22 ms·

What Is a Manifold?

283 points by Topolomancer 8y ago

mjfl 8y ago
f(x) = 0. Move along it by moving perpendicular to the gradient of f.
termiefoo 8y ago
what ain't a manifold at this point
- username90 8y ago
  Things with edges, like disks and finite lines.
  - srean 8y ago
    Nothing wrong with them, they can be manifolds.
  - deleted 8y ago
    [deleted]
  - math_and_stuff 8y ago
    Both of those are typically referred to as manifolds with boundary. The setting of Stokes Theorem is often manifolds with boundary.
    - hansen 8y ago
      Mathematicians messed up here… A manifold with boundary is not a manifold. But a manifold is a manifold with boundary (the empty set).
      - deleted 8y ago
        [deleted]
      - Topolomancer 8y ago
        I thought a manifold with a boundary would still be a manifold, but its boundary has to satisfy a dimensionality condition. For example, the 2D disk is a 2-manifold with a 1-dimensional boundary. Strictly speaking, this is a _topological manifold with a boundary_, though.
      - JadeNB 8y ago
        > Mathematicians messed up here… A manifold with boundary is not a manifold. But a manifold is a manifold with boundary (the empty set). It depends on which mathematicians! Plenty of differential geometers allow manifolds to have boundary, and say "closed manifold" (https://en.wikipedia.org/wiki/Closed_manifold https://en.wikipedia.org/wiki/Closed_manifold) to emphasise when they are dealing with a (compact) manifold without boundary (or, as you point out, really a manifold whose boundary is empty).
- deleted 8y ago
  [deleted]
- srean 8y ago
  A fractal isnt a manifold. Discrete sets are an edge case. They are considered to be 0 dimensional manifolds to make them fit, but there is a not much that manifold theory has to say about them.
  - man-and-laptop 8y ago
    > A fractal isnt a manifold Some are. A Koch curve for instance.
    - srean 8y ago
      I believe you are correct. Koch curve is a continuous curve so its topological dimension would be 1 (not its fractal dimension).
- JulianWasTaken 8y ago
  Besides the other answers, things also start to get interesting when you add some qualifiers. E.g., asking for a "differentiable manifold" now asks for manifolds with enough additional structure to do calculus on top of.
  - Topolomancer 8y ago
    Good point; I originally wanted to talk about the Poincaré conjecture as well, but then I realized that this would make the post even longer. Do you have some ideas about other interesting topics?
    - JulianWasTaken 8y ago
      To be honest it's not even my field so I know a lot less than I wish I did but yeah agreed it has no shortage of interesting things to talk about :) Poincaré Conjecture is yeah interesting to explain though to be honest I vaguely recall an article in similar style to your post that explained it. Possibly something like describing gradient descent on a manifold is interesting to this audience? Or maybe a post on Flatland? Many possibilities really on good follow ups.
    - amelius 8y ago
      > Good point Can a point be a manifold?
      - Topolomancer 8y ago
        Yes, but it's a simple one of dimension zero. There's not much to do here---every 'neighbourhood' of the point is exactly that: just the point. (not that there are different 'classes' of manifolds out there; I am not sure if the point would qualify as a Riemannian manifold, for example)
- ColinWright 8y ago
  A Y-shaped "line".
- badosu 8y ago
  f(x) = 0 (x < 1, x > 1) f(0) = 1
- Sinbe 7y ago
  Simplest example of what not a manifold is the shape of the letter “Y”. Technically for a topological space to be a manifold, every small neighborhood around each point should look the same. But “Y” has a special point at the center.
lordnacho 8y ago
Is the conclusion that a manifold is a space that looks like a lower dimensional space that it really is?
- Topolomancer 8y ago
  To some extent, yes! It needs to have a 'nice' structure, though---this is why we have the condition that is should resemble, say, an Euclidean space.
- JulianWasTaken 8y ago
  You can have a thing in some higher dimensional space that still "looks like" a thing in lower dimensional space (like the surface of the sphere in R^3 being two dimensional), but R^3 itself is a 3 dimensional manifold that "looks" 3 dimensional as well.
- mlevental 8y ago
  no. there's no requirement for lower dimension. like another person that responded to you R^3 is 3 dimensional manifold. a (smooth) manifold is a space you can do calculus on. that means you have a consistent notion of distance (so that you can figure out when points are close together, ie you approach limits). the set of charts and continuous transformations between them is what encodes this constraint.
- patrickthebold 8y ago
  Yes and no. The definition doesn't require that what you have is embedded in a higher dimensional space. Though that's a good source of examples. https://en.m.wikipedia.org/wiki/Whitney_embedding_theorem https://en.m.wikipedia.org/wiki/Whitney_embedding_theorem Basically says that many manifolds can be described as a lower dimensional submanifold of some euclidean space.
ErotemeObelus 8y ago
I think everything said here can applies to just topologies.
- dan-robertson 8y ago
  That isn’t really true. Dimension isn’t so well defined in topology but is reasonably straightforward with a manifold. The article also touched on geometry (sum of angles of a triangle). To be topological a bunch of things must change. The bug walking example doesn’t work and is demoted to an analogy. All notion of distance and direction is lost. If the article were about topology then it would probably be talking about distinguishing shapes by homotopy/homology rather than lines and walks and angles. The article also gets some details a bit wrong/fuzzy: it says you can tell if you are on a sphere by walking in a straight line infinitely far and seeing if you ever cross yourself. But this property also follows on the surface of a (rounded at the top) cone. Even if you require this property in all directions you get problems on eg a torus.
  - Topolomancer 8y ago
    Thanks for pointing out the fuzziness! I did not want to extend this further so as not to confuse readers who have no familiarity with the subject. Of course, the fact that you cross your own path at some point is _not_ indicative of living on a sphere; there are numerous other manifolds that satisfy this (not to speak of the torus, as you yourself mentioned). I will try to be more precise about this in a subsequent article!
    - ErotemeObelus 8y ago
      Don't. Precision obfuscates more than it increases resolution. You don't want to be precise. You want to increase the resolution of the key ideas being seen. This is also why parables are so popular for teaching: they're literally not true, but they resolve to an image of something that is true. Please don't introduce homotopy/homology. Lines and angles are perfect.
