nu}{\partial}_{\nu} e_{|{\hat{J}}]}^{\sigma} dx^{\mu}({\partial}_{\sigma})\\ &=\sigma^{\hat{K}}_{\mu} 2e_{[{\hat{K}}|}^{\nu}{\partial}_{\nu} e_{|{\hat{J}}]}^{\mu}\\ &=\sigma^{\hat{K}}_{\mu} e_{{\hat{K}}}^{\nu}{\partial}_{\nu} e_{{\hat{J}}}^{\mu}-\sigma^{\hat{K}}_{\mu} e_{{\hat{J}}}^{\nu}{\partial}_{\nu} e_{{\hat{K}}}^{\mu}\\ &={\partial}_{\nu} e_{{\hat{J}}}^{\nu}+e_{{\hat{J}}}^{\nu} \dfrac{1}{\sqrt{-g}} {\partial}_{\nu} \sqrt{-g}\\ &=\dfrac{1}{\sqrt{-g}} {\partial}_{\nu} (\sqrt{-g}e_{{\hat{J}}}^{\nu}). \end{array} \ \ \ \ \ (52)$

Substituting this into eq.(51), we obtain

$\displaystyle \begin{array}{rl} 2\sqrt{-g}e_{\hat{I}}(C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}}) =tot.- 2\sqrt{-g}C^{\hat{K}}{}_{{\hat{K}}{\hat{I}}} C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}}, \end{array} \ \ \ \ \ (53)$

and hence the Lagrangian becomes

$\displaystyle {\cal L} =tot.- \sqrt{-g} C^{\hat{K}}{}_{{\hat{K}}{\hat{I}}} C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}} + \dfrac{1}{4} \sqrt{-g} C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{I}}{\hat{J}}{\hat{K}}} -{\dfrac{1}{2}} \sqrt{-g}C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{J}}{\hat{K}}{\hat{I}}}. \ \ \ \ \ (54)$

8.2. Decomposing Einstein-Hilbert action

Let us write the Einstein-Hilbert action in terms of ${3+1}$ formalism. Let us first decompose ${\sqrt{-g}.}$ Note that

$\displaystyle \begin{array}{rl} n\wedge\sigma^1\wedge\sigma^2\wedge\sigma^3 &=d^4x\ \sqrt{-g}. \end{array} \ \ \ \ \ (55)$

On the other hand,

$\displaystyle \begin{array}{rl} n\wedge\sigma^1\wedge\sigma^2\wedge\sigma^3 &=Ndt\wedge(\sqrt{\gamma}dx^1\wedge dx^2\wedge dx^3)\\ &=d^4 x\ N\sqrt{\gamma}. \end{array} \ \ \ \ \ (56)$

Therefore, ${\sqrt{-g} = N\sqrt{\gamma}.}$

Next, let us try to decompose ${R.}$ Consider pull-back of coframe field. By using eq.(3)(4), we have

$\displaystyle {\phi}^*n = 0, \ \ \ \ \ (57)$

and

$\displaystyle \begin{array}{rl} \Sigma^I &\equiv{\phi}^*{\sigma}^I\\ &={\phi}^*(N^I dt + {\Lambda}_i^I dx^i)\\ &=({\phi}^*N^I)d({\phi}^*t) + ({\phi}^*{\Lambda}_i^I) d({\phi}^* x^i)\\ &=({\phi}^*{\Lambda}_i^I) dX^i. \end{array} \ \ \ \ \ (58)$

So the induced coframe field on hypersurface is given by ${(\Sigma^I).}$ With this, one can define the corresponding frame field ${(E_I)}$ such that ${\Sigma^I(E_J) = {\delta}^I_J.}$ It is easy to see that

$\displaystyle E_I = ({\phi}^*V_I^i)\frac{{\partial}}{{\partial} X^i}. \ \ \ \ \ (59)$

