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• CommentRowNumber101.
• CommentAuthorDavidRoberts
• CommentTimeOct 24th 2018

Ah, irreps, not irrational! That makes more sense.

• CommentRowNumber102.
• CommentAuthorUrs
• CommentTimeOct 24th 2018

Oh, I see. I’ll make it more explicit.

(For that reason I had, elsewhere, spelled out “irrational”, even though it’s clunky as a superscript. But so I should also spell of “irreducible”, thanks.)

• CommentRowNumber103.
• CommentAuthorUrs
• CommentTimeOct 24th 2018
• (edited Oct 24th 2018)

Richard, thanks for highlighting, now I see what David C. had in mind.

certainly one could try to construct topological K-theory over $\mathbb{F}_{1}$ by looking at $\mathbb{F}_1$-vector bundles over a topological space. The question of course is what this means. I guess it would mean the same as usual but looking at fibres as pointed sets. So one could try to see whether can one represent such a thing, and if so, that would be topological $\mathbb{F}_{1}$.

Bundles with fibers pointed sets are covering spaces, classified, for compact base, by maps to $\underset{\underset{n}{\longrightarrow}}{lim} BS_n$, and equipped with a choice of section. But the Grothendieck group of these can’t quite be stable cohomotopy, can it?

• CommentRowNumber104.
• CommentAuthorDavid_Corfield
• CommentTimeOct 24th 2018

Have you worked out the kernels of these maps, the hidden M-theoretic degrees of freedom?

• CommentRowNumber105.
• CommentAuthorDavid_Corfield
• CommentTimeOct 24th 2018

Re #103, you’d think there’d be a morphism of the whole differential cohomology hexagon, from a $\mathbb{F}_1$ version to a $\mathbb{C}$ version.

1. Re #103: thanks for the quick thoughts, Urs. On reflection, since there is an assumption that the vector spaces are finite dimensional in the definition of a vector bundle, one should be looking at fibres which are finite pointed sets. Do you still know how to classify these (apologies if the answer is obvious)?

• CommentRowNumber107.
• CommentAuthorDavid_Corfield
• CommentTimeOct 24th 2018

This popular MO question brings out good material on the construction that Urs is pointing to in #103.

• CommentRowNumber108.
• CommentAuthorUrs
• CommentTimeOct 24th 2018
• (edited Oct 24th 2018)

I realize that the comparison morphism at the end in “Part B” of “Part II” is to be improved:

The comparison morphism to be considered in the first place is of course the equivariant Boardman homomorphism

$\beta_{G_{ADE}} \;\colon\; \mathbb{S}^{4}_{G_{ADE}}( X) \longrightarrow H^4_{G_{ADE}}(X,\mathbb{Z})$

But then for $A$-type $G_A$, the ordinary equivariant cohomology is Borel equivariant, hence right induced (here). This means by adjunction that the above morphism factors as

$\beta_{G_{A}} \;\colon\; \mathbb{S}^{4}_{G_{A}}(X) \overset{\alpha}{\longrightarrow} \mathbb{S}^4(X \sslash G_A) \overset{\beta}{\longrightarrow} H^4(X \sslash G_A ,\mathbb{Z})$

where now $\alpha$ is Segal-Carlsson completion and $\beta$ is the plain Boardman homomorphism. So $\alpha$ is surjective and $\beta$ is still surjective at least as soon as ${\vert G_A \vert} \gt 4$, so the conclusion here for the A-series remains unchanged.

But previously the argument didn’t properly apply to the DE-series, whereas now it does: In this case the equivariance on $H \mathbb{Z}$ is no longer just right induced, but incorporates the orientation twist, as it should be for M-theoretic discrete torsion.

• CommentRowNumber109.
• CommentAuthorDavid_Corfield
• CommentTimeOct 24th 2018

That MO question is saying Richard’s guess in #98 that the algebraic and topological K-theories for $\mathbb{F}_1$ coincide is right, no?

• CommentRowNumber110.
• CommentAuthorUrs
• CommentTimeOct 24th 2018
• (edited Oct 24th 2018)

Let me see, I gather you are after the statement that the stable cohomotopy in degree 0 of some compact space X is the Grothendieck group of virtual $\mathbb{F}_1$-bundles over X, hence the Grothendieck group of virtual covering spaces with sections over X, hence the Grothendieck dieck group of virtual covering spaces over X.