  - senderista 8y ago
ulucs 8y ago
All is well until the poor bug finds out it's on a Klein Bottle, or worse yet RP2. Is there any way to distinguish between orientable or nonorientable surfaces when you're simply walking on it? If the bug was also 2d, I might have had some ideas hut I'm totally blanking for 3d.
- ColinWright 8y ago
  > Is there any way to distinguish between orientable or nonorientable surfaces when you're simply walking on it? Draw a circle on the surface and orient it. If the surface is non-orientable then there's a way to take a long walk, return to the circle, and find that the direction of the orientation has changed.
  - tgb 8y ago
    This depends upon the circle you chose. Nonorientable manifolds still have many orientable circles (any sufficiently small loop is orientable).
    - ColinWright 8y ago
      You confuse me. If the surface is non-orientable then I can draw a very small circle, put an orientation on it, then there's a path I can take that brings me back to find the orientation has changed. Let's be more explicit. Take a non-self-intersecting embedding of RP^2 in R^4, a point P in our RP^2, and a sufficiently small epsilon e. Then take three points on the circle of size e centred on P, and think of them as going in order, thus defining an orientation. Now take an appropriate walk around RP^2 and return to the circle. For an appropriate walk I will now find that the order of the points on my circle has reversed. That's a more precise way of saying what I originally intended, and in that context your comment doesn't make sense to me. Can you expand on it?
      - a1369209993 8y ago
        I think when they say "circle" they're talking about your path, not your orientation reference. That's probably wrong terminology, but it makes some sense if you're thinking about how the path has to be a closed loop.
joncrane 8y ago
It's funny, because in my much younger days I dropped out of college and worked as a mechanic for 3 years. In the automotive world and internal combustion World, manifold has a much different meaning, though related. Basically, a manifold means something that takes the flow of gases from a one-to-many or a many-to-one. An intake manifold takes one single entry point for air feeding the engine and splits up into a separate input for each cylinder. An exhaust manifold takes the hot exhaust from each cylinder separately and combines them into one big pipe.
- tyingq 8y ago
  That's what manifold means to me as well. Though, there are exceptions to the "one" part, like intake manifolds with two inlets (dual plane is common), exhaust manifolds with two outlets (common on inline 6 engines), and also some really odd setups: https://speedmaster79.com/media/catalog/product/cache/1/thumbnail/9df78eab33525d08d6e5fb8d27136e95/p/c/pce148.1086_2.jpg https://speedmaster79.com/media/catalog/product/cache/1/thum...
  - joncrane 8y ago
    That picture looks like an intake for a V8 and each trumpet is still for an individual cylinder; it's just that the intake ports on the heads are paired up for manufacturing and thermal efficiency.
    - tyingq 7y ago
      My point was that manifolds aren't all many to one. This one is clearly many to many.
- commandlinefan 8y ago
  I actually clicked the link expecting exactly that explanation.
  - Topolomancer 8y ago
    Sorry! It just goes to show that I am living in a filter bubble of mathematics. My intention was not to deceive you :-)
tw1010 8y ago
Love me some explanation innovation. More of this and the coding world will reach a plateau.
xelxebar 8y ago
If anyone has questions (especially technical ones), the ##math channel on freenode is pretty phenomenal. It's quite active, with quite a few grad students, across a range of fields. The best part is how friendly and helpful they are to learners. Anyway, highly recommended.
- mlevental 8y ago
  this in general accurate but I would warn people to have a thick skin in there because there are some seriously toxic people in there (TRW3W or whatever his/her name that was fairly knowledgeable but hung out in there I think purely to assert his/her superiority).
  - microcolonel 8y ago
    This is good advice for existing at all in the world. They're not toxic except in high doses, like all sorts of people.
  - hhmc 8y ago
    I still recognize that name/handle and I haven't logged into ##math for ~10 years, so I can certainly corroborate what you're saying.
krosaen 8y ago
Another connection to make: quaternions can be represented as a Lie group, which is a (smooth differentiable) manifold. This means representing 3d orientation using quaternions works better for optimization than e.g Euler angles (roll, pitch, heading) in things like SLAM.
- mlthoughts2018 8y ago
  What do you mean “works better”? Symbolically simpler once you know the requisite math behind Lie groups? In terms if computation, why would one representation be any different than the other? Unless you mean there’s a subset of certain computations where that representation is better but also a subset where it is worse? Similar to how there are cases where a Hough transform allows a more efficient calculation, but many cases where you can’t efficiently extract other calculations out of the Hough transform with effectively inverting it entirely.
  - ColinWright 8y ago
    Representing rotations via Euler angles can lead to gimbal lock[0], whereas using quaternions doesn't. The best explanation I could find quickly is this one: https://mathoverflow.net/a/95908 https://mathoverflow.net/a/95908 [0] https://en.wikipedia.org/wiki/Gimbal_lock#In_applied_mathematics https://en.wikipedia.org/wiki/Gimbal_lock#In_applied_mathema...
    - blt 8y ago
      euler angles are sort of the odd one out, a particularly bad way to represent rotations. Orthonormal matrices, quaternions, skew-symmetric tensors with exponential map -- all represent SO(3) without gimbal lock, with different pros and cons for each.
  - yiyus 8y ago
    They are more practical, specially when using a computer. Besides the gimbal lock problem mentioned in another problem, Euler angles are not easy to compose. It is also not obvious how to calculate the misorientation between two rotations with Euler angles, but it is trivial with quaternions. Other approaches have their own problems. For example, matrices are more expensive to use in a computer (normalizing a quaternion to avoid rounding errors is much faster than normalizing the equivalent matrix, and quaternions need less memory), and Rodrigues vectors, while usually very useful, present the problem of infinite values for rotations of 180 degrees (which, in my opinion, should not be a problem, but it is in many programming languages). At the end of the day, quaternions present the best compromise.
no_identd 8y ago
