text:
\begin{document}

\begin{abstract}
We use topology of configuration spaces to give a characterization
of Neuwirth--Stallings pairs $(S^5, K)$ with $\dim K = 2$.
As a consequence, we construct
polynomial map germs
$(\mathbb{R}^{6},0) \to (\mathbb{R}^{3},0)$
with an isolated singularity at the origin such
that their Milnor fibers are not diffeomorphic to a disk,
thus putting an end to
Milnor's non-triviality question.
Furthermore, for a polynomial map germ
$(\mathbb{R}^{2n},0) \to (\mathbb{R}^{n},0)$ or
$(\mathbb{R}^{2n+1},0) \to (\mathbb{R}^{n},0)$, $n \geq 3$,
with an isolated singularity at the origin,
we study the conditions under which
the associated Milnor fiber has the homotopy type
of a bouquet of spheres. We then construct,
for every pair $(n, p)$ with $n/2 \geq p \geq 2$, a
new example of a
polynomial map germ $(\mathbb{R}^n,0) \to (\mathbb{R}^p,0)$
with an isolated singularity at the origin such that
its Milnor fiber has the homotopy type of a bouquet of
a positive number of spheres.
\end{abstract}

\maketitle

\section{Introduction}\label{section1}

In the book ``Singular points of complex hypersurfaces'' \cite{Mi},
John Milnor studied the topology of complex
polynomial function germs in terms of the associated locally trivial
fiber bundles. He also showed
the existence of such structures for real
polynomial map germs with an isolated singularity as follows.

\begin{theorem}[{\cite[Theorem~11.2]{Mi}}]\label{TeoMilnor}
Let $f: (\mathbb{R}^n,0) \to (\mathbb{R}^p,0)$,
$n \geq p \geq 2$, be a polynomial map germ
with an isolated singularity at the origin. Then,
there exists an $\varepsilon_0 > 0$ such that
for all $0 < \varepsilon \leq \varepsilon_{0}$,
the complement of an open tubular neighborhood of the link $K =
f^{-1}(0) \cap S^{n-1}_{\varepsilon}$ in $S^{n-1}_{\varepsilon}$
is the total space of a smooth fiber bundle over the sphere $S^{p-1}$,
with each fiber $F_{f}$ being a smooth compact $(n-p)$-dimensional manifold
bounded by a copy of $K$, where $S^{n-1}_\varepsilon$ denotes
the sphere in $\mathbb{R}^n$ with radius $\varepsilon$
centered at the origin.
\end{theorem}

By using the conical structure of the singularity, Milnor proved
that the diffeomorphism type of the link does not change
for all $\varepsilon > 0$ small enough. Moreover, since the
origin is an isolated singularity, we have that
$0 \in \mathbb{R}^{p}$ is a regular value of
$f|S^{n-1}_{\varepsilon} : S^{n-1}_{\varepsilon} \to \mathbb{R}^{p}$.
Therefore, if the link $K=f^{-1} (0) \cap S^{n-1}_{\varepsilon}$
is not empty, then it is a smooth $(n-p-1)$-dimensional
submanifold of the sphere with trivial normal bundle. It also implies
that, for each fixed $\varepsilon$ one can find a small enough
$\delta$, $0 < \delta \ll \varepsilon$, and a closed
disk $D_{\delta}^{p}$ centered at the origin in $\mathbb{R}^{p}$
with radius $\delta$, such that the restriction map
$f: f^{-1}(D_{\delta}^{p})\cap S_{\varepsilon}^{n-1}
\to D_{\delta}^{p}$ is a smooth trivial fiber bundle,
which implies the triviality of the fibration
$$f: f^{-1}(D_{\delta}^{p} \setminus \{0\})\cap
S_{\varepsilon}^{n-1} \to D_{\delta}^{p} \setminus \{0\}.$$
By composing this with the radial projection
$\displaystyle{\pi: D_{\delta}^{p} \setminus \{0\}
\to S^{p-1}_{\delta}}$ onto the boundary sphere
and scaling to the unit sphere,
one finds that the bundle structure on a neighborhood of
the link $K$ is given by
$$\frac{f}{\|f\|}:f^{-1}(D_{\delta}^{p} \setminus \{0\})
\cap S_{\varepsilon}^{n-1} \to S^{p-1}.$$

Now, one can glue this fiber bundle with that given in
Theorem~\ref{TeoMilnor} along the common boundary
$f^{-1}(S_{\delta}^{p-1})\cap S_{\varepsilon}^{n-1}$
in a smooth way, so that we get a smooth locally trivial
fiber bundle
$$S_{\varepsilon}^{n-1}\setminus K \to S^{p-1}.$$

\begin{remark}
Following Milnor's proof of Theorem~\ref{TeoMilnor},
one sees that, although no precise information about the bundle
projection above was provided, in the real settings in general
one cannot expect that it is given by $f/\|f\|$ outside a
neighborhood of the link.

See Milnor's example \cite[p.~99]{Mi} or the following
example adapted from Milnor's one.

Consider $f:(\mathbb{R}^{3},0)\to (\mathbb{R}^{2},0)$ given by
$$f(x,y,z)=(x, x^2+yx^2+y^3+yz^2).$$
It is easy to see that the singular point set is given by
$\Sigma(f)=\{(0,0,0)\}$, and for all
$\varepsilon>0$ small enough, we have
$K=\{(0,0,\varepsilon), (0,0,-\varepsilon)\}$.
However, $f/\|f\|$ does have singular points
and therefore it is not a submersion.
\end{remark}

\begin{definition}[Looijenga \cite{Lo}]
Let $K = K^{n-p-1}$ be an oriented
submanifold of dimension $n-p-1$ of
the oriented sphere $S^{n-1}$ with trivial normal
bundle, or let $K=\emptyset$. Suppose that for some trivialization
$c: N(K) \to K \times D^p$ of a tubular neighborhood $N(K)$ of $K$,
the fiber bundle defined by the composition
$$N(K) \setminus K \stackrel{c}{\to}
K \times (D^p \setminus \{0\}) \stackrel{\pi}{\to} S^{p-1},$$
with the last projection being given by $\pi(x,y) = y/\|y\|$,
extends to a smooth fiber bundle $S^{n-1} \setminus K \to S^{p-1}$.
Then, the pair $(S^{n-1}, K^{n-p-1})$ is called a
\emph{Neuwirth--Stallings pair}, or an \emph{NS-pair} for short.
\end{definition}

According to Theorem~\ref{TeoMilnor} and the previous discussion,
for all $\varepsilon >0 $ sufficiently
small, the pair $(S^{n-1}_{\varepsilon}, f^{-1}(0)
\cap S^{n-1}_{\varepsilon})$ is
an NS-pair. In this case Looijenga called it the
\emph{NS-pair associated to the singularity}.

More recently, several generalizations of such a structure have been
obtained. For instance, in \cite{AT} the authors considered a
real analytic map germ $f:(\mathbb{R}^{n},0) \to (\mathbb{R}^{p},0)$
with non-isolated singularities at the origin and introduced
a condition which also ensures that the pair
$(S^{n-1}_{\varepsilon}, f^{-1}(0) \cap S^{n-1}_{\varepsilon})$
is an NS-pair with the bundle projection given by
$f/\|f\| : S^{n-1}_{\varepsilon} \setminus K \to S^{p-1}$,
where the link $K = f^{-1}(0) \cap S^{n-1}_{\varepsilon}$
is a smooth manifold.
It was called a \emph{higher open book structure} of the sphere $S_{\varepsilon}^{n-1}$. In \cite{ACT, ACT1}
it was shown an extension of such structures for spheres of
small and big enough radii (at infinity), but allowing
singularity in the ``binding'' $K$.
In this case, it was called a \emph{singular open book structure}
of the sphere.

As pointed out by Milnor in \cite[p.~100]{Mi}, the hypothesis of
Theorem~\ref{TeoMilnor} is so strong that examples are difficult to
find, and he posed the following question.

\begin{problem}\label{problem}
For which dimensions $n \geq p \geq 2$
do non-trivial examples exist ?
\end{problem}

Milnor did not exactly specify what ``trivial'' means here:
however, he proposed to say that a real polynomial map germ
$f: (\mathbb{R}^n,0) \to (\mathbb{R}^p,0)$ is \emph{trivial} if
the fiber $F_{f}$ of the bundle given in Theorem~\ref{TeoMilnor}
is diffeomorphic to a closed disk $D^{n-p}$.
In particular, this implies that the fibers of the
associated NS-pair are diffeomorphic to the $(n-p)$-dimensional
open disk.

\begin{remark}
For a holomorphic function germ $f:(\mathbb{C}^{n+1},0) \to
(\mathbb{C},0)$ with an isolated singularity
at the origin, it follows from \cite[Appendix B]{Mi} that the
fibers of the associated Milnor fibration are diffeomorphic to a
$2n$-dimensional disk if and only if $0$ is a non-singular point
of $f$; in fact, the function germ $f$ is trivial if and
only if the Milnor number $\mu_{f} =
\mathrm{deg}_{0}(\nabla f(z))$ is equal to zero, where
$\mathrm{deg}_{0}(\nabla f(z))$ stands for the topological
degree of the map

$$\varepsilon\frac{\nabla f}{\|\nabla f \|}:
S^{2n+1}_{\varepsilon} \to S^{2n+1}_{\varepsilon}$$
for all $\varepsilon>0$ small enough, and
$$\nabla f = \left(\frac{\partial f}{\partial z_1},
\frac{\partial f}{\partial z_2}, \ldots,
\frac{\partial f}{\partial z_{n+1}}\right).$$
\end{remark}

In \cite{CL} Church and Lamotke used results of
Looijenga \cite{Lo} and answered the above question in the following
way.

\begin{theorem}\label{T2} {\rm(a)}
For $0 \leq n-p \leq 2$, non-trivial examples occur precisely
for the dimensions $(n,p)\in \{(2,2), (4,3), (4,2)\}$.

\noindent {\rm(b)}
For $n-p \geq 4$, non-trivial examples occur for
all $(n,p)$.

\noindent {\rm(c)} For $n-p=3$, non-trivial examples occur for
$(5,2)$ and $(8,5)$. Moreover, if the $3$-dimensional Poincar\'e
Conjecture is false, then there are non-trivial examples for all
$(n,p)$. If the Poincar\'e Conjecture is true, then
all examples are trivial except $(5,2)$, $(8,5)$ and possibly $(6,3)$.
\end{theorem}

Since the Poincar\'e Conjecture has been
proved to be true, we have that for $n-p = 3$ the map $f$
can be non-trivial
only if $(n,p)\in \{(6,3), (8,5), (5,2)\}$.
Therefore, Problem~\ref{problem} has been open
uniquely for the dimension pair $(6, 3)$.

