# Morse Theory and Exotic Spheres

Hey everyone! Im 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 refference im 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 substitudes $(u',v')$ 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".

Thats where i get lost, i dont 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 cant get it to work out, i tought maybe he was using the involution $v \mapsto v^{-1}$ somehow but it doesnt add up either... Any help is much appreciated! Thanks in advance!

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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. –  Juan OS Oct 21 '10 at 6:02

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.
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! –  Juan OS Oct 21 '10 at 5:59
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! –  Juan OS Oct 21 '10 at 6:04