"Quick" etymological fact about what mathematicians in English call Manifolds: If one would translate Riemann's original German word for them, "Mannigfaltigkeit", it would translate to "manifoldyhead", or, more understandable to the speaker of Modern English: Manyfoldyhood. (Think of -head as in "Godhead", not as in "Brotherhood", in the same sense that the "ring" in "algebraic ring" refers to "ring" in the sense of "smuggler ring", not in the sense of "gold ring". The two suffixes merged in English, and have very different and rather complex etymological origins.) While this word sounds somewhat ridiculous and perhaps a bit infantile in English, it, in my opinion, conveys what we mean by them significantly better. Clifford tried to accommodate Riemann's highly specific word choice in his translation, as Jost notes: "The English of Clifford may appear somewhat old-fashioned for a modern reader. For instance, he writes “manifoldness” instead of the simpler modern translation “manifold” of Riemann’s term “Mannigfaltigkeit”. But Riemann’s German sounds likewise somewhat old-fashioned, and for that matter, “manifoldness” is the more accurate translation of Riemann’s term. In any case, for historical reasons, I have selected that translation here." Unfortunately, neither Jost (a native German speaker!) nor Clifford realized that English can and does accommodate Riemann's exact meaning directly. To make a comparison which might require quite a bit of German knowledge beyond high school education to describe in exact linguistic terms, but which native speakers should hopefully find intuitive to distinguish (I at least, do): Translating "Mannigfaltigkeit" as "Manifoldness" seems equal to mistranslating "Geheimnis" as "Secrethood" and "Geheimheit" as "Secretness", whereas the opposite pairing would yield an accurate (albeit not necessarily immediately apparent—in terms of the differences between the two—to the English native speaker, unless explicitly pointed out) translation. So much for the last suffix, but that still leaves insertion of the one before it (' * foldy * ' instead of ' * fold[( * )]' unclear.) One may observe the difference involved there by a converse example, also involving "Geheim", as well as the root word of it, "Heim", by dragging forth a rather rare and archaic—but none the less highly likely intuitive to the native speaker—word:"Geheimig". Whose meaning starkly differs from both "Geheimnis" and "Geheimheit". For further language related hijinks related to /Manyfoldyhoods/, see: A) The Dutch word for them, which would back translate to "Variety"; and B) This quote by Poincaré: "I prefer the translation of Mannigfaltigkeit by multiplicity, because the two words have the same etymological meaning. The word set is more adapted to the Mannigfaltigkeiten considered by Mr. Cantor and which are discrete. It would be less adapted to those which I consider and which are discontinuous." (As I don't speak French, I can't make much of any statements about the accuracy about the etymological claim by Poincaré, so I'll close with another Poincaré quote instead: " Mathematics is the art of giving the same name to different things." —Henri Poincaré)
braxxox 8y ago
What a great read! I've never thought about how the way we experience the surface of the earth as 2D manifold of a 3D space, and how up until recently (relatively speaking) this was unknown. It's interesting to expand on this idea and realize that maybe we are making the same mistake again, and that from our local perspective the universe is 3D, when in reality, it's a 3D manifold of a higher dimensional space. One thing I am still unclear of though, is that isn't this proven to be the case? Is it not true that we are provably living in at least a 4D space, where time is the fourth dimension? We can observe its existence but cannot move freely through it and are confined to free transformations only in 3D space? In this way, aren't we living in a 3D manifold in a 4D space? So maybe then the question is are we living in a 3D manifold of a +4D space?
- crdrost 8y ago
  Sort of, not quite. The basic background is that Einstein wrote a paper taking Lorentz more seriously than Lorentz took his own work: and that paper suggested that just maybe, when you accelerate in any given direction by an acceleration A, you see all of the clocks ahead of you some distance z tick faster by a factor A z / c², where c is the speed of light in vacuum, and behind you they tick slower with the corresponding negative z until a wall of death at z = -c² / A where clocks do not tick at all and time appears to stand still. This is in fact the only new fact that special relativity adds. Lots of people got very confused about the philosophical implications, but the mathematical implication is that time and space can be mixed together by these accelerations and must be treated as one unified geometrical entity. Einstein then went one further on the whole 4D thing, because arguably we are always accelerating in this whole gravitational field of the Earth. You have a lot of options to choose from. So this part took Einstein many many years to work out. Maybe the easiest is to say that we standing on Earth are a non-accelerating reference frame, and then anybody who falls must see a wall of death somewhere out in space. That turns out to be a very boring approach, and also wrong. What Einstein suggested instead was that you are in a non-accelerating reference frame with no wall of death if you are in free-fall, and we standing on the Earth would see a wall of death beneath our feet, except we can't see beneath our feet. But if a body were more massive, maybe we could see the wall of death from orbit. And now we have a photo of a black hole to prove it! But even before that, the essential point is that if I put a clock up somewhere high (on a tower, in a plane, or at the top of a mountain) and I am standing on the ground, then I am accelerating towards that clock relative to free-fall: so that clock must be ticking faster than my clocks are. And we have had direct observation of that “gravitational time dilation” for a long time. Mathematically, this means that we are in a manifold that looks locally 4 dimensional, in this weird way of coupling the four dimensions that couples accelerations with the ticking of clocks. We say, going back to this guy who worked out all of the mathematics before Einstein, that the manifold is locally Lorentzian, as opposed to Euclidean. But the manifold is four dimensional, not three dimensional. It has to have this coupling between time and space locally. But then globally it can have these interesting features like black holes. Now, whether we can embed this curvy universe that we inhabit into a larger dimensional flat space, is not necessarily a given. I don't know many physicists who are deeply interested in that sort of question. Certainly to