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have added this pointer:
and this one:
and this couple:
Stanley Kochman, The symplectic cobordism ring I, Memoirs of the American Mathematical Society 1980, Volume 24, Number 271 (doi:10.1090/memo/0271)
Stanley Kochman, The symplectic cobordism ring II, Memoirs of the American Mathematical Society 1982 Volume 40, Number 271 (doi:10.1090/memo/0271)
I am trying to find out:
The K3-surface should represent a nontrivial class in ΩSp4, no?
I am trying to find a reference that would admit this and provide more details.
Presumably at some stage we’d split off ’quaternionic cobordism’, e.g., to include the kind of things Laughton speaks about in Chapter 7:
We begin with some preliminaries on quaternionic cobordism,… (p. 99)
’Quaternionic towers’, etc.
The K3-surface should represent a nontrivial class in ΩSp4, no?
At the end of
it says
M4 is a generator of ΩSp4=Z,
where M4 is mentioned at the top of p. 432, and M4r earlier.
Thanks for checking.
But I cannot tell from this if K3 represents a generator, hence if that M4 is Sp-bordant to K3. I suspect it is, but then why wouldn’t anyone comment on it.
I know that the bordism class of K3 is a non-trivial element in ΩSU4, e.g. from Novikov, p. 218 (and in ΩSpin4, for that matter). I am thinking this ought to imply at least that it’s also non-trivial in ΩSp4, I suppose.
When you say K3, you mean what Novikov has as the Kummer surface K4?
(Edit this is a little bit confused, take with a grain of salt)
I think from (3.16) on page 218 one gets that K4 is a K3. Certainly c1,c2 and χ work out right. The Pontryagin class p1 is correct, up to sign, and the signature τ is also correct, up to sign.
So the only issue is the signs on p1 and τ, and I don’t know if this is just due to normalisation, or if one needs to take the reverse orientation, or what.
Hmm, no: the Pontryagin classes are independent of the orientation! So it might be something to do with a choice of basis?
And ’Kummer surface’ seems to be in some places a synonym for K3, but it’s not quite clear. A smooth quartic in ℂℙ3 is a K3, and Novikov says a “generic” quartic.
Hmm, but now I read your comment #7 more closely, this was just about ΩSU4, and extending this to ΩSp4. Man, should have slowed down…
Thanks for checking. Maybe it can indeed be decided on the characteristic classes.
More modern (in fact recent) accounts of K3 as representing an element in the SU-bordism ring are referenced at MSU, here. (These authors even mention Sp-bordism, but only to say that it remains mostly unknown.)
I was vaguely thinking that the identification SU(2)≃Sp(1) of the K3’s structure group might allow to see how its bordism class behaved as we move from SU-bordism to Sp-bordism. But one needs more info than that.
(On terminology: It seems that these days “Kummer surface” is mostly used for the singular incarnation of K3 as the orbifold 𝕋4/ℤ2 Usage seems to have been different when Novikov wrote his survey, not sure.)
Words of wisdom, perhaps, from Nige Ray (here):
Maybe MSp*, is such a problem because Sp-manifolds admit so few alternative Sp structures.
added the remark that
The canonical topological group-inclusions
Sp(k)⊂SU(2k)⊂U(2k)(quaternionic unitary group into special unitary group into unitary group) induce ring spectrum-homomorphism of Thom spectra
MSp⟶MSU⟶MUand hence corresponding multiplicative cohomology theory-homomorphisms of cobordism cohomology theories.
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