Exploiting the Linearity of the Pullback [closed]

Question

Edit: This question has been significantly revised.

Some recent developments in computational geometry (for example see http://geometry.stanford.edu//papers/fmfrmbs-obsbg-12/fmfrmbs-obsbg-12.pdf) are based on the idea of considering the pullback of a map between two manifolds. As the pullback is a linear map (insofar that it is well-defined), it is much more friendly to work with than the original map, and many tools from Hilbert spaces can now be exploited. However, the "price paid" is that now you are perhaps working in a higher dimensional space, or some information about the original map is lost.

I am very interested to learn of any other mathematics literature that may be related to these ideas or this approach. Thank you very much.

Answer 1

I should let you in a revolutionary point of view proposed by I.M. Gelfand more than seven decades ago. More precisely he observed that a compact topological $X$ space is completely determined by the algebra $C(X)$ of continuous complex valued functions on it. (This is a commutative Banach algebra, but I will not dwell on this, referring you instead to this Wikipedia article.) He noticed that the space $X$ can be identified as a set with the set of maximal ideal of $C(X)$, called the maximal spectrum of $C(X)$. This spectrum can then be equipped with a natural topology making it homeomorphic to $X$.

The point of this result is that you can read the topology of $X$ from the space of continuous functions on $X$. Moreover any Banach algebra morphism $T:C(X)\to C(Y)$ is determined by a continuous map

$$F: Y\to X. $$

More precisely $Tu= u\circ F$, $\forall u\in C(X)$.

This point of view lead to the development of schemes by Grothendieck and to the creation of non-commutative geometry by Alain Connes.

If you are interested in more refined properties of the space $X$, then you need to add additional structure to the ring of functions on $X$. If for example, $X$ is a compact submanifold of some Euclidean space $\mathbb{R}^n$, then $X$ is equipped with a Riemann metric giving it some shape (think ellipsoid vs. round sphere). The metric on $X$ defines a natural (unbounded) operator on $L^2(X)$, the Laplace-Beltrami operator in the paper you quote.

It can be proved that the metric on $X$, hence its shape, is completely determined by the spectral decomposition of $L^2(X)$ determined by this operator.