In \cite{SHMA} the authors used an extension of
Milnor-Khimshiashvili's formula proved in \cite{ADD}
(see Theorem~\ref{euler} of the present paper)
for real isolated singularity map germs
to show a manageable characterization of Church-Lamotke's results
when the Milnor fiber is $3$-dimensional, as follows.

\begin{theorem}\label{teorema 1}
Let $f:(\mathbb{R}^{n},0) \to (\mathbb{R}^{p},0)$,
$f(x)=(f_{1}(x), f_2(x), \ldots, f_{p}(x))$, be a
polynomial map germ with an
isolated singularity at the origin, and suppose $n-p=3$.
Denote by
$\mathrm{deg}_{0}(\nabla f_{1})$
the topological degree of the mapping
$$\varepsilon \frac{\nabla f_{1}}{\|\nabla f_{1}\|} :
S^{n-1}_{\varepsilon} \to
S^{n-1}_{\varepsilon},$$
where
$$\nabla f_1 = \left(\frac{\partial f_1}{\partial x_1},
\frac{\partial f_1}{\partial x_2}, \ldots,
\frac{\partial f_1}{\partial x_{n}}\right).$$

\noindent {\rm(a)} If the pair $(n,p)=(6,3)$, then the following
three are equivalent.
\begin{itemize}
\item[{\rm({i})}] $f$ is trivial.
\item[{\rm({ii})}] $\mathrm{deg}_{0}(\nabla f_{1})=0$.
\item[{\rm({iii})}] The link $K$ is connected.
\end{itemize}

\noindent {\rm(b)} If the pair $(n,p)=(8,5)$, then the
following three are equivalent.
\begin{itemize}
\item[{\rm({i})}] $f$ is trivial.
\item[{\rm({ii})}] $\mathrm{deg}_{0} (\nabla f_{1})=0$.
\item[{\rm({iii})}] The link $K$ is not empty.
\end{itemize}

\noindent {\rm(c)} If the pair $(n,p)=(5,2)$, then
the following two are equivalent.
\begin{itemize}
\item[{\rm({i})}] $f$ is trivial.
\item[{\rm({ii})}] $\pi_{1}(F_{f})=1$, i.e.\
the Milnor fiber $F_{f}$ is simply connected.
\end{itemize}
\end{theorem}

In this paper we aim to give a characterization of NS-pairs
$(S^5, K)$ with $\dim K=2$,
and use it to prove the existence of
non-trivial real polynomial map germs
$(\R^6, 0) \to (\R^3, 0)$ with an isolated
singularity at the origin, putting an end to
Problem~\ref{problem} posed by Milnor.
For this, we will use tools from configuration spaces
and a construction by Funar in \cite[Section~2.7]{Funar}.
More precisely, we first classify fiber bundles
$E^5 \to S^2$ with fiber the $3$-sphere with the interiors
of a disjoint union of
$3$-disks removed, such that the boundary
fibrations are trivial. We will show that the
isomorphism classes of such bundles are in
one-to-one correspondence with the second
homotopy group of a certain configuration space,
and that its elements correspond to a skew-symmetric
integer matrix. Then, we show that a given
fiber bundle $E^5 \to S^2$ is associated with
an NS-pair $(S^5, K)$ if and only if the skew-symmetric
matrix is unimodular.
As a consequence, we see that the number of boundary
components of a fiber is always odd.
Furthermore, this allows us to construct
a lot of non-trivial NS-pairs $(S^5, K)$, and then
the Looijenga construction \cite{Lo} leads to non-trivial
polynomial map germs with an isolated singularity.

Our second aim in this paper is to
introduce necessary and sufficient conditions
under which the Milnor fiber in the pairs of dimensions $(2n, n)$
and $(2n+1,n)$, $n \geq 3$, is, up to homotopy, a bouquet (or a wedge)
of spheres.

As applications, we give examples of polynomial map
germs $(\mathbb{R}^{n},0) \to (\mathbb{R}^{p},0)$,
$n/2 \geq p \geq 2$, such that the associated Milnor fiber
is a bouquet of a non-zero number of spheres.

Throughout the paper, the (co)homology groups are with
integer coefficients unless otherwise specified.
The symbol
``$\cong$'' denotes a diffeomorphism between
smooth manifolds or an appropriate isomorphism
between algebraic objects.

\section{Classification of bundles}\label{section2}

Let $(S^5, K^2)$ be an NS-pair, where $K^2$ is a closed
$2$-dimensional manifold embedded in the $5$-dimensional
sphere $S^5$. We have the associated
fibration $\pi : S^5 \setminus \Int N(K^2) \to S^2$, where
$N(K^2)$ denotes a closed tubular neighborhood of $K^2$
in $S^5$, and we denote by $F$ its fiber, which
is a compact $3$-dimensional manifold bounded by
a copy of $K^2$.
Since $S^5$ does not fiber over $S^2$, we have $K^2
\neq \emptyset$.
Furthermore, we have the homotopy exact sequence
$$\pi_2(S^5 \setminus \Int N(K^2)) \to \pi_2(S^2)
\to \pi_1(F) \to \pi_1(S^5 \setminus \Int N(K^2)).$$
Since $\pi$ is trivial on the boundary, it has a section,
so that the homomorphism $\pi_2(S^5 \setminus \Int N(K^2)) \to \pi_2(S^2)$
is surjective. Furthermore, $S^5 \setminus \Int N(K^2)$ is
simply connected. Therefore, the compact
$3$-dimensional manifold $F$ is also simply connected.

Then, by a standard argument, we see that $K^2 \cong \partial F$
consists of some copies of $S^2$ and that $F$ is homotopy equivalent
to a $3$-dimensional sphere with some points removed.
Then, by the solution to the
Poincar\'e Conjecture, we see that $F$ is diffeomorphic to
$S^3_{(k+1)}$ for some non-negative integer $k$,
where $S^3_{(k+1)}$ denotes the $3$-sphere with
the interiors of $k+1$ disjoint $3$-balls removed.
Therefore, $\pi$ is a smooth fiber bundle with
fiber $S^3_{(k+1)}$ such that it is trivial on the
boundary. In this section, we classify such fiber
bundles.

Let $\mathrm{Diff}(S^3)$ be the topological group of diffeomorphisms
of $S^3$. By the solution to the Smale Conjecture by Hatcher
\cite{Hatcher}, we have that $\mathrm{Diff}(S^3)$ is homotopy
equivalent to the orthogonal group $O(4)$.

Let us denote by $B^3$ the $3$-dimensional closed ball and for a
non-negative integer $k$, we denote by $\cup^{k+1} B^3$
the disjoint union of $k+1$ copies of $B^3$. We sometimes
regard $\cup^{k+1} B^3$ to
be ``standardly'' embedded in $S^3$, and we denote by
$j_{k+1}: \cup^{k+1} B^3 \to S^3$ the inclusion map.

We denote by $\mathrm{Emb}(\cup^{k+1} B^3, S^3)$ the space of all smooth
embeddings of $\cup^{k+1} B^3$ into $S^3$, not necessarily the
standard one, and by $\mathrm{Diff}(S^3, \cup^{k+1} B^3)$ the
subspace of $\mathrm{Diff}(S^3)$ consisting of those
diffeomorphisms which restrict to the inclusion map
$j_{k+1}$ on $\cup^{k+1} B^3$.
Furthermore, we denote

by $\mathrm{Diff}(S^3_{(k+1)}, \partial S^3_{(k+1)})$ the
topological group of diffeomorphisms of $S^3_{(k+1)}$ which
restrict to the identity on the boundary. Note that
$S^3_{(k+1)} = S^3 \setminus \cup^{k+1} \Int{B^3}$.

The lemma below follows from \cite[Proposition~1, p.~120]{Cerf}.

\begin{lemma}\label{lemma1}
The canonical map $\mathrm{Diff}(S^3, \cup^{k+1} B^3) \to
\mathrm{Diff}(S^3_{(k+1)}, \partial S^3_{(k+1)})$ induces
isomorphisms
$$\pi_i(\mathrm{Diff}(S^3, \cup^{k+1} B^3)) \to
\pi_i(\mathrm{Diff}(S^3_{(k+1)}, \partial S^3_{(k+1)}))$$
for all $i$.
\end{lemma}

Now consider the natural map
$$\varphi : \mathrm{Diff}(S^3) \to \mathrm{Emb}(\cup^{k+1} B^3, S^3)$$
that sends each diffeomorphism of $S^3$ to its restriction to
$\cup^{k+1} B^3$. The following is a consequence of the Cerf--Palais
fibration theorem (see \cite[Appendice]{Cerf}, \cite{Palais}).

\begin{lemma}\label{lemma:CP}
The natural map $\varphi$ as above
is the projection of a locally trivial fiber bundle with fiber
$\mathrm{Diff}(S^3, \cup^{k+1} B^3)$.
\end{lemma}

Therefore, we have the homotopy exact sequence:

\begin{eqnarray}
& & \pi_2(\mathrm{Diff}(S^3), \id) \to
\pi_2(\mathrm{Emb}(\cup^{k+1} B^3, S^3), j_{k+1}) \to
\pi_1(\mathrm{Diff}(S^3, \cup^{k+1} B^3), \id)
\nonumber \\
& \to & \pi_1(\mathrm{Diff}(S^3), \id) \to
\pi_1(\mathrm{Emb}(\cup^{k+1} B^3, S^3), j_{k+1}) \to \cdots.
\label{eqn1}
\end{eqnarray}

Let $\mathbb{F}_{k+1}(S^3)$ be the
configuration space of $k+1$ points in $S^3$. This space can be
naturally identified with $\mathrm{Emb}(\{0, 1, \ldots, k\}, S^3)$.

\begin{lemma}
The space $\mathrm{Emb}(\cup^{k+1} B^3, S^3)$ is homotopy
equivalent to $\mathbb{F}_{k+1}(S^3) \times O(3)^{k+1}$.
\end{lemma}

\begin{proof}
For a given embedding $\eta : \cup^{k+1} B^3 \to S^3$, we associate
the element of $\mathbb{F}_{k+1} (S^3)$ which sends the $i$-th
point to the $\eta$-image of the center of the $i$-th $3$-ball.
Furthermore, by
associating the normalized differential of $\eta$ at each center, we
get an element of $O(3)^{k+1}$. (Note that the tangent bundle
$TS^3$ of $S^3$ is trivial, and we fix its trivialization here.)
Then, we can show that the map $\mathrm{Emb}(\cup^{k+1} B^3, S^3) \to
\mathbb{F}_{k+1}(S^3) \times O(3)^{k+1}$ thus obtained gives a homotopy
equivalence. (For example, see
\cite[Appendice, \S5, Proposition~3]{Cerf}.)

\end{proof}

Recall that $\mathrm{Diff}(S^3) \simeq O(4)$. Furthermore, $\mathbb{F}_{k+1}(S^3)$ is $1$-connected
(see \cite{FH}, \cite[Proof of Proposition~2.30]{Funar})
and $\pi_2(O(3)) = 0$. Thus, the
exact sequence (\ref{eqn1}) turns into
\begin{eqnarray*}
& 0 & \to \pi_2(\mathbb{F}_{k+1} (S^3)) \to
\pi_1(\mathrm{Diff}(S^3, \cup^{k+1} B^3), \id)
\to \pi_1(O(4), \id)
\\
& & \to \pi_1(O(3)^{k+1}, \id) \to \cdots.

\end{eqnarray*}

Note that $\pi_1(O(4), \id) \cong \mathbb{Z}_2$ and
$\pi_1(O(3)^{k+1}, \id) \cong (\mathbb{Z}_2)^{k+1}$. By choosing the
standard embedding $j_{k+1}$ so that it is equivariant with respect
to the natural $SO(2)$-actions, we see that the homomorphism
$\pi_1(O(4), \id) \to \pi_1(O(3)^{k+1}, \id)$ sends the generator $1 \in
\mathbb{Z}_2$ to $(1, 1, \ldots, 1) \in (\mathbb{Z}_2)^{k+1}$. In
particular, it is injective. Thus, we have that the boundary
homomorphism
\begin{equation}
\pi_2(\mathbb{F}_{k+1} (S^3)) \to
\pi_1(\mathrm{Diff}(S^3, \cup^{k+1}  B^3), \id)
\label{eqn3}
\end{equation}
is an isomorphism.

By \cite{FH} and \cite[Proof of Proposition~2.30]{Funar},

we have the following important result.

\begin{lemma}\label{lemma:iso}
The homotopy group $\pi_2(\mathbb{F}_{k+1}(S^3))$ is isomorphic to
$\mathbb{Z}^{k(k-1)/2}$.
\end{lemma}

Note that, for

a smooth fiber bundle $S^3_{(k+1)} \hookrightarrow E^5 \to S^2$
with structure group $\mathrm{Diff}(S^3_{(k+1)}, \partial S^3_{(k+1)})$,
its characteristic map

is an element of
$$\pi_1(\mathrm{Diff}(S^3_{(k+1)}, \partial S^3_{(k+1)}), \id),$$
which is isomorphic to
$\pi_1(\mathrm{Diff}(S^3, \cup^{k+1} B^3), \id) \cong
\mathbb{Z}^{k(k-1)/2}$
by Lemma~\ref{lemma1}, (\ref{eqn3}) and Lemma~\ref{lemma:iso}.

In fact, given such a smooth fiber bundle
$\pi : E^5 \to S^2$,
one can consider $S^2=D_1^2 \cup D_2^2$,
where $D_i^2$, $i=1,2$, denote the $2$-dimensional
closed disk.
Since each $D_i^2$ is contractible, the restriction
$\pi : \pi^{-1} (D_i^2) \to D_i^2$
is a trivial fiber bundle with fiber $S^3_{(k+1)}$,
and we have $\pi^{-1} (D_i^2) \cong S^3_{(k+1)} \times D_i^2$.
Hence, we can recover the total space
$$E^5 = (S^3_{(k+1)} \times D_1^2) \cup_h
(S^3_{(k+1)} \times D_2^2)$$

for some diffeomorphism $h: S^3_{(k+1)} \times \partial D_2^2 \to
S^3_{(k+1)} \times \partial D_1^2$ defined by
$h(x, t)= (\alpha(t)(x), t)$, where $\alpha : S^1 = \partial D^2
\to \mathrm{Diff}(S^3_{(k+1)}, \partial S^3_{(k+1)})$
corresponds to the characteristic map. Therefore, the structure of

the fiber bundle is
completely determined by the homotopy class
$[\alpha] \in \pi_1(\mathrm{Diff}(S^3_{(k+1)},
\partial S^3_{(k+1)}), \id)$.

