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  1.  
    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeNov 22nd 2018

    am starting something here. For the moment I just want to record today’s

    • Daniel Grady, Cobordisms of global quotient orbifolds and an equivariant Pontrjagin-Thom construction (arXiv:1811.08794)

    eventually more references ought to be added

    v1, current

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeFeb 4th 2019

    I hope to write out some simple but interesting explicit examples of the equivariant Pontryagin-Thom isomorphism. I may fumble around quite a bit at the beginning.

    To warm up, I added what is possibly the simplest non-trivial example: ℤ 2\mathbb{Z}_2-equivariant maps from S 1S^1 to itself (here).

    Simplistic as it is, it is already kind of interesting in how it works.

    diff, v3, current

    • CommentRowNumber3.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 4th 2019

    There’s a typo after “Indeed, under passage to fixed points any”, where you have S 1≃{0,1}S^1 \simeq \{0,1\}.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeFeb 5th 2019

    Thanks, fixed now. And there is something strange about the example, too. Hope to sort it out…

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeFeb 11th 2019

    What is the class in π 4(S 3)≃ℤ 2\pi_4(S^3) \simeq \mathbb{Z}_2 of the map

    S ℍ ⟶ S (ℍ im) q ↦ qvq¯ \array{ S^{\mathbb{H}} &\longrightarrow& S^{\left(\mathbb{H}_{im}\right)} \\ q &\mapsto& q v \bar q }

    where v∈ℍ im∖{0}v \in \mathbb{H}_{im} \setminus \{0\} is any non-zero imaginary quaternion?

    • CommentRowNumber6.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 11th 2019

    Let me try some things on the fly… Take v=iv=i, the choice is immaterial.

    My first thought is that it should be the nontrivial class, since restricted to Sp(1)=S(ℍ)⊂S ℍSp(1) = S(\mathbb{H}) \subset S^\mathbb{H} the equator, this looks like the Hopf map H:S 3=Sp(1)→S(ℍ im)=S 2H\colon S^3 = Sp(1) \to S(\mathbb{H}_{im}) = S^2, which can be expressed as q↦qiq¯q\mapsto qi\bar{q} (I learned this from Atiyah’s book “Geometry of Yang-Mills fields”). And now that I think about it, the map you give is the suspension of HH, so yes, it is the nontrivial class.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeFeb 11th 2019

    Thanks!!

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeFeb 11th 2019

    Just for citation purposes: Where in Atiyah do you see this stated?

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeFeb 11th 2019

    But then, even Wikipedia spells this out. All right, thanks again!

    • CommentRowNumber10.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 12th 2019

    Yes, this is essentially coming from the representation Sp(1)→SO(ℍ im)Sp(1) \to SO(\mathbb{H}_{im}) by conjugation, and recognising the unit sphere in the imaginary quaternions as a homogeneous space for the action of Sp(1)Sp(1) via this rep.

    • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeFeb 16th 2019

    Now my next question is the following variant of the above:

    For G=ℤ n=⟨2πi/n⟩↪SU(2)G = \mathbb{Z}_n = \langle 2 \pi \mathrm{i}/n\rangle \hookrightarrow SU(2) a cyclic subgroup, is there any bi-pointed GG-equivariant continuous function

    S ℍ⟶S ⟨j,k⟩ S^{\mathbb{H}} \longrightarrow S^{ \langle \mathrm{j}, \mathrm{k}\rangle }

    from the representation sphere of ℍ\mathbb{H} with its canonical left action to the representation sphere of the 2d linear span of the other two imaginary quaternions, equipped with the conjugation action?

    Here by “bi-pointed” I mean that it restricts to the canonical bijection on the fixed points S 0S^0.

    I suspect there is not. Because the evident way to achieve equivariance would seem to be to send quaternions on th eleft to their conjugation action on something, as in #5, but that does’t work here with the shrunken codomain.

    I’d like to turn that into a proof that no such map can exist. But not sure yet.

    • CommentRowNumber12.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 16th 2019

    Doesn’t this reduction to a statement about the non-compactified GG-spaces? And asking there is no pointed equivariant map, linear or otherwise, between the representations? I suspect it would be easy to prove that no linear map exists, but not sure about the general case yet.

    • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeFeb 16th 2019

    Sure, we may ask equivalently for equivariant maps ff between the linear GG-spaces that take 0 to 0 and for which limr 1→∞r 2(f)\underset{r_1 \to \infty}{lim} r_2(f) is ∞\infty.

    Linear maps won’t exist by Schur’s lemma. One could see this nicely in the previous example we did, which was a quadratic map, thus reducing to zero in the linear approximation.

    Similarly for instance for G=ℤ/2G = \mathbb{Z}/2 the quadratic map (−) 2:S ℝ sgn→S ℝ(-)^2 \colon S^{\mathbb{R}_{sgn}} \to S^{\mathbb{R}} exists equivarianty (and is “bi-pointed”) but the only equivarian linear map ℝ sgn→ℝ\mathbb{R}_{sgn} \to \mathbb{R} is zero.

    • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeFeb 17th 2019

    finally added pointer to

    diff, v8, current

    • CommentRowNumber15.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 17th 2019
    • (edited Feb 17th 2019)

    Assume we have such an equivariant map (we should write μ n\mu_n for the group, since it’s the roots of unity in U(1)⊂Sp(1)U(1) \subset Sp(1)). I’m wondering what would be the induced linear map between the representations on the tangent spaces at 0∈S ℍ0\in S^\mathbb{H} and 0∈S ⟨i,k⟩0\in S^{\langle i,k\rangle}. I can’t see how those representations are anything other than ℍ\mathbb{H} and ⟨j,k⟩\langle j,k \rangle. Edit Actually I guess this linear map being 0 isn’t going to give us much.

    We also know that the fibres of the map are, generically, 2-dimensional submanifolds that intersect any μ n\mu_n-orbits only once (for nn odd) or twice (for nn-even). I was thinking what we would do if instead of μ n\mu_n we were looking at U(1)U(1). Then any nontrivial U(1)U(1)-orbit in S ℍS^\mathbb{H} is mapped 2:1 to the orbit in the other sphere. Thus in this case, the 2-dimensional generic fibres can still only intersect those circle orbits twice.

    • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeFeb 17th 2019

    Hi David,

    you seem to be arguing some point now, maybe disagreeing with something I said, but I am not sure. Could you say again what statement you are after now? I was suggestion there ought to be no bipointed μ n\mu_n-equivariant map S ℍ→S ⟨j,k⟩S^\mathbb{H} \to S^{\langle \mathrm{j}, \mathrm{k}\rangle}. Do you feel this can be proven, or that it is false?

    Assume we have such an equivariant map (we should write μ n\mu_n for the group, since it’s the roots of unity in U(1)⊂Sp(1)U(1) \subset Sp(1)). I’m wondering what would be the induced linear map between the representations on the tangent spaces at 0∈S ℍ0\in S^\mathbb{H} and 0∈S ⟨i,k⟩0\in S^{\langle i,k\rangle}. I can’t see how those representations are anything other than ℍ\mathbb{H} and ⟨j,k⟩\langle j,k \rangle.

    Right, the tangent GG-representation of the GG-representation sphere S VS^V at 00 is VV. And so if V 1V_1 and V 2V_2 are distinct linear irreps, then by Schur the only linear GG-map V 1→V 2V_1 \to V_2 is zero, which means that any bipointed differentiable GG-equivariant function f:S V 1→S V 2f : S^{V_1} \to S^{V_2} must have vanishing differential at 0.

    Which is just what we see in the examples we discussed, where ff is quadratic.

    My almost-proof that there is no bipointed equivariant S ℍ→S ⟨j,k⟩S^{\mathbb{H}} \to S^{\langle \mathrm{j}, \mathrm{k}\rangle} goes like so:

    The fact that we need to equivariantly send a left quaternion action to adjoint quaternion action “obviously” forces us to send any quaternion qq in S ℍS^{\mathbb{H}} to its conjugation action on something, and that something can “obviously” only be an imaginary quaternion, and that leads to the map S ℍ→S ℍ imS^{\mathbb{H}} \to S^{\mathbb{H}_{im}} which does however not factor through S ⟨j,k⟩S^{\langle \mathrm{j}, \mathrm{k}\rangle}. QED.

    Now how to remove the quotation marks around “obviously”? :-)

    • CommentRowNumber17.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 17th 2019
    • (edited Feb 17th 2019)

    Do you feel this can be proven, or that it is false?

