11
$\begingroup$

In unoriented cobordism there exist stable cohomology operations looking similar to Steenrod squares (they were used by Quillen to compute the unoriented cobordism ring with its formal group law defined by tensor product of bundles).

Since their properties are so similar to the properties of Steenrod squares, my questions are: do they become normally Steenrod squares in $N^*(X) \otimes_{N*} \mathbb{Z}_2 $? Do they correspond to $Sq^i \otimes id_{N^*} $ under the isomorphism $N^*(X) \simeq H^*(X) \otimes_{\mathbb{Z}_2} N^*$?

Improve this question
$\endgroup$
8
  • $\begingroup$ I'm not sure I follow. The spectrum $MO$ is a wedge of Eilenberg-MacLane spectra (that's Quillen's theorem), so all its cohomology operations come from Steenrod squares. $\endgroup$
    – Denis Nardin
    Commented Jun 23, 2016 at 16:28
  • $\begingroup$ Could it be shown more directly, e.g. that these operations act on $N^*(pt)$ trivially? $\endgroup$
    – vareps
    Commented Jun 23, 2016 at 16:48
  • 1
    $\begingroup$ Denis: that was proved long before Quillen, and is an additive statement in any case... That said, it's true that MO is actually an HF_2 algebra so it is indeed the case that all its power operations come from the Dyer-Lashof algebra for HF_2. $\endgroup$
    – Dylan Wilson
    Commented Jun 23, 2016 at 16:51
  • $\begingroup$ @DylanWilson Sorry for the misattribution (and I agree that it is an additive statement, but cohomology operations are the homotopy of the endomorphism ring anyway, so they depend only on the additive structure) Is it true it is a commutative HF_2-algebra? Do you have a reference? I thought that the map HF_2->MO was only E_2 (so no control on power operations). $\endgroup$
    – Denis Nardin
    Commented Jun 23, 2016 at 17:55
  • $\begingroup$ These Steenrod ops the OP is referring to are the power operations... So they definitely depend on the multiplicative structure, not just the additive structure. But you're right, I guess I don't know whether that map is anything more than E_2... It's likely not E_infty but I would be surprised if it wasn't H_infty (which is what we're asking). I dunno how I'd prove it... The analog for MU and BP is false I think, but with HF_2 I feel more optimistic. $\endgroup$
    – Dylan Wilson
    Commented Jun 23, 2016 at 19:57

1 Answer 1

Reset to default
7
$\begingroup$

There is a natural transformation $N^*(X)\to H^*(X;\mathbb{Z}/2)$ given by the universal Thom class, and this maps the Steenrod-tom Dieck operations in cobordism to the Steenrod squares in cohomology. A reference is Section 15 of

Tammo tom Dieck, MR 244989 Steenrod-Operationen in Kobordismen-Theorien, Math. Z. 107 (1968), 380--401.

This answers your first question, I think. As to your second question, I don't know the answer but I suspect the Steenrod-tom Dieck operations are non-trivial on $N^*(pt)$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .