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am starting something here. For the moment I just want to record today’s
eventually more references ought to be added
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: $\mathbb{Z}_2$-equivariant maps from $S^1$ to itself (here).
Simplistic as it is, it is already kind of interesting in how it works.
There’s a typo after “Indeed, under passage to fixed points any”, where you have $S^1 \simeq \{0,1\}$.
Thanks, fixed now. And there is something strange about the example, too. Hope to sort it out…
What is the class in $\pi_4(S^3) \simeq \mathbb{Z}_2$ of the map
$\array{ S^{\mathbb{H}} &\longrightarrow& S^{\left(\mathbb{H}_{im}\right)} \\ q &\mapsto& q v \bar q }$where $v \in \mathbb{H}_{im} \setminus \{0\}$ is any non-zero imaginary quaternion?
Let me try some things on the fly… Take $v=i$, the choice is immaterial.
My first thought is that it should be the nontrivial class, since restricted to $Sp(1) = S(\mathbb{H}) \subset S^\mathbb{H}$ the equator, this looks like the Hopf map $H\colon S^3 = Sp(1) \to S(\mathbb{H}_{im}) = S^2$, which can be expressed as $q\mapsto qi\bar{q}$ (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 $H$, so yes, it is the nontrivial class.
Thanks!!
Just for citation purposes: Where in Atiyah do you see this stated?
But then, even Wikipedia spells this out. All right, thanks again!
Yes, this is essentially coming from the representation $Sp(1) \to SO(\mathbb{H}_{im})$ by conjugation, and recognising the unit sphere in the imaginary quaternions as a homogeneous space for the action of $Sp(1)$ via this rep.
Now my next question is the following variant of the above:
For $G = \mathbb{Z}_n = \langle 2 \pi \mathrm{i}/n\rangle \hookrightarrow SU(2)$ a cyclic subgroup, is there any bi-pointed $G$-equivariant continuous function
$S^{\mathbb{H}} \longrightarrow S^{ \langle \mathrm{j}, \mathrm{k}\rangle }$from the representation sphere of $\mathbb{H}$ 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 $S^0$.
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.
Doesn’t this reduction to a statement about the non-compactified $G$-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.
Sure, we may ask equivalently for equivariant maps $f$ between the linear $G$-spaces that take 0 to 0 and for which $\underset{r_1 \to \infty}{lim} r_2(f)$ 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 $G = \mathbb{Z}/2$ the quadratic map $(-)^2 \colon S^{\mathbb{R}_{sgn}} \to S^{\mathbb{R}}$ exists equivarianty (and is “bi-pointed”) but the only equivarian linear map $\mathbb{R}_{sgn} \to \mathbb{R}$ is zero.
finally added pointer to
Assume we have such an equivariant map (we should write $\mu_n$ for the group, since it’s the roots of unity in $U(1) \subset Sp(1)$). I’m wondering what would be the induced linear map between the representations on the tangent spaces at $0\in S^\mathbb{H}$ and $0\in S^{\langle i,k\rangle}$. I can’t see how those representations are anything other than $\mathbb{H}$ and $\langle j,k \rangle$. 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 $\mu_n$-orbits only once (for $n$ odd) or twice (for $n$-even). I was thinking what we would do if instead of $\mu_n$ we were looking at $U(1)$. Then any nontrivial $U(1)$-orbit in $S^\mathbb{H}$ 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.
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 $\mu_n$-equivariant map $S^\mathbb{H} \to S^{\langle \mathrm{j}, \mathrm{k}\rangle}$. Do you feel this can be proven, or that it is false?
Assume we have such an equivariant map (we should write $\mu_n$ for the group, since it’s the roots of unity in $U(1) \subset Sp(1)$). I’m wondering what would be the induced linear map between the representations on the tangent spaces at $0\in S^\mathbb{H}$ and $0\in S^{\langle i,k\rangle}$. I can’t see how those representations are anything other than $\mathbb{H}$ and $\langle j,k \rangle$.
Right, the tangent $G$-representation of the $G$-representation sphere $S^V$ at $0$ is $V$. And so if $V_1$ and $V_2$ are distinct linear irreps, then by Schur the only linear $G$-map $V_1 \to V_2$ is zero, which means that any bipointed differentiable $G$-equivariant function $f : S^{V_1} \to S^{V_2}$ must have vanishing differential at 0.