have the structure that we need, it needs to have Lorentzian timelike dimensions in it, one of which we use as our time dimension. Certainly also, the people working on string theory use these extra-dimensional possibilities to solve certain mathematical inadequacies that their string theories otherwise have: but usually those dimensions are locally available, so we would see them; so there is some sort of hand-waving about how they must be curled up into such a small length scale that we cannot actually observe them. But there is certainly a branch of string theory called M-theory that I do not personally know too much about which has something to do with viewing our universe as a geometric entity in a larger space. Thankfully, the mathematics does not require this. The essential point of a manifold that the article somehow leaves out, is that I am no longer going to rely on global coordinates. I only care about local coordinates. So on the sphere, it is a two dimensional object, even though it lives in a three dimensional space, and that's because depending on where I am, I can uniquely identify points near me on the sphere by either their x & y coordinates, or their y & z coordinates, or their x & z coordinates. This fails for points that are not nearby me, because projecting a globe on to a flat surface this way will project two hemispheres onto the same point. But I can always choose one of those three and find myself in the middle of a hemisphere, and describe everything else on that hemisphere with those coordinates. So the whole point of a manifold is that I don't need global coordinates, and therefore it doesn't matter much whether or not I am embedded in a larger flat space or not.
colllectorof 8y ago
If you want more, there is an excellent course on "introductory" topology on YouTube: https://www.youtube.com/watch?v=CEXSSz0gZI4 https://www.youtube.com/watch?v=CEXSSz0gZI4 It is made by a physicist, not mathematician, and it's a great combination of informal explanations and carefully covering the actual math definitions. I feel that his approach works way better than making analogies. There is another course from Wildberger. A bit more hardcore, but also has some interesting perspectives: https://www.youtube.com/watch?v=Ap2c1dPyIVo&list=PL6763F57A61FE6FE8 https://www.youtube.com/watch?v=Ap2c1dPyIVo&list=PL6763F57A6...
amai 8y ago
So a manifold is a surface?
- chadcmulligan 8y ago
  it can be, I think of a manifold (probably naively) as the mathematical abstraction that you end up with if you start from Euclidean space and fiddle with different features - e.g. if you're in a Euclidean space how do you get from one point to another - what are the minimum requirements? What if we want calculus to work what are the minimum requirements? If we're on a surface and we walk around is there any way to tell what shape that surface is?
- hopler 8y ago
  A generalized surface in arbitrary dimensions. A common English "surface" maps naturally to a 2D manifold embedded in 3D space.
  - amai 8y ago
    Ok, so a manifold is simply a hypersurface. At least I‘m pretty sure in machine learning it means exactly that.
    - ginnungagap 8y ago
      A hypersurface is usually a codimension one object, while a manifold can have bigger codimension or even not be embedded in an higher dimensional space at all
      - amai 7y ago
        Ok, a line in 3D space can also be called a manifold? Maybe a better question is then to ask, what is not a manifold?
valw 8y ago
Nice article, it would have been nice to add a word or two about manifold embedding though.
- Topolomancer 8y ago
  I plan on adding this in a sequel :) Thanks for the suggestion!
arduinomancer 8y ago
I think the natural next question is how does a bug in R3 check if its part of a 4-dimensional manifold? This is related to people measuring the curvature of the universe right?
- improbable22 8y ago
  The article squeezes together two notions, of whether a manifold is curved, and whether it's a submanifold of some higher-dimensional one. Sub-spaces of R^3 are a useful way of generating and picturing examples of 2D manifolds. But these things exist by themselves. As the article mentions with angles of a triangle, you can tell that a 2D manifold is curved from living inside. Likewise you can tell whether a 3D manifold is curved of not, without any mention of a 4th dimension. Questions of curvature of the universe are a step harder, as we are talking about 3+1-dimensional space-time. But the slice of constant time is a 3D manifold, and as far as we can tell right now, it appears to be flat.
alcolade 8y ago
If the bug were truly mathematical, he would notice the curvature of S^1 and S^2 without walking around (since curvature is defined at each point), but that is obviously very pedantic of me and shows why I wouldn't write an article nearly as interesting as this one.
- improbable22 8y ago
  But S^1 isn't curved. At least if we mean intrinsic curvature, which is the kind that belongs to the manifold itself. S^2 is curved, but locally you can't tell it's not (say) RP^2.
  - alcolade 8y ago
    I was thinking of the Riemann curvature tensor which is just some measure of curvature at each point. If the bug were at some point of S^2 he would notice the curvature. Of course, this is assuming S^2 is getting its geometry from a certain embedding into R^3 that comes to mind. You could define different Riemann curvature tensors over S^2 that may have zero curvature in certain places, like squishing a balloon against a flat table for example. I guess my point was that topology and curvature are different things (like you pointed out with your comment about RP^2), and saying that "a sphere looks flat locally" is missing the point! Also, it didn't occur to me that this notion of curvature doesn't make sense in 1d, i.e. for S^1 like you said.
    - improbable22 8y ago
      Not only is there no intrinsic curvature in 1D, but the 2D and 3D versions are also simplified special cases. In 2D there is only scalar curvature, i.e. you don't need the whole Riemann tensor, just one number (at a given point). In 3D you need the Ricci tensor, but still not the full curvature tensor. This is still much simpler, for instance IIRC this is why you cannot have gravitational waves in 3 (meaning 2+1) dimensional spacetime. In 4D you get the full complication.
loeg 8y ago
Ah, and here I was imagining this[0] is a manifold. I must be spending too much time with PEX lately. [0]: https://www.supplyhouse.com/Bluefin-CM4-75PXZ-PXV-3-4-PEX-Crimp-x-Closed-Copper-Manifold-w-1-2-PEX-Crimp-Ball-Valves-Lead-Free-4-Outlets https://www.supplyhouse.com/Bluefin-CM4-75PXZ-PXV-3-4-PEX-Cr...
725686 8y ago
At the end of the article he mentions Jeffrey Weeks... I searched youtube and found this very interesting vid: https://www.youtube.com/watch?v=j3BlLo1QfmU https://www.youtube.com/watch?v=j3BlLo1QfmU

