\input amstex
\documentstyle{amsppt}
\input cnm.def
\input usebolditalic
\def\leftheadtext{Zeroes of Pullback Vector Fields and Fixed Point
Theory for Bodies}
\def\rightheadtext{Daniel Henry Gottlieb}
\def\br{{\Bbb R}}
\def\bz{{\Bbb Z}}
\def\Coinc{\operatorname{Coinc}}
\def\Ind{\operatorname{Ind}}
\def\mod{\operatorname{mod}}
\def\sign{\operatorname{sign}}
\def\hn{\hat N}
\def\hv{\hat V}
\def\tv{\widetilde V}
\def\ov{\overline V}
\NoBlackBoxes

\topmatter
\title Zeroes of Pullback Vector Fields and Fixed Point
Theory for Bodies  \endtitle
\author Daniel Henry Gottlieb \endauthor
\address{Purdue University and I.H.E.S.}                  

\cvol{00}
\cvolyear{1988}
\cyear{1988}
\subjclass{Primary 57R25; Secondary 57R45, 55M20, 55M25, 53A05}  % Math Subject Classifications 
\abstract{There is an old little used equation due to Marston Morse concerning
the index of vector fields.  We apply it to several different vector fields
and obtain interesting equations.  We obtain results involving the degree of
the Gauss map, fixed points for transformations of bodies in space, inductive
equations relating the Euler--Poincar\'e number of a body with the
 Euler--Poincar\'e number of plane sections, a relationship between curvature
and the index of vector fields, invariants of the singular set of a smooth 
mapping of a manifold to Euclidean space, and an extension of the Lefschetz 
Fixed Point theorem}

\endtopmatter

\document

\subheading{1. Introduction}  		% bold, run in; don't type final
                                        % punctuation

Morse's equation, which is the main tool, is equation (7). It is used to
prove several different equations, many of which are described briefly
in this introduction and more fully in the body of the paper.

To motivate the concept of the pullback vector field, which is the first item 
in the title and the first subject of the paper, we consider the following 
question.
Let $V$ be a vector field on $\br^n$. Suppose that $M^n$ is a smooth manifold
of dimension $n$, and suppose that $f:M \to \br^n$ is a smooth map. We ask the
question:~Is there a vector field $V^*$ on $M$ so that $f_*(V^*(m)) = V(f(m))$
for every $m \in M$? Here $f_*:TM \to T\br^n$ is the differential of $f$ on the
tangent bundles.

Such a vector field $V^*$ need not exist. But there is a vector field on $M$    which 
always exists which we call the pullback vector field $f^*V$. It is defined in
section 4. The index of $f^*V$ equals the index of $V^*$ when it exists. And in
in the case where $f$ restricted to
$\partial M$ is an immersion and $V$ has no zeroes on the image of $f|\partial
M$ we calculate the index in Theorem 5 which gives the following formula. 
$$
\Ind(f^*V) = \sum_i \omega_iv_i + \left(\chi(M) - \deg\hn\right)\tag1
$$
where $V$ has only isolated zeroes indexed by $i$ and $v_i$ represents the
index and $\omega_i$ the winding number of the $i^{th}$ zero. The term $\chi(M)
- \deg\hn$ is zero for the odd dimensional case, and $\chi(M)$ is the
Euler--Poincar\'e number of $M$ and $\deg\hn$ is the normal degree of the
imbedding $f:\partial M \to \br^n$. In fact, $\hn$ represents the Gauss
map or normal map. Its degree is called the normal degree or Curvatura
Integrala of an immersion. In some sense this equation is a 
generalization of the global Gauss-Bonnet Theorem.

A local way to calculate $\Ind f^*V$ yields the equation
$$
\Ind(f^*V) = \sum_i n_iv_i + \sum_j r_j\tag2
$$
where $n_i$ is the number of regular points in the inverse image of the
$i^{th}$ zero and $r_j$ represents the index of the $j^{th}$ ``latent'' zero of
$f^*V$. It is the existence of latent zeros of $f^*V$ which obstructs the
existence of $V^*$. These two formulas depend only on quantities which can be
calculated in $\br^n$. They imply a generalization of the theorem of
Haefliger,
\cite{8} which states that the normal degree of the boundary of a
codimension zero immersion $M^n \to \br^n$ is equal to $\chi(M)$.

In section 7 we use the index of vector fields to study fixed points for compact
manifolds in the same dimensional Euclidean space.
Suppose that $M^n \subset \br^n$ is a compact body in Euclidean space. If
it undergoes a transformation $f:M \to \br^n$, what points remain fixed? The
index of a vector field $V_f$ defined on $M$ using $f$ gives us a means to
devise a fixed point index in this case. Using this, we find that if $f:M \to
M$ has no fixed points on $\partial M$,
$$
\Lambda_f + \Lambda_{-\hv_f,\hn}\hn = \chi(M).\tag3
$$
Here $\Lambda_f$ is the Lefschetz number for $f$ and $\Lambda_{-\hv,\hn}$ is
the coincidence numbers of two maps. $\hn$ is the Gauss map $\partial M \to
S^{n-1}$ and $-\hv_f$ is the ``Gauss map'' for the vector field $-V_f$. Thus we
have a topological description of the difference between $\Lambda_f$ and
$\chi(X)$ for any finite complex $X$. No matter what the thickening of $X$ is,
the two Gauss maps defined on the boundary always have the same coincidence
number.