Which is just what we see in the examples we discussed, where $f$ is quadratic.
My almost-proof that there is no bipointed equivariant $S^{\mathbb{H}} \to S^{\langle \mathrm{j}, \mathrm{k}\rangle}$ 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 $q$ in $S^{\mathbb{H}}$ to its conjugation action on something, and that something can “obviously” only be an imaginary quaternion, and that leads to the map $S^{\mathbb{H}} \to S^{\mathbb{H}_{im}}$ which does however not factor through $S^{\langle \mathrm{j}, \mathrm{k}\rangle}$. QED.
Now how to remove the quotation marks around “obviously”? :-)
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: $\mu_2 = \{\pm 1\}$? It has trivial action on $S^{\langle j,k\rangle}$ but acts on the equator of $S^\mathbb{H}$ as the antipodal map and then extends to the obvious thing. Any $\mu_{2k}$-equivariant map would give one of these, so this is at least a potential obstruction to all the even cases.
Given the $\mu_2$-equivariant map, we’d get a ordinary bipointed map $\Sigma\mathbb{RP}^3\to S^2$. Obstruction theory might have something to say about this.
OK, so assuming (for the sake of contradition) there is some $\mu_2$-equivariant map $e\colon S^{\mathbb{H}} \longrightarrow S^{ \langle \mathrm{j}, \mathrm{k}\rangle }$, then this map factors into bipointed maps
$S^{\mathbb{H}} \longrightarrow \Sigma\mathbb{RP}^3 \longrightarrow S^{ \langle \mathrm{j}, \mathrm{k}\rangle } .$Up to homotopy there are only two maps $S^4 \to S^2$: the trivial one, and the composite $\eta \circ \Sigma \eta$ for $\eta$ 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 $e$ would have to be nullhomotopic. I would think that $S^4 \to \Sigma\mathbb{RP}^3$ is not null-homotopic. It may be possible to push this line of argument further, but it’s late here :-)
Thanks, David!! I’ll think about it.
Ah, here’s a construction of such a map! Recall that $\Sigma pt \simeq [0,1]$ and consider it as a bipointed space. Pick a path in $S^2$ from $0$ to $\infty$, hence a map $\Sigma pt \to S^2$. Then compose this with $S^4 \to \Sigma \mathbb{RP}^3 \to \Sigma pt$ to get a bipointed map $S^4 \to S^2$ factoring through the quotient $S^4 \to \mathbb{RP}^2$, hence a $\mu_2$-equivariant map on the representation spheres $S^\mathbb{H} \to S^{ \langle \mathrm{j}, \mathrm{k}\rangle }$.
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 $n=2$ does a $\mu_n$-equivariant bipointed map exist.
Hmm… Now I’m really off to bed.
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 $\mu_2$ should act trivially on $S^{\langle \mathrm{j}, \mathrm{k}\rangle}$. This doesn’t sound right to me: The nontrivial element acts by rotation by $\pi$ in the $(\mathrm{j},\mathrm{k})$-plane, hence it rotates our 2-sphere by $\pi$ 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 $S^{\mathbb{H}} \to S^{\langle i,j\rangle}$, to which it does not apply, like to so many other examples, due to its extremely strong assumptions…)
$\mu_2 = \{1,e^{i\pi}\}$, no? Try conjugating $a j + b k$ by $e^{i\pi}$ and see what happens ;-)
I did calculate a matrix representative of the action of $e^{2\pi i k/n}$, and it’s rotation by 4$\pi i k/n$.
Sorry, my bad, you are right, of course. I was thinking about the $Z/2$ down in $SO(3)$, but you are of course using the correct $Z/2$ up in $SU(2)$, which covers $1$ downstairs.
And now it’s me who is off to bed. See you tomorrow!
But so we need to exclude that case of $\mathbb{Z}/2$ 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 $S^\mathbb{H}$ for the case of the MK6) to the representation sphere of – and that’s the key point here - the orthogonal complement $V-V^G$ of the trivial rep inside $V$.
Hmm, will have to think on that, but not too much at present, with other things on.
