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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeNov 25th 2020

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeNov 25th 2020

and this one:

• D. M. Segal, On the symplectic cobordism ring, Commentarii Mathematici Helvetici 45, 159–169 (1970) (doi:10.1007/BF02567323)
• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeNov 25th 2020

and this couple:

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeNov 25th 2020

I am trying to find out:

The K3-surface should represent a nontrivial class in $\Omega^{Sp}_4$, no?

I am trying to find a reference that would admit this and provide more details.

• CommentRowNumber5.
• CommentAuthorDavid_Corfield
• CommentTimeNov 25th 2020

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.

• CommentRowNumber6.
• CommentAuthorDavid_Corfield
• CommentTimeNov 26th 2020

The K3-surface should represent a nontrivial class in $\Omega^{Sp}_4$, no?

At the end of

• R. E. Stong, Some Remarks on Symplectic Cobordism, (JSTOR)

it says

$M^4$ is a generator of $\Omega^{Sp}_{4}= Z$,

where $M^{4}$ is mentioned at the top of p. 432, and $M^{4r}$ earlier.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeNov 26th 2020
• (edited Nov 26th 2020)

Thanks for checking.

But I cannot tell from this if K3 represents a generator, hence if that $M^4$ 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 $\Omega^{SU}_4$, e.g. from Novikov, p. 218 (and in $\Omega^{Spin}_4$, for that matter). I am thinking this ought to imply at least that it’s also non-trivial in $\Omega^{Sp}_4$, I suppose.

• CommentRowNumber8.
• CommentAuthorDavid_Corfield
• CommentTimeNov 26th 2020

When you say $K 3$, you mean what Novikov has as the Kummer surface $K^4$?

• CommentRowNumber9.
• CommentAuthorDavidRoberts
• CommentTimeNov 26th 2020
• (edited Nov 27th 2020)

(Edit this is a little bit confused, take with a grain of salt)

I think from (3.16) on page 218 one gets that $K^4$ is a K3. Certainly $c_1,c_2$ and $\chi$ work out right. The Pontryagin class $p_1$ is correct, up to sign, and the signature $\tau$ is also correct, up to sign.

So the only issue is the signs on $p_1$ and $\tau$, 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?

• CommentRowNumber10.
• CommentAuthorDavidRoberts
• CommentTimeNov 26th 2020

And ’Kummer surface’ seems to be in some places a synonym for K3, but it’s not quite clear. A smooth quartic in $\mathbb{CP}^3$ is a K3, and Novikov says a “generic” quartic.

Hmm, but now I read your comment #7 more closely, this was just about $\Omega_4^{SU}$, and extending this to $\Omega_4^{S p}$. Man, should have slowed down…

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeNov 27th 2020
• (edited Nov 27th 2020)

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 $SU(2) \simeq Sp(1)$ 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 $\mathbb{T}^4/\mathbb{Z}_2$ Usage seems to have been different when Novikov wrote his survey, not sure.)

• CommentRowNumber12.
• CommentAuthorDavid_Corfield
• CommentTimeNov 27th 2020

Words of wisdom, perhaps, from Nige Ray (here):

Maybe $M S p_{\ast}$, is such a problem because $S p$-manifolds admit so few alternative $S p$ structures.

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeJan 23rd 2021

The canonical topological group-inclusions

$Sp(k) \;\subset\; SU(2k) \;\subset\; U(2k)$ $M Sp \;\longrightarrow\; M SU \;\longrightarrow\; M \mathrm{U}$

(from MSp to MSU to MU)

and hence corresponding multiplicative cohomology theory-homomorphisms of cobordism cohomology theories.