added the following pointers:

Discussion of supersymmetric Hopf fibrations:

A. P. Balachandran, G. Marmo, B.-S. Skagerstam and A. Stern, section 9.3 of

*Gauge Symmetries and Fibre Bundles*, Lect. Notes in Physics 188, Springer-Verlag, Berlin, 1983 (arXiv:1702.08910)Simon Davis, section 3 of

*Supersymmetry and the Hopf fibration*(doi:10.4995/agt.2012.1623)

This makes much more sense now. Thank you @Urs, @Charles and @Dylan.

]]>Actually, Sugawara proves everything you need: he shows that if $X$ is a *grouplike $H$-space” (i.e., a unital multiplication such that left multiplication by any element is an equivalence, and likewise for right multiplication), then the homotopy fiber of $h\colon X\star X\to S(X)$ over a particular point of $S(X)$ is equivalent to $X$. Since $S(X)$ must be connected whenever $X$ is non-empty, this means that the homotopy fiber over any point of $S(X)$ is equivalent to $X$.

The HoTT proof is some sort of instantiation of the following observation: we have a map from the diagram

$X \xleftarrow{\pi_1} X\times X \xrightarrow{\pi_2} X$to the diagram

$* \leftarrow X \rightarrow *$where the map in the middle column $\mu\colon X\times X\to X$ is multiplication. The “grouplike” condition lets you prove that both commutative squares you get are homotopy pullbacks. Then “descent” applies to identify the fibers of the induced map from the homotopy colimit of the first diagram to the homotopy colimit of the second diagram. (Actually, the multiplication doesn’t even need to be unital.)

]]>@Urs I have no problem with the case for $S^0$, perhaps my question is better phrased as “when is the hopf construction a fibration”? The problem I have for $S^0$ is that the article reads like the hopf construction gives you that it is a fibration, fortunately, as you said twice it is easy to see that this is actually a fibration for $S^0$. I just want to clear up the conditions needed for the hopf construction to give a fibration.

]]>I have made a brief edit to record Charles’ observations from #18 and #19 here. No time for more.

]]>I may never have understood what your question actually is. You seemed to claim some problem for the case $S^0$, where I showed you that one gets the double cover map all right, which is as fibration as it gets.

]]>@Urs and both time I agree, what I am confused about is why this doesn’t work in HoTT. I am not at all disputing what you have written both times about $S^0$. The nlab article on Hopf construction, attributes Theorem 3.1 to Sugawara. Unless I am mistaken, Charles has just said that Sugawara does not show this is a quasifibration, which is what Stasheff’s book and the article claim. Is this not a problem?

]]>I have spelled it out twice now, in some detail here. Both times you come back just re-iterating the question.

]]>The reason I brought it up is because in HoTT we have “0-connectedness” floating around. I was further confused by Stasheff’s book mentioning the condition. When I was learning Hopf fibrations in Algebraic topology I don’t ever remember using the hopf construction to make them. I only really found out about it when learning HoTT. The mathoverflow question hasn’t helped in the slightest, though I can’t blame it since my question wasn’t exactly well written. I am inclined to believe that it does work for $S^0$ though I am made slightly uncomfortable about what Charles has said.

Now I know it isn’t exactly a good motivation to think this way because of HoTT, but this still poses the question of why it is needed in HoTT. In the HoTT proof that the hopf construction gives a fibration, an essential part of which is showing left and right multiplication in the H-space is an equivalence. For this one “needs” 0-connectedness.

Overall I couldn’t care less if the hopf construction works for $S^0$, since the real hopf fibration is so easy to give explicitly in HoTT and AT. The only thing confusing me is the hopf construction working for $S^0$. And on this I am unfortunately still confused.

]]>Just to say that I have recorded the simple example from #7 (of working out the real Hopf fibration as the special case of the Hopf construction for $X = \mathbb{Z}/2$) to the entry *real Hopf fibration*: here

I’m guessing Stasheff is referring Theorem 4 (p. 119) of

Sugawara, Masahiro.

On a condition that a space is an H-space.Math. J. Okayama Univ. 6 (1957), 109–129.

This basically shows that the fiber of $h\colon X\star X\to SX$ over one of the cone points of $SX$ (which is in fact homeomorphic to $X$) is weakly equivalent (by the tautological map) to the homotopy fiber of $h$. It does *not* show that *every* fiber of $h$ is weakly equivalent (via the tautological map) to the corresponding homotopy fiber, so it does not show that $h$ is a quasifibration.

