Skip to main content
14 events
when toggle format what by license comment
Feb 23, 2016 at 17:18 answer added Tyler Lawson timeline score: 6
Feb 23, 2016 at 12:12 comment added Fernando Muro @SebastianGoette You're right, I somehow missed the track of the fact that we have a tensor product.
Feb 23, 2016 at 12:09 comment added Sebastian Goette @FernandoMuro I don't understand your comment (or maybe I don't understand the question correctly). $G=\{1\}$ gives the algebraic Künneth formula at $E_2$, which holds without Tor terms because $\mathbb C$ is a field. Note however that the tensor product of singular cochain complexes is not the singular cochain complex of the product, so the example at the end is misleading.
Feb 23, 2016 at 11:36 comment added Fernando Muro The answer is not positive in general, not even for $G$ finite. Take $G=\{1\}$. You can also construct more involved examples, of course.
Feb 23, 2016 at 11:30 history edited Sebastian Goette CC BY-SA 3.0
Corrected Borel construction near the very end.
S Feb 23, 2016 at 8:30 history suggested Ali Taghavi CC BY-SA 3.0
I add a tag
Feb 23, 2016 at 8:19 review Suggested edits
S Feb 23, 2016 at 8:30
Nov 24, 2014 at 0:35 history edited Yeping Zhang CC BY-SA 3.0
added 1 character in body
Nov 23, 2014 at 21:39 history edited Yeping Zhang CC BY-SA 3.0
added 12 characters in body
Nov 23, 2014 at 21:34 comment added Yeping Zhang It is tensor product over $\mathbb{C}$, and yes, i wrote $\oplus$ by mistake, thanks a lot.
Nov 23, 2014 at 21:33 history edited Yeping Zhang CC BY-SA 3.0
added 12 characters in body
Nov 23, 2014 at 21:31 comment added John Pardon Is this tensor product over $\mathbb C$ or over $\mathbb C[G]$? (I assume you must mean $\otimes$ instead of $\oplus$ in the definition of the product complex).
Nov 23, 2014 at 21:25 review First posts
Nov 23, 2014 at 21:39
Nov 23, 2014 at 21:23 history asked Yeping Zhang CC BY-SA 3.0