\section{Characterization of NS-pairs}\label{section3}

For a non-negative integer $k$, let
\begin{equation}
S^3_{(k+1)} \hookrightarrow E^5 \stackrel{\pi}{\rightarrow} S^2
\label{eq:bundle}
\end{equation}
be a smooth fiber bundle
such that its restriction to the boundary
$\partial S^3_{(k+1)} \hookrightarrow \partial E^5 \rightarrow S^2$
is a trivial bundle.
In this section, we characterize such fiber
bundles that arise from an NS-pair $(S^5, K)$ with
$K \cong \cup^{k+1}S^2$.

We start by gluing the trivial bundle
$\cup^k (B^3 \times S^2) \to S^2$ along the $k$ boundary
components to get the $B^3$-fibration
\begin{equation}
\widetilde{\pi} : Y = E^5 \cup (\cup^k (B^3 \times S^2)) \rightarrow S^2.
\label{eq:tilde-psi}
\end{equation}
This fibration is trivial, since the structure
group $\mathrm{Diff}(B^3, \partial B^3)$
is contractible by Hatcher's solution to the Smale Conjecture
\cite{Hatcher}, where $\mathrm{Diff}(B^3, \partial B^3)$ denotes
the space of those diffeomorphisms of $B^3$ which fix
$\partial B^3$ pointwise.
Therefore, the total space $Y$ of the fibration (\ref{eq:tilde-psi})
is diffeomorphic to $B^3 \times S^2$.
Then, by gluing $B^3_0 \times S^2$ to $Y = B^3 \times S^2$ by the map
$\partial B^3_0 \times S^2 \to \partial B^3 \times S^2$ given by
$(x, y) \mapsto (y, x)$, we get the sphere $S^5$,
where $B^3_0$ is a copy of the closed $3$-dimensional ball.
Set $S^2_0 = x_0 \times S^2$, where $x_0$ is the
center of $B^3_0$, and we write $N(S^2_0) = B^3_0 \times S^2$,
which is identified with the closed tubular neighborhood
of $S^2_0$ in $S^5$.

To fix the notation we write $\cup^{k+1} B^3 = \cup^k_{i=0} B^3_i$,
and denote by $x_i$ the
center of $B^3_i$, where we consider
$\cup^k B^3 = \cup_{i=1}^k B^3_i$.
We also write $S^2_i = x_i \times S^2$, $i = 1, 2, \ldots, k$.
Note that the $2$-spheres $S^2_i$, $i = 0, 1, \ldots, k$, are all
embedded in $S^5$ in a standard way. Furthermore,
each of $S^2_i$, $i = 1, 2, \ldots, k$, has
linking number $\pm 1$ with $S^2_0$.
In the following discussions, we orient
$S^2_i$, $i = 0, 1, 2, \ldots, k$, in such a way that
the linking number of $S^2_i$ with $S^2_0$
is equal to $+1$, $i = 1, 2, \ldots, k$.

For $y \in S^2$,
we have $\cup^k_{i=1} (B^3_i \times y) \subset
\widetilde{\pi}^{-1}(y) \cong B^3$.
Therefore, to each $y \in S^2$ we can naturally associate an element of
the $k$-point configuration space
$\mathbb{F}_k (\Int B^3) \cong \mathbb{F}_k (\mathbb{R}^3)$.
This defines a \emph{classifying map}
$c : S^2 \to \mathbb{F}_k(\mathbb{R}^3)$.

Then, we have the following.

\begin{lemma}\label{lemma:cl}
The isomorphism classes of the
fibrations as in \textup{(\ref{eq:bundle})}
are in one-to-one correspondence with
$\pi_2(\mathbb{F}_k(\mathbb{R}^3)) \cong \Z^{k(k-1)/2}$.
The correspondence is given by associating
the homotopy class of the classifying map $c$.
\end{lemma}

Recall that according to \cite[Lemma~2.31]{Funar},
a fiber bundle (\ref{eq:bundle}) corresponds to
the element $(\mathrm{lk}(S^2_i, S^2_j))_{1 \leq i < j \leq k}
\in \Z^{k(k-1)/2}$ in the above correspondence,
where $\mathrm{lk}$ denotes the linking number in $S^5$, and we fix
orientations of $S^2_i$, $i = 1, 2, \ldots, k$, and $S^5$.

\begin{proof}[Proof of Lemma~\textup{\ref{lemma:cl}}]
As has been seen in Section~\ref{section2}, the isomorphism
classes of the bundles in question are in one-to-one
correspondence with $\pi_2(\mathbb{F}_{k+1} (S^3)) \cong
\pi_1(\mathrm{Diff}(S^3, \cup^{k+1}  B^3), \id)$.
On the other hand, it is known that
$\pi_2(\mathbb{F}_k(\mathbb{R}^3))$ is naturally
isomorphic to $\pi_2(\mathbb{F}_{k+1} (S^3))$
(see \cite[p.~38]{FH}).

Recall the locally trivial fiber bundle
$$\mathrm{Diff}(S^3, \cup^{k+1} B^3) \stackrel{\iota}{\longrightarrow}
\mathrm{Diff}(S^3) \stackrel{\varphi}{\longrightarrow}
\mathrm{Emb}(\cup^{k+1} B^3, S^3)$$
of Lemma~\ref{lemma:CP},
where $\iota$ is the natural inclusion map.
For the homotopy class $[\alpha] \in \pi_1(
\mathrm{Diff}(S^3_{(k+1)}, \partial S^3_{(k+1)}))
\cong \pi_1(\mathrm{Diff}(S^3, \cup^{k+1} B^3))$
of the characteristic map,
its $\iota_*$-image vanishes in
$\pi_1(\mathrm{Diff}(S^3))$, so that there exists
a continuous map $\tilde{\alpha} : D^2 \to \mathrm{Diff}(S^3)$
which extends $\iota \circ \alpha : S^1 \to  \mathrm{Diff}(S^3)$.
Then, the homotopy class of $\varphi \circ \tilde{\alpha} : D^2 \to
\mathrm{Emb}(\cup^{k+1} B^3, S^3)$ in
$\pi_2(\mathrm{Emb}(\cup^{k+1} B^3, S^3))
\cong \pi_2(\mathbb{F}_{k+1}(S^3)) \cong
\pi_2(\mathbb{F}_k(\R^3))$ is the class
corresponding to $[\alpha]$ by the
isomorphism (\ref{eqn3}). By construction,
this coincides with the homotopy class of the
classifying map $c$.

This completes the proof.
\end{proof}

Now, we have the following natural question.

\begin{problem}
Which elements of $\pi_2 (\mathbb{F}_k(\mathbb{R}^3))
\cong \mathbb{Z}^{k(k-1)/2}$ correspond to an NS-pair?
\end{problem}

We answer this question in our main result in this section, as follows.

\begin{theorem} \label{main-theorem}
The fiber bundle $S^3_{(k+1)}\hookrightarrow E^5 \stackrel{\pi}{\to} S^2$ as
in \textup{(\ref{eq:bundle})} arises from
an NS-pair if and only if
$\mathrm{det} \left(\mathrm{lk}(S_i^2,S_j^2)\right)_{1 \leq i,j \leq k}
= \pm 1$, where $\mathrm{lk}(S_i^2,S_i^2) = 0$ for all $1 \leq i \leq k$
by convention.
\end{theorem}

Note that $\left(\mathrm{lk}(S_i^2,S_j^2)\right)_{1 \leq i,j \leq k}$
is a $k \times k$ skew-symmetric integer matrix.

The rest of this section is devoted to the proof
of the above theorem.

For a fiber bundle (\ref{eq:bundle}), let $F$
be its fiber. We fix the trivialization of the boundary
fibration, and we write
$\partial E^5 = \cup^k_{i=0} (K_i \times S^2)$, where
$K_i \cong S^2$ are the boundary components of $F \cong
S^3_{(k+1)}$ and are
oriented in such a way that the cycle represented by $K_0$
is homologous to the sum of the cycles represented by $K_i$,
$i=1, 2, \ldots, k$. Let $X^5 =
E^5 \cup (\cup^k_{i=0} (K_i \times B^3))$
be the closed $5$-dimensional manifold obtained by
gluing $E^5$ and $\cup^k_{i=0} (K_i \times B^3)$ along their
boundaries
in such a way that the natural projection
$$\cup^k_{i=0} (K_i \times (B^3 \setminus \{0\}))
\to S^2$$
extends to a smooth fibration
$X^5 \setminus K \to S^2$, where
$K = \cup^k_{i=0} (K_i \times \{0\})$.
Note that in this notation, $K_i$ is identified with
$\partial B^3_i$, $i = 1, 2, \ldots, k$,
and $K_0$ is identified with $\ast \times S^2
\subset \partial B^3_0 \times S^2$.
We warn the reader that the way that $K_i \times B^3$
are attached to $E^5$ is very different from
that for the construction of $Y \subset S^5$ in (\ref{eq:tilde-psi}).

The theorem is a consequence of Lemmas~\ref{lem: 0} and \ref{lem: 1}
below.

\begin{lemma} \label{lem: 0}
The fiber bundle $\pi$ \textup{(\ref{eq:bundle})}
arises from an NS-pair if and only if
$X^5$ is homotopy equivalent to $S^5$.
\end{lemma}

The above lemma is a consequence of the well-known
fact that every homotopy $5$-sphere is standard \cite{Smale}.

Since we see easily that $X^5$ is simply connected,
it suffices to study the homology group $H_2(X^5)$.
Now consider the following piece of the Mayer-Vietoris
exact sequence:
$$H_2 (\cup^k_{i=0} (K_i \times \partial B^3))
\stackrel{\rho}{\to} H_2(E^5) \oplus H_2(\cup^k_{i=0} (K_i \times B^3))
\to H_2(X^5)\to 0,$$
where the homomorphism $\rho = (i_{1*}, -i_{2*})$ is induced by
the inclusions $i_{1}: \cup^k_{i=0} (K_i \times \partial B^3) \to E^5$
and $i_{2}: \cup^k_{i=0} (K_i \times \partial B^3) \to
\cup^k_{i=0} (K_i \times B^3)$.

Figure~\ref{fig: S0completa} helps us to understand the images
of the elements of $H_2(\cup^k_{i=0} (K_i \times \partial B^3))$
by the homomorphism $\rho$. Note that this depicts the
situation in $S^5$ and not in $X^5$.

\begin{figure}[t]
\psfrag{S0}{$S_0^2$}
\psfrag{S1}{$S_1^2$}
\psfrag{S2}{$S_2^2$}
\psfrag{Sk}{$S_k^2$}
\psfrag{K0}{$K_0$}
\psfrag{K1}{$K_1$}
\psfrag{K2}{$K_2$}
\psfrag{Kk}{$K_k$}
\psfrag{m0}{$\mu_0$}
\psfrag{m1}{$\mu_1$}
\psfrag{m2}{$\mu_2$}
\psfrag{mk}{$\mu_k$}
\includegraphics[width=0.95\linewidth,height=0.4\textheight,
keepaspectratio]{fig1-2.eps}
\caption{Situation in $S^5$}
\label{fig: S0completa}
\end{figure}

In order to describe the homomorphism $\rho$,
let us fix bases of the homology groups.
In the following, for a cycle $z$, we denote
by $[z]$ the homology class represented by $z$.
First, we have
$$H_2(\cup^k_{i=0} (K_i \times \partial B^3)) \cong
\oplus^k_{i=0} H_2(K_i \times \partial B^3)$$
and each $H_2(K_i \times \partial B^3) \cong \Z \oplus \Z$
is generated by $[K_i \times \ast]$ and
$[y_i \times \partial B^3]$, where $y_i \in K_i$,
$i = 0, 1, \ldots, k$. Furthermore, we have
$$H_2(\cup^k_{i=0} (K_i \times B^3)) \cong
\oplus^k_{i=0} H_2(K_i \times B^3)$$
and each $H_2(K_i \times B^3) \cong \Z$
is generated by $\delta_i  =
[K_i \times \ast]$, $\ast \in B^3$, $i = 0, 1, \ldots, k$.