What Is a Manifold?

283 points by Topolomancer 8y ago

mjfl 8y ago

f(x) = 0. Move along it by moving perpendicular to the gradient of f.

termiefoo 8y ago

what ain't a manifold at this point

username90 8y ago
Things with edges, like disks and finite lines.
- srean 8y ago
  Nothing wrong with them, they can be manifolds.
- deleted 8y ago
  [deleted]
- math_and_stuff 8y ago
  Both of those are typically referred to as manifolds with boundary. The setting of Stokes Theorem is often manifolds with boundary.
  - hansen 8y ago
    Mathematicians messed up here… A manifold with boundary is not a manifold. But a manifold is a manifold with boundary (the empty set).
    - deleted 8y ago
      [deleted]
    - Topolomancer 8y ago
      I thought a manifold with a boundary would still be a manifold, but its boundary has to satisfy a dimensionality condition. For example, the 2D disk is a 2-manifold with a 1-dimensional boundary. Strictly speaking, this is a _topological manifold with a boundary_, though.
    - JadeNB 8y ago
      > Mathematicians messed up here… A manifold with boundary is not a manifold. But a manifold is a manifold with boundary (the empty set). It depends on which mathematicians! Plenty of differential geometers allow manifolds to have boundary, and say "closed manifold" (https://en.wikipedia.org/wiki/Closed_manifold https://en.wikipedia.org/wiki/Closed_manifold) to emphasise when they are dealing with a (compact) manifold without boundary (or, as you point out, really a manifold whose boundary is empty).
deleted 8y ago
[deleted]
srean 8y ago
A fractal isnt a manifold. Discrete sets are an edge case. They are considered to be 0 dimensional manifolds to make them fit, but there is a not much that manifold theory has to say about them.
- man-and-laptop 8y ago
  > A fractal isnt a manifold Some are. A Koch curve for instance.
  - srean 8y ago
    I believe you are correct. Koch curve is a continuous curve so its topological dimension would be 1 (not its fractal dimension).
JulianWasTaken 8y ago
Besides the other answers, things also start to get interesting when you add some qualifiers. E.g., asking for a "differentiable manifold" now asks for manifolds with enough additional structure to do calculus on top of.
- Topolomancer 8y ago
  Good point; I originally wanted to talk about the Poincaré conjecture as well, but then I realized that this would make the post even longer. Do you have some ideas about other interesting topics?
  - JulianWasTaken 8y ago
    To be honest it's not even my field so I know a lot less than I wish I did but yeah agreed it has no shortage of interesting things to talk about :) Poincaré Conjecture is yeah interesting to explain though to be honest I vaguely recall an article in similar style to your post that explained it. Possibly something like describing gradient descent on a manifold is interesting to this audience? Or maybe a post on Flatland? Many possibilities really on good follow ups.
  - amelius 8y ago
    > Good point Can a point be a manifold?
    - Topolomancer 8y ago
      Yes, but it's a simple one of dimension zero. There's not much to do here---every 'neighbourhood' of the point is exactly that: just the point. (not that there are different 'classes' of manifolds out there; I am not sure if the point would qualify as a Riemannian manifold, for example)
ColinWright 8y ago
A Y-shaped "line".
badosu 8y ago
f(x) = 0 (x < 1, x > 1) f(0) = 1
Sinbe 7y ago
Simplest example of what not a manifold is the shape of the letter “Y”. Technically for a topological space to be a manifold, every small neighborhood around each point should look the same. But “Y” has a special point at the center.

lordnacho 8y ago

Is the conclusion that a manifold is a space that looks like a lower dimensional space that it really is?

Topolomancer 8y ago
To some extent, yes! It needs to have a 'nice' structure, though---this is why we have the condition that is should resemble, say, an Euclidean space.
JulianWasTaken 8y ago
You can have a thing in some higher dimensional space that still "looks like" a thing in lower dimensional space (like the surface of the sphere in R^3 being two dimensional), but R^3 itself is a 3 dimensional manifold that "looks" 3 dimensional as well.
mlevental 8y ago
no. there's no requirement for lower dimension. like another person that responded to you R^3 is 3 dimensional manifold. a (smooth) manifold is a space you can do calculus on. that means you have a consistent notion of distance (so that you can figure out when points are close together, ie you approach limits). the set of charts and continuous transformations between them is what encodes this constraint.
patrickthebold 8y ago
Yes and no. The definition doesn't require that what you have is embedded in a higher dimensional space. Though that's a good source of examples. https://en.m.wikipedia.org/wiki/Whitney_embedding_theorem https://en.m.wikipedia.org/wiki/Whitney_embedding_theorem Basically says that many manifolds can be described as a lower dimensional submanifold of some euclidean space.

ErotemeObelus 8y ago

I think everything said here can applies to just topologies.

dan-robertson 8y ago
That isn’t really true. Dimension isn’t so well defined in topology but is reasonably straightforward with a manifold. The article also touched on geometry (sum of angles of a triangle). To be topological a bunch of things must change. The bug walking example doesn’t work and is demoted to an analogy. All notion of distance and direction is lost. If the article were about topology then it would probably be talking about distinguishing shapes by homotopy/homology rather than lines and walks and angles. The article also gets some details a bit wrong/fuzzy: it says you can tell if you are on a sphere by walking in a straight line infinitely far and seeing if you ever cross yourself. But this property also follows on the surface of a (rounded at the top) cone. Even if you require this property in all directions you get problems on eg a torus.
- Topolomancer 8y ago
  Thanks for pointing out the fuzziness! I did not want to extend this further so as not to confuse readers who have no familiarity with the subject. Of course, the fact that you cross your own path at some point is _not_ indicative of living on a sphere; there are numerous other manifolds that satisfy this (not to speak of the torus, as you yourself mentioned). I will try to be more precise about this in a subsequent article!
  - ErotemeObelus 8y ago
    Don't. Precision obfuscates more than it increases resolution. You don't want to be precise. You want to increase the resolution of the key ideas being seen. This is also why parables are so popular for teaching: they're literally not true, but they resolve to an image of something that is true. Please don't introduce homotopy/homology. Lines and angles are perfect.
- senderista 8y ago

ulucs 8y ago

All is well until the poor bug finds out it's on a Klein Bottle, or worse yet RP2. Is there any way to distinguish between orientable or nonorientable surfaces when you're simply walking on it? If the bug was also 2d, I might have had some ideas hut I'm totally blanking for 3d.