We say that $f:M^n \to \br^n$ is {\it transverse} if the line from $m$ to
$f(m)$ is not tangent to $\partial M$ at $m$. Then we see that transverse maps
must have a fixed point if $\chi(M) \not\equiv 0 (\mod 2)$. Indeed if $M$ is
even dimensional, a transverse map has a fixed point if $\chi(M) \not= 0$.
These methods imply Dold's generalization of the Brouwer fixed point theorem
found in \cite{3}. If $f:D^n \to \br^n$ so that either $x - f(x)$ never 
points normally inward, or else never points normally outward, where $x \in 
\partial D^n = S^{n-1}$, then $f$ must have a fixed point.

In section 8 we consider
another type of function, the restriction to $M$ of the
projection of $\br^n$ to a plane $\Pi^k$ of dimension $k$. Then the study of
the index of $V_f$ results in an equation
$$
\chi(M) = (-1)^{n-k}\chi(\Pi \cap M) + \sum_i\text{ convexity numbers.}\tag4
$$
The convexity numbers depend on the curvature of $\partial M$ and the position
of $\Pi^k$ and normal lines from $\partial M$ to $\Pi$. They maybe related to 
the
convexity numbers defined in a more general context in \cite{1}.

For example, if\quad $\Pi^0$ is a point in $\br^3$, then $k = 0$ and $n = 3$. 
Then we have
$$
\chi(M) = - \chi\left(\Pi^0 \cap M\right) + \sum_ic_i\tag5
$$
where $c_i$ is calculated as follows in the generic case. The point $\Pi^0$
lies on a set of lines normal to $\partial M$ so that as one moves along the
line from $\partial M$ to $\Pi$, one enters into the body $M$. These lines are
indexed by $i$. Now let us imagine that $\Pi$ moves along some normal $\ell_i$.
When $\Pi = x_i \in \partial M$ we let $c_i = +1$. As $\Pi$ moves along the
normal and crosses a point which is the center of curvature for one of the
principal curves of curvature for $x_i$, then the $c_i$ should change sign.
Note that $\chi(\Pi \cap M) = 0$ or $1$ according to whether $\Pi$ is outside
or inside $M$.

Thus for example, if $\Pi^0$ is inside a diffeomorphic $3$--ball, the sum total
of principal radii of curvature it crosses as it moves toward $\partial B$
along all exiting normals must be even. If $\Pi^0$ were outside, it must be
odd.

Another example when $k = 1$ and $n = 2$. Then $\Pi$ is a line in the plane.
Then (4) reduces to
$$
\chi(M) = -\chi(\Pi \cap M) + \Sigma c_i.\tag6
$$
In this case note that $\chi(\Pi \cap M)$ is the number of components of the
line in $M$. Also $c_i$ is computed as follows. Let $i$ index those lines in
the plane which are normal to $\Pi$ and to $\partial M$, and so that traveling
from $\partial M$ towards $\Pi$, one initially enters into $M$. Then if
$\partial M$ curves towards $\Pi$ at $\chi_i$, let $c_i = +1$ and $c_i = -1$ if
$\partial M$ curves away.

In the case of the plane $(k=2)$ in $\br^3$, the $i$ indexes lines normal to
$\Pi$ and $\partial M$ which enter $M$ as one moves from $\partial M$ to $\Pi$.
Then the $c_i$ should equal the sign of the curvature at $x_i$.

A different relationship involving Euler-Poincar\'e numbers and curvature is
given in theorem 15. We give a very special consequence of it here. Suppose 
a line intersects a solid torus in two segments. Then there are four points
on the boundary torus with negative curvature whose tangent planes all contain 
the line.

Our methods are based on the following index formula. Let $M$ denote a compact
manifold, with or without boundary $\partial M$. Let $V$ be a tangent vector
field on $M$. Suppose that $V$ has no zeroes on $\partial M$. There is an
integer associated to $V$ called the index of $V$, denoted $\Ind(V)$ here. This
index satisfies the following equation, \cite{11}, \cite{12},
\cite{9}, \cite{3}, \cite{4},
$$
\Ind(V) + \Ind(\partial_-V) = \chi(M).\tag7
$$

Here $\chi(M)$ denotes the Euler--Poincar\'e number of $M$, and $\partial_-V$
is a vector field defined on part of $\partial M$ as follows. Let $\partial_-M$
be the subset of $\partial M$ containing all $m$ so that $V(m)$ is pointing
into the manifold. Then $\partial_-V$ is the vector field given by first
restricting $V$ to $\partial M$, and then projecting the vectors to $\partial
M$ so as to get a vector field tangent to $\partial M$, and finally restricting
$\partial V$ to $\partial_-M$.

This equation could almost serve as a definition of the index of a vector
field. All the properties of the index should be derivable from it. Yet, for
the past sixty years, it has largely been ignored. What we do here is look at
various situations in topology where a vector field can be defined, and study the
different interpretations of the least familiar term, $\Ind(\partial_-V)$.  In
the next section, we will show that $\Ind(\partial_-V)$ equals a coincidence
number under some circumstances.