On the other hand, we have
$$E^5 = S^5 \setminus (\cup^k_{i=0} \Int N(S^2_i)),$$
where $N(S^2_i) = B^3_i \times S^2$ is the closed tubular
neighborhood of $S^2_i$ in $S^5$,
and $S^2_i = x_i \times S^2$, $i = 1, 2, \ldots, k$, are
so-called ``Hopf duals'' to $S^2_0$.
Therefore, by Alexander duality we have
$$H_2(E^5) \cong H^2 (\cup^k_{i=0} \Int N(S^2_i)).$$
Since $N(S^2_i) = B^3_i \times S^2$, we can take the
generators $\mu_i = [\partial B^3_i \times \ast]
\in H_2(E^5)$, $\ast \in S^2$, and $H_2(E^5)$
is freely generated by $\mu_i$, $i = 0, 1, \ldots, k$.
Here, we orient $\mu_i$ in such a way that the
linking number of $\mu_i$ with $S^2_i$ is equal to $+1$.
Observe that $\mu_i = i_{1*}([K_i \times \ast])$ for
$i=1, 2, \ldots, k$, and

$\delta_i = i_{2*}([K_i \times \ast])$
for $i=0, 1, \ldots, k$. Therefore, the images of the generators
by the homomorphism $\rho$ can be written as follows.
\begin{eqnarray*}
\rho([K_0 \times \ast]) & = & \mu_1 + \mu_2 + \cdots +
\mu_k - \delta_0, \\
\rho([K_i \times \ast]) & = & \mu_i - \delta_i \quad (1\leq i \leq k), \\
\rho([y_0 \times \partial B^3]) & = & \mu_0 + 0, \\
\rho([y_i \times \partial B^3]) & = &
\sum_{0 \leq j \leq k, j \neq i} \mathrm{lk}(S^2_i, S^2_j)\mu_j + 0
\quad (1 \leq i \leq k).
\end{eqnarray*}

Therefore, with respect to the above bases,
the homomorphism $\rho$ is represented by the following matrix:
\begin{equation*}
R = \left(
\begin{array}{ccccccccl}
0 & 0 & \cdots & 0 &  & 1 & 1 & \cdots & 1 \\
1 & 1 & \cdots & 0 & & 0 & a_{11} & \cdots & a_{1k} \\
\vdots & \vdots & \ddots & \vdots &  & \vdots &
\vdots & \ddots & \vdots \\
1 & 0 & \cdots & 1 &  & 0 & a_{k1} & \cdots & a_{kk}\\
 &  &  &  &  & &  &  & \\
-1 & 0 & \cdots & 0 &  & 0 & 0 & \cdots & 0\\
0 & -1 & \cdots & 0 &  & 0 & 0 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots &  & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & -1 &  & 0 & 0 & \cdots & 0 \\

\end{array}
\right),

\end{equation*}
where $a_{ij} = \mathrm{lk}(S^2_j , S^2_i)$, $i \neq j$,
$1 \leq i,j\leq k$ and $a_{ii}=0$. Observe that $a_{ij}=-a_{ji}$.

\begin{lemma} \label{lem: 1}
The $5$-dimensional manifold $X^5$ is homotopy
equivalent to $S^5$ if and only if $\mathrm{det}\,R =\pm 1$.
\end{lemma}

\begin{proof}
If $X^5$ is homotopy equivalent to $S^5$, then
its second homology group must vanish and
therefore the homomorphism $\rho$ must
be an epimorphism, which implies that $\mathrm{det}\,R =\pm 1$.

Conversely, if $\rho$ is an isomorphism, by the
above Mayer-Vietoris exact sequence, we have
$H_2(X^5) = 0$. Then by Poincar\'e duality,
we see that $X^5$ has the homology of $S^5$.
Then, a standard argument in
algebraic topology shows that $X^5$
is homotopy equivalent to $S^5$.

This completes the proof of Lemma~\ref{lem: 1},
and hence Theorem~\ref{main-theorem} has been proved.
\end{proof}

Since a skew-symmetric integer matrix has determinant
$\pm 1$ only if its size is even, we have
the following.

\begin{corollary}
If the fiber bundle $S^3_{(k+1)}\hookrightarrow E^5 \to S^2$
as in \textup{(\ref{eq:bundle})} arises from an NS-pair,
then $k$ must be even.
\end{corollary}

\begin{remark} \label{lobs}
In \cite{Lo} Looijenga showed how to use the connected sum of NS-pairs
to construct new ones. In fact, he proved that given
an NS-pair $(S^n, K^{n-p-1})$ with fiber $F$,
there exists a polynomial map germ $f: (\mathbb{R}^{n+1},0)
\to (\mathbb{R}^{p+1},0)$
with an isolated singularity at the origin such that
the associated NS-pair is isomorphic to the connected sum
$$(S^n, K^{n-p-1}) \sharp ((-1)^{n-1}S^n, (-1)^{n-p}K^{n-p-1})$$
with fiber being diffeomorphic to the interior of
$\overline{F} \natural (-1)^{n-p}\overline{F}$, where
``$\natural$'' means the connected sum along the boundary.

For further details the reader is referred to \cite[p.~421]{Lo}.
\end{remark}

The following proposition follows from the remark above and the previous result.

\begin{proposition}\label{non-trivial}
For every even integer $k \geq 0$, there exists
an NS-pair $(S^5, L_{k+1})$ with $L_{k+1}$ being
diffeomorphic to the disjoint union of $k+1$ copies of $S^2$,
and there exists a polynomial map germ $f:(\mathbb{R}^6, 0) \to (\mathbb{R}^3,0)$
with an isolated critical point at $0$ such that the
associated NS-pair is isomorphic to
$(S^5, L_{k+1} \sharp (-L_{k+1}))$.
In particular, $L_{k+1} \sharp (-L_{k+1})$ consists of $2k+1$
connected components.
\end{proposition}

\begin{proof}
First note that for each positive even integer
$k$, there exists a skew-symmetric integer matrix
of determinant $\pm 1$. (For example,
consider the direct sum of the matrix
$$
\begin{pmatrix}
0 & 1 \\ -1 & 0
\end{pmatrix}
$$
and its copies.)
Then, by the above argument,
there exists an NS-pair $(S^5, L_{k+1})$ corresponding
to that matrix. Now, one can just apply Looijenga's
construction explained above in Remark~\ref{lobs}.
\end{proof}

\begin{corollary}
Given a real polynomial map germ as in
Proposition~\textup{\ref{non-trivial}} with $k > 0$,
the fiber of the associated Milnor fibration is not
diffeomorphic to a disk.
\end{corollary}

This answers to Milnor's non-triviality question,
Problem~\ref{problem},
for the dimension pair $(6, 3)$.

\section{A generalization to higher dimensions}\label{section4}

We can generalize the construction of Section~\ref{section3}
in higher dimensions as follows, in order to obtain
new non-trivial examples of real polynomial map
germs with an isolated singularity.

Let $n \geq 3$ be an integer.
For a non-negative integer $k$, let
$S^n_{(k+1)}$ denote the $n$-dimensional
sphere $S^n$ with the interior of the disjoint
union of $k+1$ copies of the $n$-dimensional disks removed.
In this section, we will construct a smooth fiber bundle
$$S^n_{(k+1)} \hookrightarrow E^{2n-1} \stackrel{\pi}{\to} S^{n-1}$$
such that the restriction to the boundary
$$\partial S^n_{(k+1)} \hookrightarrow \partial E^{2n-1}
\stackrel{\pi}{\to} S^{n-1}$$
is a trivial bundle and that it arises from
an NS-pair $(S^{2n-1}, K^{n-1})$.

Let $A = (a_{ij})$ be a $k \times k$ integer matrix
which is $(-1)^n$-symmetric such that the
diagonal entries all vanish.
Let $S_0 \cong S^{n-1}$ be a trivially embedded oriented
$(n-1)$-sphere in $S^{2n-1}$. Then, there exist
mutually disjoint
smoothly embedded oriented $(n-1)$-spheres
$S_i$ in $S^{2n-1}$, $i = 1, 2, \ldots, k$,
such that
\begin{itemize}
\item[$(1)$] $S_i$ do not intersect $S_0$,
\item[$(2)$] $S_i$ have linking number $+1$ with $S_0$,
\item[$(3)$] the linking number $\mathrm{lk}(S_i, S_j)
= a_{ij}$, $i \neq j$.
\end{itemize}
Such embeddings do exist (for example,
see \cite{H3}).
Note that then
$$E^{2n-1} = S^{2n-1} \setminus \cup_{i=0}^k
\Int N(S_i)$$
naturally fibers over $S^{n-1}$ in such a way that
the restriction to the boundary is trivial.
(More precisely, consider
the associated sub-fibration of the
trivial fiber bundle $S^{2n-1} \setminus \Int N(S_0)
\cong B^n \times S^{n-1} \to S^{n-1}$.)
Then, by the same construction as in Section~\ref{section3},
we obtain an object $(X^{2n-1}, K^{n-1})$, where
$X^{2n-1}$ is a $(2n-1)$-dimensional smooth closed manifold,
$X^{2n-1} \setminus \Int{N(K^{n-1})}$ is diffeomorphic to $E^{2n-1}$, and it
fibers over $S^{n-1}$ with fiber $S^n_{(k+1)}$ in such a way
that the projection map is compatible with a trivialization of
the closed tubular neighborhood $N(K^{n-1})$.
Then, we have the following.

\begin{lemma}
The manifold $X^{2n-1}$ is a homotopy $(2n-1)$-sphere if and only if
$\mathrm{det}\,A = \pm 1$.
\end{lemma}

\begin{proof}
We see easily that $E^{2n-1}$, and hence $X^{2n-1}$
is $(n-2)$-connected. Thus, $X^{2n-1}$ is a homotopy
$(2n-1)$-sphere if and only if $H_{n-1}(X^{2n-1})$ vanishes.
Then, an argument using a Mayer-Vietoris exact sequence
as in the previous section leads to the desired result.
\end{proof}

Combining this with the Looijenga construction (Remark~\ref{lobs}),
we have the following.

\begin{corollary}\label{cor:hdim}
Let $n \geq 3$ be an integer.
For every positive integer $k$ with $k \equiv 1 \pmod 4$,
there exists an NS-pair $(S^{2n-1}, L_k)$ with $L_k$ being
diffeomorphic to the disjoint union of $k$ copies of $S^{n-1}$,
and there exists a polynomial map germ $f:(\mathbb{R}^{2n}, 0) \to (\mathbb{R}^n,0)$
with an isolated singularity at $0$ such that the
associated NS-pair is isomorphic to
$(S^{2n-1}, L_k \sharp (-1)^{n}L_k)$.
In particular, $L_k \sharp (-1)^{n}L_k$ consists of $2k-1$
connected components.
\end{corollary}

Note that the associated Milnor fiber is
diffeomorphic to $S^n_{(2k-1)}$ and is homotopy
equivalent to the bouquet of $2k-2$ copies of the
$(n-1)$-sphere.

\begin{proof}[Proof of Corollary~\textup{\ref{cor:hdim}}]
Set $\ell = (k-1)/2$, which is a nonnegative even integer.
There exists an $\ell \times \ell$ $(-1)^n$-symmetric integer matrix $A$
with determinant $\pm 1$. By Haefliger \cite{H3},
there exists an embedding of $\cup_{i=0}^k S^{n-1}_i$ into
$S^{2n-1}$ such that each component is embedded trivially,
that $S^{n-1}_i$ links with $S_0$ once for all $i > 0$,
and that the linking matrix for $\cup_{i=1}^\ell S^{n-1}_i$
coincides with $A$, where each $S^{n-1}_i$ is a copy
of $S^{n-1}$ and the linking number $\mathrm{lk}(S^{n-1}_i,
S^{n-1}_i) = 0$, $i = 1, 2, \ldots, \ell$, by
convention. Then by the above construction,
we get the homotopy sphere $X^{2n-1}$ in which the disjoint
union of $\ell + 1$ copies of the $(n-1)$-spheres is embedded.
Then, the connected sum $X^{2n-1} \sharp (-X^{2n-1})$ is
diffeomorphic to $S^{2n-1}$ by \cite{KM}, since $n \geq 3$.
Therefore, by the connected sum construction,
we get an NS-pair $(S^{2n-1}, L_k)$, where $L_k$
is diffeomorphic to the disjoint union of $2\ell+1 = k$
copies of $S^{n-1}$.

