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The hardest thing for me to grok trying to learn enough AT for TDA (but: I mean really learn, not just understand in a loose Medium-blog kind of way) was the ke

The hardest thing for me to grok trying to learn enough AT for TDA (but: I mean really learn, not just understand in a loose Medium-blog kind of way) was the ker/im definition of homology. I pestered people on IRC for days, and just worked hard to stretch my brain enough to see that it applies in many "small" cases. If you're coming from topology to AT and can grok the ker/im stuff, TDA should be easy peasy to you. I flunked (I think I scraped a couple of points on appeal by sophistry and desperation so I passed, but really) real analysis, so I never moved into topology beyond skimming books to "learn about" key concepts. You simply can't have too much math in life. I'm having a kid soon and I struggle between what seems truer to my core values (let the kid have a childhood and then personal inclinations; advise but not push; if he wants to do the gifted kid thing, yay) and what seems practical advice (learn as many human languages as you can, at least English, French and German; learn math deeply, suffering along the way).

spekcular 6y ago
Did anyone suggest thinking of ker/im as "boundaries modulo cycles"? That is, homology detects "holes," loosely speaking, and a hole is exactly something that you can draw a cycle around, where that cycle is not the boundary of some disk (in 2 dimensions, for example). To be even more concrete, if I cut out a disk from a piece of paper, I've created a cycle (the circle bounding the disk) that is not the boundary of any disk in the piece of paper (because I cut it out). Admittedly, this gets a little funkier in higher dimensions, but for 2 dimensions, it's a good guide. (I think Hatcher's book has a discussion of how to make this idea precise in higher dimensions.)
- prionassembly 6y ago
  It was the "modulo" I had trouble with (the idea of quotient spaces). I understood well how kernels corresponded to cycles (people on IRC made me do lots of calculations with chain complexes both in Z/2 and R), and had some idea of groups (or, at least, of when something is a semigroup and not a group...) but didn't have an internal feeling of what quotient groups were.
  - jvvw 6y ago
    Do you grok equivalence classes because quotients (of anything) are basically sets of equivalence classes? And on the parenting front, luckily you start with a baby not a ten year old so you have lots of chance to grow as a parent before making those sorts of decisions - a lot of parenting is deciding when to be directive/non-directive and when to take charge based on your superior understanding of the world (not letting your toddler walk into the road!) and when to let them learn from their own mistakes.
- Y_Y 6y ago
  This is just the same as my own conception, then again I also used (and can highly recommend) Hatcher. He kindly provides a pdf on his webpage here: https://pi.math.cornell.edu/~hatcher/AT/ATpage.html https://pi.math.cornell.edu/~hatcher/AT/ATpage.html
- bikenaga 6y ago
  Small correction - it's "cycles mod boundaries". In general, a quotient object R/I is read "R mod I" (top mod bottom). What's a cycle? It's something in the kernel of the boundary map. To be in the kernel means the boundary map takes it to 0. So heuristically, cycles are "things with no (zero) boundary". For the quotient cycles/boundaries to make sense - for any quotient to make sense - the bottom should be contained in the top. Like in Z/6Z, the integers mod 6, the "6Z" (the integers which are a multiple of 6) is contained in Z (the integers). In this case, for "cycles/boundaries" to make sense, a boundary should be a cycle - that is, a boundary should have no boundary. This is (again heuristically) what the definition of a boundary map (d^2 = 