# Why no cosmological constant in momentum constraint?

In the ADM formalism of general relativity, one decomposes the Einstein equations in (3+1) dimensions. More explicitely, if the Einstein equations are given by

$$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}=0$$

then one introduces a coordinate system in which $$g$$ can be written as

$$g_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu}=(-N^{2}+X^{i}X_{i})\mathrm{d}t^{2}+2X_{i}\mathrm{d}x^{i}\mathrm{d}t+h_{ij}\mathrm{d}x^{i}\mathrm{d}x^{j}$$

and writes down the equations for $$\mu\nu=00,0i,ij$$, which yields

$$\partial_{t}k_{ij}=N\bigg[\mathrm{Ric}(h)_{ij}+\mathrm{tr}(k)k_{ij}-2k_{ik}{k^{k}}_{j}-\Lambda h_{ij}\bigg]-\nabla_{i}\nabla_{j}N+X^{k}\nabla_{k}k_{ij}+k_{ik}\nabla_{j}X^{k}+k_{kj}\nabla_{i} X^{k}$$ $$k_{ij}k^{ij}-\mathrm{tr}(k)^{2}-\mathrm{Scal}(h)=2\Lambda$$ $$\partial_{i}\mathrm{tr}(k)-\nabla_{j}{k_{i}}^{j}=0$$

where $$k_{ij}$$ is the second fundamental form determined by

$$\partial_{t} h_{ij}=-2Nk_{ij}+2\nabla_{(i}X_{j)}$$

Now, I am wondering why there is no cosmological constant in the last equation $$\partial_{i}\mathrm{tr}(k)-\nabla_{j}{k_{i}}^{j}=0$$. I mean, the Einstein equations are equivalent to $$R_{\mu\nu}=\Lambda g_{\mu\nu}$$. A straight-forward computation by using the Christoffel symbols coming from the metric above yields

$$R_{0i}=\frac{1}{N}(\partial_{i}\mathrm{tr}(k)-\nabla_{j}{k_{i}}^{j})$$

so I would expect an equation like

$$\partial_{i}\mathrm{tr}(k)-\nabla_{j}{k_{i}}^{j}=N\Lambda X_{i}$$

which however is wrong, apparently (at least in all books/papers it is without cosmological constant). What am I doing wrong?