Suppose a Lie group $K$ acts freely, properly and isometrically on a Riemannian manifold $(M,g)$, which is equipped with a $K$-invariant measure. Then $(M/K,\check{g})$ is a Riemannian manifold, where the metric $\check{g}$ and the measure descend from the metric and measure on $M$ in the natural way.
Question: Can one interpret a bounded linear operator $T$ on $L^2(M/K)$ as a bounded linear operator $\tilde{T}$ on $L^2(M)$ that is invariant under the action of $K$?
Thoughts: What I mean by being invariant under the action of $K$ is that the action of $T$ on $L^2(M)$ commutes with action of $k$ on $L^2(M)$ (defined by $kf(x):=f(k^{-1}x)$) for every $k\in K$. One can check that a $K$-invariant operator on $L^2(M)$ always descends to an operator on $L^2(M/K)$. I wonder if there is a reference that addresses the issue of lifting an operator the other way.