0) means. Consider your disk/circle analogy. The circle is a boundary (the boundary of the disk). Does the circle have a "boundary"? No - the boundary of a boundary is 0. When the quotient cycles/boundaries is nonzero, you have nontrivial homology. Think of nontrivial homology as "having a hole". To see an example of this, take the plane but remove the origin (0, 0). Consider your circle again (say the unit circle x^2 + y^2 = 1). It's still a cycle, because its boundary is 0 (no boundary). But it's not the boundary of "something" (where "something" means roughly "something nice" - it's hard to explain this better without getting technical). So the cycle represented by the circle represents a nontrivial homology class (in the homology of plane minus origin).
WoahNoun 6y ago
Yea, my own mental picture of homology and cohomology tends to default to DeRham (co)homology or the axiomatic approach versus the combinatorial approaches that are normally taught (I'm also bad at combinatorics). I'm going to be hand wavey here, but in DeRham, the "boundary" operator is basically just the derivative. So the DeRham homology of a manifold (the ker/im) are the functions that differentiate to zero (locally constant functions) / functions that can be integrated (eg functions that are the derivative of some function). You get left the functions that are "interesting" and define the geometry of the manifold in some way. Since manifolds are patched together pieces of Euclidean space, locally constant functions don't need to be globally constant which is why the quotient operation isn't trivially zero. If you've never seen quotient rings/groups/topologies before though, I can see how ker/im would be a major trip-up no matter the homology domain. The crazy part of algebraic topology (to me) is that all these homology theories are isomorphic. The DeRham homology and the simplicial homology give you the same homology group despite not having much to do with each other at first glance. (Stoke's theorem provides the isomorphism).
- prionassembly 6y ago
  Wikipedia is wonderful at giving you the handwavey picture, and then you can even skim some more advanced papers with handwavey ideas (e.g. there's a rigorous theory of PDEs in graphs relating the Khirchoff/incidence matrix to the de Rham operator, and then developing a theory of how graph Laplacians are really the Laplacians from calculus). But I kind of have a BDSM relationship with mathematics. I'm not happy until it makes me suffer.
- auntienomen 6y ago
  Wait until you learn about modern homotopy theory. It turns out that basically every cohomology theory can be described as homotopy classes of maps into some space. The standard cohomology groups with integer coefficients H^n(X; Z), those are just homotopy classes of maps from X into the Eilenberg-MacLane K(Z,n). Want K-theory instead? No problem. That cohomology group K(X) is homotopy classes of maps [X, Z x BU], where Z is the integers and BU is the classifying space for the unitary group. It turns out essentially all cohomology theories are represented this way. h(X) = [X,S_h], where [] is homotopy classes of maps and S_h is a topological space called the spectrum of the cohomology theory. Even wackier, the various algebra operations you can do on cohomology classes are mirrored by / derived from algebraic operations on the spectra. Which means that much of the machinery of algebraic geometry can be used to study cohomology theories.
Y_Y 6y ago
I'm sold. I'm not very good at maths, but modding the image of ∂ₙ by the kernel of ∂ₙ₊₁ is something I "understand" in the Von Neumann sense. That is to say I've become "used to it" and now it seems to me utterly obvious. (I accept Dunning-Kruger objections may apply.) Anyway, where is this Topological Data Analysis that's going to be easy for me?
- prionassembly 6y ago
  My tip is to look for slide decks before diving into papers.