\subheading{2. Coincidence numbers}

We want to show that $\Ind(\partial_-V)$ equals a certain coincidence number.
In this section we study the local situation.

Let $\alpha:D^{n-1} \to \br^n$ be a smooth imbedding of an $n-1$ dimensional
disk into $n$--dimensional Euclidean space. Suppose that $V$ is a vector field
on $\br^n$ which is not zero on the image of $D$ in $\br^n$. That is,
$$
V(\alpha(m)) \not= \overset\rightarrow\to0 \in T_{\alpha(m)} (\br^n)\text{
where }m \in D.
$$

Let $N$ be a vector field on $\alpha(D)$ of unit length normal to $\alpha(D)$.
It will be necessary to choose orientations of $D$ and $\br^n$ in order to use
coincidence numbers. But since the index of a vector field does not depend upon
the choice of orientations, it is necessary to be careful. We relate the
orientations of $D^{n-1}$ and $\br^n$ as follows.

Choose a basis of the tangent space at $m$, denoted $T_m(D)$. Call the basis
$b_1,\dots,b_{n-1}$ and give it the ordering which agrees with the orientation
of $D$. Then $\left\{\alpha_*\left(b_1\right),\dots,\alpha_*\left(b_{n-1}
\right), N(\alpha(m))\right\}$ is a basis for
$T_{\alpha(m)} \left(\br^n\right)$. Then we let the ordering
$\left\{\alpha_*\left(b_1\right),\dots,\alpha_*\left(b_{n-1}\right), 
N(\alpha(m))\right\}$ represent the orientation of $T_{\alpha(m)} 
\left(\br^n\right)$.

Now, using $N$, we want to project $V|\alpha(D)$ down onto a vector field $V'$
which is tangent to $\alpha(D)$. We define $V'$ by 
$$
V'(\alpha(m)) = V(\alpha(m)) - [V(\alpha(m)) \bullet N(\alpha(m))]
N(\alpha(m)).\tag8
$$
Or more abbreviated, $V' = V - (V\bullet N)N$, where $\bullet$ represents the 
dot product. Next we pullback $V'$ to a vector field $V^*$ using $\alpha$. We
define $V^*$ on $D$ by $V^*(m) = \alpha^{-1}_*\left(V'(\alpha(m))\right)$. This
is well defined since $\alpha$ is an imbedding.

The vector field $N$ gives rise to a Gauss map or normal map $\hn:D \to
S^{n-1}$. This is defined by parallel translating the normal vector
$N(\alpha(m))$ to a vector based at the origin of $\br^n$. Then the tip of this
vector lies on the unit sphere. We introduce the concept of the {\it Gauss map}
$\hv$ for any non zero vector field. Then $\hv:D \to S^{n-1}$ is given by
making $V$ of unit length and parallel translating the vector to the origin,
and looking at its end point. Thus
$$
m \mapsto V(\alpha(m)) \mapsto \frac{V(\alpha(m))}{\Vert V(\alpha(m))\Vert}
\mapsto \hv(m)
$$
where $\hv(m)$ a parallel vector of unit length based at the origin.

Now $V^*$ is related to $\hv$ and $\hn$ as follows. The point $m_0 \in D$ is a
coincidence of $\hv$ and $\hn$, that is $\hn\left(m_0\right) =
\hv\left(m_0\right)$, if and only if $N\left(m_0\right) \bullet
V\left(m_0\right) > 0$ and $V^*\left(m_0\right) = \overset\rightarrow\to0 \in
T_{m_0}(D)$. Now this is obvious. What is not so obvious, and needs a careful
argument, is that for isolated coincidence points $m_0$, the index of $V^*$ at
$m_0$ is equal to the coincidence number, $\Coinc(\hn,\hv)$.

We follow the excellent, and perhaps unique, account of coincidence
\cite{13}, see chapter 6. Suppose we have two {\it oriented} manifolds $M^n$
and $N^n$. Suppose that $f,g:M \to N$ are two maps with an isolated coincidence
point $m_0$. Then the coincidence number of $f$ and $g$ at $m_0$, denoted
$\Coinc\left(f,g;m_0\right)$, is defined as follows. Choose two homeomorphisms
of the unit disk $D \subset \br^n$, denote them by $h:D \to M$ and $k:D \to N$.
We choose $h$ and $k$ so that $h(0) = m_0$ and $k(0) = f\left(m_0\right) =
g\left(m_0\right)$. Also we want the image of $h(D) \subset M$ under both $f$
and $g$ to be contained in $k(D)$, so that $k^{-1}fh$ and $k^{-1}gh$ are
defined. Also we choose $h$ so that $h(D)$ contains only one coincidence point,
$m_0$. And last, but very important, we choose $h$ and $k$ so that $k^{-1}h:
D\to D$ preserves orientation. Now define
$$
\varphi:S^{n-1} = \partial D \overset{h}\to\longrightarrow M
\overset{(f,g)}\to\longrightarrow N \times N 
@>k^{-1} \times k^{-1}>> D \times D \overset{F}\to\longrightarrow D - 
\overset\rightarrow\to0\tag9
$$
where $F(x,y) = \frac12 (y-x)$. In other words,
$$
\varphi(m) = \frac12 \left(k^{-1}gh(m) - k^{-1}fh(m)\right) \in D -
\overset\rightarrow\to0.\tag10
$$
Then the degree of $\varphi$ is defined to be $\Coinc\left(f,g;m_0\right)$. If
$M_1 = M_2$ and the identity is orientation preserving, then
$\Coinc\left(f,1;m_0\right)$ is the usual fixed point index. This definition is
independent of the choices made. Also the definition is stable under homotopies
of $f$ and $g$ such that no coincidence point of $f_t$ and $g_t$, for any $t$,
lies on the boundary of $h(D)$. However, if the orientation of either $M$ or
$N$ is reversed, then the sign of the coincidence number changes. And if the
roles of $f$ and $g$ are reversed, then
$$
\Coinc(f,g) = (-1)^n \Coinc(g,f)\text{ where }n = \dim M.\tag11
$$

