Given a function $p(x), x\in \mathbb{R}^F$ I am able to calculate the gradient $\nabla_x\ p(x)$.
Let's consider the subspaces
$$ \mathbb S_x = \{x\in\mathbb{R}^F\ |\ x[f]=o[f] \text{ for some } f\in\{1,...,F\} \} $$
where $o \in \mathbb{R}^F$, $x[f]$ means a the $f$-th entry of $x$, and
$$ \mathbb S_y = \{ y\in\mathbb{R}^N \} \subset \mathbb{R}^F $$
for $N\in\mathbb N, 0<N<F$, where we chose the same dimensions for $\mathbb S_y$ that are not fixed to $o[f]$ in $\mathbb S_x$. (I hope it is clear what I mean.)
In words: I slice $\mathbb{R}^F$ at a particular position and I want to look at this (hyper) surface. There I want to calculate the gradient $\nabla_y\ p(y)$. What I do know is the gradient $\nabla_x\ p(x)$, which of course has a too large dimension for $\nabla_y\ p(y)$.
Sorry for defining everything so non-publication-ready, but it would take a while to figure out how to define it so exactly. Anyway:
Question: Can I simply take those dimensions of $\nabla_x\ p(x)$ that correspond to the dimensions in $\nabla_y\ p(y)$ or do I need to recalculate $p(y)$ somehow? For example if $\nabla_x\ p(x) = (1,2)$, can I say $\nabla_y\ p(y) = (1)$?
To make it more clear:
The 2-D Gaussian is $p(x)$. The surface / subspace I am interested in, is where it is cut away in the picture. Please note, that I am not interested in the entire projection of the 2-D Gaussian into the 1-D cut subspace, but the cut, as it is displayed in the picture, so the maxima of the 2-D Gaussian is lower than the maxima of the subspace 1-D Gaussian.
In opposite to the picture, I only cut parallel to the dimensionality axis.
I hope it is clear now, but please ask if not.
You have a function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ and you are defining a new function $g:\mathbb{R}^{n-1} \rightarrow \mathbb{R}$ by $$g(x_1, \ldots, x_{n-1}) = f(x_1, …, x_{n-1}, c)$$ for some given constant $c \in \mathbb{R}$. So, with $x=(x_1, …, x_n) \in \mathbb{R}^n$ and $y=(y_1, …, y_{n-1}) \in \mathbb{R}^{n-1}$: \begin{align} \nabla f(x)& = [\partial f(x)/\partial x_1 , …, \partial f(x)/\partial x_n] \\ \nabla g(y) &= [\partial f(y,c)/\partial x_1, …, \partial f(y,c)/\partial x_{n-1}] \end{align} So, yes, you just consider the relevant dimensions of the original function $f$.