Meaning and freedom of slicing conditions for 3+1 relativity In 3+1 relativity, it is often stated that there is freedom of choice in the gauge functions, being the lapse $\alpha(t,x^{i})$ and the shift $\beta^{i}(t,x^{i})$.  However, going by the definitions of the 3+1 quantities it doesn't seem so freely specifiable, at least in terms of how they change over time.
By foliation of the spacetime into spacelike hypersurfaces, we obtain the unit normal vectors to the hypersurfaces.
$$n_{\mu}=(-\alpha,0,0,0)\quad n^{\mu}=\left(\frac{1}{\alpha},\frac{-\beta^{i}}{\alpha}\right)$$
This also represents the 4-velocity of the Eulerian observers.
Next, there is the definition of the extrinsic curvature
$$K_{\mu\nu} \equiv -\left(\nabla_{\mu}n_{\nu} + n_{\mu}n^{\alpha}\nabla_{\alpha}n_{\nu}\right)$$
If one takes the trace of the curvature, one obtains
$$K = -\nabla_{\mu}n^{\mu}$$
Expanding,
$$K = -\nabla_{0}\frac{1}{\alpha} + \nabla_{i}\frac{\beta^{i}}{\alpha}$$
$$\partial_{t}\alpha = \alpha^{2}K - \alpha^{2}\nabla_{i}\frac{\beta^{i}}{\alpha}$$
But this presents an evolution equation for the lapse!
How can the slicing conditions be arbitrary if there is an evolution equation for the lapse that is derived purely from the definitions of the 3+1 quantities?  Also, for hyperbolic slicings (with zero shift) it is often given as
$$\partial_{t}\alpha = -\alpha^{2}f(\alpha)K$$
For harmonic slicing (f=1), where did the negative sign come from?
 A: I don't think the way you take the covariant derivatives is correct.
I will show that, in fact, there is no time derivative of the lapse in the trace of extrinsic curvature.
We'll be using the four-dimensional point of view, in which fields living on $\mathbb{R} \times \Sigma$ are treated as defined on the whole spacetime manifold $\mathcal{M}$.
The extensions of tensor fields on $\Sigma$ to $\mathcal{M}$ we call spatial, as their contractions with the normal vector field $\mathbf{n}$ vanish, and the action of the orthogonal projector $\gamma^{\mu}_{\;\;\nu}$ on these fields is an identity. For example, $\beta$ is defined to be a spatial vector field that quantifies the difference between normal observers and coordinate time observers:
$$ \partial_{t} = \alpha n + \beta, \\[0.5em]
 g(\beta, n) = 0.$$
From the latter, knowing that $n_{\mu}=(-\alpha, 0,0,0)$, we can derive:
$$ 0= \beta^{\mu}n_{\mu} = \beta^{t}n_{t} + \beta^{i}n_{i}= \beta^{t}n_{t}.$$
And therefore the temporal component of the shift vector is identically zero $$ \beta^{t}=0.$$
CONTENT OF THE EDIT:
Contract $K_{\mu\nu}=-\nabla_{\nu}n_{\mu} -a_{\mu}n_{\nu}$ with the inverse 4-metric:
$$ K=K_{\mu\nu}g^{\mu\nu} = -g^{\mu\nu}\nabla_{\nu}n_{\mu} = \nabla_{\mu}n^{\mu} = -\partial_{\mu}n^{\mu} - \Gamma^{\mu}_{\mu\lambda}n^{\lambda}.$$
Here $a_{\mu}n_{\nu}g^{\mu\nu}=0$, since $a_{\mu}=n^{\lambda}\nabla_{\lambda}n^{\mu}.$
Now let us introduce the notation: $g$, $\gamma$ are the determinants of the 4 and 3-metrics, respectively, satisfying the relation $\sqrt{-g}=\alpha \sqrt{\gamma}$ (ref 1, eq. 4.55) and we will use the fact that:
$$\Gamma^{\mu}_{\mu\lambda}=\frac{1}{2}g^{\mu\nu}g_{\mu\nu,\lambda}=\frac{1}{\sqrt{-g}}\frac{\partial \sqrt{-g}}{\partial x^{\lambda}}=\frac{1}{\alpha\sqrt{\gamma}}\frac{\partial (\alpha\sqrt{\gamma})}{\partial x^{\lambda}}, $$
as well as:
$$ n^{\mu}=(\frac{1}{\alpha}, \frac{-\beta^{i}}{\alpha})$$
Expanding the expression of interest:
$$-\partial_{\mu}n^{\mu} - \Gamma^{\mu}_{\mu\lambda}n^{\lambda} = -\partial_{t}n^{t} - \partial_{i}n^{i} - \frac{1}{\alpha\sqrt{\gamma}}\partial_{\lambda}(\alpha\sqrt{\gamma}) n^{\lambda} =\\   
\frac{1}{\alpha^{2}}\partial_{t}\alpha - \partial_{i}n^{i} -\frac{n^{\lambda}}{\sqrt{\gamma}}\partial_{\lambda}\sqrt{\gamma} -\frac{1}{\alpha}n^{\lambda}\partial_{\lambda}\alpha = \\
\frac{1}{\alpha^{2}}\partial_{t}\alpha - \partial_{i}n^{i} - \frac{n^{\lambda}}{\sqrt{\gamma}}\partial_{\lambda}\sqrt{\gamma} - \frac{1}{\alpha}n^{t}\partial_{t}\alpha - \frac{1}{\alpha}n^{i}\partial_{i}\alpha=\\
\frac{1}{\alpha^{2}}\partial_{t}\alpha - \partial_{i}n^{i} - \frac{n^{\lambda}}{\sqrt{\gamma}}\partial_{\lambda}\sqrt{\gamma} - \frac{1}{\alpha^{2}}\partial_{t}\alpha - \frac{1}{\alpha}n^{i}\partial_{i}\alpha$$
As you see, the time derivatives of the lapse cancel out.
Going further:
$$
 -\partial_{i}n^{i} - \frac{n^{\lambda}}{\sqrt{\gamma}}\partial_{\lambda}\sqrt{\gamma}  - \frac{1}{\alpha}n^{i}\partial_{i}\alpha =\\
 \partial_{i}(\frac{\beta^{i}}{\alpha}) - \frac{n^{t}}{\sqrt{\gamma}}\partial_{t}\sqrt{\gamma} - \frac{n^{i}}{\sqrt{\gamma}}\partial_{i}\sqrt{\gamma}  + \frac{1}{\alpha^{2}}\beta^{i}\partial_{i}\alpha=\\
 \frac{1}{\alpha}\partial_{i}\beta^{i} - \frac{1}{\alpha^{2}}\beta^{i}\partial_{i}\alpha - \frac{1}{\alpha\sqrt{\gamma}}\partial_{t}\sqrt{\gamma} + \frac{\beta^{i}}{\alpha\sqrt{\gamma}}\partial_{i}\sqrt{\gamma}  + \frac{1}{\alpha^{2}}\beta^{i}\partial_{i}\alpha=\\
\frac{-1}{\alpha}\Big{[} \frac{1}{\sqrt{\gamma}}\partial_{t}\sqrt{\gamma} - \frac{1}{\sqrt{\gamma}}\partial_{i}(\sqrt{\gamma}\beta^{i})\Big{]}=\\
\frac{-1}{\alpha}\Big{[} \frac{1}{\sqrt{\gamma}}\partial_{t}\sqrt{\gamma} - D_{i}\beta^{i} \Big{]}$$
Where the last expression involving the shift vector is just an alternative formula for the divergence of a vector field.
We can get to the expression we have just derived by using an an alternative (but equivalent) formula for the extrinsic curvature tensor.
$$ K_{ij} = \frac{-1}{2\alpha}(\mathcal{L}_{m}\gamma)_{ij}=\frac{-1}{2\alpha}\big{[}\partial_{t}\gamma_{ij} - (\mathcal{L}_{\beta}\gamma)_{ij}  \big{]}.$$
Upon contraction and using $(\mathcal{L}_{\beta}\gamma)_{ij}= D_{i}\beta_{j} + D_{j}\beta_{j}$:
$$ K=K_{ij}\gamma^{ij} = \frac{-1}{\alpha}\Big{(} \frac{1}{\sqrt{\gamma}} \frac{\partial \sqrt{\gamma}}{\partial t} - D_{i}\beta^{i}\Big{)} .$$
An easier derivation could be made from the start, based on the fact that $\mathbf{K}$ is a spatial tensor, and that $g^{\mu\nu}= \gamma^{\mu\nu} + n^{\mu}n^{\nu}$.
Contracting as before:
$$ K=K_{\mu\nu}g^{\mu\nu} = K_{\mu\nu}\gamma^{\mu\nu}=K_{ij}\gamma^{ij} = -\gamma^{ij}\nabla_{j}n_{i}$$
and once again, no time derivative of the lapse is present in the end.
In the above we use the fact that $\gamma^{t\mu}=0$ is zero, as it is a spatial tensor with an upper temporal index $\gamma^{\mu\nu}n_{\nu}=0$. Additionally, we use the fact that $n_{i}=0$ (but not the lowered-index version covariant derivative of its covariant derivative, $(\nabla_{j}n)_{i}$, mind you!
This $\gamma$ is, by the way, the extension of the 3-metric $\gamma_{ij}$ to a spatial tensor field in 4-dimensional spacetime.
As for the derivations of different slicings:

*

*to obtain maximal slicing prescription for the lapse, you should contract the evolution equation for $K_{\mu\nu}$ and obtain an evolution equation for its trace; then, you enforce for all times $t$ the vanishing of $K$, i.e. $K=0$ and $\partial_{t}K=0$, thereby obtaining an elliptic equation for the lapse (ref 1, eq. 4.63, 4.64, 6.90, 9.15); the maximal slicing prescription requires that at each time step, the elliptic equation for the lapse on the hypersurface is solved for before we evolve further


*to obtain the harmonic slicing evolution equation, we demand that the time coordinate of our foliation is a harmonic function: $\Box_{g}t=0$. (ref 1, section 9.2.3); in there, the time derivative of lapse appears in a natural way and the derivation answers your question regarding the minus sign, I think; you can generalize this prescription and obtain a family of slicings based on different choices of the $f(\alpha)$ function
The arbitrariness in either of these choices is now evident:

*

*for maximal slicing, you choose that your foliation has $K=0$ everywhere, and you use a no-longer-an-evolution-equation to fix your lapse; inverting the reasoning, you pick such a lapse that will guarantee $K=0$ for each hypersurface


*for harmonic slicing, the choice that the time coordinate of the slicing is a harmonic function is totally arbitrary
For details I refer you to the 3+1 Formalism and Bases of Numerical Relativity, specifically Chapter 9 for details on the lapse conditions.
DESCRIPTION OF THE EDIT: I owe you an apology - the answer was not correct. In fact, both parts of the full expression (the one with the Christoffel symbol as well) contribute to the absence of the temporal derivative of the lapse. From one part we get $\sim \partial_{t}\alpha$ and from the other $\sim - \partial_{t}\alpha$, so these cancel out.
In particular, I was wrong stating that $g^{t \mu} = \frac{\beta^{\mu}}{\alpha^{2}}$. This concerns the $g^{ti}$, but not $g^{tt}$. In the quoted part, you can even see that $g^{tt}=\frac{-1}{\alpha^{2}}$.
The fact that $\beta^{t}=0$ is of course true, but does not come into play in this calculation in the end. Also, one must be careful. The orthogonality of a tensor field does not mean that its temporal component vanishes. When the one of the indices of a spatial tensor field component is $t$, the component will be zero by virtue of this contraction: $T_{\;\;\;\mu\nu}^{\alpha\beta}n_{\alpha}=0$ (since $n_{i}=0$). However, for a covariant index, this is not longer the case, easily visible even in the case of $\beta$.
$$ 0 = \beta_{\mu}n^{\mu} = \beta_{t}n^{t} + \beta_{i}n^{i}$$
which leads to:
$$\beta_{t}=g_{t\mu}\beta^{\mu}=g_{ti}\beta^{i} = \beta_{i}\beta^{i}$$