Then, the closure of Lie bracket gives

$\displaystyle \ [E_I, E_J] = \tilde C^K{}_{IJ}E_K. \ \ \ \ \ (60)$

One may wish to determine ${\tilde C^K{}_{IJ}}$ in terms of ${C^{\hat{K}}{}_{{\hat{I}}{\hat{J}}}.}$ So one would try to pushforward the above equation. However, there is no pushforward on vector fields — only pushforward on vectors are allowed. So let us first evaluate the above equation at a point ${p}$ on the hypersurface. This gives

$\displaystyle [(E_I)_p, (E_J)_p] = \tilde C^K{}_{IJ}(p)(E_K)_p. \ \ \ \ \ (61)$

Then

$\displaystyle {\phi}_*([(E_I)_p, (E_J)_p]) = \tilde C^K{}_{IJ}(p){\phi}_*((E_K)_p). \ \ \ \ \ (62)$

For any vectors ${u_p, v_p}$ on the hypersurface, the identity ${{\phi}_*([u_p,v_p]) = [{\phi}_*u_p,{\phi}_*v_p]}$ applies. Let us also compute

$\displaystyle \begin{array}{rl} {\phi}_*(E_I)_p &= {\phi}_*\left(V_I^i({\phi}(p))\left(\dfrac{{\partial}}{{\partial} X^i}\right)_p\right)\\ &= V_I^i({\phi}(p)){\phi}_*\left(\dfrac{{\partial}}{{\partial} X^i}\right)_p\\ &= V_I^i({\phi}(p))\left(\dfrac{{\partial}}{{\partial} x^i}\right)_{{\phi}(p)}\\ &= (e_I)_{{\phi}(p)}. \end{array} \ \ \ \ \ (63)$

Putting everything together, one obtains

$\displaystyle \ [(e_I)_{{\phi}(p)},(e_J)_{{\phi}(p)}] =\tilde C^K{}_{IJ}(p)(e_K)_{{\phi}(p)}. \ \ \ \ \ (64)$

Let us compare with

$\displaystyle \ [e_I,e_J] = C^K{}_{IJ}e_K + C^0{}_{IJ}u. \ \ \ \ \ (65)$

But since ${e_I = V_I^i{\partial}_i,}$ the Lie bracket with themselves do not produce ${{\partial}_t.}$ So one can conclude that ${C^0{}_{IJ} = 0,}$ and that

$\displaystyle \tilde C^K{}_{IJ}(p) = C^K{}_{IJ}({\phi}(p)). \ \ \ \ \ (66)$

Therefore, eq.(61) becomes

$\displaystyle \ [E_I,E_J] = ({\phi}^*C^K{}_{IJ})E_K. \ \ \ \ \ (67)$

With this equation as a starting point, one may assume that it is possible to have definitions and identities corresponding to the ones from the previous subsection. Then eventually, one would obtain Ricci scalar on the hypersurface in the form

$\displaystyle \tilde R =2E_I({\phi}^*C_J{}^{JI}) + \dfrac{1}{4} {\phi}^*C_{IJK}{\phi}^*C^{IJK} -{\dfrac{1}{2}} {\phi}^*C_{IJK}{\phi}^*C^{JKI} + {\phi}^*C_J{}^{JK}{\phi}^*C_I{}^I{}_{K}. \ \ \ \ \ (68)$

Consider

$\displaystyle \begin{array}{rl} \tilde R(p) &=2(E_I)_p({\phi}^*C_J{}^{JI}) + \dfrac{1}{4} ({\phi}^*C_{IJK})(p)({\phi}^*C^{IJK})(p)\\ &\qquad-{\dfrac{1}{2}} ({\phi}^*C_{IJK})(p)({\phi}^*C^{JKI})(p) + ({\phi}^*C_J{}^{JK})(p)({\phi}^*C_I{}^I{}_{K})(p)\\ &=2{\phi}_*(E_I)_p(C_J{}^{JI}) + \dfrac{1}{4} C_{IJK}({\phi}(p))C^{IJK}({\phi}(p))\\ &\qquad-{\dfrac{1}{2}} C_{IJK}({\phi}(p))C^{JKI}({\phi}(p)) + C_J{}^{JK}({\phi}(p))C_I{}^I{}_{K}({\phi}(p))\\ &=2(e_I)_{{\phi}(p)}(C_J{}^{JI}) + \dfrac{1}{4} C_{IJK}({\phi}(p))C^{IJK}({\phi}(p))\\ &\qquad-{\dfrac{1}{2}} C_{IJK}({\phi}(p))C^{JKI}({\phi}(p)) + C_J{}^{JK}({\phi}(p))C_I{}^I{}_{K}({\phi}(p))\\ &\equiv ({}^{(3)}R)({\phi}(p)), \end{array} \ \ \ \ \ (69)$