ColinWright 8y ago
> Is there any way to distinguish between orientable or nonorientable surfaces when you're simply walking on it? Draw a circle on the surface and orient it. If the surface is non-orientable then there's a way to take a long walk, return to the circle, and find that the direction of the orientation has changed.
- tgb 8y ago
  This depends upon the circle you chose. Nonorientable manifolds still have many orientable circles (any sufficiently small loop is orientable).
  - ColinWright 8y ago
    You confuse me. If the surface is non-orientable then I can draw a very small circle, put an orientation on it, then there's a path I can take that brings me back to find the orientation has changed. Let's be more explicit. Take a non-self-intersecting embedding of RP^2 in R^4, a point P in our RP^2, and a sufficiently small epsilon e. Then take three points on the circle of size e centred on P, and think of them as going in order, thus defining an orientation. Now take an appropriate walk around RP^2 and return to the circle. For an appropriate walk I will now find that the order of the points on my circle has reversed. That's a more precise way of saying what I originally intended, and in that context your comment doesn't make sense to me. Can you expand on it?
    - a1369209993 8y ago
      I think when they say "circle" they're talking about your path, not your orientation reference. That's probably wrong terminology, but it makes some sense if you're thinking about how the path has to be a closed loop.

joncrane 8y ago

It's funny, because in my much younger days I dropped out of college and worked as a mechanic for 3 years. In the automotive world and internal combustion World, manifold has a much different meaning, though related. Basically, a manifold means something that takes the flow of gases from a one-to-many or a many-to-one. An intake manifold takes one single entry point for air feeding the engine and splits up into a separate input for each cylinder. An exhaust manifold takes the hot exhaust from each cylinder separately and combines them into one big pipe.

tyingq 8y ago
That's what manifold means to me as well. Though, there are exceptions to the "one" part, like intake manifolds with two inlets (dual plane is common), exhaust manifolds with two outlets (common on inline 6 engines), and also some really odd setups: https://speedmaster79.com/media/catalog/product/cache/1/thumbnail/9df78eab33525d08d6e5fb8d27136e95/p/c/pce148.1086_2.jpg https://speedmaster79.com/media/catalog/product/cache/1/thum...
- joncrane 8y ago
  That picture looks like an intake for a V8 and each trumpet is still for an individual cylinder; it's just that the intake ports on the heads are paired up for manufacturing and thermal efficiency.
  - tyingq 7y ago
    My point was that manifolds aren't all many to one. This one is clearly many to many.
commandlinefan 8y ago
I actually clicked the link expecting exactly that explanation.
- Topolomancer 8y ago
  Sorry! It just goes to show that I am living in a filter bubble of mathematics. My intention was not to deceive you :-)

tw1010 8y ago

Love me some explanation innovation. More of this and the coding world will reach a plateau.

xelxebar 8y ago

If anyone has questions (especially technical ones), the ##math channel on freenode is pretty phenomenal. It's quite active, with quite a few grad students, across a range of fields. The best part is how friendly and helpful they are to learners. Anyway, highly recommended.

mlevental 8y ago
this in general accurate but I would warn people to have a thick skin in there because there are some seriously toxic people in there (TRW3W or whatever his/her name that was fairly knowledgeable but hung out in there I think purely to assert his/her superiority).
- microcolonel 8y ago
  This is good advice for existing at all in the world. They're not toxic except in high doses, like all sorts of people.
- hhmc 8y ago
  I still recognize that name/handle and I haven't logged into ##math for ~10 years, so I can certainly corroborate what you're saying.

krosaen 8y ago

Another connection to make: quaternions can be represented as a Lie group, which is a (smooth differentiable) manifold. This means representing 3d orientation using quaternions works better for optimization than e.g Euler angles (roll, pitch, heading) in things like SLAM.

mlthoughts2018 8y ago
What do you mean “works better”? Symbolically simpler once you know the requisite math behind Lie groups? In terms if computation, why would one representation be any different than the other? Unless you mean there’s a subset of certain computations where that representation is better but also a subset where it is worse? Similar to how there are cases where a Hough transform allows a more efficient calculation, but many cases where you can’t efficiently extract other calculations out of the Hough transform with effectively inverting it entirely.
- ColinWright 8y ago
  Representing rotations via Euler angles can lead to gimbal lock[0], whereas using quaternions doesn't. The best explanation I could find quickly is this one: https://mathoverflow.net/a/95908 https://mathoverflow.net/a/95908 [0] https://en.wikipedia.org/wiki/Gimbal_lock#In_applied_mathematics https://en.wikipedia.org/wiki/Gimbal_lock#In_applied_mathema...
  - blt 8y ago
    euler angles are sort of the odd one out, a particularly bad way to represent rotations. Orthonormal matrices, quaternions, skew-symmetric tensors with exponential map -- all represent SO(3) without gimbal lock, with different pros and cons for each.
- yiyus 8y ago
  They are more practical, specially when using a computer. Besides the gimbal lock problem mentioned in another problem, Euler angles are not easy to compose. It is also not obvious how to calculate the misorientation between two rotations with Euler angles, but it is trivial with quaternions. Other approaches have their own problems. For example, matrices are more expensive to use in a computer (normalizing a quaternion to avoid rounding errors is much faster than normalizing the equivalent matrix, and quaternions need less memory), and Rodrigues vectors, while usually very useful, present the problem of infinite values for rotations of 180 degrees (which, in my opinion, should not be a problem, but it is in many programming languages). At the end of the day, quaternions present the best compromise.

