# “The” Milnor Fibration, and Why-You-Should-Care.

I’m back!  After a long period of laziness, I’m back.  Mainly, because the past week, I’ve been kicking myself in the ass for losing basically all my notes over the past few months, and I have to present at the math department’s new seminar in singularity theory.  Aren’t I smart?

The topic?  A really useful technique/object used to study the topology of (complex) analytic spaces, called the Milnor fibration.  The reason for the quotes in the title is simple: there are several objects and manifold manifestations of this so-called “Milnor fibration.”  Hence, I’ll do my best to introduce “the” main idea, and hopefully walk both myself and whatever readers are out there through its different forms.

Let’s start simple, with an innocent question:  “Given a complex analytic hypersurface $Y$ in a complex manifold $M$, and a point $p \in Y$, how does $Y$ ‘sit inside’ $M$ at $p$?”

Since we only care about the local behavior (here) of $Y$ at $p$ in $M$, without any loss of generality, we can rephrase this as follows:  $M = \mathcal{U}$ is a connected open neighborhood of $p = 0 \in \mathbb{C}^{n+1}$, $Y = V(f) := f^{-1}(0)$ for some (non-constant) complex analytic function germ $f : (\mathcal{U},0) \to (\mathbb{C},0)$.  In this form, our innocent question becomes less hand-wavy:

“How does $V(f)$ (topologically) ‘sit inside’ $\mathcal{U}$ at $0$?”

Wtf do we mean by “sit inside”?  We want to know about the “topological type” (i.e., the homeomorphism class of the pair $(V(f),\{0\})$), but more importantly, we want to know how $V(f)$ is embedded as a subset of $\mathcal{U}$.  For this, we mean the local, ambient, topological type of $V(f)$ in $\mathcal{U}$ at 0, given by the datum of a triple $(\mathcal{W},\mathcal{W} \cap V(f),\{0\})$, where $\mathcal{W}$ is an open neighborhood of 0 in $\mathcal{W}$.  In our case, it suffices to consider triples of the form $(\overset{\circ}{B}_\varepsilon, \overset{\circ}{B}_\varepsilon \cap V(f), \{0\})$, where $\overset{\circ}{B}_\varepsilon$ denotes the open ball centered at the origin in $\mathbb{C}^{n+1} \cong \mathbb{R}^{2n+2}$.  For the sake of horribly cramped notation, I’ll suppress the dimension, and hope you can keep up.

From general topological stuff, if we can determine the structure of a triple $(B_\varepsilon, B_\varepsilon \cap V(f),\{0\})$, we can understand $(\overset{\circ}{B}_\varepsilon, \overset{\circ}{B}_\varepsilon \cap V(f), \{0\})$.

Now, the question becomes:

“What is the topological type of the triple $(B_\varepsilon, B_\varepsilon \cap V(f),\{0\})$?”

But wait, it gets better (due to a result of Milnor):

Theorem (Milnor):

For $\varepsilon > 0$ sufficiently small, there is a homeomorphism of triples $(B_\varepsilon, B_\varepsilon \cap V(f),\{0\}) \cong (Cone(S_\varepsilon),Cone(S_\varepsilon \cap V(f)), \{0\})$.  Moreover, the topological type of this triple is independent of $\varepsilon$ (for $\varepsilon$ sufficiently small).

Where “Cone” is the topological cone. That is, if $X$ is a topological space, $Cone(X) := X \times [0,1] / X \times \{0\}$.  This is just a fancy way of saying “all the interesting topological behavior occurs on the boundary of the ball.”  So…if we can figure out how this space $K := S_\varepsilon \cap V(f)$ (called the Real Link of $V(f)$ at 0) is embedded inside $S_\varepsilon$, then we’ve (more or less) answered our question.  It’s important to note that the real link of $V(f)$ at 0 is actually a well-defined object, since the topological type of the above triple is independent of epsilon, when epsilon is chosen “small enough”.

This is basically what Milnor strove to answer with his Fibration.  But, you’ll have to wait for the sequel to see that in action.

Before I leave,  I wanted to talk about some nice cases/descriptions of the real link of $V(f), K$:

Lemma:

Let $X$ be a real analytic subset of $\mathbb{R}^{n+1}$  containing the origin, such that $X - \{0\}$ is a real analytic submanifold of $\mathbb{R}^{n+1}$.  Then, for $\varepsilon > 0$ sufficiently small, $X$ transversely intersects the $n$-sphere $S_\varepsilon.$

proof:

Suppose, for the sake of contradiction, that no such $\varepsilon$ exists.  Then, for all $\varepsilon > 0$, there is some $p \in S_\varepsilon \cap X$ such that

$T_p S_\varepsilon + T_p X \neq T_p \mathbb{R}^{n+1}$

Let $r(x) = \|x\|^2$ be the (real analytic) “norm-squared” function on $\mathbb{R}^{n+1}$, so that $r^{-1}(\varepsilon^2) = S_\varepsilon$.  In addition, since the origin is the only critical point of $r$, $T_p S_\varepsilon = T_p r^{-1}(r(p))$.  Then, the above implies $p \in Z := \Sigma(r|_X)$ (where $\Sigma(r|_X)$ is the critical locus of $r$, restricted to $X$, which is a real analytic subset of $\mathbb{R}^{n+1}$).  Since we get such a $p$ for all $\varepsilon > 0$, it follows that $0 \in \overline{Z}$, so we can apply the Curve Selection Lemma: there exists a real analytic curve $\gamma : [0,\eta) \to \mathbb{R}^{n+1}$ such that $\gamma(0) = 0$, and for $t \in (0,\eta)$, $\gamma(t) \in Z$.  What then?

When in doubt, take the derivative of something: for $t \neq 0$,

$r(\gamma(t))' = d_{\gamma(t)} r(\gamma(t)') = 0$

since $\gamma(t)' \in \text{Ker}(d_{\gamma(t)}r)$.  That is, $r(\gamma(t)) = \text{ constant}$.  BUT, $r(\gamma(0)) = 0$, and $r \circ \gamma$ is a real analytic function, so we actually get $r(\gamma(t)) = 0$ for ALL $t \in [0,\eta)$.  BUT, $r(\gamma(t)) =0$ iff $\gamma(t) = 0$, CONTRADICTION!

And the lemma follows.

Done.

So…here, if 0 is an isolated critical point of $f$, this tells us that (for small enough $\varepsilon > 0$), that the real link $K$ is a smooth submanifold of $S_\varepsilon$, of real dimension $2n-1$.

More later.

## Author: brianhepler

I'm a second-year math postdoc at the University of Wisconsin-Madison, and I think math is pretty neat. Especially the more abstract stuff. It's really hard to communicate that love with the general population, but I'm going to do my best.