Clearly something close is true, and I feel dumb for not seeing it right away, but: Why is that true exactly?

• CommentRowNumber111.
• CommentAuthorRichard Williamson
• CommentTimeOct 24th 2018
• (edited Oct 24th 2018)

Hmm, wouldn’t the statement be that $\pi_{0}$ of the sphere spectrum coincides with something, where ’something’ is roughly (I guess would actually be a product with $\mathbb{Z}$ of) the classifying space for $\mathbb{F}_{1}$-vector bundles? We have that $\pi_{0}$ of the sphere spectrum is $\mathbb{Z}$. The classifying space for covering spaces with fibres with $n$ elements would appear to be, as you say, $BS_{n}$, the classifying space of the symmetric group with $n$ elements. Presumably we put these together by taking the homotopy colimit of $BS_{0} \rightarrow BS_1 \rightarrow \ldots$ where the maps are induced by inclusions of symmetric groups, although I’m not 100% sure yet why this works. Anyhow, the question is whether this homotopy colimit is contractible (we have to have this if we are to end up with $\mathbb{Z}$ after taking a product with $\mathbb{Z}$!) [Edit: this sentence does not seem correct, see #113. It seems that $\pi_0$ being $\mathbb{Z}$ is the correct thing to look for]. I suppose it is well-known whether or not this is the case? Indeed, I suppose that it can’t be if it is the correct classifying space, as it would imply that all covering spaces with finite fibres are trivial. Thus one of three things must be true: it is not the correct classifying space; algebraic $K$-theory of $\mathbb{F}_{1}$ does not agree with $\mathbb{F}_{1}$-topological K-theory; there is something else wrong with this story!

• CommentRowNumber112.
• CommentAuthorRichard Williamson
• CommentTimeOct 24th 2018
• (edited Oct 24th 2018)

Although in retrospect, I think I may have got my wires crossed when I suggested that they might be the same.

At the level of $\pi_{0}$, having just a comparison morphism (not necessarily isomorphism) from algebraic to topological, which there should at least be, is immediate, since $\mathbb{Z}$ is initial in abelian groups.

• CommentRowNumber113.
• CommentAuthorRichard Williamson
• CommentTimeOct 24th 2018
• (edited Oct 24th 2018)

On the other hand, and maybe that is what David C was getting at, it seems that $BS_{n}$ might be homotopic to the zero-th space of the sphere spectrum (for any n?). Which would suggest that $\mathbb{F}_{1}$-topological K-theory does agree with $K(\mathbb{F}_{1})$, as long as one ignores the taking of product with $\mathbb{Z}$ that I was referring to (or maybe this product with $\mathbb{Z}$ doesn’t matter in this case, i.e. maybe one will then end up, i.e. at the level of abelian groups, with a tensor product of two copies of $\mathbb{Z}$, i.e. just $\mathbb{Z}$, as desired)?

2. If it is true that the two coincide, that would beautiful, and be exactly the kind of thing that is supposed to happen over $F_1$: the topological/analytical becomes algebraic!

• CommentRowNumber115.
• CommentAuthorUrs
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

Here is where I get stuck:

For topological K-theory, the fact that $B U \times \mathbb{Z}$ (with $B U = \underset{\underset{n}{\longrightarrow}}{\lim} B U(n)$) classifies virtual vector bundles over compact (and Hausdorff) spaces $X$ crucially depends on the fact that every vector bundle on a compact space is the direct summand of a trivial vector bundle (this lemma). Namely, it implies that every virtual vector bundle $(V^+, V^-)$ over $X$ is equivalent to one whose anti-bundle part is trivial, $(W,\mathbb{C}^n)$. Pairs of this form clearly are classified by the Cartesian product $B U \times \mathbb{Z}$ of the classifying space for plain vector bundles and that for integers, which is $B U \times \mathbb{Z}$.

Now for $\mathbb{F}_1$-bundles = covering spaces with section, it is no longer true that each is a direct summand of a trivial covering space.

If so, then the classifying space for virtual $\mathbb{F}_1$-bundles is not $B \Sigma \times \mathbb{Z}$ (with $B \Sigma = \underset{\underset{n}{\longrightarrow}}{\lim} \Sigma_n$).