Then, applying Looijenga's construction,
we get the desired conclusion.
\end{proof}

\section{Bouquet theorems for real isolated singularities}

It is known that for a holomorphic function germ
$f: (\mathbb{C}^{n+1}, 0) \to (\mathbb{C}, 0)$
with an isolated singularity at the origin, the Milnor fiber
$F_{f}$ has the homotopy type of a bouquet (or a wedge) of $n$-dimensional spheres.
For real polynomial map germs with an isolated singularity,
we cannot expect, in general, such a bouquet theorem, which
can be seen as follows.

By Zeeman's twist spinning construction \cite{Z},
one can construct an NS-pair $(S^4, K^2)$ such that
the fundamental group of the fiber
is not a free group. Then Looijenga's construction
leads to a non-trivial polynomial map germ
$(\R^5, 0) \to (\R^2, 0)$ with an isolated singularity
at the origin such that the Milnor fiber does not
have a free fundamental group. Consequently, the Milnor
fiber is not homotopy equivalent to a bouquet of spheres.

\begin{remark}\label{spinning}
In the following,
in order to get examples in higher dimensions,
we use the spinning construction due to Artin
\cite{Artin}. For completeness, let us recall the
construction.
Let $(S^m, K^k)$ be an NS-pair
with $K^k \neq \emptyset$ and $\pi : S^m
\setminus K^k \to S^{m-k-1}$ the associated
fibration. We denote the fiber of the fibration
$S^m \setminus \Int N(K^k) \to S^{m-k-1}$ by $F^{k+1}$,
where $N(K^k)$ is the closed tubular neighborhood of $K^k$
in $S^m$.
We take a point $q \in K^k$ and a small
$m$-disk neighborhood $D$ in $S^m$ such that
$(D, D \cap K^k)$ is diffeomorphic to the
standard disk pair $(D^m, D^k)$ and that
$\pi$ restricted to $D \setminus (D \cap K^k)$
is equivalent to the standard fibration $D^m \setminus
D^k \to S^{m-k-1}$.
Then, we consider the quotient space
of $(S^m \setminus \Int D, K^k \setminus (\Int D \cap K^k))
\times S^1$, where for each $x \in \partial D$,
the points of the form $(x, t)$ are identified
to a point for all $t$. This kind of a construction
is called the \emph{spinning}.
The resulting pair gives $(S^{m+1}, \tilde{K}^{k+1})$,
where $\tilde{K}^{k+1}$ is a smoothly embedded submanifold
of $S^{m+1}$ of dimension $k+1$. By construction,
there exists a fibration $\tilde{\pi} : S^{m+1}
\setminus \tilde{K}^{k+1} \to S^{m-k-1}$
which restricts to $\pi$ on $(S^m \setminus (\Int D \cup K^k))
\times \{t\}$
for each $t \in S^1$. It is straightforward to see
that $(S^{m+1}, \tilde{K}^{k+1})$ is an NS-pair.
We call it the \emph{spun}
of the NS-pair $(S^m, K^k)$.
Note that the fiber $\tilde{F}^{k+2}$
of the fibration $S^{m+1} \setminus
\Int N(\tilde{K}^{k+1}) \to S^{m-k-1}$
is diffeomorphic to the $(k+2)$-dimensional
manifold obtained from $F^{k+1} \times S^1$
by identifying, for each $x \in \Delta^k$, the points
of the form $(x, t)$ to a point for all $t$, where
$\Delta^k$ is a $k$-dimensional
disk embedded in $\partial F^{k+1}$ (near $q$).
Note that the fundamental groups of $S^m \setminus K^k$
and $S^{m+1} \setminus \tilde{K}^{k+1}$ are isomorphic,
and that $F^{k+1}$ and $\tilde{F}^{k+2}$ also
have isomorphic fundamental groups.
\end{remark}

Let $(S^4, K^2)$ be an NS-pair such that the fiber
has non-free fundamental group.
Then, applying once the spinning construction explained
above to $(S^4, K^2)$, one gets a non-trivial example in dimension
$(6,2)$ such that the Milnor fiber is not homotopy equivalent to
a bouquet of spheres.
Performing such procedures inductively one can construct
examples in all pairs of dimensions $(n, 2)$, $n \geq 5$,
such that the Milnor fiber is not homotopy
equivalent to a bouquet of spheres.

In this section we give sufficient conditions to
guarantee that the real Milnor fiber is
homotopy equivalent to a bouquet of spheres of
the same dimension, or of different dimensions.

Throughout this section we consider $f: (\mathbb{R}^{n}, 0)
\to (\mathbb{R}^{p}, 0)$, $n > p \geq 2$,
a polynomial map germ with an isolated singularity
at the origin and the Milnor fibration (the ``Milnor tube"),
$$f: f^{-1} (S_{\delta}^{p-1}) \cap D_{\varepsilon}^{n}
\to S_{\delta}^{p-1},$$
where $0 < \delta \ll \varepsilon \ll 1$.
We denote by $F_{f}$ its fiber and by
$\beta_{j}=\rk H_{j}(F_{f})$ its $j$-th Betti number.

Consider $\pi :(\mathbb{R}^{p},0) \to (\mathbb{R}^{p-1},0)$,
$p \geq 3$, the germ of the canonical projection. Clearly,
the composition map germ $G=\pi \circ f :(\mathbb{R}^{n},0)
\to (\mathbb{R}^{p-1},0)$ also has an isolated singularity
at the origin and thus we have two fibrations:
$$f: f^{-1} (S_{\delta}^{p-1}) \cap D_{\varepsilon}^{n}
\to S_{\delta}^{p-1},$$
and
$$G: G^{-1} (S_{\delta}^{p-2}) \cap D_{\varepsilon}^{n}
\to S_{\delta}^{p-2}.$$

In \cite{DA} it was shown the relationship between the
Milnor fibers $F_{f}$ and $F_{G}$.

It is worth pointing out that the results in
\cite{DA} hold in a more general setting which includes
the case of non-isolated singularities.
Nevertheless, in the special case of an isolated singularity,
it provides a positive answer to a conjecture stated by
Milnor in \cite[p.~100]{Mi} as follows:

\begin{theorem}[\textup{\cite{DA}}] \label{proj}
Let $f:(\mathbb{R}^{n},0)\to (\mathbb{R}^{p},0)$, $n > p \geq 2$,
be a polynomial map germ with an isolated singularity at the
origin and set $G=\pi \circ f :(\mathbb{R}^{n},0) \to
(\mathbb{R}^{p-1},0)$. Then, the Milnor fiber $F_{G}$ of $G$
is homeomorphic to $F_{f}\times [0,1]$, where for $p=2$ the Milnor fiber
of $G$ is, by definition, the intersection of a sufficiently
small closed ball centered at the origin and the
inverse image of a regular value
sufficiently close to the origin. In particular,
the Milnor fibers $F_f$ and $F_G$ have the same homotopy type.
\end{theorem}

In \cite[Chapter~11]{Mi}, Milnor provided information concerning
the topology of the fiber $F_{f}$. It was proved in
Lemma~11.4 that if $n < 2(p-1)$, then the Milnor fiber
is necessarily contractible. It also follows from the
first paragraph of the proof that for $n > p \geq 2$ in general,
if the link is not empty, then the fiber $F_{f}$ is
$(p-2)$-connected, i.e., $\pi_{i}(F_{f})=0$, $i=0,1, \ldots, p-2$.

In \cite{ADD} the authors proved formulae relating
the Euler characteristic of the Milnor fiber and the
topological degree of the gradient mapping of the coordinate
functions, which extends Milnor's formula for
complex function germs with an isolated singularity
(see \cite[p.~64]{Mi}) and Khimshiashvili's formula \cite{Khim}
for isolated singularity real analytic function germs, as follows.

\begin{theorem}[\cite{ADD}]\label{euler}
Let $f:(\mathbb{R}^{n},0) \to (\mathbb{R}^{p},0)$, $n > p \geq 2$,
be a polynomial map germ with an isolated singularity at the origin,
and consider
$$f(x)=(f_{1}(x), f_2(x), \ldots, f_{p}(x)),$$
an arbitrary representative of the germ. Denote by
$\mathrm{deg}_{0}(\nabla f_{i}(x))$, for $i=1, 2, \ldots, p$,
the topological degree of the map
$\displaystyle{\varepsilon\frac{\nabla f_{i}}{\|
\nabla f_{i} \|}:S^{n-1}_{\varepsilon} \to S^{n-1}_{\varepsilon}}$,
for $\varepsilon > 0$ small enough.
\begin{itemize}
\item[(i)] If $n$ is even, then
$\chi (F_{f}) = 1-\mathrm{deg}_0 \nabla f_1$.
Moreover, we have
$$\mathrm{deg}_0 \nabla f_{1}= \mathrm{deg}_0 \nabla f_{2}
= \cdots= \mathrm{deg}_0 \nabla f_{p}.$$
\item[(ii)] If $n$ is odd, then
$\chi (F_{f}) = 1$. Moreover, we have
$\mathrm{deg}_0 \nabla f_{i}=0$ for $i=1, 2, \ldots, p$.
\end{itemize}
\end{theorem}

In particular, from item (ii) above it follows that
if the source space is odd-dimensional, then the
fiber can never be homotopy equivalent to
a bouquet of a positive number
of spheres of the same dimension.

In the following subsections, we consider the dimension
pairs $(2n, n)$ and $(2n+1, n)$, and study
conditions for a Milnor fiber to have the homotopy type
of a bouquet of spheres. We also
study the dimension pairs $(2n, p)$ and $(2n+1, p)$
with $2 \leq p \leq n$ using the composition with a
projection.

\subsection{The case of $(\mathbb{R}^{2n}, 0) \to (\mathbb{R}^n, 0)$}
\label{even}

Consider $f: (\mathbb{R}^{2n}, 0) \to (\mathbb{R}^n, 0)$,
$n\geq 2$, a polynomial map germ with an isolated
singularity at the origin. Note that $S^{2n-1}$
does not smoothly fiber over $S^{n-1}$. Hence, in this case
$F_{f}$ is an $n$-dimensional compact orientable
manifold with non-empty boundary and $\pi_{i}(F_{f})=0$
for $i=0,1, \ldots, n-2$.

Since $\partial F_{f} \neq \emptyset$, we have
$H_n(F_{f}) = 0$.
Moreover, since $F_{f}$ is
orientable, the homology $H_{n-1}(F_{f})$ is torsion free.

Then, by Theorem~\ref{euler}, item (i),
we have
$\beta_{n-1}=(-1)^{n}\mathrm{deg}_{0}(\nabla f_{1})$.

Furthermore, in the special case $n=2$, the fibers are compact
connected surfaces with non-empty boundary, so that
they have the homotopy type of a bouquet of $1$-dimensional
spheres (circles). Furthermore, for $n=3$, we have
seen in Section~\ref{section2}
that the fibers are diffeomorphic to
$S^3_{(k+1)}$ for some non-negative integer $k$,
and hence they are homotopy equivalent to
a bouquet of $2$-spheres.
Therefore, we
may assume that $n \geq 4$.
Note that if $\mathrm{deg}_{0}(\nabla f_{1}) = 0$,
then the Milnor fiber is contractible.

It follows from the Hurewicz theorem that the
Hurewicz homomorphism
$$\rho_{n-1} : \pi_{n-1} (F_{f}) \to  H_{n-1} (F_{f})
\cong \mathbb{Z}^{\beta_{n-1}}$$
is an isomorphism. Then, for each generator
$\gamma_i \in H_{n-1}(F_{f}) \cong \mathbb{Z}^{\beta_{n-1}}$
there exists a continuous map $\varphi_i : S^{n-1} \to F_f$,
$i = 1, 2, \ldots, \beta_{n-1}$, such that
$\gamma_i = \rho_{n-1} ([\varphi_i])= (\varphi_i)_* ([S^{n-1}])$,
where $[S^{n-1}] \in H_{n-1} (S^{n-1})\cong \mathbb{Z}$ is
the fundamental class (given by
the natural orientation of $S^{n-1}$).
Therefore, we have the continuous map
$$\varphi:\displaystyle{\bigvee^{\beta_{n-1}} S^{n-1} \to F_{f}}$$
obtained by the wedge of the maps $\varphi_i: S^{n-1} \to F_{f}$, for
$i= 1, 2, \ldots, \beta_{n-1}$, which is a homotopy
equivalence by the Whitehead theorem.