spekcular 6y ago

Did anyone suggest thinking of ker/im as "boundaries modulo cycles"? That is, homology detects "holes," loosely speaking, and a hole is exactly something that you can draw a cycle around, where that cycle is not the boundary of some disk (in 2 dimensions, for example). To be even more concrete, if I cut out a disk from a piece of paper, I've created a cycle (the circle bounding the disk) that is not the boundary of any disk in the piece of paper (because I cut it out). Admittedly, this gets a little funkier in higher dimensions, but for 2 dimensions, it's a good guide. (I think Hatcher's book has a discussion of how to make this idea precise in higher dimensions.)

prionassembly 6y ago
It was the "modulo" I had trouble with (the idea of quotient spaces). I understood well how kernels corresponded to cycles (people on IRC made me do lots of calculations with chain complexes both in Z/2 and R), and had some idea of groups (or, at least, of when something is a semigroup and not a group...) but didn't have an internal feeling of what quotient groups were.
- jvvw 6y ago
  Do you grok equivalence classes because quotients (of anything) are basically sets of equivalence classes? And on the parenting front, luckily you start with a baby not a ten year old so you have lots of chance to grow as a parent before making those sorts of decisions - a lot of parenting is deciding when to be directive/non-directive and when to take charge based on your superior understanding of the world (not letting your toddler walk into the road!) and when to let them learn from their own mistakes.
Y_Y 6y ago
This is just the same as my own conception, then again I also used (and can highly recommend) Hatcher. He kindly provides a pdf on his webpage here: https://pi.math.cornell.edu/~hatcher/AT/ATpage.html https://pi.math.cornell.edu/~hatcher/AT/ATpage.html
bikenaga 6y ago
Small correction - it's "cycles mod boundaries". In general, a quotient object R/I is read "R mod I" (top mod bottom). What's a cycle? It's something in the kernel of the boundary map. To be in the kernel means the boundary map takes it to 0. So heuristically, cycles are "things with no (zero) boundary". For the quotient cycles/boundaries to make sense - for any quotient to make sense - the bottom should be contained in the top. Like in Z/6Z, the integers mod 6, the "6Z" (the integers which are a multiple of 6) is contained in Z (the integers). In this case, for "cycles/boundaries" to make sense, a boundary should be a cycle - that is, a boundary should have no boundary. This is (again heuristically) what the definition of a boundary map (d^2 = 0) means. Consider your disk/circle analogy. The circle is a boundary (the boundary of the disk). Does the circle have a "boundary"? No - the boundary of a boundary is 0. When the quotient cycles/boundaries is nonzero, you have nontrivial homology. Think of nontrivial homology as "having a hole". To see an example of this, take the plane but remove the origin (0, 0). Consider your circle again (say the unit circle x^2 + y^2 = 1). It's still a cycle, because its boundary is 0 (no boundary). But it's not the boundary of "something" (where "something" means roughly "something nice" - it's hard to explain this better without getting technical). So the cycle represented by the circle represents a nontrivial homology class (in the homology of plane minus origin).

WoahNoun 6y ago

Yea, my own mental picture of homology and cohomology tends to default to DeRham (co)homology or the axiomatic approach versus the combinatorial approaches that are normally taught (I'm also bad at combinatorics). I'm going to be hand wavey here, but in DeRham, the "boundary" operator is basically just the derivative. So the DeRham homology of a manifold (the ker/im) are the functions that differentiate to zero (locally constant functions) / functions that can be integrated (eg functions that are the derivative of some function). You get left the functions that are "interesting" and define the geometry of the manifold in some way. Since manifolds are patched together pieces of Euclidean space, locally constant functions don't need to be globally constant which is why the quotient operation isn't trivially zero. If you've never seen quotient rings/groups/topologies before though, I can see how ker/im would be a major trip-up no matter the homology domain. The crazy part of algebraic topology (to me) is that all these homology theories are isomorphic. The DeRham homology and the simplicial homology give you the same homology group despite not having much to do with each other at first glance. (Stoke's theorem provides the isomorphism).

prionassembly 6y ago
Wikipedia is wonderful at giving you the handwavey picture, and then you can even skim some more advanced papers with handwavey ideas (e.g. there's a rigorous theory of PDEs in graphs relating the Khirchoff/incidence matrix to the de Rham operator, and then developing a theory of how graph Laplacians are really the Laplacians from calculus). But I kind of have a BDSM relationship with mathematics. I'm not happy until it makes me suffer.
auntienomen 6y ago
Wait until you learn about modern homotopy theory. It turns out that basically every cohomology theory can be described as homotopy classes of maps into some space. The standard cohomology groups with integer coefficients H^n(X; Z), those are just homotopy classes of maps from X into the Eilenberg-MacLane K(Z,n). Want K-theory instead? No problem. That cohomology group K(X) is homotopy classes of maps [X, Z x BU], where Z is the integers and BU is the classifying space for the unitary group. It turns out essentially all cohomology theories are represented this way. h(X) = [X,S_h], where [] is homotopy classes of maps and S_h is a topological space called the spectrum of the cohomology theory. Even wackier, the various algebra operations you can do on cohomology classes are mirrored by / derived from algebraic operations on the spectra. Which means that much of the machinery of algebraic geometry can be used to study cohomology theories.

Y_Y 6y ago

I'm sold. I'm not very good at maths, but modding the image of ∂ₙ by the kernel of ∂ₙ₊₁ is something I "understand" in the Von Neumann sense. That is to say I've become "used to it" and now it seems to me utterly obvious. (I accept Dunning-Kruger objections may apply.) Anyway, where is this Topological Data Analysis that's going to be easy for me?

prionassembly 6y ago
My tip is to look for slide decks before diving into papers.