$$ \newcommand{\R}{\mathbb{R}} \newcommand{\s}{\mathring{\Delta}^n} \newcommand{\topspace}{\R^{n+1}_{>0}} \newcommand{\P}{\mathcal{P}} \newcommand{\O}{\mathcal{O}} $$
Quotient
Let $M$ be a smooth manifold endowed with some equivalence relation and let $\Phi: M \rightarrow S $ be the associated quotient map. Assume that the quotient space $S$ is also a manifold (this is the case, for example, if the equivalence relation arises from a Lie group action on $M$ which is free and proper, by the Quotient Manifold Theorem).
For example, consider
$$ \begin{split} \Phi : \, & M = \topspace \rightarrow S = \s \\ & \mu \mapsto x = \frac{\mu}{\mu_0 + \dots + \mu_n} \end{split} $$
where $\s$ is the interior of the standard $n$-dimensional simplex $$ \s = \{ x \in \topspace : x_0 + \dots + x_n = 1 \} $$ Now if we have a vector field $X$ in $M$ we can try to find a vector field $Y$ on $S$ that is $\Phi$-related to $X$. Nor its existence nor its uniqueness are granted, but looking for it makes sense.
In the example above, we want that
$$ Y(x) = d_{\mu}\Phi(X_\mu) \quad \text{for any } \mu \in \Phi^{-1}(x) $$ This is well defined if we start from a vector field $\Phi$ that is $1$-homogeneous (namely $X(\lambda \mu) = \lambda \, X(\mu)$ for any $\lambda > 0, \mu \in \topspace$). In such case, the vector field on the quotient space $\s$ is $$ Y(x) = X(x) - x \left( \, X_0(x) + \dots + X_n(x) \, \right) $$
Similarly and more simply, if we have a smooth function or a 1-form on the quotient space that we can always pull it back along $\Phi$, obtaining rescp. a smooth function or a 1-form on $M$.
Riemannian projection
Now my question is whether these ideas can be recast in a setting in which the way to go from the ambient space to the subspace is a Riemannian projection rather than a quotient.
Endow $M$ with a Riemannian metric $g$ and assume $S$ is an orientable submanifold of $M$. Then we can always obtain a vector field $\P(X)$ on $S$ from a vector field $X$ on $M$ by means of the orthogonal projection $\P$ induced by $g$: $$ \P(X) = X - \frac{g(X, n)}{g(n, n)}n $$ where $n$ is a vector field nowhere vanishing on $S$ and normal to $S$, that exists since $S$ is orientable.
Questions
- Can this be generalized to orthogonally project higher rank contravariant tensor fields, i.e. rank $(k, 0)$, from $M$ to $S$? I have the feeling that if the projection operator can be written in matrix form, like $\P(X) = \O{X}$ for some matrix-valued function $\O$, then the orthogonal projection of a bivector $\pi$ (i.e. a rank $(2,0)$ tensor field) could be $\P(\pi) = \O{\pi}\O^{t}$.
- Can this procedure be "inverted" to "pull-back" smooth functions and 1-forms from $S$ to $M$ in a $g$-dependent manner?