@Dylan: I don’t know where Sugawara’s proof is found: there are three papers of his in the references of Stasheff’s book, and I’m not going to hunt them all down.

Suppose $X$ is a grouplike $H$-space, with multiplication $\mu$. Let $f\colon X\times X\to X$ be any morphism in the relative mapping space $\mathrm{Map}_{X\vee X/}(X\times X,X)$ in the path component of $\mu$. Every such $f$ determines a group-like $H$-space structure on $X$, and if we believe the theorem all fibers of all the Hopf constructions $h_f\colon X\star X\to SX$ must be weakly equivalent to $X$. The map $f$ can be found inside $h_f$ (i.e., $h_f$ restricts to $f$ along inclusions $X\times X\subset X\star X$ and $X\subset S(X)$), so in particular *all the fibers of all the maps $f$ over all points of $X$ must be weakly equivalent to $X$*. This seems unlikely.

Dold-Lashof prove that a *different* construction gives a quasifibration; this other construction, applied to a group-like $H$-space, gives a map homotopic to the Hopf construction as usually defined.

I don’t see why the Hopf construction applied to an $H$-space $X$ must necessarily produce a quasifibration, even if we assume $\pi_0X$ is a group or even just $X$ connected. (It is certainly true that the *homotopy fiber* of $h\colon X\star X\to SX$ will be equivalent to $X$ under these hypotheses; I just don’t see why the *actual fibers* of $h$ should also be equivalent to $X$, which is what necessarily must happen if $h$ is a quasifibration.)

added brief mentioning (here) of the suspensions of the Hopf fibrations generating the corresponding stable homotopy groups of spheres

]]>@Urs I am confused what you mean. Is it in the definition of H-space that left-multiplication is a homotopy equivalence? I thought a H-space was just a unital magma in the homotopy category.

]]>$X$ must be an H-space. As it says in the entry.

]]>I have asked a question on MO hopefully we will get some correct answers rather than my vague thoughts.

]]>In general it is not correct to say the hopf construction works for any $X$ but for a $0$-connected $X$ or at least $\pi_0(X)$ being a group. When these are CW-complexes it seems quasifibrations become fibrations due to Dold. I am not so familar with proving left multiplication in a H-space is a weak homotopy equivalence however, especially with the connectedness condition weakened.

]]>Actually ignore what I said above that is complete nonsense. I found that X is 0-connected can be weakened to $\pi_0(X)$ is a group, so this is in fact correct. On page 4. The reason I got confused is because the multiplcation maps $X\times X \to X$ should be weak homotopy equivalences $X \simeq X$ when given one argument. It seems the book claims this can be proven on the weaker assumption that $\pi_0(X)$ is a group. Which means the articles are kind of correct, but only when this condition holds.

]]>Sorry, which problem do you think occurs?

]]>I think the problem occurs because we get a quasifibration not necesserily a fibration. I am not so sure however.

]]>Hm, I had another look at the Definition as stated in the entry here and plugged in $S^0 \simeq \mathbb{Z}/2$ for $X$. From, this I do seem to get the real Hopf fibratoipn $S^1 \overset{\cdot 2}{\longrightarrow} S^1$ all right.

Let’s go through it in detail:

So on the left we have four copies of the interval, labeled in $\mathbb{Z}/2 \times \mathbb{Z}/2$, while on the right we have two copies, labeled by $\mathbb{Z}/2$.

As we run once around on the left (with coordinates in $[0,1] \times (\mathbb{Z}_2 \times \mathbb{Z}_2)$)

$(0,(0,0)) \to (1,(0,0)) \sim (1,(1,0)) \to (0,(1,0)) \sim (0,(1,1)) \to (1,(1,1)) \sim (1,(0,1)) \to (0,(0,1)) \sim (0,(0,0))$we run on the right (with coordinates in $[0,1] \times \mathbb{Z}_2$)

$(0,(0+0 = 0)) \to (1,(0 + 0 = 0)) \sim (1,(1 + 0 = 1)) \to (0,(1 + 0 = 1)) \sim (0,(1+1 = 0)) \to (1,(1+1 = 0)) \sim (1,(0 + 1 = 1)) \to (0,(0 + 1 = 1)) \sim (0,(0 + 0 = 0))$That’s twice around the circle on the right, for once on the left.

Seems right to me. But let me know if I am missing something.

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