\proclaim{Lemma 1} \quad If $m_0$ is an isolated coincidence of $\hn$ and $\hv$,
then $\Ind\left(V^*\right) = \Coinc\left(\hn,\hv;m_0\right)$.
\endproclaim

\demo{Proof} Without loss of generality, we can assume that the imbedding
$\alpha:D^{n-1} \to \br^n$ takes the unit disk $D$ in $\br^{n-1}$ into $\br^n$
so that the coincidence point $m_0 = \overset\rightarrow\to0$. Assume also that
$N(0) = (0,0,\dots,0,1)$, so that $\hv(0) = \hn(0)$ is the ``north pole'' of
$S^{n-1} \subset \br^n$. We can assume also that $\alpha(D)$ is small enough so
that the images of $\hv$ and $\hn$ lie in a small neighborhood of the north
pole in $S^{n-1}$. Let $\pi:S^{n-1} \to D^{n-1}$ be the projection. It gives a
homeomorphism between the northern hemisphere and the equitorial disk. In our
calculation of $\Coinc(\hn,\hv)$, the role of $h$ will be played by the
identity and the role of $k$ by $\pi^{-1}$. The choice of orientations is given
by choosing an orientation of $\br^n$. This induces orientations on the
northern hemisphere and on $D^{n-1}$ and $\pi$ is orientation preserving. The
outward pointing normal to $S^{n-1}$ induces the orientation on $S^{n-1}$.

Let $\alpha_t:D \to \br^n$ be an isotopy of embeddings so that $\alpha_t(0)$ is
fixed. Then we have normal fields $N_t(m)$ for each $t \in I$, and each $N_t$
is defined on $\alpha_t(D)$. We choose the isotopy $\alpha_t$ so that $N_t(0) =
N(0)$. More precisely, $N_t(m) \in T_{\alpha_t(m)}(\br^n)$ so that $\Vert
N_t(m)\Vert = 1$ and $N_t(m)$ is orthogonal to $\left(\alpha_t\right)_*
\left(T_m(D)\right)$. Finally we assume that $\alpha_1 =$ translation of $D$,
(so that $0$ goes to $N(0)$ in the $(n-1)$--dimensional hyperplane passing
through $N(0)$ and orthogonal to $N(0)$). So $\alpha_t$ can be thought of as a
deformation of $\alpha(D)$ which flattens it out into the tangent hyperplane at
$\alpha(0)$. We assume that $\alpha_1$ translates $D$ to the unit disk tangent
to the north pole.

Now we define a family of vector fields $V_t$, each one defined on the
corresponding subspace $\alpha_t(D)$ of $\br^n$, by letting
$$
V_t(m) = [V(\alpha(m))\bullet N(m)] N_t(m) +
\left(\alpha_t\right)_*\left(V^*(m)\right)\qquad \in T_{\alpha(m)} (\br^n).\tag12
$$
Then we have $(\alpha_t)_*(V^*) = V_t$ for every $t$.

