I'm new to manifolds and in my Computer Graphics class, we briefly explored the topic superficially through visual examples. From my intuitive understanding, we should be able to place a Euclidean grid on a manifold's surface or it should appear flat locally. However, I'm puzzled by how a pyramid with sharp edges and corners can still be a manifold. When we zoom in on the edges or corners, we only see flat areas along the faces of the pyramid, in any other direction it doesn't look flat. What is the difference between this case and two pyramids sharing a vertex (as an hourglass shape), which we classified as non-manifold?

I'd appreciate if someone can explain what I'm missing in my intuitive understanding.

  • $\begingroup$ here and here? $\endgroup$
    – FShrike
    Commented Sep 25, 2023 at 21:36
  • $\begingroup$ @FShrike the answers to the first link are likely too complicated for the OP to gain an intuitive understanding from, and the second link doesn't address the question of two touching pyramids is not a manifold $\endgroup$
    – Carlyle
    Commented Sep 25, 2023 at 22:06
  • $\begingroup$ For embedded subsets $S$, we can directly classify if they are a $C^k$ manifolds by asking if every point is locally the graph of a $C^k$ function. Think of square in $\mathbb{R}^2$. At a corner, you look like the function $|x|$ which is a continuous (even Lipschitz) function. So in particular, it's a $C^0$ manifold, but not a $C^1$ manifold. $\endgroup$
    – Mr. Brown
    Commented Sep 25, 2023 at 22:20

2 Answers 2


The surface of a pyramid is not a smooth manifold, for exactly the reasons that you have perceived: as you zoom in on particular points, the image of that surface does not ever get close to flat.

Nonetheless, the surface of a pyramid is a topological manifold. For example, at each point on an edge of the pyramid, as we zoom in it looks like two half-discs meeting at a dihedral angle, and by a non-smooth yet continuous function we can unbend that angle to get two half-discs meeting along a segment and forming a single flat disc. We can do something similar at a vertex of the pyramid, where three or more "angular" discs meet, stretching them an unbending their angles until we again form a single flat disc, and again the functions needed will be continuous, but they will not be smooth.

  • 2
    $\begingroup$ Well, topologically, a pyramid is homeomorphic to a sphere (for example by some central projection); so in principle the manifold could me made into a smooth manifold. It's just that it's not a smooth submanifold of $\mathbb{R}^3$. $\endgroup$ Commented Sep 25, 2023 at 21:40
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    $\begingroup$ Yes, and indeed every topological 2-manifold can be made into a smooth manifold. I didn't think that would help the OP's desire for missing intuition, however. $\endgroup$
    – Lee Mosher
    Commented Sep 25, 2023 at 21:42
  • $\begingroup$ @LeeMosher thank you for great intuitive explanation! But if I have two intersecting perpendicular rectangles in 3D (shaped like "T"), do I understand correctly that "unbending" along the intersecting edge by laying one rectangle on top of the other rectangle is not an option? $\endgroup$
    – Aid
    Commented Sep 25, 2023 at 21:59
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    $\begingroup$ The more formal point is that I am describing a non-smooth homeomorphism between a neighborhood of the surface of the pyramid and an open set in a flat plane. Homeomorphisms must be bijections. Laying one rectangle on top of the other is not a bijection. $\endgroup$
    – Lee Mosher
    Commented Sep 25, 2023 at 23:02

Instead of thinking of "local flatness at each point" as an intuitive definition of a manifold, you can think of it in this, slightly more accurate way:

A manifold is some object that has the property that at each point on the surface of this object, you can take some disc cutout from a very stretchy material so that you can place the centre of this stretchy disc on the chosen point without cutting or tearing the disc, and deform the disc to fit "nicely" on the surface of the object, without any parts of this disc not being stuck to the surface of the object, and such that the disc doesn't overlap itself anywhere

This is probably the closest you will get to the formal definition of a manifold, and it allows you to see that the pyramid is a manifold, since even at the sharp points of the pyramid, you can simply place the stretchy disc, and stick it smoothly to the faces that surround that point, and since the disc is so stretchy there are no creases and it doesn't need to get torn. However if you try to do this with the two pyramids with their points stuck together, the only way to even get the centers of the disc and this meeting point to align is to first fold the disc in half, an then bring the centre of the disc to the meeting point, but then if you wanted to smooth out the disc around this meeting point, you would need to wrap it around the pyramids, and you might be able to imagine why no matter how small you choose the disc, it would always overlap with itself if you get close enough to the meeting point. This of course does not constitute a proof since we don't know that there aren't other ways to make the centres meet or smooth out a potential disc, but it should at least give a bit more intuition.

I'll also give a brief explanation of why this intuitive definition is a good starting point for someone that isn't familiar enough with Topology:

The formal definition of a manifold is that it is a set $S$ together with a topology on $S$ such that every point in $S$ has a neighbourhood which is homeomorphic to some open subset of $\mathbb{R}^n$.

For your purposes it suffices to think of a topology on a set as a way to determine the closeness of points in $S$ to each other, which induces some sort of "geometric" structure on the set. A neighbourhood of point is then just some subset of $S$ containing that point, which is "localised" around that point with respect to this "closeness" that we mentioned earlier. In our pyramid example, the neighbourhoods of the pointy tip of the pyramid were all those points on the surface of the pyramid within a certain distance of the pointy tip. The open subset of $\mathbb{R}^n$ was the disc, i.e. a subset of $\mathbb{R}^2$, and the condition that The neighborhood be homeomorphic to this disc was captured in everything else: The disc being forced to be smoothed out on the surface corresponds to the definition of homeomorphic requiring a function from the disc to the neighborhood of the pointy tip (The function has to map every point in the disc somewhere so the disc cannot stick out, or not touch the surface anywhere). The description of the disc being stretchy, but not being allowed to tear corresponds to the requirement that the function from the disc to the neighborhood should be continuous. And the requirement that the disc should not overlap anywhere corresponds to requiring that the function sending the disc to the neighborhood should be a bijection. There was just one requirement that we did not build into the intuitive definition, which is that the inverse function, sending the neighborhood to the disc, should also be continuous, but I couldn't think of a nice visual interpretation of this.


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