no_identd 8y ago

"Quick" etymological fact about what mathematicians in English call Manifolds: If one would translate Riemann's original German word for them, "Mannigfaltigkeit", it would translate to "manifoldyhead", or, more understandable to the speaker of Modern English: Manyfoldyhood. (Think of -head as in "Godhead", not as in "Brotherhood", in the same sense that the "ring" in "algebraic ring" refers to "ring" in the sense of "smuggler ring", not in the sense of "gold ring". The two suffixes merged in English, and have very different and rather complex etymological origins.) While this word sounds somewhat ridiculous and perhaps a bit infantile in English, it, in my opinion, conveys what we mean by them significantly better. Clifford tried to accommodate Riemann's highly specific word choice in his translation, as Jost notes: "The English of Clifford may appear somewhat old-fashioned for a modern reader. For instance, he writes “manifoldness” instead of the simpler modern translation “manifold” of Riemann’s term “Mannigfaltigkeit”. But Riemann’s German sounds likewise somewhat old-fashioned, and for that matter, “manifoldness” is the more accurate translation of Riemann’s term. In any case, for historical reasons, I have selected that translation here." Unfortunately, neither Jost (a native German speaker!) nor Clifford realized that English can and does accommodate Riemann's exact meaning directly. To make a comparison which might require quite a bit of German knowledge beyond high school education to describe in exact linguistic terms, but which native speakers should hopefully find intuitive to distinguish (I at least, do): Translating "Mannigfaltigkeit" as "Manifoldness" seems equal to mistranslating "Geheimnis" as "Secrethood" and "Geheimheit" as "Secretness", whereas the opposite pairing would yield an accurate (albeit not necessarily immediately apparent—in terms of the differences between the two—to the English native speaker, unless explicitly pointed out) translation. So much for the last suffix, but that still leaves insertion of the one before it (' * foldy * ' instead of ' * fold[( * )]' unclear.) One may observe the difference involved there by a converse example, also involving "Geheim", as well as the root word of it, "Heim", by dragging forth a rather rare and archaic—but none the less highly likely intuitive to the native speaker—word:"Geheimig". Whose meaning starkly differs from both "Geheimnis" and "Geheimheit". For further language related hijinks related to /Manyfoldyhoods/, see: A) The Dutch word for them, which would back translate to "Variety"; and B) This quote by Poincaré: "I prefer the translation of Mannigfaltigkeit by multiplicity, because the two words have the same etymological meaning. The word set is more adapted to the Mannigfaltigkeiten considered by Mr. Cantor and which are discrete. It would be less adapted to those which I consider and which are discontinuous." (As I don't speak French, I can't make much of any statements about the accuracy about the etymological claim by Poincaré, so I'll close with another Poincaré quote instead: " Mathematics is the art of giving the same name to different things." —Henri Poincaré)

braxxox 8y ago

What a great read! I've never thought about how the way we experience the surface of the earth as 2D manifold of a 3D space, and how up until recently (relatively speaking) this was unknown. It's interesting to expand on this idea and realize that maybe we are making the same mistake again, and that from our local perspective the universe is 3D, when in reality, it's a 3D manifold of a higher dimensional space. One thing I am still unclear of though, is that isn't this proven to be the case? Is it not true that we are provably living in at least a 4D space, where time is the fourth dimension? We can observe its existence but cannot move freely through it and are confined to free transformations only in 3D space? In this way, aren't we living in a 3D manifold in a 4D space? So maybe then the question is are we living in a 3D manifold of a +4D space?

crdrost 8y ago
Sort of, not quite. The basic background is that Einstein wrote a paper taking Lorentz more seriously than Lorentz took his own work: and that paper suggested that just maybe, when you accelerate in any given direction by an acceleration A, you see all of the clocks ahead of you some distance z tick faster by a factor A z / c², where c is the speed of light in vacuum, and behind you they tick slower with the corresponding negative z until a wall of death at z = -c² / A where clocks do not tick at all and time appears to stand still. This is in fact the only new fact that special relativity adds. Lots of people got very confused about the philosophical implications, but the mathematical implication is that time and space can be mixed together by these accelerations and must be treated as one unified geometrical entity. Einstein then went one further on the whole 4D thing, because arguably we are always accelerating in this whole gravitational field of the Earth. You have a lot of options to choose from. So this part took Einstein many many years to work out. Maybe the easiest is to say that we standing on Earth are a non-accelerating reference frame, and then anybody who falls must see a wall of death somewhere out in space. That turns out to be a very boring approach, and also wrong. What Einstein suggested instead was that you are in a non-accelerating reference frame with no wall of death if you are in free-fall, and we standing on the Earth would see a wall of death beneath our feet, except we can't see beneath our feet. But if a body were more massive, maybe we could see the wall of death from orbit. And now we have a photo of a black hole to prove it! But even before that, the essential point is that if I put a clock up somewhere high (on a tower, in a plane, or at the top of a mountain) and I am standing on the ground, then I am accelerating towards that clock relative to free-fall: so that clock must be ticking faster than my clocks are. And we have had direct observation of that “gravitational time dilation” for a long time. Mathematically, this means that we are in a manifold that looks locally 4 dimensional, in this weird way of coupling the four dimensions that couples accelerations with the ticking of clocks. We say, going back to this guy who worked out all of the mathematics before Einstein, that the manifold is locally Lorentzian, as opposed to Euclidean. But the manifold is four dimensional, not three dimensional. It has to have this coupling between time and space locally. But then globally it can have these interesting features like black holes. Now, whether we can embed this curvy universe that we inhabit into a larger dimensional flat space, is not necessarily a given. I don't know many physicists who are deeply interested in that sort of question. Certainly to have the structure that we need, it needs to have Lorentzian timelike dimensions in it, one of which we use as our time dimension. Certainly also, the people working on string theory use these extra-dimensional possibilities to solve certain mathematical inadequacies that their string theories otherwise have: but usually those dimensions are locally available, so we would see them; so there is some sort of hand-waving about how they must be curled up into such a small length scale that we cannot actually observe them. But there is certainly a branch of string theory called M-theory that I do not personally know too much about which has something to do with viewing our universe as a geometric entity in a larger space. Thankfully, the mathematics does not require this. The essential point of a manifold that the article somehow leaves out, is that I am no longer going to rely on global coordinates. I only care about local coordinates. So on the sphere, it is a two dimensional object, even though it lives in a three dimensional space, and that's because depending on where I am, I can uniquely identify points near me on the sphere by either their x & y coordinates, or their y & z coordinates, or their x & z coordinates. This fails for points that are not nearby me, because projecting a globe on to a flat surface this way will project two hemispheres onto the same point. But I can always choose one of those three and find myself in the middle of a hemisphere, and describe everything else on that hemisphere with those coordinates. So the whole point of a manifold is that I don't need global coordinates, and therefore it doesn't matter much whether or not I am embedded in a larger flat space or not.