Now we consider $\hn_t$ and $\hv_t$. For any $t$, the maps $\hn_t$ and $\hv_t$
have only one coincidence point, at $0$, since $V_t = V^*$ has only one zero,
at $0$. That being the case,
$$
\Coinc(\hn,\hv) = \Coinc(\hn_t,\hv_t) = \Coinc(\hn_1,\hv_1).\tag13
$$
Recall that $\hv_1(m)$ is parallel to $N(\overset\rightarrow\to0)$. Thus
$$
V_1(m) = [V(\alpha(m)) \bullet N(m)] N_1(m) + \alpha_{1_*}
\left(V^*(m)\right).\tag14
$$
In addition we have, where $v_0$ denotes $v_m$ translated to the origin, 
$$
\gathered
\hn_1(m) =\text{ north pole; }\hv_1(m) =\\
\frac{\left([V(\alpha(m))\bullet N(m)] N(\overset\rightarrow\to0) +
\left(\alpha_{1_*}(V^*(m))\right)_0\right)}{\left(\left(V(\alpha(m)) \bullet
N(m)\right)^2 +
\left\Vert\alpha_{1_*}\left(V^*(m)\right)\right\Vert^2\right)^\frac12}.
\endgathered\tag15
$$
Recall that $\pi:${\it Northern hemisphere} $\rightarrow D$ was the projection
and $\pi \circ \alpha_1 =$ identity. Then $\pi \circ \hn_1 =
\overset\rightarrow\to0$ and $\pi \circ \hv_1(m) = k(m)
\left(\alpha_{1^*}(V^*(m))\right)_0 = k(m) (V^*(m))_0$ where $k(m) =
\left[\left(V(\alpha(m)) \bullet N(m)\right)^2 + \left\Vert \alpha_{1_*}
\left(V^*(m)\right)\right\Vert^2\right]^{-\frac12} > 0$. Now by (10), the
coincidence number $\Coinc(\hn_1,\hv_1)$ is the degree of $\varphi:\partial
D\to \br^{n-1} - \overset\rightarrow\to{0}$, where
$$
\gathered
\varphi(m) = \frac12 \left(\pi\circ\hv_1(m) - \pi \circ \hn_1(m)\right) =\\
\frac12 k(m) \left(\alpha_{1_*}(V^*(m))\right)_0 \in \br^{n-1}
\endgathered
\tag16
$$
On the other hand, the index of a vector field $V$ is given by the degree of
the map
$$
\psi:\partial D \to \br^{n-1} - 0\text{ where }\psi(m) - (V(m))_0.\tag17
$$
Thus we have shown that
$$
\Coinc(\hn,\hv) = \Coinc(\hn_1,\hv_1) = \deg\varphi = \Ind \left(\frac{k(m)}2
V^*\right).\tag18
$$
Now $\frac{k}2: D \to \br_+$ is homotopic to the constant $1$, and since this
homotopy does not introduce new zeroes, we see that $\Ind\left(\frac{k(m)}2
V^*\right) = \Ind(V^*)$. This proves the lemma.
\enddemo

\subheading{3. Codimension one immersions}

In this section we consider the following situation. Let $C^{n-1}$ be a smooth
closed oriented manifold. Suppose that $\alpha:C^{n-1} \to \br^n$ is a smooth
immersion. Let $V$ be a vector field on $\br^n$ which is not zero on the image
$\alpha(C^{n-1}) \subset \br^n$. We choose an orientation of $\br^n$. Now for
each point $m \in C$, we choose a vector $N(m) \in T_{\alpha(m)}(\br^n)$ such
that $N(m)$ has unit length and $N(m)$ is orthogonal to the tangent space of
$\alpha(C)$ at $\alpha(m)$. Now $N$ induces an orientation on $C$ exactly as in
the last section. Note that $N$ is not a vector field on $\alpha(C)$ since      $\alpha$ need not
be injective. Now for each $m \in C$, we define a vector $V'(m) \in
T_{\alpha(m)} (\br^n)$ by
$$
V'(m) = - [V(\alpha(m)) \bullet N(m)] N(m) + V(\alpha(m)).
$$

Thus $V'(m) \bullet N(m) = 0$, so $V'(m)$ is tangent to $\alpha(C)$ at
$\alpha(m)$. Again $V'$ does not define a vector field on $\alpha(C)$ .         We define a vector
field $V^*$ on $C$ by the equation
$$
\alpha_*(V^*(m)) = V'(m).
$$
Now $V^*$ is well defined since $\alpha$ is an immersion, so $\alpha_*$ is
injective, and $V'(m)$ is in the image of $\alpha_*$.

We denote by $V^*_+$, the vector field $V^*$ restricted to the open subspace of
$C$ such that $V(\alpha(m)) \bullet N(m) < 0$. Note that every zero of $V^*$ is
either in $V^*_+$ or $V^*_-$ since $V(\alpha(m))$ is never zero. Thus
$$
\Ind V^*_+ + \Ind V^*_- = \Ind(V^*) = \chi(C).\tag20
$$

We recall here some more coincidence theory, suppose $M_1$ and $M_2$ are both
closed, oriented manifolds of dimension $n$. Suppose $f$ and $g$ are maps from
$M_1$ to $M_2$. Suppose that $f$ and $g$ have a finite number of coincidence
points, $M_i$. Then we introduce the Lefschetz trace
$$
\Lambda_{f,g} = \sum_j (-1)^j tr(g^!f^*)_j.\tag21
$$
Here $f^*:H^j(M_2) \to H^j(M_1)$ is the induced map in cohomology and
$g^!:H^j(M_1) \to H^j(M_2)$ is the umkehr map, defined by $g^! = D^{-1}g_*D$
where $D:H^j(M) \to H_{n-j}(M)$ is the Poincar\'e duality map. Also
$tr(f^*g^!)_j$ denotes the trace of $f^*g^!:H^j(M_1) \to H^j(M_1)$.

Note that $g^!$ reverses sign when the orientation of either $M_1$ or $M_2$ is
reversed. If $M_1 = M_2$ and the orientations are chosen so that the identity
$1$ preserves orientation, then $\Lambda_{f,1} = \Lambda_f$ where $\Lambda_f$
is the usual Lefschetz number. Now the main result of coincidence theory is
$$
\Lambda_{f,g} = \sum_i\Coinc(f,g;m_i).\tag22
$$
This generalizes the result for fixed points.

