In the first few pages of his book Introduction to Smooth Manifolds, Lee writes:

But for more sophisticated applications, it is an undue restriction to require smooth manifolds to be subsets of some ambient Euclidean space.

A way to motivate why we develop tools to do calculus in manifolds is through examples in physics. For instance, one might want to do calculus on the function that maps the surface of the earth to its temeprature on the real line, we might want to study electromagnetic properties of a torus, and so forth. In each of these cases, it is easy to study these surfaces as subsets (or submanifolds) of Euclidean space.

What are some "sophisticated applications" of manifolds outside of math - as Lee writes - that do not allow us to work with manifolds that are embedded in Euclidean space?

Edit: I'm obviously looking for examples other than the one Lee gives himself (and perhaps the most popular here): Looking at space-time as a four dimensional manifold, where it doesn't make sense to embed it in an ambient space.

  • 4
    $\begingroup$ I am not sure what Prof. Lee actually meant; but think of examples from physics. For example, in relativity we think of space-time as a 4-dimensional manifold. And, it would not make sense to think of spacetime as embedded into an ambient space. What would be “physical meaning” of any such ambient space? $\endgroup$
    – Raghav
    Commented Mar 22, 2020 at 7:55
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    $\begingroup$ What did you mean "outside of math"? $\endgroup$
    – IAmNoOne
    Commented Mar 22, 2020 at 8:01
  • $\begingroup$ @IAmNoOne Just that the "sophisticated application" should not be a problem strictly within math itself, i.e. it is easy for me to agree with the idea that we can have purely abstract structures that can be studied through manifolds without embedding, but the same isn't true when I think of, say, physics. $\endgroup$
    – gtoques
    Commented Mar 22, 2020 at 9:06
  • $\begingroup$ Nowhere in Solid Mechanics - or elsewhere in classical physics - there exist manifolds that are curved into nothingness. It's typical for General Relativity, as far as I know. $\endgroup$ Commented Apr 25, 2021 at 9:55

3 Answers 3


In my opinion, the point here is that one can think of manifolds intrinsically, without need of seeing them as subsets of some Euclidean space, and this provides a much cleaner theory.

Indeed, any smooth manifold can be smoothly embedded into Euclidean space, as proven by the Whitney embedding theorem (and other results if you impose additional structure, e.g. the Nash embedding theorem).


A fundamental application for which it is undue to consider a manifold as a subset of some Euclidean space is the whole field of Riemannian geometry (as you know), but also the whole field of Pseudoriemannian geometries, like Lorentzian Geometry.

An example is the hyperbolic model with the Minkowski metric. The Minkowski metric is a particular kind of Lorentzian metric, meaning a metric that is no more positive definite, and may have positive and negative eigenvalues. The Minkowski metric has three positive eigenvalues and one negative. This has enormously different properties than the standard rules of calculus in Euclidean spaces: for example for some vectors it even holds a reverse Cauchy-Schwartz inequality and reverse triangular inequality, in the opposite direction of the usual. This has lots of implications in special relativity, giving rise to the famous cone structure. If we considered an hyperboloid in the Euclidean space (meaning with the standard positive definite scalar product), then we would never be able to describe these physical phenomena.


Another source of examples is quotient manifolds. See for example the book by Absil, Mahony and Sepulchre, which studies numerical algorithms to solve optimization problems on Riemannian manifolds. The idea is as follows:

Say you want to find a minimizer of $f \colon M \to \mathbb{R}$, where $M$ is some Riemannian manifold. It may be the case that there is some Lie group $G$ which acts on $M$ and for which $f$ is invariant, i.e., $f(g \cdot x) = f(x)$ for all $g\in G$ and all $x \in M$. In words: there may be some symmetries at play.

A typical example is: let $M$ be the Stiefel manifold (matrices whose columns are orthonormal), and let $f(X) = \mathrm{trace}(X^\top\! A X)$ for some given symmetric matrix $A$. The minimizers of $f$ are all the matrices whose columns form an orthonormal basis for a left-most invariant subspace of $A$ (essentially, eigenvectors associated to the smallest eigenvalues of $A$).

You can see that this optimization problem is invariant under rotations of the basis formed by the columns of $X$. Specifically, let $G$ be the orthogonal group (of size corresponding to the number of columns of $X$). Then, $f(XQ) = f(X)$ for all $Q\in G$ and all $X \in M$. In words: it doesn't matter which basis we pick; it only matters which subspace it spans.

This makes it quite natural to ask: could I study/solve the problem on the quotient space instead, that is, on $\tilde M = M / G$ where the equivalence classes are the orbits of the group action?

That quotient set is 1-to-1 with the Grassmann manifold (linear subspaces of given dimension), which is indeed a manifold in its own right, though it's not "embedded" in an obvious way. This holds more generally, provided the group action is free, proper and smooth (Lee's book includes a theorem to that effect). So, to make sense of the claim that "the quotient set is a smooth manifold", it seems you do need the general machinery of charts, atlases etc.

To be fair though, in the case of the Grassmann manifold (with Riemannian metric inherited from Stiefel through the quotient map), you can also view it as a Riemannian submanifold that consists of orthogonal projectors: one for each subspace. So, if Grassmann is all you need, you still don't need the general formalism.. but (in my opinion) it's still a very potent way of thinking about it, providing useful intuitions also beyond that particular case.

(I also discuss these things in Chapter 9 of my book.)


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