How closely should the energy components of ab initio SCF results at a stationary nuclear configuration preserve the virial theorem?
I hope I am not mistaken about this, but if I'm wrong then I'll learn so no big deal.

I believe this is a result of the fact that the electrons do not experience a true $r^{-1}$ potential due to electron-electron repulsion. Namely, we should expect that the electrons will repel each other which will cause the apparent attraction to the nuclei to be weaker than $r^{-1}$.

In other words, for a general potential,
$$
V(r)=r^{n}
$$
we have,
$$
\frac{dV(r)}{dr}=nr^{n-1}
$$

Because the Virial theorem can be stated as,
$$
\langle T\rangle=\frac12\langle\vec{r}\cdot\nabla V\rangle
$$
But for a potential which is less than coulombic, we have,
$$
\langle T\rangle=-\frac{n}{2}\langle V\rangle=-nE_{tot}
$$

That is to say, the electrons in water are feeling an effective potential of the form $V(r)=r^{-.99787}$ or so.

Or, in the language of the book you linked, the trial wavefunction of the form being tried for the hydrogen atom does not evaluate to a parameter $\zeta=1$.

Again, I am not completely confident that I nailed this answer, but for systems that have no obvious analytic form of the potential energy, there is no reason I see that we have should have $\langle T\rangle=-E_{tot}$ be exactly true. That is not to say the virial theorem is not satisfied, it certainly is, only that it is not as clean as we are used to.



I was a bit loose with some constants in my equations, please forgive me :)