where

$\displaystyle {}^{(3)}R =2e_I(C_J{}^{JI}) + \dfrac{1}{4} C_{IJK}C^{IJK} -{\dfrac{1}{2}} C_{IJK}C^{JKI} + C_J{}^{JK}C_I{}^I{}_{K}. \ \ \ \ \ (70)$

We see from its definition in eq.(69) that essentially the Ricci scalar ${\tilde R}$ on the hypersurface is the pull-back of ${{}^{(3)}R.}$ I think this statement is simply heuristic. While geometrical meaning of ${\tilde R}$ is clear, the geometrical meaning of ${{}^{(3)}R}$ is not so clear.

By comparing eq.(48) with eq.(70), we see that ${{}^{(3)}R}$ is a part of ${R.}$ So the Lagrangian would read

$\displaystyle {\cal L} = N\sqrt{{\gamma}}({}^{(3)}R +\cdots), \ \ \ \ \ (71)$

where ${\cdots}$ are to be determined. Let us rewrite ${N\sqrt{{\gamma}}{}^{(3)}R}$ by using the similar trick to the rewriting of ${\sqrt{-g}R.}$ Consider

$\displaystyle \begin{array}{rl} 2N\sqrt{{\gamma}}e_I(C_J{}^{JI}) &=tot.-{\partial}_i(2N\sqrt{{\gamma}}V_I^i)C_J{}^{JI}\\ &=tot.-2N{\partial}_i(\sqrt{{\gamma}}V_I^i)C_J{}^{JI} -2\sqrt{{\gamma}}V_I^iC_J{}^{JI}{\partial}_iN \end{array} \ \ \ \ \ (72)$

By imitating the calculation of eq.(52), one obtains

$\displaystyle {\partial}_i(\sqrt{{\gamma}}V_I^i) = \sqrt{{\gamma}}C^K{}_{KI}. \ \ \ \ \ (73)$

Consider

$\displaystyle \begin{array}{rl} C^0{}_{0I} &={\sigma}^0([e_0,e_I])\\ &=n([u,e_I])\\ &=n([-g^{-1}n,e_I])\\ &=Ndt\left(\left[\dfrac{1}{N}{\partial}_t - \frac{N^i}{N}{\partial}_i,V_I^j{\partial}_j\right]\right)\\ &=Ndt\left(-V_I^j{\partial}_j\dfrac{1}{N}{\partial}_t + \cdots\right)\\ &=\dfrac{1}{N}V^j_I{\partial}_j N, \end{array} \ \ \ \ \ (74)$

where ${\cdots}$ in penultimate step represents terms in ${{\partial}_i}$ directions. Putting everything together, eq.(72) becomes

$\displaystyle 2N\sqrt{{\gamma}}e_I(C_J{}^{JI}) =tot.-2N\sqrt{{\gamma}}C^K{}_{KI}C_J{}^{JI} -2N\sqrt{{\gamma}}C_J{}^{JI}C^0{}_{0I}. \ \ \ \ \ (75)$

So from eq.(70), we have

$\displaystyle N\sqrt{{\gamma}}{}^{(3)}R =tot.+N\sqrt{{\gamma}} \left(-C^K{}_{KI}C_J{}^{JI} -2C_J{}^{JI}C^0{}_{0I} + \dfrac{1}{4} C_{IJK}C^{IJK} -{\dfrac{1}{2}} C_{IJK}C^{JKI}\right). \ \ \ \ \ (76)$

