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Aug 18, 2018 at 18:12 comment added darij grinberg @JoeSilverman: Nice observation!
Aug 18, 2018 at 17:40 comment added Joe Silverman Regarding Volker Strehl's identity, it seems to be true for any function, not just $\mu$, although presumably taking $\mu$ has some application. Thus let $f:\mathbb{N}\to\mathbb{Q}$ be any function. Then in $\mathbb Q[[a,b,z]]$, we have the formal identity $$ \prod_{k\ge1} \left(\frac{1}{1-az^k}\right)^{\frac{1}{k}\sum_{d\mid k} f(d)b^{k/d}} = \exp\left( \sum_{d=1}^\infty\frac{f(d)}{d} \sum_{i,j=1}^\infty \frac{a^ib^j}{ij} z^{ijd}\right). $$ so symmetry in $a$ and $b$ is clear. Proof: take logs of both sides, use the series for $\log(1-t)^{-1}$, and flip the order of series.
Aug 18, 2018 at 15:09 history edited darij grinberg CC BY-SA 4.0
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Dec 15, 2013 at 5:14 comment added Ryan Reich The Tor symmetry is basically just that $M \otimes N \cong N \otimes M$, and you take the derived functors of both sides. Generalizing, any and all nice properties of (co)homology groups would seem to be mysterious symmetries if you consider the definition to be messing around with projective or injective modules, and not something more intrinsic like derived functors.
S Dec 14, 2013 at 21:37 history answered darij grinberg CC BY-SA 3.0
S Dec 14, 2013 at 21:37 history made wiki Post Made Community Wiki by darij grinberg