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Mar 1, 2010 at 19:54 comment added Don Stanley I see a problem since the internal differential in the bar construction could land in terms in $T(V)\otimes T(V)\otimes T(V)$ that are not in $T(V)\otimes V\otimes T(V)$. For example if $V$ is free on basis $a,b$ and $db=a^2$.
Mar 1, 2010 at 19:50 vote accept Don Stanley
Mar 1, 2010 at 19:50
Mar 1, 2010 at 19:44 vote accept Don Stanley
Mar 1, 2010 at 19:50
Mar 1, 2010 at 14:54 comment added Mariano Suárez-Álvarez Indeed. The complex is just a subcomplex of the bar construction.
Mar 1, 2010 at 14:45 comment added Ben Webster Presumably the just the usual differential on the tensor product of compelxes.
Mar 1, 2010 at 14:13 comment added Don Stanley What is the differential on the $T(V)\otimes V\otimes T(V)$ (or $T(V)\otimes V\otimes L(V)$) term?
Mar 1, 2010 at 14:05 comment added Mariano Suárez-Álvarez You can take the total complex of the resolution I constructed above: that is a free $T(V)$-module with a differential. The map induced by my map $T(L)\otimes L(V)\to L(V)$ is then a quasi-iso.
Mar 1, 2010 at 13:58 comment added Don Stanley Slightly less good would be to have a $T(V)$ dg-module quasi-isomorphic to $L(V)$ that was free as a $T(V)$ module (like the bar construction).
Mar 1, 2010 at 13:53 comment added Don Stanley For the differential I mean that $T(V)$ is a differential graded algebra and $L(V)$ is a differential graded module over it. Then it would be great to have $L(V)$ as the cone on a map in the category of $T(V)$ modules (as in your resolution), but I guess that's not possible.
Mar 1, 2010 at 13:39 comment added Don Stanley This is great. Thanks a lot. I need to think about it for a bit.
Mar 1, 2010 at 13:26 history edited Mariano Suárez-Álvarez CC BY-SA 2.5
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Mar 1, 2010 at 13:21 history answered Mariano Suárez-Álvarez CC BY-SA 2.5