By considering eq.(54) and eq.(76), let us compute ${{\cal L} - N\sqrt{{\gamma}}{}^{(3)}R.}$ For this, consider term-by-term:

$\displaystyle \begin{array}{rl} - N\sqrt{{\gamma}}( C^{\hat{K}}{}_{{\hat{K}}{\hat{I}}} C_{\hat{J}}{}^{{\hat{J}}{\hat{I}}}-C^K{}_{KI}C_J{}^{JI}) &=- N\sqrt{{\gamma}}( C^0{}_{0I} C_0{}^{0I}+C^K{}_{K0}C_J{}^{J0}+2C^K{}_{KI}C_0{}^{0I}), \end{array} \ \ \ \ \ (77)$

$\displaystyle \begin{array}{rl} -{\dfrac{1}{2}} N\sqrt{{\gamma}}\left(C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{J}}{\hat{K}}{\hat{I}}}-C_{IJK}C^{JKI}\right) &=-{\dfrac{1}{2}} N\sqrt{{\gamma}}\left(C_{IJ0}C^{J0I}+C_{00K}C^{0K0}\right)\\ &={\dfrac{1}{2}} N\sqrt{{\gamma}}\left(C_{IJ0}C^{JI0}+C_{00K}C^{00K}\right), \end{array} \ \ \ \ \ (78)$

$\displaystyle \begin{array}{rl} \dfrac{1}{4} N\sqrt{{\gamma}} \left(C_{{\hat{I}}{\hat{J}}{\hat{K}}}C^{{\hat{I}}{\hat{J}}{\hat{K}}}-C_{IJK}C^{IJK}\right) &=\dfrac{1}{2} N\sqrt{{\gamma}} \left(C_{IJ0}C^{IJ0}+C_{00K}C^{00K}\right). \end{array} \ \ \ \ \ (79)$

In these calculations, we made use of the fact that ${C^0{}_{IJ} = 0.}$ By combining eq.(77)(79), we obtain

$\displaystyle {\cal L} - N\sqrt{{\gamma}}{}^{(3)}R = N\sqrt{{\gamma}} \left(C^K{}_{K0}C^J{}_{J0} + C_{IJ0}C^{(IJ)0}\right). \ \ \ \ \ (80)$

Let us compute

$\displaystyle \begin{array}{rl} C_{IJ0} &={\sigma}_I([e_J,u])\\ &=-{\Lambda}_{Ii}\left(\dfrac{1}{N}V_J^j{\partial}_j N^i + u[V_J^i]\right). \end{array} \ \ \ \ \ (81)$

So

$\displaystyle \begin{array}{rl} 2C_{(IJ)0} &=-{\Lambda}_{Ii}\left(\dfrac{1}{N}V_J^j{\partial}_j N^i + u[V_J^i]\right) -{\Lambda}_{Ji}\left(\dfrac{1}{N}V_I^j{\partial}_j N^i + u[V_I^i]\right)\\ &\\ &=-{\gamma}_{ij}u[V_I^i V_J^j]-V^m_I V_J^l\dfrac{2}{N}{\partial}_{(m}N^i{\gamma}_{l)i}\\ &\\ &=V^m_I V_J^l\left(u({\gamma}_{ml})-\dfrac{2}{N}{\partial}_{(m}N^i{\gamma}_{l)i}\right)\\ &\\ &\equiv 2V^m_I V_J^l K_{ml}. \end{array} \ \ \ \ \ (82)$

This gives

$\displaystyle C_{IJ0}C^{(IJ)0} =-[K^2], \ \ \ \ \ (83)$

where the bracket ${[.]}$ stands for trace over spatial indices ${i,j,\cdots.}$ So

$\displaystyle {\cal L} = N\sqrt{{\gamma}} \left({}^{(3)}R+[K]^2 - [K^2]\right). \ \ \ \ \ (84)$

