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Jun 19, 2014 at 22:23 vote accept Lisa
Jun 19, 2014 at 22:09 answer added Misha timeline score: 14
Jun 19, 2014 at 22:06 history edited Lisa CC BY-SA 3.0
Whoops, I got the arrows reversed (though since $G$ is finite and we are working over $\mathbb{Q}$ the invariants and coinvariants are equal, so really there no error)
Jun 19, 2014 at 20:48 review First posts
Jun 19, 2014 at 20:58
Jun 19, 2014 at 20:31 history asked Lisa CC BY-SA 3.0