Thus we have proved the following:

\begin{proposition} \label{wedge-even}
Let $f : (\R^{2n}, 0) \to (\R^n, 0)$ be a polynomial
map germ with an isolated singularity at the origin, $n \geq 2$.
Given $f(x)=(f_{1}(x), f_2(x), \ldots, f_{n}(x))$, a
representative of the germ $f$, we have the following.
\begin{itemize}
\item[(i)] $\beta_{n-1} = (-1)^n \mathrm{deg}_{0}(\nabla f_{1})$.
\item[(ii)] The Milnor fiber $F_{f}$ has the homotopy type of
a bouquet of $(n-1)$-dimensional spheres
$$\displaystyle {\bigvee^{\beta_{n-1}} S^{n-1}},$$
where it means a point when $\beta_{n-1}=0$.
\end{itemize}
\end{proposition}

For $n \geq 4$, it follows from Theorem~\ref{T2}, item (c),
that in all pairs of dimensions $(2n,n)$ there exist non-trivial
examples.

However, these non-trivial examples due to
Church--Lamotke \cite{CL} have contractible Milnor fibers
(with non-simply connected links).
On the other hand, according to our
construction in Section~\ref{section4} together with
Theorem~\ref{proj},
we get the following.

\begin{corollary}\label{bouquet}

For each pair of dimensions $(2n, p)$, $2 \leq p \leq n$,
there exists a real isolated singularity polynomial map germ
$(\mathbb{R}^{2n}, 0)\to (\mathbb{R}^p, 0)$ such that the
Milnor fiber is, up to homotopy, a bouquet of $(n-1)$-dimensional
spheres with the number of
spheres equal to $|\mathrm{deg}_{0}(\nabla f_{1})| > 0$,
where $f(x)=(f_{1}(x), f_2(x), \ldots, f_p(x))$.
\end{corollary}

\subsection{The case of $(\mathbb{R}^{2n+1}, 0) \to (\mathbb{R}^n, 0)$}
\label{odd}

Consider now $f: (\mathbb{R}^{2n+1}, 0) \to (\mathbb{R}^n, 0)$,
$n\geq 3$, a polynomial map germ with an isolated
singularity at the origin. In this case, the Milnor fiber $F_{f}$
is an $(n+1)$-dimensional compact orientable manifold with
non-empty boundary and is
$(n-2)$-connected. Then, $H_{n+1}(F_{f})=0$,
$H_{n}(F_{f})$ is torsion free and, by Theorem~\ref{euler},
$\beta_{n}=\beta_{n-1}$.
Suppose that $H_{n-1}(F_{f})$ is torsion free. Then, we have $H_{n-1}
(F_f)\cong \mathbb{Z}^{\beta_{n-1}} \cong H_n (F_f)$. By Hurewicz
theorem, the Hurewicz homomorphisms
$$\rho_{n-1} : \pi_{n-1} (F_{f}) \to H_{n-1} (F_{f}) \cong \mathbb{Z}^{\beta_{n-1}}$$
and
$$\rho_n : \pi_n (F_{f}) \to H_n (F_{f}) \cong \mathbb{Z}^{\beta_{n-1}}$$
are surjective. Then, by an argument similar to that
used in the case $(2n, n)$, we can construct a homotopy equivalence
$$\varphi : \left(\displaystyle{\bigvee^{\beta_{n-1}}S^{n-1}}\right) \vee
\left(\displaystyle{\bigvee^{\beta_{n}}S^{n}}\right)
\to F_f.$$
Thus, we have proved the following result.

\begin{proposition}\label{wedge-odd}
Let $f:(\mathbb{R}^{2n+1},0)\to (\mathbb{R}^{n},0)$, $n\geq 3$,
be a real isolated singularity polynomial map germ.
Then, the $(n-1)$-th homology $H_{n-1}(F_{f})$ of the
Milnor fiber is torsion free if and only if $F_f$
has the homotopy type of
a bouquet of spheres of the form
$$\left(\displaystyle{\bigvee^{\beta_{n-1}}S^{n-1}}\right)
\vee
\left(\displaystyle{\bigvee^{\beta_{n-1}}S^{n}}\right)
= \bigvee^{\beta_{n-1}} (S^{n-1} \vee S^n),$$
where it means a point when $\beta_{n-1}=0$.

\end{proposition}

According to our
construction in Section~\ref{section4} together with
Theorem~\ref{proj} again, we get the following.

\begin{corollary}\label{bouquet2}

For each pair of dimensions $(2n+1, p)$, $2 \leq p \leq n$,
there exists a real isolated singularity polynomial map germ
$(\mathbb{R}^{2n+1}, 0)\to (\mathbb{R}^p, 0)$ such that the
Milnor fiber is, up to homotopy, a bouquet of $\ell$ copies
of the $n$-dimensional
sphere and $\ell$ copies of the $(n-1)$-dimensional sphere
with $\ell > 0$.
\end{corollary}

\begin{proof}
For $n \geq 3$, this is a consequence of
Proposition~\ref{wedge-odd}.
For $n = 2$, we start with a non-trivial
fibered knot $(S^3, K)$. Then, its spun $(S^4, \tilde{K})$
is a non-trivial fibered $2$-knot,
and its fiber is obtained by spinning a positive
genus surface with boundary. Therefore, the fiber
of $(S^4, \tilde{K})$ has the homotopy
type of a bouquet of a positive number of circles
and $2$-spheres. This completes the proof.
\end{proof}

\subsection{Application to $k$-stairs maps}
Given a polynomial map germ $f:(\mathbb{R}^n, 0)\to (\mathbb{R}^q, 0)$,
$n \geq q \geq 1$,
with an isolated singularity at the origin, we say that a map
germ $F: (\mathbb{R}^{n}, 0) \to (\mathbb{R}^{p}, 0)$,
$1 \leq q \leq p$, is a
\emph{$(p-q)$-stairs map for $f$} if there exist germs of
polynomial functions $g_{j}: (\mathbb{R}^{n}, 0) \to (\mathbb{R}, 0)$,
$q+1 \leq j  \leq p$, such that $F(x)=(f(x), g_{q+1}(x), g_{q+2}(x),
\ldots, g_{p}(x))$ has an isolated singularity at the origin.
If $p=q$, then by definition, we have $F(x)=f(x)$ and $f$
is its own $0$-stairs map.

\begin{corollary}
Let $f:(\mathbb{R}^{n},0)\to (\mathbb{R}^{p},0)$,
$n/2 \geq p \geq 2$,
be a polynomial map germ with an isolated singularity
at the origin. Then we have the following.
\begin{itemize}
\item [(i)] If $n$ is even and $f$ admits a
$(n/2-p)$-stairs map, then the Milnor fiber
is homotopy equivalent to a bouquet of $(n/2-1)$-dimensional spheres.
\item [(ii)] Suppose $n$ is odd and $H_{k}(F_{f})$ is torsion free
for $k=(n-1)/2-1$, where $F_f$ denotes the Milnor fiber.
If $f$ admits a $((n-1)/2-p)$-stairs map,
then the Milnor fiber is homotopy equivalent to a bouquet of
$k$- and $(k+1)$-dimensional spheres, where the numbers of
spheres are the same.
\end{itemize}
\end{corollary}

\begin{proof}
Just apply Propositions~\ref{wedge-even}, \ref{wedge-odd}, and
Theorem~\ref{proj}.
\end{proof}

We do not know whether or not the bouquet structure in
the fiber characterizes the existence of such $k$-stairs maps
for $k \geq 1$.

\section*{Acknowledgements}

The authors would like to express their sincere gratitude to
Edivaldo Lopes dos Santos and Denise de Mattos
for stimulating discussions and invaluable comments.

The first author would like to thank very much the Fapesp grant 2013/23443-5
and CNPq/PQ-2 309819/2012-1.
The second author would like to thank very much all Fapesp support given during
the development of the PhD project, grant number 2012/12972-4.
The third author has been supported in part by JSPS KAKENHI Grant Number
23244008, 23654028.
The fourth author would like to thank CNPq/PDJ grant 502638/2012-5 during
the post-doc in ICMC/USP.


TITLE: Friction up a slope?
QUESTION [1 upvotes]: I'm confused by this question:

...
After resolving forces, I got $T=20g\sin{60} + 0.4\cdot 20g\cos{60} = 209N $
However, this is the answer for part B. After looking through solutions, the minimum tension required is when the frictional force is acting up the slope, lowering the amount of force needed to keep it up.
But, when would the frictional force be acting up the slope? This makes no sense to me. If the force of tension is upwards, surely the friction will always act oppositionally, downwards.
If anything, wouldn't the minimum effect of friction be 0, when the tension is equal to the component of mass parallel to the slope?
Making $T_{\text{min}} = 20g\sin{60}$.

$ F_{\text{friction}} \le \mu F_{\text{normal-reaction}}$

REPLY [0 votes]: Friction opposes relative motion.
The box trying to move up or down the slope will be opposing friction. the direction of friction will change depending on what is happening.


TITLE: How to apply Borel-Cantelli Lemma to find if something converges almost surely to $0$.
QUESTION [6 upvotes]: Suppose that $Z_1,Z_2,...$ are random variables with $Z_n\sim\mathrm{Exp}(1)$.  (We do not assume that these random variables are independent.)  Show that $Z_n/\big(\ln^2(n)\big)$ converges to $0$ almost surely.
To be honest I am not really sure where to start with this problem , I believe I should start with the lim(sup) of the given function summed up. But after that I have no idea.
Help would be appreciated, thanks.

REPLY [4 votes]: It suffices to show that $P\Big(\bigcap_{m\geq 1} \bigcup_{N\geq 1}\bigcap_{n\geq 1} \big[ Z_n/\big(\ln^2(n)\big)\leq \frac 1m\big]\Big) = 1$.
In turn, it suffices to show that $\forall m\in \mathbb N\setminus\{1\}$, $P\Big( \bigcup_{N\geq 1}\bigcap_{n\geq 1} \big[ Z_n/\big(\ln^2(n)\big)\leq \frac 1m\big]\Big) = 1$, which is the same as
$P\Big(\liminf_n [Z_n/\big(\ln^2(n)\big)\leq \frac 1m]\Big)=1$, or $P\Big(\limsup_n [Z_n/\big(\ln^2(n)\big)>\frac 1m]\Big)=0$.
The last equality follows from Borel-Cantelli. I leave the details to you, but it suffices to show that for fixed $m$, the series $\sum_{n\geq 1} \exp(-\frac{\ln^2(n)}{m})$ is convergent. Note that for $n$ large enough, $\frac{\ln(n)}m \geq 2$ and compare with $\sum_{n\geq 1} \frac 1{n^2}$.


text:
\section{Reduction from Proper to Improper Learning}\label{sec:proper-to-improper}
In this section, we develop blackbox reductions from proper to improper learning under Massart noise. This means that we show how,
given query access to any classifier $h$ which achieves 0-1 errer upper bounded by $L$, how to generate a \emph{proper} classifier $\vx \mapsto \sgn(\langle \vw, \vx \rangle)$ which achieves equally good 0-1 loss in polynomial time and using a polynomial number of samples from the model. The significance of this is that it allows us to use a much more powerful class to initially learn the classifier, which may be easier to make robust to the Massart noise corruption, and still extract at the end a simple and interpretable halfspace predictor. Schemes along these lines, which use a more powerful learner to train a weaker one, are referred to as knowledge distillation or model compression procedures (see e.g. \cite{hinton2015distilling}), and sometimes involve the use of ``soft targets'' (i.e. mildly non-blackbox access to the teacher). Our results show that the in the Massart setting, there is in fact a simple way to perform blackbox knowledge distillation which is extremely effective.

The key observation we make is that such a reduction is actually \emph{implied} by the existence of an $\eta + \epsilon$ Massart proper learner based upon implementation of a gradient or separating hyperplane oracle, as developed in Section~\ref{sec:proper-learning}. The reason is that if we view an improper hypothesis as a ``teacher'' and a halfspace $\sgn(\langle \vw, \rvx \rangle)$ parameterized by $\vw$ in the unit ball as a ``student'', as long as the student is inferior to the teacher we can make progress by focusing on the region where the teacher and student disagree. On this region, the student has accuracy lower than $50\%$, so by appealing to the previously developed separation oracle for the proper learner, we can move $w$ towards $w^*$ (in the case of the gradient-based learner) or otherwise zoom in on $w^*$ (for the cutting-plane based learner).