    I was trying some obvious things to see if I could get an easy contradiction from assuming such a map exists. :-)

    How about we consider the easiest case: μ 2={±1}\mu_2 = \{\pm 1\}? It has trivial action on S ⟨j,k⟩S^{\langle j,k\rangle} but acts on the equator of S ℍS^\mathbb{H} as the antipodal map and then extends to the obvious thing. Any μ 2k\mu_{2k}-equivariant map would give one of these, so this is at least a potential obstruction to all the even cases.

    Given the μ 2\mu_2-equivariant map, we’d get a ordinary bipointed map Σℝℙ 3→S 2\Sigma\mathbb{RP}^3\to S^2. Obstruction theory might have something to say about this.

    • CommentRowNumber18.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 17th 2019
    • (edited Feb 17th 2019)

    OK, so assuming (for the sake of contradition) there is some μ 2\mu_2-equivariant map e:S ℍ⟶S ⟨j,k⟩e\colon S^{\mathbb{H}} \longrightarrow S^{ \langle \mathrm{j}, \mathrm{k}\rangle }, then this map factors into bipointed maps

    S ℍ⟶Σℝℙ 3⟶S ⟨j,k⟩. S^{\mathbb{H}} \longrightarrow \Sigma\mathbb{RP}^3 \longrightarrow S^{ \langle \mathrm{j}, \mathrm{k}\rangle } .

    Up to homotopy there are only two maps S 4→S 2S^4 \to S^2: the trivial one, and the composite η∘Ση\eta \circ \Sigma \eta for η\eta the Hopf fibration. Somehow I doubt this factorises like this, but I haven’t a reference and can’t quickly see how to prove it. So I conjecture that ee would have to be nullhomotopic. I would think that S 4→Σℝℙ 3S^4 \to \Sigma\mathbb{RP}^3 is not null-homotopic. It may be possible to push this line of argument further, but it’s late here :-)

    • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeFeb 17th 2019

    Thanks, David!! I’ll think about it.

    • CommentRowNumber20.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 17th 2019
    • (edited Feb 17th 2019)

    Ah, here’s a construction of such a map! Recall that Σpt≃[0,1]\Sigma pt \simeq [0,1] and consider it as a bipointed space. Pick a path in S 2S^2 from 00 to ∞\infty, hence a map Σpt→S 2\Sigma pt \to S^2. Then compose this with S 4→Σℝℙ 3→ΣptS^4 \to \Sigma \mathbb{RP}^3 \to \Sigma pt to get a bipointed map S 4→S 2S^4 \to S^2 factoring through the quotient S 4→ℝℙ 2S^4 \to \mathbb{RP}^2, hence a μ 2\mu_2-equivariant map on the representation spheres S ℍ→S ⟨j,k⟩S^\mathbb{H} \to S^{ \langle \mathrm{j}, \mathrm{k}\rangle }.

    Thus there is at least one case that does exist, and we don’t get the obstruction I envisaged. However, it’s possible that only for n=2n=2 does a μ n\mu_n-equivariant bipointed map exist.

    Hmm… Now I’m really off to bed.

    • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeFeb 17th 2019
    • (edited Feb 17th 2019)

    Sorry, I haven’t been following closely. Now that I am chasing back through your comments I get stuck in #17, where you say that μ 2\mu_2 should act trivially on S ⟨j,k⟩S^{\langle \mathrm{j}, \mathrm{k}\rangle}. This doesn’t sound right to me: The nontrivial element acts by rotation by π\pi in the (j,k)(\mathrm{j},\mathrm{k})-plane, hence it rotates our 2-sphere by π\pi around its axis thtough 0 and ∞\infty.

    Am I misreading your comment? But it’s great that you are thinking about this. Don’t let me stop you. :-)

    Myself, I got sidetracked today by seeing how much tom Dieck’s theorem (here) may be of help (generally, not for the particular question of S ℍ→S ⟨i,j⟩S^{\mathbb{H}} \to S^{\langle i,j\rangle}, to which it does not apply, like to so many other examples, due to its extremely strong assumptions…)

    • CommentRowNumber22.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 17th 2019
    • (edited Feb 17th 2019)

    μ 2={1,e iπ}\mu_2 = \{1,e^{i\pi}\}, no? Try conjugating aj+bka j + b k by e iπe^{i\pi} and see what happens ;-)

    I did calculate a matrix representative of the action of e 2πik/ne^{2\pi i k/n}, and it’s rotation by 4πik/n\pi i k/n.

