Return to Answer

added 518 characters in body

Source Link

edited Feb 15, 2013 at 19:11

30.4k
3
96
140

Dylan, I am afraid that you are missing something. It is true that each element of the group gives a map of spectral sequences. But look at the fibers and my comment about isotropy groups above. If you construct the Serre spectral sequence in the usual way, there is a non-equivariantly inconsequential choice of base point in the base space. But that is no longer inconsequential equivariantly. You do not get to collate your actions one element of the group at a time into an algebraically meaningful action of the group on "the" spectral sequence. If you try to express the action of $G$ on the $E_2$ term algebraically, you will see the point. Unfortunately for calculations, the "right" way to think about the question is to switch to Bredon cohomology, say with the constant integer coefficient system. Then there is an equivariant Serre spectral sequence, developed by Moerdijk and Svensson: The equivariant Serre spectral sequence. Proc. Amer. Math. Soc. 118 (1993), no. 1, 263–278. While there has been more recent theoretical work, there have been few if any serious calculations with it.

Edit: Dylan, in answer to your question below, the incorporation of varying fixed point data (as in the base space here) is just the kind of thing Bredon cohomology is designed for. For a simple but serious use of Bredon cohomology to obtain information about ordinary cohomology see "A generalization of Smith theory" (#57 on my web page). For a nice exposition and a worked example of the equivariant Serre spectral sequence, see Megan Shulman's "Equivariant spectral sequences for local coefficients" arXiv:1005.0379

Dylan, I am afraid that you are missing something. It is true that each element of the group gives a map of spectral sequences. But look at the fibers and my comment about isotropy groups above. If you construct the Serre spectral sequence in the usual way, there is a non-equivariantly inconsequential choice of base point in the base space. But that is no longer inconsequential equivariantly. You do not get to collate your actions one element of the group at a time into an algebraically meaningful action of the group on "the" spectral sequence. If you try to express the action of $G$ on the $E_2$ term algebraically, you will see the point. Unfortunately for calculations, the "right" way to think about the question is to switch to Bredon cohomology, say with the constant integer coefficient system. Then there is an equivariant Serre spectral sequence, developed by Moerdijk and Svensson: The equivariant Serre spectral sequence. Proc. Amer. Math. Soc. 118 (1993), no. 1, 263–278. While there has been more recent theoretical work, there have been few if any serious calculations with it.

Dylan, I am afraid that you are missing something. It is true that each element of the group gives a map of spectral sequences. But look at the fibers and my comment about isotropy groups above. If you construct the Serre spectral sequence in the usual way, there is a non-equivariantly inconsequential choice of base point in the base space. But that is no longer inconsequential equivariantly. You do not get to collate your actions one element of the group at a time into an algebraically meaningful action of the group on "the" spectral sequence. If you try to express the action of $G$ on the $E_2$ term algebraically, you will see the point. Unfortunately for calculations, the "right" way to think about the question is to switch to Bredon cohomology, say with the constant integer coefficient system. Then there is an equivariant Serre spectral sequence, developed by Moerdijk and Svensson: The equivariant Serre spectral sequence. Proc. Amer. Math. Soc. 118 (1993), no. 1, 263–278. While there has been more recent theoretical work, there have been few serious calculations with it.

Edit: Dylan, in answer to your question below, the incorporation of varying fixed point data (as in the base space here) is just the kind of thing Bredon cohomology is designed for. For a simple but serious use of Bredon cohomology to obtain information about ordinary cohomology see "A generalization of Smith theory" (#57 on my web page). For a nice exposition and a worked example of the equivariant Serre spectral sequence, see Megan Shulman's "Equivariant spectral sequences for local coefficients" arXiv:1005.0379

Source Link

created Feb 15, 2013 at 16:37

30.4k
3
96
140

Dylan, I am afraid that you are missing something. It is true that each element of the group gives a map of spectral sequences. But look at the fibers and my comment about isotropy groups above. If you construct the Serre spectral sequence in the usual way, there is a non-equivariantly inconsequential choice of base point in the base space. But that is no longer inconsequential equivariantly. You do not get to collate your actions one element of the group at a time into an algebraically meaningful action of the group on "the" spectral sequence. If you try to express the action of $G$ on the $E_2$ term algebraically, you will see the point. Unfortunately for calculations, the "right" way to think about the question is to switch to Bredon cohomology, say with the constant integer coefficient system. Then there is an equivariant Serre spectral sequence, developed by Moerdijk and Svensson: The equivariant Serre spectral sequence. Proc. Amer. Math. Soc. 118 (1993), no. 1, 263–278. While there has been more recent theoretical work, there have been few if any serious calculations with it.