Now, that need not be fatal for the statement in question, since, as stated in that MO question, the degree-0 space of $\mathbb{S}$ is in fact not quite the space $B \Sigma \times \mathbb{Z}$, but is its Quillen plus-construction.

So the usual argument for the classifying space of complex topological K-theory breaks in two places when one passes to $\mathbb{F}_1$-topological K-theory.

Now it could maybe be that these two failures somehow cancel each other and so that the final statement remains the same after all. I am not saying that this is not the case, but I also don’t see why it would be the case.

• CommentRowNumber116.
• CommentAuthorRichard Williamson
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

I think there are three points which I am not sure about, which might have some intersection with yours! The first is the role of the pointedness. I saw some reference which said that covering spaces of degree $n$ are classified by $B\Sigma_{n}$. No sections were mentioned: I don’t whether they are needed and were just omitted, or whether they change anything. The second is whether it is correct that one can conclude from this that $B\Sigma$ is the classifying space for covering spaces with any finite degree. The third is whether we need to take a product with $\mathbb{Z}$: it seems that everything would work out without that product if we ignore sections.

One thing that might resolve all three points here is if the sections do change things, so that $B\Sigma$ is no longer quite correct, and so that the product with $\mathbb{Z}$ is needed again. The other way to resolve them might be that the sections are irrelevant, but that the product with $\mathbb{Z}$ ’does nothing’.

• CommentRowNumber117.
• CommentAuthorUrs
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

The sections/pointedness becomes irrelevant to the extent that only isomorphisms matter, because isomorphisms need to preserve the section/basepoint and hence reduce to isomorphisms of the remaining non-pointed bit.

The fact that $B \Sigma \coloneqq \underset{\underset{n}{\longrightarrow}}{\lim} B \Sigma_n$ classifies finite covering spaces over compact base spaces $X$ uses that compact objects may be taken inside the filtered colimit. (Slight subtlety here, since compact topological spaces are the compact objects only up to some fine print, but if we are really mainly interested in compact manifolds, then all is fine.)

So all that is fine. I think. To my mind, the subtlety begins when asking for the Grothendieck group of these finite covering spaces, as indicated in #115.

• CommentRowNumber118.
• CommentAuthorRichard Williamson
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

Thanks! Regarding the following…

It is no longer true that each is a direct summand of a trivial covering space

…I think we need to be a bit careful. First, we need to say what a trivial $\mathbb{F}_{1}$-bundle is. Not completely obvious, because one needs to say what $\mathbb{F}_{1}^{n}$ is for $n \gt 1$, or something similar to this. And secondly we need to be careful about what we mean by ’direct sum’. I guess it is going to amount to a coproduct on each fibre. Did you already have in mind how to make sense of these two things?

• CommentRowNumber119.
• CommentAuthorDavid_Corfield
• CommentTimeOct 25th 2018

I wonder if it’s been worked out in the finite field case. Presumably one meets the same issue that a bundle need not be a direct summand of a trivial covering space.

Would another approach be to think of the differential K-theory for $\mathbb{F}_1$?

• CommentRowNumber120.
• CommentAuthorDavid_Corfield
• CommentTimeOct 25th 2018

Re #118, most people take vector spaces over $\mathbb{F}_1$ as pointed sets. Then a trivial bundle over $X$ is just a finite number of copies of $X$ (with one copy designated as pointed).

• CommentRowNumber121.
• CommentAuthorRichard Williamson
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

My guess for what a trivial $\mathbb{F}_{1}$-bundle is would be something like: a product $X \times A$ (together with projection down to $X$) for some finite set $A$ with the discrete topology. I guess with this definition, the lemma that is needed will be true: [Edit: gave, too quickly, some nonsensical argument here].

(Edit: I did not see David C’s two comments in #119 and #120 before writing the above. It seems that David’s definition of trivial bundle agrees with mine, which is good! So it remains to be seen whether my argument for why the required lemma should hold goes through.)

• CommentRowNumber122.
• CommentAuthorUrs
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

Yeah, as David C. says, it seems to me the nature of $\mathbb{F}_1$-vector bundles in this discussion here is fixed by the context to be pointed-set-bundles = covering-spaces-with-section. That’s what allows us to ask for a relation to the Barratt-Priddy-Quillen theorem in the first place.