9. Hamiltonian

Let us now Legendre transform the Lagrangian to get Hamiltonian. However, instead of keeping on working with frame field, let us now switch to using metric. This means that we will write ${{\cal L}}$ in terms of ${{\gamma}_{ij}, N, N^i.}$ Note that there is only time derivative on ${{\gamma}_{ij},}$ but no time derivatives on ${N, N^i.}$ The time derivative on ${{\gamma}_{ij}}$ only appears through ${K_{ij},}$ which can be expressed as

$\displaystyle \begin{array}{rl} K_{ml} &={\dfrac{1}{2}}\left(u({\gamma}_{ml}) - \dfrac{2}{N}{\partial}_{(m}N^i{\gamma}_{l)i}\right)\\ &\\ &=\dfrac{1}{2N}\left(\dot{\gamma}_{ml} - N^i{\partial}_i{\gamma}_{ml} - 2{\partial}_{(m}N^i{\gamma}_{l)i}\right). \end{array} \ \ \ \ \ (85)$

So

$\displaystyle \begin{array}{rl} \dfrac{{\delta}{\cal L}}{{\delta}\dot{\gamma}_{ij}} &=2N\sqrt{{\gamma}}([K]{\gamma}^{ml} - K^{ml})\dfrac{{\delta} K_{ml}}{{\delta}\dot{\gamma}_{ij}}\\ &\\ &=\sqrt{{\gamma}}([K]{\gamma}^{ij} - K^{ij}). \end{array} \ \ \ \ \ (86)$

Therefore, conjugate momenta for ${{\gamma}_{ij}}$ are given by

$\displaystyle {\zeta}^{ij} \equiv \sqrt{{\gamma}}([K]{\gamma}^{ij} - K^{ij}). \ \ \ \ \ (87)$

Let us next trade ${\dot{\gamma}_{ij}}$ for ${{\zeta}^{ij}.}$ For this, we make simple observations that

$\displaystyle \ [{\zeta}]= 2\sqrt{{\gamma}}[K],\qquad [{\zeta}^2] = {\gamma}([K]^2 + [K^2]),\qquad {\zeta}^{ij}K_{ij} = \sqrt{{\gamma}}([K]^2 - [K^2]) \ \ \ \ \ (88)$

Therefore,

$\displaystyle \begin{array}{rl} {\cal H} &={\zeta}^{ij}\dot{\gamma}_{ij}-{\cal L}\\ &\\ &=2N{\zeta}^{ij}K_{ij} + {\zeta}^{ij}(N^k{\partial}_k{\gamma}_{ij} + 2{\partial}_{(i}N^k{\gamma}_{j)k}) -{\cal L}\\ &\\ &=N\sqrt{{\gamma}}([K]^2 - [K^2]) + {\zeta}^{ij}(N^k{\partial}_k{\gamma}_{ij} + 2{\partial}_{(i}N^k{\gamma}_{j)k}) -N\sqrt{{\gamma}}{}^{(3)}R\\ &\\ &=N\left(-\sqrt{{\gamma}}{}^{(3)}R+\dfrac{1}{2\sqrt{{\gamma}}}[{\zeta}]^2 - \dfrac{1}{\sqrt{{\gamma}}}[{\zeta}^2]\right) + {\zeta}^{ij}(N^k{\partial}_k{\gamma}_{ij} + 2{\partial}_{(i}N^k{\gamma}_{j)k})\\ &\\ &=N\left(-\sqrt{{\gamma}}{}^{(3)}R+\dfrac{1}{2\sqrt{{\gamma}}}[{\zeta}]^2 - \dfrac{1}{\sqrt{{\gamma}}}[{\zeta}^2]\right) + N^k({\zeta}^{ij}{\partial}_k{\gamma}_{ij} - 2{\partial}_i{\zeta}^i{}_k) + {\partial}_i(2 N^k{\zeta}^i{}_k), \end{array} \ \ \ \ \ (89)$

where indices of ${{\zeta}^{ij}}$ are raised/lowered by ${{\gamma}^{ij}, {\gamma}_{ij}.}$ It is convenient to define a covariant derivative which is compatible with ${{\gamma}_{ij}.}$ Let us note this operation by using the symbol ${|.}$ From the definition eq.(87), ${{\zeta}^{ij}}$ is not a tensor density on the hypersurface. The corresponding tensor is ${{\zeta}^{ij}/\sqrt{{\gamma}}.}$ A direct computation gives

