﻿What is “smooth” in smooth manifolds? Why do we care about smoothness in the first place?


Let’s look at it from a lense of isomorphism. In other words can you morph it from one to the other. If yes then those two things are equal at that level of isomorphism.


Level 1 is isomorphic at sets. Do you have a bijection between set A and set B.
Level 2 is isomorphic as topological spaces. Do you have a homeomorphism between set A and set B.
Level 3 is isomorphic as a smooth manifold. Is there a diffeomorphism between set A and set B.


Different examples of A and B will fail at different levels.


Consider a sphere and a potato. This pair will be equal for level 1, level 2, and level 3.


Level 1 you can get the answer smartly because the existence of a bijection is entirely dedicated by the cardinality of the sets, which in this case is equal (both uncountably infinite). Because both sets are uncountably infinite, they are the same as sets. There’s nothing to differentiate them. So to wrap up, the mental trick here is to ask “do A and B have the same cardinality?” to know if they’re equal or not.


Level 2 you can’t get the answer smartly because topological spaces don’t have a defining classification in terms of something. The category is simply too large. There are too many topological spaces. So the proof that the sphere and potato is a bit more manual, but there is still a mental trick you can imagine. You need to mentaly confirm or deny the existence of a homeomorphism between A and B. Pretty much, can you imagine forward and inverse mapping every point to each other in a way that continuity is preserved, or said more elegantly, can you imagine deforming object A into object B continuously, which is to say without any tear, or filling in any holes. This is equivalent to the classic introduction of topology as a donut and sphere are equal because you can morph one into the other like this
  

So in this case I can clearly imagine a sphere and potato being equal because I can imagine a continuous transformation of one into the other. Hence the homeomorphism exists.


Level 3 is isomorphic as a smooth manifold and that’s also I believe not immediately trivial, but some neat tricks arise. First off you can imagine a diffeomorphism between A and B meaning you can mould A into B, except this time creases don’t go away and must be preserved. So a cube is not diffeomorphic to a sphere for example, because the corners are creased. The idea is that you need the existence of a diffeomorphism map.
Let me talk about what smoothness means in the first place.


Smoothness of a curve means that all of the charts in the atlas agree about the curve’s smoothness, yes/no, where smoothness in each chart is defined the euclidean way (since charts are euclidean anyways). This is to say that all charts are compatible.


Let me restart.


A manifold becomes a smooth manifold when it’s equipped with a smooth atlas.


A smooth atlas is a C-infinity atlas, which means that it’s a selection of charts that are all C-infinity compatible with each other. Which means for each intersection between two charts, the chart transition map is C-infinity, which is to say smooth. So we’ve stated smoothness on a manifold in terms of smoothness on R^n, which we already know how to define, which is great.


Pretty much another way to say it is that each chart must agree with each other about the differentiability or non-differentiability of a curve on a manifold. There’s nothing objective about it, it’s just that all charts agree.


The manifold must basically contain a mechanism for judging whether a curve is differentiable or not (duh), and this mechanism must be valid, in all edge cases, which funnily in this case literally can mean edge of chart, where two charts intersect.


OK but not really. What a smooth manifold exactly requires is that each chart transition map on shared neighborhoods is smooth. More specifically, the chart coordinate transition functions must be differentiable in R^1, which is very intuitive to us.


I wonder: can this be stated in terms of changing frames of references, and making sure that the change of reference frame doesn’t break smoothness of an observed curve? I think so.


A frame of reference is essentially a chart of the universe. So when a smooth manifold says that chart transition maps must be smooth, it’s saying that the function responsible for the change of frames of reference is smooth.


* Lin alg
* Tensors
* Bundles
* Tangent spaces
* (Tangent tensor bundles)


What is linear algebra? Linear algebra is the study of vector spaces and their structure preserving maps, which is to say linear maps.


Why are linear maps “structure preserving”? Why is the linearity on the + and x considered necessary to preserve structure? Well it all has to do with vector spaces. Imagine A is guaranteed to be a vector space, and you have a map f. That map f might be messed up. What restrictions do you have to put on f so that the image of A under the map f, which we will call B, is a vector space.


Well it turns out we must restrict f to be linear. Quite neat.


At the start of linear algebra, we start with vector spaces, and we ask “is this a vector space?”, “how about this one, is it a vector space?”, asking these kinds of questions for odd samples. We come up with algorithms like for example saying that a subset of R^n will be a vector space given certain conditions (closed under +, x, and contains 0).


The next thing we do is talk about linear independence and span, so that we can talk about basis, so that we can talk about dimension of a vector space, because it turns out that vector spaces are classified by dimension. Which means you can immediately know if there exists an invertible linear map between V and W, depending on if they have the same dimension. If not then that map does not exist. This can be quite useful to replace one kind of vector space with another you’re more familiar with if they’re identical under the eyes of the category.


Next we have a more detailed look at linear maps, 


The classification of surfaces is determined by how many donut holes are in the surface.