colllectorof 8y ago

If you want more, there is an excellent course on "introductory" topology on YouTube: https://www.youtube.com/watch?v=CEXSSz0gZI4 https://www.youtube.com/watch?v=CEXSSz0gZI4 It is made by a physicist, not mathematician, and it's a great combination of informal explanations and carefully covering the actual math definitions. I feel that his approach works way better than making analogies. There is another course from Wildberger. A bit more hardcore, but also has some interesting perspectives: https://www.youtube.com/watch?v=Ap2c1dPyIVo&list=PL6763F57A61FE6FE8 https://www.youtube.com/watch?v=Ap2c1dPyIVo&list=PL6763F57A6...

amai 8y ago

So a manifold is a surface?

chadcmulligan 8y ago
it can be, I think of a manifold (probably naively) as the mathematical abstraction that you end up with if you start from Euclidean space and fiddle with different features - e.g. if you're in a Euclidean space how do you get from one point to another - what are the minimum requirements? What if we want calculus to work what are the minimum requirements? If we're on a surface and we walk around is there any way to tell what shape that surface is?
hopler 8y ago
A generalized surface in arbitrary dimensions. A common English "surface" maps naturally to a 2D manifold embedded in 3D space.
- amai 8y ago
  Ok, so a manifold is simply a hypersurface. At least I‘m pretty sure in machine learning it means exactly that.
  - ginnungagap 8y ago
    A hypersurface is usually a codimension one object, while a manifold can have bigger codimension or even not be embedded in an higher dimensional space at all
    - amai 7y ago
      Ok, a line in 3D space can also be called a manifold? Maybe a better question is then to ask, what is not a manifold?

valw 8y ago

Nice article, it would have been nice to add a word or two about manifold embedding though.

Topolomancer 8y ago
I plan on adding this in a sequel :) Thanks for the suggestion!

arduinomancer 8y ago

I think the natural next question is how does a bug in R3 check if its part of a 4-dimensional manifold? This is related to people measuring the curvature of the universe right?

improbable22 8y ago
The article squeezes together two notions, of whether a manifold is curved, and whether it's a submanifold of some higher-dimensional one. Sub-spaces of R^3 are a useful way of generating and picturing examples of 2D manifolds. But these things exist by themselves. As the article mentions with angles of a triangle, you can tell that a 2D manifold is curved from living inside. Likewise you can tell whether a 3D manifold is curved of not, without any mention of a 4th dimension. Questions of curvature of the universe are a step harder, as we are talking about 3+1-dimensional space-time. But the slice of constant time is a 3D manifold, and as far as we can tell right now, it appears to be flat.

alcolade 8y ago

If the bug were truly mathematical, he would notice the curvature of S^1 and S^2 without walking around (since curvature is defined at each point), but that is obviously very pedantic of me and shows why I wouldn't write an article nearly as interesting as this one.

improbable22 8y ago
But S^1 isn't curved. At least if we mean intrinsic curvature, which is the kind that belongs to the manifold itself. S^2 is curved, but locally you can't tell it's not (say) RP^2.
- alcolade 8y ago
  I was thinking of the Riemann curvature tensor which is just some measure of curvature at each point. If the bug were at some point of S^2 he would notice the curvature. Of course, this is assuming S^2 is getting its geometry from a certain embedding into R^3 that comes to mind. You could define different Riemann curvature tensors over S^2 that may have zero curvature in certain places, like squishing a balloon against a flat table for example. I guess my point was that topology and curvature are different things (like you pointed out with your comment about RP^2), and saying that "a sphere looks flat locally" is missing the point! Also, it didn't occur to me that this notion of curvature doesn't make sense in 1d, i.e. for S^1 like you said.
  - improbable22 8y ago
    Not only is there no intrinsic curvature in 1D, but the 2D and 3D versions are also simplified special cases. In 2D there is only scalar curvature, i.e. you don't need the whole Riemann tensor, just one number (at a given point). In 3D you need the Ricci tensor, but still not the full curvature tensor. This is still much simpler, for instance IIRC this is why you cannot have gravitational waves in 3 (meaning 2+1) dimensional spacetime. In 4D you get the full complication.

loeg 8y ago

Ah, and here I was imagining this[0] is a manifold. I must be spending too much time with PEX lately. [0]: https://www.supplyhouse.com/Bluefin-CM4-75PXZ-PXV-3-4-PEX-Crimp-x-Closed-Copper-Manifold-w-1-2-PEX-Crimp-Ball-Valves-Lead-Free-4-Outlets https://www.supplyhouse.com/Bluefin-CM4-75PXZ-PXV-3-4-PEX-Cr...

725686 8y ago

At the end of the article he mentions Jeffrey Weeks... I searched youtube and found this very interesting vid: https://www.youtube.com/watch?v=j3BlLo1QfmU https://www.youtube.com/watch?v=j3BlLo1QfmU