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Feb 15 at 1:07 history edited Tony Huynh
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Nov 24, 2021 at 15:02 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jul 27, 2021 at 15:01 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jun 27, 2021 at 14:22 history edited Tony Huynh
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Jun 26, 2021 at 0:08 answer added Tony Huynh timeline score: 2
Jun 25, 2021 at 22:05 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Feb 25, 2021 at 22:04 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jan 26, 2021 at 20:45 comment added H A Helfgott @FedorPetrov Calling it $H^1$ makes it clear why I said "holey", and also why it should have anything to do with cycles.
Jan 26, 2021 at 17:57 comment added Fedor Petrov Why not to formulate it as $|E|-|V|=d-1$?
Jan 26, 2021 at 17:16 answer added John Machacek timeline score: 0
Jan 26, 2021 at 17:11 comment added H A Helfgott Isomorphic. Just changed it to $\cong$.
Jan 26, 2021 at 17:11 history edited H A Helfgott CC BY-SA 4.0
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Jan 26, 2021 at 17:09 comment added Wojowu What does $\sim$ mean here?
Jan 26, 2021 at 17:01 history edited H A Helfgott CC BY-SA 4.0
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Jan 26, 2021 at 16:56 comment added H A Helfgott I obtain the graph $\Gamma$ I care about by starting with a tree $T$, partitioning its vertices into $d$ equivalence classes - with adjacent vertices being non-equivalent - and take the quotient of the graph by the equivalence relation. I assume that gives me a fully general $\Gamma$, and so knowing that $\Gamma$ arises in this way doesn't help?
Jan 26, 2021 at 16:55 history asked H A Helfgott CC BY-SA 4.0