$\displaystyle \left(\dfrac{{\zeta}^i{}_{k}}{\sqrt{{\gamma}}}\right)_{|i} ={\partial}_i\left(\dfrac{{\zeta}^i{}_k}{\sqrt{{\gamma}}}\right) + \dfrac{1}{\sqrt{{\gamma}}}{\partial}_j\sqrt{{\gamma}}\dfrac{{\zeta}^j{}_k}{\sqrt{{\gamma}}} - {\dfrac{1}{2}}\dfrac{{\zeta}^{ij}}{\sqrt{{\gamma}}}{\partial}_k{\gamma}_{ij}. \ \ \ \ \ (90)$

So

$\displaystyle {\zeta}^i{}_{k|i} ={\partial}_i{\zeta}^i{}_k - {\dfrac{1}{2}}{\zeta}^{ij}{\partial}_k{\gamma}_{ij}. \ \ \ \ \ (91)$

Therefore,

$\displaystyle \begin{array}{rl} {\cal H} &=N\left(-\sqrt{{\gamma}}{}^{(3)}R+\dfrac{1}{2\sqrt{{\gamma}}}[{\zeta}]^2 - \dfrac{1}{\sqrt{{\gamma}}}[{\zeta}^2]\right) -2 N^k {\zeta}^i{}_{k|i} + {\partial}_i(2 N^k{\zeta}^i{}_k). \end{array} \ \ \ \ \ (92)$

Let us ignore the last term which is a total derivative, and label the quantities

$\displaystyle {\cal H}_0\equiv-\sqrt{{\gamma}}{}^{(3)}R+\dfrac{1}{2\sqrt{{\gamma}}}[{\zeta}]^2 - \dfrac{1}{\sqrt{{\gamma}}}[{\zeta}^2], \ \ \ \ \ (93)$

$\displaystyle {\cal H}_k\equiv-2 {\zeta}^i{}_{k|i}. \ \ \ \ \ (94)$

Therefore,

$\displaystyle \begin{array}{rl} {\cal H} &=N{\cal H}_0 + N^k{\cal H}_k. \end{array} \ \ \ \ \ (95)$

Pure gravity theory is a constrained system. By carrying out the constrained analysis, it can be seen that ${{\cal H}_0}$ and ${{\cal H}_k}$ are first class constraints. These constraints generate diffeomorphism transformation. Furthermore, by counting the degrees of freedom from constrained analysis, pure gravity theory has ${2}$ degrees of freedom. This is related to two polarisation modes of gravitational waves: “cross mode” and “plus mode”. Details on constrained analysis of pure gravity theory will likely appear in some future posts.

In some future posts, I might also revisit some discussions in this post by using alternative point of view. This is to allow the cross-checks of the calculations and the arguments.

One thought on “Basic setup of 3+1 formalism of General Relativity”

1. Elegant!
That swapping of \partial_t and dt is an interesting idea. However, as you pointed out that \partial_t is in the directions of vectors associated with integral curves parametrised by time coordinate, in other words, directions of the time axis. The direction of time axis’ seems to have no direct relation with the hypersurfaces that foliate the spacetime, except through the normal dt. Is the perpendicular direction of the t-axis geometrically interesting?

Another interesting alternative decomposition of the metric is the NRG parametrisation developed in these papers:https://arxiv.org/abs/0712.4116 and https://arxiv.org/abs/1011.6024
Interestingly, the ADM decomposition is regarded as a KK reduction in spatial direction, while the alternative, NRG decomposition, is regarded as a temporal KK reduction and thus useful in some applications, e.g. post-Newtonian calculations.

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