• CommentRowNumber123.
• CommentAuthorUrs
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

Now I have expanded the last slides in Part II of my slides (here) as indicated in #108

• CommentRowNumber124.
• CommentAuthorDavid_Corfield
• CommentTimeOct 25th 2018

Re #121, but if $X$ is the circle, and the bundle includes a component which is the boundary of a Moebius band, then one can’t add in another bundle to trivialize.

• CommentRowNumber125.
• CommentAuthorRichard Williamson
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

Re #122: yes, I agree on that too.

Re #124: yes, I agree now that my too-quick suggestion in #121 that the lemma is true is probably nonsense! Interesting…! It seems reasonable that somebody might have studied what the Grothendieck group of finite covering spaces is already, but I wouldn’t know where to look.

For the circle, the finite coverings correspond to $\mathbb{N}$, so that the Grothendieck construction produces $\mathbb{Z}$. Not sure how to do something to $B\Sigma$ to give the right answer in this case.

• CommentRowNumber126.
• CommentAuthorDavid_Corfield
• CommentTimeOct 25th 2018

The reason I latched on to that MO question is that the third instantiation of the entity is

The group completion of $B(\sqcup_n \Sigma_n)$, where $\Sigma_n$ is the symmetric group on $n$ letters, and $B(\sqcup_n \Sigma_n)$ is given the structure of a topological monoid via the block addition map $\Sigma_n\times \Sigma_m\to \Sigma_{n+m}$.

Is that group completion not enough for the relevant topological K-theory?

• CommentRowNumber127.
• CommentAuthorUrs
• CommentTimeOct 25th 2018
• (edited Oct 25th 2018)

This “group completion” is the $\infty$-theoretic group completion, given in degree 0 by forming the loop space $\Omega B_{\sqcup} ( Set_{fin} )$ (here). We are trying to see why or why not the connected components of the mapping space into this $\Omega B_{\sqcup} Set_{fin}$ is in bijection to the actual group completion of the additive monoid of isomorphism classes of covering spaces of $X$:

$\pi_0 Maps(X, \Omega B_{\sqcup} Set_{fin}) \;\overset{???}{\simeq}\; K( CoveringSpaces(X)/\sim, \sqcup )$

I don’t see how to go about relating the two sides of this. (Which doesn’t imply that it’s wrong, but I just don’t see why it would be right.)

My best guess for how to try to prove this would be to relate $\Omega B_{\sqcup} Set_{fin}$ to $B \Sigma \times \mathbb{Z}$ and then try to proceed as in the argument for actual topological K-theory. But that best guess fails, it seems, for the two problems pointed out in #115.

So therefore I am still stuck.

• CommentRowNumber128.
• CommentAuthorDavid_Corfield
• CommentTimeOct 25th 2018

Ah, OK.

So let’s test out the circle as base, and ignore the pointed section part of the cover. Then any $n$-cover is indexed by a cycle shape in a permutation on $n$-elements. So that’s a multiset of positive natural numbers adding to $n$.

So in terms of virtual covers, this should amount to one of Loeb’s hybrid sets of positive natural numbers, at least a finite one, in other words a finitely-supported map $\mathbb{N}^+ \to \mathbb{Z}$.

• CommentRowNumber129.
• CommentAuthorRichard Williamson
• CommentTimeOct 26th 2018
• (edited Oct 26th 2018)

Looking at the circle is a good idea. I discussed this at the end of #125: I thought one just (up to isomorphism) has a single cover for each integer $n \geq 1$, namely $S^{1} \rightarrow S^{1}$ given by $z \rightarrow z^{n}$? Is this wrong? If it is correct, the Grothendieck group is $\mathbb{Z}$. And this leads me to…

Regarding #127: if we look at the case of the circle, does $B\Sigma \times \mathbb{Z}$ work? I.e. do homotopy classes of maps from $S^{1}$ to this give $\mathbb{Z}$?

I wonder if one can make sense of $\Omega \Sigma$, and if so whether this would work instead of $B\Sigma \times \mathbb{Z}$ (initially, as a consequence of mis-remembering the parallel with the case of complex $K$-theory, I wrote $\Omega B\Sigma$, which definitely is wrong, e.g. it gives the wrong answer for the circle). In any case, one will have to consider something like this if one is to obtain a spectrum, i.e. if one wants something which works in all degrees.