In other words, in context of the $\vw$ versus $S$ game introduced in the previous section, blackbox access to an improper learner massively increases the strength of the $S$ player. We make the preceeding arguments formal in the remainder of this section.

\subsection{Boosting Separation Oracles using a Good Teacher}
Given blackbox access to a ``teacher'' hypothesis $h$ with good 01 loss, in order to produce an equally good proper learner we simply run one of our proper learning algorithms replacing the original separation oracle $\mathcal{O}(\vw)$ with the following ``boosted'' separation oracle $\textsc{BoostSeparationOracle}(h,\vw,\mathcal{O})$. Remarkably, even if $\eta = 0.49$ and the original separation oracle can only separate $\vw^*$ from $\vw$ with $01$ loss greater than $49\%$, if we have black-box query access to hypothesis $h$ achieving error rate $1\%$, the resulting boosted oracle can suddenly distinguish between $\vw^*$ and all $\vw$ achieving error greater than $1\%$.
\begin{algorithm}\caption{\textsc{BoostSeparationOracle}$(h,\vw,\mathcal{O})$}
\DontPrintSemicolon
Let $\mathcal{R} = \{\vx : h(\vx) \ne \sgn(\langle \vw, \rvx\rangle)\}$.\;
Define $\calD_{\calR}$ to be $\calD$ conditional on $\rvx \in \calR$. Note that given sampling access to $\calD$, distribution $\calD_{\calR}$ is efficiently sampleable using rejection sampling.\;
Return the output of $\mathcal{O}(\vw)$ on distribution $\calD_{\calR}$.
\end{algorithm}
\begin{lemma}
Suppose that the joint distribution $(\rvx, Y)$ follows the $\eta$-Massart halfspace model, $h : \mathbb{R}^d \to \{\pm 1\}$ and $\vw \in \mathbb{R}^d$ a vector such that
\[ \err(h) - \err(\vw) \ge \epsilon \]
for $\epsilon > 0$. Suppose $\calO$ is an oracle which with probability $1 - \delta/2$, access to $m$ samples from any Massart distribution $\mathcal{D}'$, and input $\vw'$ such that $\Pr[\mathcal{D}']{\sgn(\iprod{\vw', \rvx}) = Y} < 1/2$ outputs a separating hyperplane $\vg'$ with $\|\vg'\| \le 1$ such that $\langle \vw^* - \vw', -\vg' \rangle \ge \gamma$ for $\gamma \ge 0$. Then $\textsc{BoostSeparationOracle}(h,\vw,\calO)$ returns with probability at least $1 - \delta$ a vector $\vg$ with $\|\vg\| \le 1$ such that
\[ \langle \vw^* - \vw, -\vg \rangle \ge \gamma \]
and has sample complexity $O\left(\frac{m + \log(2/\delta)}{\epsilon}\right)$.
\end{lemma}
\begin{proof}
Observe that
\[ \epsilon \le \err(h) - \err(\vw) \le \Pr{\rvx \in \mathcal{R}} \]
since $h$ and $\sgn(\iprod{\vw,\rvx})$ agree perfectly outside of $\mathcal{R}$. Furthermore, restricted to $\mathcal{R}$ it must be the case that $\Pr{\sgn(\iprod{\vw,\rvx}) = Y \mid \rvx \in \calR} < 1/2$, otherwise it would be the case that $\err(h) > \err(w)$ which contradicts the assumption. Therefore, as long as sampling from $\mathcal{D}$ produces at least $m$ samples landing in $\mathcal{R}$ the result follows from the guarantee for $\calO$; by Bernstein's inequality (Theorem~\ref{thm:bernstein}), $O(\frac{m + \log(2/\delta)}{\epsilon})$ samples will suffice with probability at least $1 - \delta/2$, so taking the union bound gives the result.
\end{proof}
\subsection{Proper to Improper Reductions}
By combining \textsc{BoostSeparationOracle} with any of our proper learners, we get an algorithm for converting any improper learner into a proper one. We state the formal results here: in all of these results we assume that $h$ is a hypothesis which we are given oracle access to.
\begin{algorithm}\caption{\textsc{FilterTronDistiller}$(\epsilon,\eta,\delta,\lambda,T)$}
\DontPrintSemicolon
Let $\vw_1$ be an arbitrary vector in the unit ball.\;
Build a empirical distribution $\hat{\calH}$ from $m=\Omega(\log(T/\delta)/\epsilon^2)$ samples (to use as a test set).\;
\For{$t = 1$ to $T$}{
    \If{$\hat{\err}(\vw_t) < \eta + \epsilon/2$}{Return $\vw_t$.\;}
    \Else{
    Let $\beta_t = \frac{1}{\sqrt{t}}$.\;
    Let $\vg_t = \textsc{BoostSeparationOracle}(h,\vw_t,\textsc{FindDescentDirection}(\cdot,\epsilon/6,\delta/2T,\lambda))$.\;
    Let $\vw_{t + 1} = \frac{\vw_t - \eta_t \vg_t}{\max(1,\|\vw_t - \eta_t \vg_t\|)}$.\;
    }
}
\end{algorithm}
\begin{theorem}\label{thm:filtertron-distiller}
Suppose that $(\rvx, Y)$ is an $\eta$-Massart halfspace model with margin $\gamma$ and $\eta < 1/2$. Suppose that $\epsilon > 0$ such that $\eta + \epsilon/6 \le 1/2$.
With probability at least $1 - \delta$, Algorithm~\textsc{FilterTron} combined with \textsc{BoostSeparationOracle} and oracle hypothesis $h$, returns $\vw$ such that
\[ \err(\vw) \le \err(h) + \epsilon \]
when $T = O\left(\frac{1}{\gamma^2 \epsilon^2}\right)$ and $\lambda = \eta + \epsilon/6$. The algorithm runs in total sample complexity $n = O\left(\frac{\log(2/\delta\gamma\epsilon)}{\gamma^4 \epsilon^6}\right)$ and runtime $O(nd + n\log n)$.
\end{theorem}
Using the cutting plane variant, as in Theorem~\ref{thm:filtertron-vaidya}, gives a matching result with improved sample complexity $O\left(\frac{d\log(2/\delta\gamma)}{\gamma^2\epsilon^4}\right)$. Finally, we state the result for halfspaces without a margin assumption.
\begin{theorem}\label{thm:distill-general}
Suppose that $(\rvx, Y)$ is an $\eta$-Massart halfspace model and $\rvx$ is supported on vectors of bit-complexity at most $b$. There exists an algorithm (the ellipsoid method combined with \textsc{BoostSeparationOracle} applied to \textsc{GeneralHalfspaceOracle}) which runs in time $poly(d,b,1/\epsilon,1/\delta)$ and sample complexity $\tilde{O}\left(\frac{\poly(d,b)}{\epsilon^4}\right)$ which outputs $\vw$ such that
\[ \err(\vw) \le \err(h) + \epsilon \]
with probability at least $1 - \delta$.
\end{theorem}
\subsection{Applications of the Proper-to-Improper Reduction}
An interesting feature of the proper to improper reduction for Massart noise is that it allows us to use our toolbox of very general learners (e.g. boosting, kernel regression, neural networks) to fit an initial classifier, while nevertheless allowing us to output a simple halfspace in the end. In this section, we describe a number of interesting applications of this idea.

To start, we give the first proper learner achieving error better than $\eta + \epsilon$ for Massart halfspaces over the hypercube, i.e. $\rvx \sim Uni(\{\pm 1\}^d)$; in fact, we show that error $\textsf{OPT} + \epsilon$ is achievable for any fixed $\epsilon > 0$ in polynomial time. (This is a stronger guarantee except when $\eta$ is already quite small.) We start with the following improper learner for halfspaces in the \emph{agnostic} model \cite{kalai2008agnostically}, which is based on $L_1$ linear regression over a large monomial basis:
\begin{theorem}[Theorem 1 of \cite{kalai2008agnostically}]\label{thm:kkms-halfspace}
Suppose that $(\rvx,Y)$ are jointly distributed random variables, where the marginal law of $\rvx$ follows $Uni(\{\pm 1\}^d)$ and $Y$ is valued in $\{\pm 1\}$. Define
\[ \textsf{OPT} = \min_{\vw^* \ne 0} \Pr{\sgn(\langle \vw^*, \rvx \rangle) \ne Y}. \]
For any $\epsilon > 0$, there exists an algorithm with runtime and sample complexity $\poly(d^{1/\epsilon^4}, \log(1/\delta))$ which with probability at least $1 - \delta$, outputs a hypothesis $h$ such that
\[ \Pr{h(\rvx) \ne Y} \le \textsf{OPT} + \epsilon. \]
\end{theorem}
No matching proper learner is known in the agnostic setting.
Based on results in computational complexity theory \cite{applebaum2008basing}, there should not be a general reduction from proper to improper learning of halfspaces. This means that there is no apparent way to convert $h(\rvx)$ into a halfspace with similar performance; it's also possible that $h$ performs significantly better than $\textsf{OPT}$, since it can fit patterns that a halfspace cannot.

If we make the additional assumption that $(\rvx,Y)$ follows the Massart halfspace model, it turns out that all of these issues go away. By combining Theorem~\ref{thm:kkms-halfspace} with our Theorem~\ref{thm:distill-general}, we immediately obtain the following new result:
\begin{theorem}\label{thm:kkms-app}
Suppose that $(\rvx,Y) \sim \mathcal{D}$ follows an $\eta$-Massart halfspace model with true halfspace $\vw^*$, and the marginal law of $\rvx$ is $Uni(\{\pm 1\}^d)$.
For any $\epsilon > 0$, there exists an algorithm (given by combining the algorithms of Theorem~\ref{thm:kkms-halfspace} and Theorem~\ref{thm:distill-general}) with runtime and sample complexity $\poly(d^{1/\epsilon^4}, \log(1/\delta))$ which with probability at least $1 - \delta$, outputs $\vw$ such that
\[ \Pr{\sgn(\langle \vw, \rvx \rangle) \ne Y} \le \textsf{OPT} + \epsilon. \]
(Recall that $\textsf{OPT} = \Pr{\sgn(\langle \vw^*, \rvx \rangle) \ne Y}$.)
\end{theorem}

As applications of the same distillation-based meta-algorithm, we note a few other interesting results for learning halfspaces in the Massart model which follow by combining an improper agnostic learner with our reduction:
\begin{enumerate}
    \item Theorem~\ref{thm:kkms-app} remains true with the weaker assumption that $\rvx$ is drawn from an arbitrary permutation-invariant distribution over the hypercube \cite{wimmer2010agnostically}.
    \item For learning Massart conjunctions (or disjunctions) over an arbitrary permutation-invariant distribution over the hypercube, there is an algorithm with runtime and sample complexity can be improved to $d^{O(\log(1/\epsilon))}$ by combining the improper learner of \cite{feldman2015agnostic} with our reduction.
    \item Suppose that $\rvz$ is a $d$-dimensional random vector with independent coordinates. Suppose that for each $i$ from $1$ to $d$, $\rvz_i$ is valued in set $\mathcal{Z}_i$ and $|\mathcal{Z}_i| = O(\poly(d))$. A \emph{linear threshold function} is a halfspace in the one-hot encoding $\rvx$ of $\rvz$, i.e. a function of the form
    \[ \vz \mapsto \sgn\left(\sum_{i = 1}^n w_i(\vz_i)\right) \]
    where each $w_i$ is an arbitrary function.
    Then combining our reduction with the result of \cite{blais2010polynomial} shows that we can properly learn Massart linear threshold functions in runtime and sample complexity $\poly(d^{O(1/\epsilon^4)},\log(1/\delta))$.
    \item If we furthermore assume $|\mathcal{Z}_i| = O(1)$ for all $i$, then the same result as 3.\ holds for $\rvz$ drawn from a mixture of $O(1)$ product measures \cite{blais2010polynomial}.
\end{enumerate}
All of these algorithms can be implemented in the SQ framework.
Interestingly, the result 2.\ above (for learning Massart conjunctions) matches the SQ lower bound over $Uni(\{\pm 1\}^d)$ observed in Theorem~\ref{thm:sq_main}, so this result is optimal up to constants in the exponent.

config: arxiv


TITLE: Prove that $OM$ is dividing $\angle IOR$ into two.
QUESTION [1 upvotes]: The square $ORIT$ is a oblong . it is given that $TR=IM$, prove that $OM$ is dividing $\angle IOR$ into two equal angles.
added my try in the pic:

Please I'm stuck a few hours need help.

