Constraints in path integral and the Lagrange multiplier I was reading some references on the slave-particle approach to the Kondo problem and Anderson model.
It is known that the slave-particle is introduced in the large Hubbard $U$ limit of the system so that the Hilbert space for the localized orbital (e.g., $f$ electron) on a single site only include: $\{ | 0 \rangle, |\uparrow \rangle, |\downarrow\rangle \}$.
The constraint $n_f \le 1$ then becomes $$b^\dagger b + \sum_{\sigma} s^\dagger_\sigma s_\sigma =1$$ (here $b$ is boson and $s_\sigma$ is fermion). In the path integral, people say the measure of the integration becomes:

$$\int D(s^\dagger,s)D(b^\dagger,b)\delta(b^\dagger b + \sum_{\sigma} s^\dagger_\sigma s_\sigma-1)$$

and a Lagrange multiplier is then introduced which replace the Dirac delta function above by: $$\int D\lambda \ e^{i\lambda(b^\dagger b + \sum_{\sigma} s^\dagger_\sigma s_\sigma-1)}.$$
Here I would like to ask how does the imaginary time dependence entered into the constraint? Because originally we just have a constraint of the particle number operators, which can be transformed into a delta function at a specific time, say, at $\tau = 0$.
 A: The constraint $\delta\left(z^{2}-1\right)$ is actually added for any time and space, rather than just $\tau=0$, thus, it need to be written as:
$$\prod_{x, \tau} \delta\left(|z(x, \tau)|^{2}-1\right)$$
since
$$\delta\left(|z(x, \tau)|^{2}-1\right)=\int_{-\infty+i c}^{+\infty+i c} d \lambda(x, \tau) e^{-i \lambda(\tau, x)\left(|z(x, \tau)|^{2}-1\right)}$$
thus,
$$\prod_{x, \tau} \delta\left(|z(x, \tau)|^{2}-1\right)=\prod_{x, \tau}\left(\int_{-\infty+i c}^{+\infty+i c} d \lambda(x, \tau) e^{-i \lambda\left(|z(x, \tau)|^{2}-1\right)}\right)=\int \mathcal{D}[\lambda] e^{-i \int d^{D} x \lambda(x, \tau)\left(|z(x, \tau)|^{2}-1\right)}$$
where
$$\int \mathcal{D}[\lambda]=\prod_{x, \tau} \int_{-\infty+i c}^{+\infty+i c} d \lambda(x, \tau)$$
Now, imaginary time dependence entered into the constraint $\lambda$. CH.8.1 of Polyakov also gives this detail.
A: The delta function can be defined in the following way:
$$2 \pi \delta(x)=\int e^{i \lambda x }d\lambda$$
Now, we can represent the path integral as a multidimensional integral and integrate it using the expression above.
A: Yes. You need a field $\lambda(x)$ in @Alexandro Nikolaenko's answer above. One lambda for each site or point $x$.
$$
\int d[\lambda(x)]e^{\int dx \lambda(x)(b^\dagger(x) b(x)+...)} 
$$
