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Aug 24, 2019 at 18:15 history edited Nicolas Boerger CC BY-SA 4.0
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Aug 24, 2019 at 6:53 comment added user43326 Since for any $Z/2[Z/2]$-module $M$ there is a splitting $M=F\oplus T$, computing $H^0$ is actually equivalent to computing the rest.
Aug 23, 2019 at 21:23 comment added Dylan Wilson for the dual Steenrod algebra itself it is an unsolved problem (as far as I know) to even compute H^0, let alone the rest.
Aug 23, 2019 at 21:09 history asked Nicolas Boerger CC BY-SA 4.0