Suppose I have a set of $k$ points $\{x\_1,x\_2,\ldots,x\_k\}$ in $\mathbb{R}^n$ that I can project into $\mathbb{R}^m$ with the linear operator $\mathcal{P}$, with $\alpha, \beta, \ldots$ parameters of the projection operator. Is there a well known best method for determining the class of functions $f:(\mathbb{R}^m)^k \rightarrow \mathbb{R}$ such that $f(\mathcal{P}x\_1,\mathcal{P}x\_2,\ldots,\mathcal{P}x\_k)$ is invariant under changes in $\alpha, \beta, \ldots$?


As an example, a situation I'm interested in has $\mathcal{P}: \mathbb{R}^3 \rightarrow \mathbb{R}^2, x \mapsto A[\phi, \theta, \eta]x$, where:


$A[\phi, \theta, \eta] = \eta \left(
\begin{matrix}
 \cos\phi\cos\theta & \sin\phi\cos & \sin\theta \\
 -\sin\phi & \cos\phi & 0
\end{matrix}\right)$


i.e. the projection of points in 3-D space onto the plane with unit normal $(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$, with an additional scaling parameter of $\eta$.


Setting derivatives of $f$ with respect to the parameters equal to zero yields a set of equations with no obvious general solution, and a feeling that I'm missing a certain set of tools.



Is there a class of functions acting on a set of projected points that remain invariant under changes in projection parameters?
I can think of (at least) two ways of interpreting this question.


**First:** You are given some specific list of $k$ points $x\_1$, $x\_2$, ..., $x\_k$ in $\mathbb{R}^n$, and you want to detect whether $k$ points $y\_1$, ..., $y\_k$ in $(\mathbb{R}^m)^k$ could be a linear projection of the original $k$ points. 


**Solution:** Compute the vector space of all $(c\_1,\ldots, c\_k)$ such that $\sum c\_i x\_i=0$. For a basis of this space, whether or not $\sum c\_i y\_i=0$. If so, the $y\_i$ are a projection, if not, they are not.


**Second:** You want to construct functions $f$ on $(\mathbb{R}^m)^k$ such that, if $(y\_1, \ldots, y\_k)$ and $(y'\_1, \ldots, y'\_k)$ are both projections of the same points $(x\_1, \ldots, x\_k)$, then $f(y\_1, \ldots, y\_k) = f(y'\_1, \ldots, y'\_k)$.


**Solution:** There are no nonconstant examples. I will show that we must have $f(y'\_1, \ldots, y'\_k) = f(y\_1,\ldots,y\_k)$ for any $y$ and $y'$ in $(\mathbb{R}^m)^k$.


For notational simplicity, I'll take $n=m+1$. Let $y^j\_i$ be the vector in $\mathbb{R}^m$ whose first $j$ coordinates are that first $j$ coordinates of $y\_i$ and whose last $m-j$ coordinates are the last $m-j$ coordinates of $y'\_i$. So $y^0\_i = y'\_i$ and $y^m\_i = y\_i$.


Let $x^j\_i = \mathbb{R}^{m+1}$ be the vector whose first $j$ coordinates are those of $y\_i$ and whose last $m+1-j$ coordinates are those of $y'\_i$. Then $(y\_1^j, \ldots, y\_k^j)$ and $(y\_1^{j+1}, \ldots, y\_k^{j+1})$ are both projections of $(x^j\_1, \ldots, x^j\_k)$. So
$$f(y\_1^0,\ldots,y\_k^0) = f(y\_1^1, \ldots, y\_k^1) = \cdots = f(y\_1^m, \ldots, y\_k^m)$$
and $f(y'\_1, \ldots, y'\_k) = f(y\_1,\ldots,y\_k)$.


I can also think of other possibilities, like that you have a data bank of a few billion options for $x$, and you want to find the best match, but this seems like enough for now.