\proclaim{Lemma 2}\quad Suppose that $C^{n-1}$ is immersed into $\br^n$ as 
above, so that $V^*$ has only a finite number of zeroes. Then
$$
\Ind V^*_+ = \Lambda_{\hn,\hv} = \deg\hv + (-1)^{n-1} \deg \hn,\text{ if }n-1 >
0.
$$
\endproclaim

\demo{Proof} Recall that $\hn$ and $\hv$ are the Gauss maps from $C^{n-1} \to
S^{n-1}$ given by taking the vector associated to $m$, making it of unit
length, and translating it to the origin. Now
$$
\Lambda_{\hn,\hv} = \sum_i \Coinc\left(\hn,\hv;m_i\right) = \sum_i
\Ind\left(V^*_+;m_i\right) = \Ind V^*_+
$$
where $m_i$ are the isolated coincidence points of $\hn$ and $\hv$, which
correspond to those zeroes of $V^*$ that are in $V^*_+$. This follows by
applying equation (22) and Lemma 1. From equation (21) we see that
$$
\aligned
\Lambda_{\hn,\hv} &= tr\left(\hv^!\hn^*\right)_0 + (-1)^{n-1}
tr\left(\hv^!\hn^*\right)_{n-1}\\
&= \deg \hv + (-1)^{n-1} \deg\hn.\endaligned
$$
\enddemo

\proclaim{Lemma 3} $\Ind V^*_- = \Lambda_{-\hv,\hn} = - \deg\hv + \deg\hn$, if
$n - 1 > 0$.
\endproclaim

\demo{Proof} We apply lemma 2 to the vector field $-V$ and obtain
$$
\Ind(-V)^*_+ = \Lambda_{\hn,-\hv} = \deg (-\hv) + (-1)^{n-1}\deg\hn.
$$
Now $(-V)^*_+ = - (V^*_-)$, so $\Ind V^*_- = (-1)^{n-1} \Ind(-(V^*_-)) =
(-1)^{n-1}\Ind V^*_+$. Also $\Lambda_{\hn,-\hv} = (-1)^{n-1}
\Lambda_{-\hv,\hn}$. And $\deg(-\hv) = (-1)^n \deg \hv$ since $- \hv$ is just
$\hv$ composed with the antipodal map. Substitution into the equation gives the
desired result.

Now adding the equations of Lemmas 2 and 3 and using equation (20) we obtain
$$
\chi(C) = \left(1 + (-1)^{n-1}\right) \deg \hn.
$$
When $C$ is odd dimensional we obtain nothing, but for even dimensional $C$ we
see that
$$
\deg \hn = \frac12 \chi(C)\text{ for }\dim C > 0\text{ even.}\tag23
$$
This was originally discovered by H.~Hopf. Hopf called $\deg \hn$
the {\it curvature Integrala}. Another name is {\it normal degree}. Milnor
\cite{10} and Bredon--Koscinski \cite{2} showed that for odd dimensional
connected $C$ the normal degree can take on any value if and only if $C$ is
parallelizable. If $C$ is connected and not parallelizable, then the normal
degree may take on any odd value, but no even values.

We can give a nice formula for $\deg \hv$. To do that we need to introduce the
concept of the {\it winding number} of $\alpha:C^{n-1} \to \br^n$ about $p \in
\br^n - \alpha(C)$. This is a generalization of the usual notion for closed
paths in the plane.
\enddemo

\demo{Definition} Choose orientations of $C^{n-1}$ and $\br^n$. Then the {\it
winding number} $w(\alpha,p)$ is the degree of the map $C^{n-1} \to \br^n - p
\approx S^{n-1}$.

In the case where $\alpha$ is an immersion, we can give a very geometric way to
calculate the winding number. The choice of orientations will give rise to the
``vector field'' $N(m)$ of normals at $\alpha(m)$. Now consider a path from $p$
out towards infinity which crosses $\alpha(C)$ transversally. For each
crossing, assign a $+1$ or a $-1$ according to whether the path crosses into
the ``side'' of $\alpha(C)$ that $N(m)$ points to or not. The sum of these
$\pm1$'s is the winding number.

Now suppose that $\alpha:C^{n-1} \to \br^n$ is an immersion as before and
suppose that the vector field $V$ on $\br^n$ has isolated zeroes $p_i$. Let
$\omega_i = W(\alpha,p_i)$, the winding number about the $i^{th}$ zero. Let
$v_i = \Ind(V,p_i)$, the index of $V$ at the $i^{th}$ zero. Then
\enddemo

\proclaim{Lemma 4}
$$
\deg \hv = \sum_i \omega_iv_i.
$$
\endproclaim

\demo{Proof} The fact that $C^{n-1}$ immerses in $\br^n$ implies that it bounds
a $\Pi$--manifold $M$. That is, $M$ is parallelizable and $\partial M = C$. So
we extend the immersion $\alpha$ to a map $f:M \to \br^n$. We can always do
this since $\br^n$ is contractible, and we can do it so that $f$ is smooth and
so that the zeroes, $p_i$, of $V$ are regular points. We think of the vector
field $V$ on $\br^n$ as a map $\tv:\br^n \to \br^n$ given by $\tv(p) =
(V(p))_0$. Now consider $\tv \circ f: M^n \to \br^n$. Restricted to $\partial
M$, the map is:
$$
\tv \circ \alpha:\partial M \to \br^n - 0 \subset \br^n.
$$

