I'm finally at the end of Milnor's "On manifolds homeomorphic to the 7-sphere", and I stumbled upon something I cant figure out...

For those with the reference I'm talking about "lemma 5", it goes something like this, you have two $\mathbb{S}^3$ bundles over $\mathbb{S}^4$, we want to obtain the total space of this bundle, so you glue them via the transition function, one can think of this as having a pair of copies of $(\mathbb{R}^4 \setminus \{0\}) \times \mathbb{S}^3$ and gluing them by identifiying $(u,v) \mapsto (u',v')=(u / \|u\|^2, u^hvu^j/\|u\|)$ where $u$ and $v$ are quaternions, so far so good, now Milnor states that if $h+j =1$ then this manifold is a $7$-sphere, his reason is that the function $f(x) = \mathfrak{R}(v)/(1+\|u\|^2)^{1/2}$ is a morse function, this with the "first" coordinate chart, for the second he defines $u'' = u'(v')^{-1}$ and substitutes $(u',v')$ for $(u'',v')$ stating that the function $f$ is now given by $\mathfrak{R}(u'')/(1+\|u''\|^2)^{1/2}$. He then says "It is easily verified that f has only two critical points (namely $(u,v) = \pm (0,1)$) and that these are nondegenerate".

That's where I get lost; I don't understand his change of coordinates $(u',v') \mapsto (u'',v')$, nor why he states the function is now the one stated... I tried developing the algebra but I can't get it to work out, I thought maybe he was using the involution $v \mapsto v^{-1}$ somehow but it doesn't add up either...

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    $\begingroup$ I believe the answer to my question is well outlined by Greg; Milnor is defining the function via the coordinate charts, and $u''$ is not a coordinate change but rather a way to simplify notation. Adressing the remarks Greg makes about how Milnor came up with this function, I believe you can get a more general picture by reading the latter article "Differentiable structures on spheres", where Milnor uses rotation groups to construct higher dimensional examples. $\endgroup$
    – Juan OS
    Oct 21, 2010 at 6:02

1 Answer 1


Milnor didn't explain the formula as much as maybe he should have, but the point is that the real part of a unit-length quaternion is invariant under both conjugation and inversion. Let $$r = ||u|| \qquad \hat{u} = u/r,$$ so that $$v' = \hat{u}^h v \hat{u}^j \qquad u' = \hat{u}r \qquad ||u'|| = ||u''|| = 1/r.$$ Thus $$v' \hat{u}^{-1} = \hat{u}^h v \hat{u}^{-h}$$ is conjugate to $v$. Thus $$\mathfrak{R}(v) = \mathfrak{R}(v'\hat{u}^{-1}) = \mathfrak{R}(\hat{u} (v')^{-1}).$$ The first equality is conjugation, the second one is inversion. So then you get $$\frac{\mathfrak{R}(v)}{\sqrt{1 + ||u||^2}} = \frac{\mathfrak{R}(v)}{\sqrt{1+r^2}} = \frac{\mathfrak{R}(\hat{u} (v')^{-1})}{\sqrt{1+r^2}} = \frac{\mathfrak{R}((\hat{u}/r) (v')^{-1})}{\sqrt{1+r^{-2}}} = \frac{\mathfrak{R}(u'')}{\sqrt{1+||u''||^2}}.$$ Note that, although $(u'',v')$ certainly is a valid parameterization of the second chart, it's enough to think of $u''$ as a convenient function rather than part of a coordinate frame.

The question now in my mind is, how did Milnor think of this algebra? I do not know the answer. Maybe he started with a round 4-sphere with its quaternionic Hopf fibration, and the elementary Morse function that consists of one of the coordinates in $\mathbb{R}^8$. You immediately get that there are two critical points (the north and south pole) and that they lie on the same Hopf fiber, since opposite points on a sphere always do. Apparently this Morse function fits together in a similar way for all of these 3-sphere bundles over the 4-sphere.

  • $\begingroup$ Thanks Greg, I realized this a couple of hours ago, conjugation is the key to it all, however, the Morse function doesnt piece together in quite the same way for all bundles, the fact that $h+j=1$ is crucial for in order for the function to be well defined, since otherwise the real part of $v$ would not be preserved! $\endgroup$
    – Juan OS
    Oct 21, 2010 at 5:59
  • $\begingroup$ In fact, if $h+j \neq \pm 1$ then the total space is not homeomorphic to the sphere, so the Morse function cannot have only two critical points! Thanks again! $\endgroup$
    – Juan OS
    Oct 21, 2010 at 6:04
  • $\begingroup$ @ Greg: In Milnor's Collected Papers, vol. 3 there is an introductory section entitled "How these papers came to be written" where he outlines the thought process that led him to the exotic spheres. It may not completely answer your question, but it is a nice window into his thinking. $\endgroup$ Feb 2, 2012 at 13:35

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