In modern chemical theory we deal with Chemical Equilibria. In the case of the decomposition of water:
$$\text{H}_2\text{O}(g) \leftrightarrows \text{H}_2(g)+\frac12 \text{O}_2(g)$$
We define an Equilibrium Constant $K$:
$$K=\frac{[\text{H}_2]\times[\text{O}_2]}{[\text{H}_2\text{O}]}$$
where the bracketed quantities are concentrations.
Nernst teaches us that:
$$\Delta G=-RT\ln K\tag{1}$$
where $\Delta G$ is the change in Gibbs Free Energy (left to right).
But from $(1)$ we glean that:
$$K=\exp{\left(-\frac{\Delta G}{RT}\right)}\tag{2}$$
This reference gives the Standard Giggs Free Energy of the Decomposition of water as:
$$\Delta G=+229\,\mathrm{kJ/mol}$$
This being $> 0$ , the value for $K$ will be $\ll 1$ and the values for $[\text{H}_2]$ and $[\text{O}_2]$ very small.
From $(2)$ can be gleaned that as $T$ increases, $K$ becomes larger (assuming $\Delta G$ in temperature invariant, which is not $100 \,\text{%}$ true) and the concentrations on the Right hand side also increase (as expected).
However, this is not a practical way of splitting water into hydrogen and oxygen because the moment you lower the temperature again, the equilibrium 'shifts back' to the left and the hydrogen and oxygen recombine to water.
The only way to avoid this recombination would be to physically separate the hydrogen and oxygen at the thermolysis high temperature. There are some well-known thermochemistry systems that rely on that method but thermolysis of water just isn't one of them.