Now $\tv$ can be homotopied to $\tv_1$ by $\tv_t(p) = k_t(p)V(p)$ where $k_t(p)
> 0$ is chosen so that $\Vert\tv_1(p)\Vert \leq 1$ for all $p \in \br^n$ and
$\Vert\tv_1(p)\Vert = 1$ for $p \in \alpha(\partial M)$. Then $\tv_1 \circ
\alpha = \hv$ and $\tv_t \circ \alpha:\partial M \to \br^n - 0$. So $\tv_1
\circ f$ can be thought of as a map from $M \to D$ which takes $\partial M$
into $\partial D$ by a  map of degree equal to $\deg \hv$. This degree can also
be calculated locally by looking at the inverse image of a small neighborhood
of $0 \in \br^n$ under $\tv_1 \circ f$. The inverse image at $0$ of $\tv_1$ is
just the zeroes of $V$, namely the $p_i$. And the inverse images of a $p_i$ are
points $m_{ij} \in M$. Now each $m_{ij}$ has a little neighborhood that maps
diffeomorphically onto a small neighborhood of $p_i$. Thus locally, the map
$\tv_1 \circ f$ restricted to a neighborhood of $m_{ij}$ has degree $\pm v_i$
depending on whether $f$ preserves or reverses the orientation of the
neighborhood. Then
$$
\deg\hv = \sum_{i,j} \pm v_i = \sum_i \omega_iv_i.
$$
\enddemo
\demo{Remark} For the results of this section there is no stipulation that the
manifold $C$ is connected. However, lemmas 2 and 3 are not true for $C$ of
dimension zero since $H_0(S^0) \napprox \bz$.
\enddemo

\subheading{4. Codimension zero maps}

Let $f:M^n \to \br^n$ be a smooth map from a compact Riemannian manifold $M$
into $\br^n$. Let $V$ be a vector field on $\br^n$. Let $\langle\ ,\ \rangle_R$
the Riemannian metric on $\br^n$.

\demo{Definition} the {\it pullback vector field} $f^*V$ on $M$ is defined by
$$
\langle f^*V(m),v_m\rangle_M = \langle V(f(m)),f_*(v_m)\rangle_R.
$$
This $f^*V$ can be thought of as the dual to the pullback of a $1$--form. Our
objective is to prove the following precise statement of equation (1).
\enddemo

\proclaim{Theorem 5} Suppose that $f$ is a smooth map which restricts to an
ambient immersion $\alpha$ on $\partial M$. We assume that $\partial M$ is
orientable and $n > 1$, and no zeroes of $V$ are in the image of $\partial M$.
Then
$$
\Ind(f^*V) = \sum_i \omega_iv_i + \chi(M) - \deg\hn.
$$
\endproclaim

The condition that $\alpha$ is an ambiant immersion on $\partial M$ means that
the singular set of $f$ is disjoint from $\partial M$. Equivalently, $f$ is an
immersion on a neighborhood of $\partial M$. Here $N(m)$ is defined as before
as the unit normal to $\alpha(\partial M)$ at $\alpha(m)$, pointing in the
outward direction. Then a choice of an orientation on $\br^n$ induces, by means
of $N(m)$, an orientation of each component of $\partial M$. Note that this
orientation of $\partial M$ may not be consistent with any orientation of $M$,
indeed $M$ may not be orientable. But $\partial M$ must be. With this
convention, we can apply the results of the last section.

Now since $\alpha_*:T_m(M) \to T_{\alpha(m)} (\br^n)$ is an isomorphism, we may
define vector fields $\alpha^{-1}_*N$ and $\alpha^{-1}_*V$ on $T(M)|\partial
M$. We also have a vector field $f^*N$ on $\partial M$.

\proclaim{Lemma 6} The vector field $f^*N$ is an outward pointing normal vector
field on $\partial M$. The vector field $\alpha^{-1}_*N$ points outwards.
\endproclaim

\demo{Proof} Let $v_m$ be any vector tangent to $\partial M$ at $m$. Then
$$
\langle f^*N(m),v_m\rangle_M = \langle N(m),\alpha_*(v_m)\rangle_R = 0.
$$
So $f^*N$ is a normal vector field. Now
$$
\langle f^*N(m),\alpha^{-1}_*N(m)\rangle_M = \langle
N(m),\alpha_*\left(\alpha^{-1}_*N(m)\right)\rangle = 1 > 0.
$$
We learn from this that $f^*N$ and $\alpha^{-1}_*N$ both point out of the same
side. Since $N$ points outwards locally in $\br^n,\ \alpha^{-1}_*N$ must point
outwards also in $M$. Given any outward pointing vector field on $\partial M$,
we may project any vector field on $\partial M$ down to a vector field tangent
to $\partial M$. Recall the definition of $V^*$ as in \S3. Thus $V^*(m) =
\alpha^{-1}_*V(\alpha(m)) - k\alpha^{-1}_*(N(m))$ for some number $k$, chosen
so that $V^*(m)$ is tangent to $\partial M$. Thus we say that $V^*$ is the
projection of $\alpha^{-1}_*V$ onto $\partial M$ by means of the outward
pointing vector field $\alpha^{-1}_*N$. Similarly, the notation
$\partial(\alpha^{-1}_*V)$ represents the projection of $\alpha^{-1}_*V$ onto
$\partial M$ by means of the outward pointing normal vector field at $\partial
M$. Recall that is $f^*N$. Finally $\partial f^*V$ is the projection of $f^*V$
onto $\partial M$ by means of $f^*N$.
\enddemo