    • CommentRowNumber23.
    • CommentAuthorUrs
    • CommentTimeFeb 17th 2019

    Sorry, my bad, you are right, of course. I was thinking about the Z/2Z/2 down in SO(3)SO(3), but you are of course using the correct Z/2Z/2 up in SU(2)SU(2), which covers 11 downstairs.

    And now it’s me who is off to bed. See you tomorrow!

    • CommentRowNumber24.
    • CommentAuthorUrs
    • CommentTimeFeb 17th 2019
    • (edited Feb 17th 2019)

    But so we need to exclude that case of ℤ/2\mathbb{Z}/2 for the purpose of classifying those transversal contributions that we talked about. These, by decree, are given by bipointed equivariant maps from a tubular neighbourhood of the fixed locus restricted to some brane in that fixed locus (which gives us S ℍS^\mathbb{H} for the case of the MK6) to the representation sphere of – and that’s the key point here - the orthogonal complement V−V GV-V^G of the trivial rep inside VV.

    • CommentRowNumber25.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 18th 2019

    Hmm, will have to think on that, but not too much at present, with other things on.

    • CommentRowNumber26.
    • CommentAuthorUrs
    • CommentTimeFeb 18th 2019

    Now back to work. Sorry again for that. I was mentally always envisioning the rotation action on the right, which is what mattered to me most in the logic of things, but it made me speak incorrectly, ignoring the double cover lift.

    But, once you have time, we should talk about that question that you ended up looking at, too: What can be said about equivariant Cohomotopy from S ℍS^{\mathbb{H}} to spheres S nS^n without group action?

    if we take the ℤ/2\mathbb{Z}/2-action (the correct one that you used, given by multiplication of quaternions by ±1\pm 1) and n=4n = 4 on the right then we get bipointed maps from each even-graded polynomial in the quaternion qq and its conjugate q¯\bar q, I suppose. So for instance q↦a⋅q 2+b⋅qq¯+c⋅q¯ 2q \mapsto a \cdot q^2 + b \cdot q \bar q + c \cdot {\bar q}^2 going S ℍ→S 4S^{\mathbb{H}} \to S^4.

    These guys need attention, too, I see now some examples where they appear. I’ll try to develop my notes on this further. Then hopefully by the time you have other duties out of the way, we could come back to this afresh.

    • CommentRowNumber27.
    • CommentAuthorUrs
    • CommentTimeFeb 19th 2019
    • (edited Feb 19th 2019)

    Have now finally re-instantiated that simplest non-trivial example of π ℝ sgn(S ℝ sgn)\pi^{\mathbb{R}_{sgn}}\left( S^{\mathbb{R}_{sgn}}\right) (here), now as a special case and illustration of tom Dieck’s equivariant Hopf degree theorem

    diff, v10, current

    • CommentRowNumber28.
    • CommentAuthorUrs
    • CommentTimeFeb 19th 2019
    • (edited Feb 19th 2019)

    Now I’d like to use that equivariant Hopf degree theorem, with its information on π V(S V)\pi^V\left( S^V\right) (here) to say more about π V(S W)\pi^V\left( S^W\right) when V≠WV \neq W, notably for that case W=ℍW = \mathbb{H}, V=ℍ imV = \mathbb{H}_{im} that we discussed above.

    So there we had one element

    [Ad (−)(i)]∈π ℍ im(S ℍ) \left[ Ad_{(-)}(\mathrm{i}) \right] \;\in\; \pi^{\mathbb{H}_{im}}\left( S^{\mathbb{H}}\right)

    and that was, admittedly, the only non-zero element I could see at all. But composition of representatives gives an action

    π V(S V)×π V(S W) ⟶ π V(S W) ([f],[g]) ↦ [f∘g] \array{ \pi^V \left( S^V\right) \times \pi^{V}\left( S^W\right) &\longrightarrow& \pi^{V}\left( S^W\right) \\ ([f],[g]) &\mapsto& [f\circ g] }

    of π V(S V)\pi^V \left( S^V\right) on π V(S W)\pi^V \left( S^W\right), and from looking at the underlying (non-equivariant) classes we know at least that any [c]∈π V(S V)[c] \in \pi^V \left( S^V\right) with deg(c {e})≠0deg\left(c^{\{e\}}\right) \neq 0 acts freely on π V(S W)\pi^{V}\left( S^W\right) this way, which generally gives lots of new elements in π V(S W)\pi^{V}\left( S^W\right).

