= Canonical momentum of the spatial metric
{title2=$\pi^{ij}=\sqrt h(K^{ij}-Kh^{ij})$}
With the convention $K_{ij}=(\dot h_{ij}-D_iN_j-D_jN_i)/(2N)$ for <extrinsic curvature>, differentiation of the gravitational Lagrangian gives $\pi^{ij}=\sqrt h(K^{ij}-Kh^{ij})$ in units $16\pi G=1$. It is a weight-one <tensor density>. In three spatial dimensions, $\pi=-2\sqrt hK$, so
$$
K^{ij}=h^{-1/2}(\pi^{ij}-\tfrac12\pi h^{ij}).
$$
The <lapse function> and <shift vector> have no time derivatives in this action and consequently have vanishing conjugate momenta. Changing the sign convention for extrinsic curvature changes the corresponding momentum relation; it must not be silently combined with this formula.
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