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 $S^{\mathbb{H}}$ to spheres $S^n$ without group action?
if we take the $\mathbb{Z}/2$-action (the correct one that you used, given by multiplication of quaternions by $\pm 1$) and $n = 4$ on the right then we get bipointed maps from each even-graded polynomial in the quaternion $q$ and its conjugate $\bar q$, I suppose. So for instance $q \mapsto a \cdot q^2 + b \cdot q \bar q + c \cdot {\bar q}^2$ going $S^{\mathbb{H}} \to S^4$.
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.
Now I’d like to use that equivariant Hopf degree theorem, with its information on $\pi^V\left( S^V\right)$ (here) to say more about $\pi^V\left( S^W\right)$ when $V \neq W$, notably for that case $W = \mathbb{H}$, $V = \mathbb{H}_{im}$ that we discussed above.
So there we had one element
$\left[ Ad_{(-)}(\mathrm{i}) \right] \;\in\; \pi^{\mathbb{H}_{im}}\left( S^{\mathbb{H}}\right)$and that was, admittedly, the only non-zero element I could see at all. But composition of representatives gives an action
$\array{ \pi^V \left( S^V\right) \times \pi^{V}\left( S^W\right) &\longrightarrow& \pi^{V}\left( S^W\right) \\ ([f],[g]) &\mapsto& [f\circ g] }$of $\pi^V \left( S^V\right)$ on $\pi^V \left( S^W\right)$, and from looking at the underlying (non-equivariant) classes we know at least that any $[c] \in \pi^V \left( S^V\right)$ with $deg\left(c^{\{e\}}\right) \neq 0$ acts freely on $\pi^{V}\left( S^W\right)$ this way, which generally gives lots of new elements in $\pi^{V}\left( S^W\right)$.
Now it would be nice if this kind of argument could be used to not just explore parts of $\pi^{V}\left( S^W\right)$, but to fully exhaust and describe it. Hm…
spelled out also the example (here) of
$\array{ \pi^{\mathbb{H}} \left( S^{\mathbb{H}} \right)^{\{0,\infty\}/} &\simeq& {\left\vert G\right\vert} \cdot \mathbb{Z} + 1 &\simeq& {\left\vert G\right\vert} \cdot \mathbb{Z} &\simeq& \mathbb{Z} \\ \left[ S^{\mathbb{H}} \overset{c}{\longrightarrow} S^{\mathbb{H}} \right] &\mapsto& deg\left( c^{ \{e\} }\right) &\mapsto& deg\left( c^{ \{e\} }\right) - 1 &\mapsto& \big( deg\left( c^{ \{e\} }\right) - 1 \big)/ {\left\vert G\right\vert} }$I finally dawns on me that tom Dieck’s construction of the isomorphism
$\mathbb{S}_G(\ast) \simeq A(G)$(between the stable equivariant cohomotopy of the point and the Burnside ring) reduces for finite groups $G$ to the following beautiful statement:
The H-marks of a virtual permutation on the right is the degree at $H$-fixed points of the corresponding representative on the left:
$\array{ \mathbb{S}_G(\ast) \simeq & \underset{\longrightarrow_{\mathrlap{V}}}{\lim} \left( \pi_0 Maps\left( S^V, S^V\right)^G \right) &\underoverset{\phi}{\simeq}{\longrightarrow}& A(G) \\ & c &\mapsto& \phi(c) }$with
$deg\left(c^H\right) \;=\; \left\vert \left(\phi(c)\right)^H \right\vert$for all $H$. This relation fully determines $\phi$, since by the equivariant Hopf degree theorem the classes of maps $S^V \overset{c}{\to} S^V$ are uniquely fixed by their $H$-degrees, as $H$-ranges, and since virtual $G$-sets $\phi(c)$ are uniquely fixed by their $H$-Burnside marks, as $H$ varies.
For $G=\mathbb{Z}/2$, what are the classes of the equivariant maps
$[S^{ 5_{sgn}} \longrightarrow S^{ 3 + 1_{sgn}}]$given as the smash product of
$\array{ S^{ \mathbb{H} } &\longrightarrow & S^{3} \\ q & \mapsto & q\cdot i \cdot \bar q }$with
$n\cdot(-) \colon S^{1_{sgn}} \longrightarrow S^{1_{sgn}}$?
Forgetting the equivariance, these classes are 0 or 1 in $\mathbb{Z}/2$, depending on whether $n$ is even or odd, respectively.
Now not forgetting the equivariance, is the $\mathbb{Z}$ worth of windings $n$ on the $S^{1_{sgn}}$ factor retained?
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