    Now it would be nice if this kind of argument could be used to not just explore parts of π V(S W)\pi^{V}\left( S^W\right), but to fully exhaust and describe it. Hm…

    • CommentRowNumber29.
    • CommentAuthorUrs
    • CommentTimeFeb 19th 2019
    • (edited Feb 19th 2019)

    spelled out also the example (here) of

    π ℍ(S ℍ) {0,∞}/ ≃ |G|⋅ℤ+1 ≃ |G|⋅ℤ ≃ ℤ [S ℍ⟶cS ℍ] ↦ deg(c {e}) ↦ deg(c {e})−1 ↦ (deg(c {e})−1)/|G| \array{ \pi^{\mathbb{H}} \left( S^{\mathbb{H}} \right)^{\{0,\infty\}/} &\simeq& {\left\vert G\right\vert} \cdot \mathbb{Z} + 1 &\simeq& {\left\vert G\right\vert} \cdot \mathbb{Z} &\simeq& \mathbb{Z} \\ \left[ S^{\mathbb{H}} \overset{c}{\longrightarrow} S^{\mathbb{H}} \right] &\mapsto& deg\left( c^{ \{e\} }\right) &\mapsto& deg\left( c^{ \{e\} }\right) - 1 &\mapsto& \big( deg\left( c^{ \{e\} }\right) - 1 \big)/ {\left\vert G\right\vert} }

    diff, v12, current

    • CommentRowNumber30.
    • CommentAuthorUrs
    • CommentTimeFeb 20th 2019

    I finally dawns on me that tom Dieck’s construction of the isomorphism

    𝕊 G(*)≃A(G) \mathbb{S}_G(\ast) \simeq A(G)

    (between the stable equivariant cohomotopy of the point and the Burnside ring) reduces for finite groups GG to the following beautiful statement:

    The H-marks of a virtual permutation on the right is the degree at HH-fixed points of the corresponding representative on the left:

    𝕊 G(*)≃ lim⟶ V(π 0Maps(S V,S V) G) ⟶ϕ≃ A(G) c ↦ ϕ(c) \array{ \mathbb{S}_G(\ast) \simeq & \underset{\longrightarrow_{\mathrlap{V}}}{\lim} \left( \pi_0 Maps\left( S^V, S^V\right)^G \right) &\underoverset{\phi}{\simeq}{\longrightarrow}& A(G) \\ & c &\mapsto& \phi(c) }

    with

    deg(c H)=|(ϕ(c)) H| deg\left(c^H\right) \;=\; \left\vert \left(\phi(c)\right)^H \right\vert

    for all HH. This relation fully determines ϕ\phi, since by the equivariant Hopf degree theorem the classes of maps S V→cS VS^V \overset{c}{\to} S^V are uniquely fixed by their HH-degrees, as HH-ranges, and since virtual GG-sets ϕ(c)\phi(c) are uniquely fixed by their HH-Burnside marks, as HH varies.

    • CommentRowNumber31.
    • CommentAuthorUrs
    • CommentTimeFeb 28th 2019
    • (edited Feb 28th 2019)

    For G=ℤ/2G=\mathbb{Z}/2, what are the classes of the equivariant maps

    [S 5 sgn⟶S 3+1 sgn] [S^{ 5_{sgn}} \longrightarrow S^{ 3 + 1_{sgn}}]

    given as the smash product of

    S ℍ ⟶ S 3 q ↦ q⋅i⋅q¯ \array{ S^{ \mathbb{H} } &\longrightarrow & S^{3} \\ q & \mapsto & q\cdot i \cdot \bar q }

    with

    n⋅(−):S 1 sgn⟶S 1 sgn n\cdot(-) \colon S^{1_{sgn}} \longrightarrow S^{1_{sgn}}

    ?

    Forgetting the equivariance, these classes are 0 or 1 in ℤ/2\mathbb{Z}/2, depending on whether nn is even or odd, respectively.

    Now not forgetting the equivariance, is the ℤ\mathbb{Z} worth of windings nn on the S 1 sgnS^{1_{sgn}} factor retained?

1 to 31 of 31

  1.