REPLY [3 votes]: $\triangle OIM$ is isosceles, $\angle IOM = \angle IMO$
Also since $TM \parallel OR$,  from opposite interior angles, $\angle ROM = \angle IMO$,
From here you have the result.


TITLE: Calculate the residue of $\frac{z^{3/2}\log(z)}{(1+z^{2})^{4}}$ at $z = i$.
QUESTION [2 upvotes]: To find the residue of $$\frac{z^{3/2}\log(z)}{(1+z^{2})^{4}} \qquad \qquad(1)$$ at $z = i$, this is what I've thought:
Express this as a partial fraction: The problem therein is that decomposition of (1) yields something like
$$P_{1}(\frac{1}{z-i}) + P_{2}(\frac{1}{z+i}) + g(z)$$ where
$g(z)$ is an analytic function and $P_{j}(z)$ is as polynomial of finite degree. The $g(z)$ term exists, since (1) is not a rational function, in which case (provided also that the denominator's degree exceeds that of the numerator), the partial fraction decomposition is some summation of $P_{j}$s.
Thus, the inclusion of $g(z)$ makes me a bit nervous to try and solve this.
One could express (1) as
$$ \frac{\frac{z^{3/2}\log(z)}{(z+i)^{4}}}{{(z-i)^{4}}}$$
and then take a third derivative of the numerator, evaluate this result at $i$, and then divide by $3!$, but that is also messy.
And I have no idea how one might express the laurent series of (1).
Is there any elegant way to do this?

REPLY [1 votes]: The easiest approach would be to note that
$$f(z) = \frac{z^{3/2} \log (z)}{(z+i)^4} * \frac{1}{(z-i)^4}=g(z) \frac{1}{(z-i)^4}$$
where $g(z)$ is analytic and non-zero near $z=i$. Thus there is a pole of order $4$.
To find the residue, use the following theorem:

Theorem:
  If $f(z)$ has a pole of order $m$ at $z_0$, then
  $$R [ f,z_0 ] = \lim_{z \to z_0} \frac{1}{( {m - 1} )!} \frac{d^{m - 1}}{dz^{m - 1}} ( ( z - z_0 )^m f( z )).$$

Thus $$R[f, i] = \lim_{z \to i} \frac{1}{3!}\frac{d^3}{dz^3} \frac{z^{3/2} \log (z)}{(z+i)^4}.  $$ There will be some messiness with nested and repeated quotient and product rules, but it's "just" a Calc I problem from here.


text:
\setcounter{chapter}{-1}

\chapter{Fundamentals and preliminaries}
\label{sec:Fundamentals-and-preliminaries}

The magic of Hilbert space resides in the following innocent-looking axioms:
\begin{enumerate}
  \item A complex vector space \sH.
  \item A complex-valued function on $\sH \times \sH$, denoted by $\la \cdot,\cdot\ra$ and satisfying
  \begin{enumerate}[(a)]
    \item For every $v \in \sH$, $\la v, \cdot\ra : \sH \to \bC$ is linear.
    \item For every $v_1,v_2 \in \sH$, $\la v_1,v_2\ra = \cj{\la v_2,v_1\ra}$.
    \item For every $v \in \sH$, $\la v,v\ra \geq 0$, with equality if and only if $v=0$.
  \end{enumerate}
  \item Under the norm defined by $\|v\|_\sH := \la v,v\ra^{1/2}$, \sH is complete.
\end{enumerate}

\begin{exm}[Square-summable sequences]
  For a \bC-valued function $v$ on the integers, let $\|v\|_2 := \left(\sum_{n \in \bZ} |v(n)|^2\right)^{1/2}$. Then
  \[\ell^2(\bZ) := \{v:\bZ \to \bC \suth \|v\|_2 < \iy\}\]
  is a Hilbert space.
\end{exm}

\begin{exm}[Classical $L^2$-spaces]
  For a measurable \bC-valued function $v$ on a measure space $(X,\gm)$, let $\|v\|_2 := \left(\int_X |v(x)|^2 \,d\gm(x) \right)^{1/2}$. Then
  \[L^2(\gm) := \{v:X \to \bC \suth \|v\|_2 < \iy\}\]
  is a Hilbert space.
\end{exm}


TITLE: Does the pressure inside a rigid container change with velocity?
QUESTION [0 upvotes]: Say you take a parcel of air and place it inside a rigid container, like a glass bottle. Then put that bottle on a rocket sled and accelerate it to very high speed. Since the air inside the bottle now has a lot of kinetic energy will the pressure inside the bottle change?

REPLY [1 votes]: There are two answers to your question, as follows. We'll assume the long axis of the bottle is aligned with the direction of the sled's travel.
1) While the bottle on the sled is accelerating, it is as if the bottle is sitting in a powerful gravitational field where "up" points in the direction of travel. The gas molecules then get compacted down into the "down" end of the bottle, leaving fewer of them in the "up" end. You then get a density gradient along the length of the bottle and if you put a series of pressure gauges along the length of the bottle, you'd find more pressure in the "down" end than in the "up" end.
2) When the motor on the sled burns out and stops producing thrust, the gas molecules are then coasting along freely and redistribute themselves evenly along the length of the bottle, the density gradient disappears, and all the pressure gauges will read the same- as if the bottle were not moving. The kinetic energy of their collective motion does not affect their random vibrations, which give rise to the pressure (and temperature) inside the bottle.
This holds true for sled speeds much smaller than the speed of light. Different reasoning must be applied in that case, which I invite the experts here to furnish.


TITLE: Question on discontinuities when denominator has roots but factors out with numerator.
QUESTION [1 upvotes]: Two questions and the answers that I came across have shaken my fundamental concepts of continuity.
They are as follows:
$$f(x) = \frac{4-x^2}{4x-x^3}$$
$$h(x) = \frac{(x^2-4)(x-1)}{x^2-a}$$
a. $f(x)$ after factorization reduces to simply $1/x$.
Does that mean it only has one single point of discontinuity at $x =0$  ?
Or will all the points where the denominator equals to 0 which are $x = -2 , \ 0 ,\  2$  be considered as points of discontinuity?
b. The second question about $h(x)$ asks for the number of values of $a$ for which the function $h(x)$ will be continuous at all values of $x$ ? The answer is given to be that exactly one value of $a$ exists such that the function will be continuous at all $x$. How is that possible ? That means they are not including the points where denominator goes to $0$.
Do we always include points where denominator = $0$ in points of discontinuity regardless of whether they get factored out by the denominator or not ? or is that not the case ?

REPLY [1 votes]: $f(x)$ will be discontinuous at all values where the denominator evaluates to 0 because it is not defined at those values. Remember the definition of continuity at a point $a$ is that $f(x)$ is equal to the limit as $x $approaches $ a$.
For you second question, $h(x)$ will always be continuous (for real values of $x$) if and only if $a $is negative (because then the denominator will never reach 0). In contrast, if a is positive, the quadratic on the bottom will certainly have two roots, or if it is 0, it will have one. Obviously this is only considering real numbers...


TITLE: Questions on Borsuk-Ulam theorem in 2d
QUESTION [0 upvotes]: I was reading this proof of Borsuk-Ulam theorem for $\Bbb S^2$ and have a couple of questions just to clear my understanding.

follows from the fact that an image of an equator under a particular map to $\Bbb S^1$ is not null-homotopic. Just for my understanding, does it mean that we proved the following result along the lines of the proof of BUT?

There are no odd functions $g:\Bbb S^2\to \Bbb S^1$, that is those sending antipodal points to antipodal ones.

Since this is a course on algebraic topology, I thought about a purely topological formulation of BUT. That seems to imply that any map $f:X\to \Bbb R^2$ is guaranteed to assign the same value to two different points of $X$ if the latter is homeomorphic to $\Bbb S^2$. Am I correct? If this is the case, would BUT follow directly from that purely topological statement?

The proof also uses the fact that $\Bbb S^2$ has a trivial fundamental group, though I did not find any formal proof of that statement in the book. What is the easiest way to show this?

REPLY [0 votes]: 1 It is a trivial consequence of Borsuk-Ulam. Such a map can be considered as a map $\mathbb{S}^2\rightarrow \mathbb{R}^2$, and so it would necessarily identify two antipodal points. A map identifying two antipodal points can't be odd.
2 Yes, you are correct, Borsuk-Ulam implies what you said. But I don't think that you can deduce BUT from that statement alone. That statement tells that any map $\mathbb{S}^2\rightarrow \mathbb{R}^2$ identifies two points, but to go from that to the antipodal thing is the whole idea of BUT, and you would need to repeat Borsuk-Ulam's proof I'm afraid.
3 The fundamental group for the sphere may be calculated by Seifert-Van Kampen theorem. If you decompose $\mathbb{S}^2$ as the union of two open sets $U, V$, then $\pi_1(\mathbb{S}^2) = \pi_1(U)*_{\pi_1(U\cap V)} \pi_1(V)$, it is a "free amalgamated product". Choosing $U, V$ to be homeomorphic to the plane (for example $U$ is the sphere minus one pole, and $V$ the sphere minus the other pole), their fundamental groups are trivial and so it is their product.


TITLE: What are Bounds in Branch Bound
QUESTION [0 upvotes]: I am looking at a video to understand branch and bound more. Specifically, the one found at this link: (21) Ch06-03 Branch and Bound Method (B&B) - YouTube.
I have two questions:

Around 9:44, I'm not sure why the upper bound for node 3 to 10.25, but kept at 11.02 for node 2. Shouldn't the expected upper bound be the same for all nodes at one level? Based on this article https://www.gurobi.com/resource/mip-basics/, the upper bound for node 3 should be 10.25.
To quote the article:
"There are two additional important values we need to introduce to complete our description of branch-and-bound. First observe that, once we have an incumbent, the objective value for this incumbent, assuming the original MIP is a minimization problem, is a valid upper bound on the optimal solution of the given MIP. That is, we know that we will never have to accept an integer solution of value higher than this value. Somewhat less obvious is that, at any time during the branch-and-bound search we also have a valid lower bound, sometimes call the best bound. This bound is obtained by taking the minimum of the optimal objective values of all of the current leaf nodes."

I am dealing with a minimization problem rather than a maximization one as shown in the video. So in this case, would my Upper Bound be the best integer solution found so far and would the lower bound will be best value for the relaxed problem or best value overall i.e. considers both the objective value for the relaxed solution and feasible solutions?

Any recommended books for optimization in particular branch and bound?

REPLY [1 votes]: For a minimization problem:

The lower bound for each branch-and-bound node arises from solving the linear programming relaxation at that node.
The minimum of lower bounds (across all active leaf nodes) is a lower bound on the optimal objective value of the original problem.
Every integer feasible solution provides an upper bound on the optimal objective value of the original problem.

For a maximization problem:

The upper bound for each branch-and-bound node arises from solving the linear programming relaxation at that node.
The maximum of upper bounds (across all active leaf nodes) is an upper bound on the optimal objective value of the original problem.
Every integer feasible solution provides a lower bound on the optimal objective value of the original problem.


TITLE: Integral $\int_{-\infty}^0 e^{(-3i+\omega)t} $
QUESTION [0 upvotes]: Let's say I am integrating this function: $e^{(-3i+\omega)t}$ from $t=-\infty $ to $t=0$
[Note: $\omega$ is just a constant]
The same function could be rewritten in this form(i believe?) : $e^{-(3i-\omega)t}$
Knowing that $e^{-\infty}$ will give me $0$, I am confused by the fact integrating the same function over the same period give me two differnt answers.
Case 1:
$\int_{-\infty}^0 e^{(-3i+\omega)t} = \dfrac{e^{(-3i+\omega)t}}{(-3i+\omega)}\bigg|_{-\infty}^0 = 0$ $(e^{-\infty}=0)$
Case 2:
$\int_{-\infty}^0 e^{-(3i-\omega)t} = -\dfrac{e^{-(3i-\omega)t}}{(3i-\omega)}\bigg|_{-\infty}^0 = \infty$ $(e^{\infty}=\infty)$
I believe I am doing something wrong? What am i missing?

REPLY [1 votes]: Actually neither of your results are correct, since $e^{(\omega-3i)t}=1$ when $t=0$. So your integral should be (and the negative infinity term does go to $0$ as mentioned in the other answers): $$\int_{-\infty}^0e^{(\omega-3i)t}\,dt=\frac{1}{\omega-3i}$$