• CommentRowNumber130.
• CommentAuthorDavid_Corfield
• CommentTimeOct 26th 2018

One normally takes a cover as connected, so then covers of the circle would correspond to positive integers as you say. But I don’t see any reason to impose the connected constraint here. Above a point we have a finite set, then passage around the circle permutes the elements. There’s no need for there to be a single cycle in the permutation, so the cover need not be connected.

3. Ah yes, OK, I agree that one should drop connectedness. Interesting. But is the result we are appealing to about $B\Sigma_{n}$ classifying covers of degree $n$ using connectedness? If not, then everything is out of the window!

• CommentRowNumber132.
• CommentAuthorUrs
• CommentTimeOct 26th 2018

Not sure if I followed what you are discussing now, but maps to $B \Sigma_n$ clearly classify $n$-sheeted covers, not necessarily connected (aka $\Sigma_n$-principal bundles, whence the notation)

4. OK, good! Then I think I am also stuck for the moment, except for the suggestion that it might be worth contemplating what $\Omega \Sigma$ could mean.

• CommentRowNumber134.
• CommentAuthorUrs
• CommentTimeOct 26th 2018

I have now a better example for the comparison morphism to ordinary cohomology in the second part of Part II:

Namely, globalizing the orbifold for an MK6-singularity to a compact space, we have a (singular) K3-surface $K3$ (e.g. Aspinwall 96, p. 23), and for comparison to heterotic 4d QFT we are to consider compactification on $K3 \times T^2$.

Now a good folklore result to compare to here is Garcia-Uranga 05, who argue (in spirit but not in detail following DMW00) that the quantization of the sugra 4-flux in ordinary integral cohomology on $K3 \times T^2$ matters and must be correct.

Hence for plausibility of “Hypothesis H” (as in the slides) we want that the Boardman homomorphism

$\mathbb{S}^4( K3 \times T^2 ) \longrightarrow H^4( K3 \times T^2 , \mathbb{Z} )$

has small cokernel, where I may ignore the equivariance now, since the cohomology of singular K3-s is that of non-singular ones.

But Arlettaz’ estimate for $coker(\beta)$ implies in this case (this example) that $2 coker(\beta) = 0$, hence that the cokernel either vanishes or is “as small as possible without vanishing”.

Now since, moreover, the integral cohomology of $K3 \times T^2$ itself is torsion-free (this Prop. togeter with the Kunneth isomorphism) this means in particular that the 4-classes $a$ in the image of $\beta$ vanish mod 2, which implies (albeit being stronger) the “integral equation of motion” $Sq^3(a) = 0$ which DMW00 argue is necessary for M-theory charges to match the K-theory classification.

• CommentRowNumber135.
• CommentAuthorUrs
• CommentTimeOct 26th 2018

In fact, if the differential $d_3$ in the AHSS for stable cohomotopy coincides on the relevant entry $H^4(X,\pi^0 = \mathbb{Z})$ with $Sq^3$ (as I suppose it must if it is non-vanishing at all), then the cokernel of $\beta$ will actually equal the cokernel of the index-2 subgroup in DMW00, section 4 of C-fields satisfying their “integral equation of motion”.

• CommentRowNumber136.
• CommentAuthorDavid_Corfield
• CommentTimeOct 27th 2018

Any sign yet of interesting M-theoretic degrees of freedom from the kernel of $\beta \circ \alpha$?

• CommentRowNumber137.
• CommentAuthorUrs
• CommentTimeOct 27th 2018
• (edited Oct 27th 2018)

Yes. There is an upcoming result by Dan Grady and one by Vincent Braunack-Mayer which show that the kernel is remarkable. Dan’s could become available by next week, or else real soon, Vincent’s will take a little longer to be ready for public consumption, but, if it works all out like it seems to, it will range real deep. Until these are officially public I should not say too much, though. But you’ll be the first to be alerted.

5. Hi Urs, sounds very exciting! For someone like myself who knows little to nothing about this, would it be possible to summarise the context of what might be being demonstrated here?

• CommentRowNumber139.
• CommentAuthorUrs
• CommentTimeOct 27th 2018
• (edited Oct 27th 2018)

Let’s see, what would you like to know. The broad impact of the project, or the specific results of Dan and Vincent that I hinted at? The latter I feel I should not circulate before their main authors feel ready, but, roughly (and now it seems I am doing it anyway, shame on me):

Dan has established the nature of the isomorphism that Pontryagin-Thom collapse induces between equivariant cohomotopy and cobordism classes inside fixed point loci. His results imply in particular that full equivariant cohomotopy in degree 4 sees the $(2,0)$-supersymmetric M5 brane inside an MK6 singularity (while the rational shadow of it that we studied previously in 1805.05987 can only see the $(1,0)$-supersymmetric $\tfrac{1}{2}M5$ being the intersection of an MK6 with an MO9).

