You Could Have Invented Spectral Sequences (2006) [pdf] 63 points by espeed 6 months ago | hide | past | web | favorite | 7 comments

 Though I have to say that [chain complexes](https://en.wikipedia.org/wiki/Chain_complex) and their [homotopy](https://en.wikipedia.org/wiki/Homotopy_category_of_chain_com...) are somewhat more difficult than the structure you need to understand monads. You can even explain monads purely in programming terms, without reference to category theory.
 Here are some illustrations of chain complexes:Homology for Normal Humans http://isomorphismes.tumblr.com/post/127950269154/graded-cha...
 Top of page 2: "Here is a simple example. Suppose we have a chain complex..."And right there I'm lost because the paper hasn't explained what a chain complex is, and I don't understand the notation. I guess I couldn't have invented spectral sequences.
 I think this was supposed to be, "if you have an undergraduate education in mathematics along with the basics of algebraic topology, you could have invented spectral sequences".The prototypical example of a chain complex is a triangulation of an n-dimensional surface. You can take each n-dimensional triangulation into its (n-1)-dimensional boundary, with the property that the boundary of a boundary is empty. Roughly, this means that you have a sequence of maps on "chains" (prototypically, triangulations)`````` d_{n} d_{n-1} d_{2} d_{1} C_{n} -----> C_{n-1} ---------> ... -------> C_{1} -------> C_0 `````` with the property that d_{k_1}(d_{k})(x) = 0 for all x and all k (or more abbreviated, d^2 = 0), the boundary of a boundary is zero.
 Yeah, it assumes basic algebraic topology (which is not a mild assumption). He states at the top that he is assuming familiarity with homology groups, and chain complexes are used to define homology groups.
 Well, as my PhD advisor once told me, those who don't understand homology are doomed to reinvent it...So maybe you'll build your own chain complexes if left to your own devices long enough.

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