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Timeline for intuition for hochschild homology

Current License: CC BY-SA 4.0

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Nov 12, 2018 at 15:22 history edited John Pardon CC BY-SA 4.0
added caveat relevant if k is not a field
Jan 2, 2018 at 12:07 vote accept Zbigniew
Dec 30, 2017 at 2:45 comment added Pedro This boils down to the fact that if $A$ is $k$-projective, then the bar resolution is a honest projective resolution of $A$ as a bimodule, and so doing $M\otimes B(A,A)\otimes N$ works to compute Tor. There is a silly cancellation (two $A$s for the two tensors over $A$) that gives the complex in this post.
Dec 30, 2017 at 2:45 comment added Pedro @QiaochuYuan One usually defines Hochschild (co)homology using Hochschild's complex, which ends up computing a relative Tor (or Ext). When the algebra is flat (or projective) over the base, you get usual Tor (or Ext) --no hypothesis should be necessary on the modules. I guess it merely depends on what you want to call HH, but I would follow Hochschild's original definition. (...)
Dec 30, 2017 at 1:11 comment added Qiaochu Yuan John is omitting the differential. Also I think this complex only computes the right thing if $M, N, A$ are all flat over the base commutative ring.
Dec 30, 2017 at 1:08 comment added darij grinberg Wait. Are you saying the Tors are literally the homology of a chain complex of tensor products? Of actual tensor products over $\mathbb {Z} $ or whatever base ring we are working over? Or is the tensor product inside the direct sum some sort of shorthand or metaphor?
Dec 30, 2017 at 0:24 comment added Qiaochu Yuan It means the derived tensor product: stacks.math.columbia.edu/tag/09LP
Dec 27, 2017 at 10:46 comment added Zbigniew Thanks for this nicely explanation but I have some confusion: What did you mean by $B\otimes_A^{\mathbb L}{}$ I do not meet this notation before.
Dec 27, 2017 at 10:40 vote accept Zbigniew
Dec 31, 2017 at 15:08
Dec 26, 2017 at 16:04 history answered John Pardon CC BY-SA 3.0