Vincent seems to have a proof that the Goodwillie-Taylor tower of the full cocycle space of equivariant cohomotopy in degree-4 on 4-manifolds $X$ produces the system of graph complexes on $X$ glued along the Ran space of $X$, and an unproven but reasonable argument that the operadic action of the Goodwillie derivatives of the identity on this give the action of the Grothendieck-Teichmueller group on the graph complexes. In summary this says something like that homotopy-theoretic perturbation theory applied to our hypothesized M-theoretic cohomology theory (“hypothesis H” last/first slides here) produces hallmark structures of perturbative interacting quantum field theory on 4d spacetimes.

• CommentRowNumber140.
• CommentAuthorUrs
• CommentTimeOct 27th 2018
• (edited Oct 27th 2018)

Regarding the last point I should add: Since this is about the full cocycle space $Maps(X_4, S^4)$, it does know about the cohomotopy of $Y_{11}$, too, as soon as we have a Kaluza-Klein-fibration or Cartesian product $Y_{11} \simeq X_4 \times K_7$, via $Maps(K_7, Maps(X_4, S^4)) \simeq Maps(X_4 \times K_7, S^4)$.

Curiously, this is the perspective on higher dimensional field theories as lower dimensional field theories “with values in” other low dimensional field theories as in the AGT correspondence.

• CommentRowNumber141.
• CommentAuthorDavid_Corfield
• CommentTimeOct 27th 2018

Coo! So into some rich terrain:

Grothendieck predicted that the GT group is closely related to the absolute Galois group. Maxim Kontsevich later conjectured its action on certain space of quantum field theories and outlined its motivic aspects.

Back to physics meets number theory!

Just so I can see some earlier comments of yours in the same location:

51, Hence we find that as we come to 10d from 11d, the twisted KU-coefficients are but one summand in the Goodwillie linearization of the dimensionally reduced M-brane coefficients. So there is more in the M-brane coefficients, in some sense. The big open question is: what is this beyond the rational approximation.

86, We are seeing the unstable spheres as the true coefficients in M-theory. From there we may think of passage to stabilization as just the first order approximation in the Goodwillie-Taylor tower, and then the identification of K-theory as just the first order approximation of that in the chromatic filration.

• CommentRowNumber142.
• CommentAuthorDavid_Corfield
• CommentTimeOct 27th 2018

Re #140, presumably dimensions should sum in $Y_{11} \simeq X_4 \times K_6$.

• CommentRowNumber143.
• CommentAuthorUrs
• CommentTimeOct 27th 2018

Thanks, yes. Fixed now.

6. Re #139: thanks Urs! I think the last sentence was the kind of thing I was looking for! But could you maybe spell out the significance of this to a lay person? Is there a particular open question this addresses, for instance?

• CommentRowNumber145.
• CommentAuthorUrs
• CommentTimeOct 29th 2018

Is there a particular open question this addresses?

Sure. Let me quote Moore 14:

We still have no fundamental formulation of “M-theory” - the hypothetical theory of which 11-dimensional supergravity and the five string theories are all special limiting cases. Work on formulating the fundamental principles underlying M-theory has noticeably waned. $[...]$. If history is a good guide, then we should expect that anything as profound and far-reaching as a fully satisfactory formulation of M-theory is surely going to lead to new and novel mathematics. Regrettably, it is a problem the community seems to have put aside - temporarily. But, ultimately, Physical Mathematics must return to this grand issue.

• CommentRowNumber146.
• CommentAuthorDavid_Corfield
• CommentTimeOct 30th 2018

Re #139, does Dan’s result give us a glimpse of Theory X, or 6d (2,0)-superconformal QFT, then?

• CommentRowNumber147.
• CommentAuthorUrs
• CommentTimeNov 2nd 2018

Not in itself yet, as it is a (equivariantly-)topological statement. But once we bring in the SuGra equations of motion by requiring (equivariant) super-torsion freedom, something should happen.

