I am trying to learn about manifold learning techniques; a family of methods in machine learning. According to this idea, there is a low ($d$) dimensional, hidden space where the real data generation mechanism lies, which has $d$ degrees of variability. But we observe the data in a high ($m$) dimensional space where $m > d$. There is a function $f:\mathbb{R^d} \to \mathbb{R^m}$ which is called embedding function which takes the data from low dimensional hidden space and maps to the high dimensional observable one as $x_i = f(\tau_i) + \epsilon_i$, where $\epsilon_i$ is a noise term. The aim here is to learn about the function $f$. All of these ideas are summed up in the following slide:

enter image description here What I don't get exactly in this kind of method is the "embedding function" $f$. It is said that this function maps a $d$ dimensional space to a $d$ dimensional manifold in a higher, $m$ dimensional space. Is it a mathematical fact that such a function $f:\mathbb{R^d} \to \mathbb{R^m}$ should always generate a $d$ dimensional manifold in its range space? I think it is not the case, since such a function can be a much more general one, mapping its input to irrelevant locations in its range.

So is it just an assumption of the approach that this $f$ function is well behaving, in the sense that it maps a low dimensional space more or less to a manifold of the same dimension in a high dimensional observation space? Or is it a mathematical fact? How should I interpret this?

  • 2
    $\begingroup$ I corrected your notation: instead of writing $f: \mathbb R^d \mapsto \mathbb R^m$ one should write $f: \mathbb R^d \to \mathbb R^m$. The symbol $\mapsto$ is used when you write $f: x \mapsto y$. $\endgroup$
    – amoeba
    Mar 24, 2015 at 10:04

1 Answer 1


It is indeed an assumption that the function $f$ is "well behaving".

If the function $f:\mathbb R^d \to \mathbb R^m$ were allowed to be arbitrary, then its image $f[\mathbb R^d] \subset \mathbb R^m$ in the target space could have any number of dimensions between $0$ and $m$. The image can e.g. be the whole target space $\mathbb R^m$ (dimensionality $m$), a single point (dimensionality zero), any weird subspace with fractional fractal dimensionality, or whatever.

But if the function $f$ is smooth (i.e. it is continuous and has continuous derivatives of all orders; mathematicians write $f \in C^\infty$), then the image of $f$ will be a differentiable manifold of dimensionality $d$. I think this assumption is implicit in the figure you provided, because the image of $f$ is displayed there as a nice curly surface which is obviously supposed to be "smooth".

Perhaps it is enough that $f$ is continuously differentiable (i.e. $f \in C^1$), which is a weaker requirement (as it assumes nothing about higher derivatives).

  • $\begingroup$ Probably continuously differentiable is sufficient for many (most?) purposes. This should allow for a coherent definition of distance. $\endgroup$
    – GeoMatt22
    Aug 31, 2016 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.