• CommentRowNumber148.
• CommentAuthorUrs
• CommentTimeNov 22nd 2018
• (edited Nov 22nd 2018)

Now the article alluded to in #139 above is out:

The bulk of the article establishes a nice equivariant Pointryagin-Thom isomorphism between equivariant cohomotopy and cobordism classes of submanifolds inside orbifold fixed points.

In section 4 the application to the M5-brane locus, that I indicated in #139 above, is discussed.

So earlier in Equivariant homotopy and super M-branes (schreiber) we showed that rational equivariant cohomotopy sees the MK6-brane, but its only way to see the M5-brane domain wall inside the MK6 is by intersecing the latter with an orientifold, which cuts down supersymetry by half and hence yields the $\mathcal{N} = (1,0)$ M5-brane locus only.

Dan establishes that beyond the rational approximation, equivariant cohomotopy in $RO$-degree $V$ classifies cobordism classes of submanifolds inside the $G$-fixed points (inside the orbifold singularities) of codimension $dim(V^G)$ relative to the ambient fixed point locus.

This means that if we measure M-brane charge with the representation sphere $S^{\mathbb{H}_{adj}}$, where the space of quaternions $\mathbb{H}$ is equipped with the adjoint action by unit quaternions $\simeq SU(2)$ (as for the M2-brane!) and consider a D- or E-series finite subgroup $G \subset SU(2)$, then equivariant cohomotopy detects codimension-1 sub-branes inside the MK6.

This result should extend straightforwardly to super-geometry, where it should say that inside the $\mathbb{R}^{6,1\vert \mathbf{16}}$-supermanifold of the MK6, we still see a codimension-1 (hence codimension-$(1\vert 0)$!)-sub-supermanifold, hence an $\mathbb{R}^{5,1\vert \mathbf{8} + \mathbf{8}}$-submanifold, hence the $\mathcal{N}= (2,0)$-M5-brane locus.

• CommentRowNumber149.
• CommentAuthorDavid_Corfield
• CommentTimeNov 22nd 2018

Great stuff! Does establishing that extension to super-geometry come next? And then perhaps integrating the equations of motion, #147?

• CommentRowNumber150.
• CommentAuthorUrs
• CommentTimeNov 22nd 2018

The supergeometric version is pretty immediate if one uses the equivalence between supermanifolds and suitable presheaves of ordinary manifolds on the site of superpoints. But somebody needs to sit down and write it out cleanly.

• CommentRowNumber151.
• CommentAuthorUrs
• CommentTimeJan 25th 2019

We are finalizing a review that sums up the whole rational story of this thread here:

(This is going to appear as our contribution to the proceedings of the Durham symposium on the topic, a while back.)

• CommentRowNumber152.
• CommentAuthorUrs
• CommentTimeJul 17th 2019
• (edited Jul 25th 2019)
• CommentRowNumber153.
• CommentAuthorUrs
• CommentTimeJul 24th 2019
• (edited Jul 24th 2019)

… and now with considerably strengthened main theorem (summary here)

• CommentRowNumber154.
• CommentAuthorDavid_Corfield
• CommentTimeJul 25th 2019

Looks intriguing. Here are some typos

superymmetric; non-pertrubative; hmotopy; super-valume; cocyle; intersting; an hence; MO9-projectin

• CommentRowNumber155.
• CommentAuthorUrs
• CommentTimeJul 25th 2019

Thanks! Fixed now.

• CommentRowNumber156.
• CommentAuthorDavid_Corfield
• CommentTimeJul 30th 2019

Utterly trivial, but I keep tripping on your ’quest’ sentences when I start reading the article

In the quest to mathematically formulating M-theory,…

is grammatically awkward. Perhaps, “In the quest for the mathematical formulation of M-theory”

In the quest for shedding some light in that direction,…

the ’quest’ here is too weak. They have aims like the Holy Grail. Perhaps, “Looking to shed some light in that direction,…”

• CommentRowNumber157.
• CommentAuthorUrs
• CommentTimeJul 31st 2019

Thanks!

This, and a bunch of other little things, now improved (hopefully) in the latest version (here).

• CommentRowNumber158.
• CommentAuthorDavid_Corfield
• CommentTimeJul 31st 2019

prinicples; has be systematically obtained; gauged-of-gauge; cohomoloy; correspong; geoemtry