\proclaim{Lemma 7} $\Ind V^*_- = \Ind \partial_-(\alpha^{-1}_*V) = \Ind
\partial_-(f^*V)$.
\endproclaim

\demo{Proof} Here the subscript minus denotes the restrictions of the vector
fields to the portion of the boundary where the original field pointed inwards.
Now the vector fields $N_t = (1-t) (\alpha^{-1}_*N) + t(f^*N)$ always point
outward for $t \in I$, since $\langle N_t,f^*N\rangle > 0$ and $f^*N$ is an
outward pointing normal. The homotopy gives rise to a homotopy $V^*_t$ of
vector fields tangent to $\partial M$ induced by projecting $\alpha^{-1}_*V$
down to $\partial M$ by means of $N_t$. If $V^*(m) = V^*_0(m)$ were tangent to
$\partial M$, it would remain constant for all $t$. Thus $(V^*_t)_-$ is a
vector field on the same open subspace of $\partial M$ for all $t$, and no
zeroes of $V^*_t$ will occur on the frontier of this inward pointing region.
Hence $\Ind(V^*_t)_-$ is a constant and we have the first equality from $\Ind
V^*_- = \Ind(V^*_t)_- = \Ind \partial_-(\alpha^{-1}_*V)$. Next we consider the
vector fields
$$
V_t = t(\alpha^{-1}_*V) + (1-t) (f^*V).
$$
This homotopy on the boundary can be extended to a homotopy $\ov_t$ of $f^*V$
over $M$. Now on the boundary $V_t$ is never zero. This can be seen since
$$
\langle f^*V(m), \alpha^{-1}_*V(\alpha(m))\rangle_M = \langle V(\alpha(m)),
V(\alpha(m))\rangle > 0.
$$
Thus $\langle V_t,f^*V\rangle_M > 0$. Hence the homotopy $\ov_t$ never has a
zero on $\partial M$. Hence $\Ind f^*V = \Ind \ov_t = \Ind \ov_1$. Now by
equation (7),
$$
\Ind(\ov_t) + \Ind(\partial_-\ov_t) = \chi(M).
$$
So $\Ind(\partial_-(f^*V)) = \Ind(\partial_-(\alpha^{-1}_*V))$, proving the
second inequality.
\enddemo

\demo{Proof of theorem 5}
$$
\aligned
\Ind f^*V &= \chi(M) - \Ind(\partial_-f^*V)\\
&= \chi(M) - \Ind(V^*_-)\\
&= \chi(M) - (-\deg\hv + \deg\hn)\\
&= \chi(M) + \Sigma \omega_iv_i - \deg \hn.\endaligned
$$
\enddemo

\demo{Remark} Note that there is no choice of orientations in theorem 5. The
outward pointing normals are all that are needed to calculate $\omega_i$ or
$\deg \hn$. In the proof, careful attention was paid to how the orientations
were chosen. If the orientation of $\br^n$ is reversed, our conventions assure
us that the orientations of $\partial M$ and $S^{n-1}$ are reversed. If we
changed our convention so that $\left\{N,b_1,\dots,b_n\right\}$ determines the
orientation instead of $\left\{b_1,\dots,b_n,N\right\}$, the consistent use of
this would change both the orientations of $\partial M$ and $S^{n-1}$ and so we
still have the correct equation. The sign changes always cancels out.

We say that a vector field $V^*$ on $M$ {\it lifts} a vector field $V$ on
$\br^n$ if $f_*(V^*(m)) = V(m)$ for all $m \in M$.
\enddemo

\proclaim{Proposition 8} If $V^*$ lifts $V$, then $\Ind V^* = \Ind f^*V$.
\endproclaim

\demo{Proof} Consider the homotopy of vector fields
$$
V_t = tV^* + (1-t)f^*V\text{ for }t \in I.
$$
Note that if $m$ is not a zero of $V^*$ and $f^*V$, then $m$ is not a zero of
$V_t$. Since both $V^*$ and $f^*V$ are never zero on $\partial M$, we see that
$V_t$ is never zero on $\partial M$. Hence $\Ind V_0 = \Ind V_1$.
\enddemo

\proclaim{Corollary 9} If $f:M^n \to \br^n$ has a lifted vector field $V^*$,
where $V$ has no zeroes, then $\deg \hn = \chi(M)$. 
\endproclaim

\demo{Proof} $0 = \Ind V^* = \Ind f^*V = \chi(M) - \deg \hn$.

If $f:M^n \to \br^n$ were an immersion, every vector field on $\br^n$ has a
lifting $V^*$. Hence we have the theorem of Haefliger \cite{8}.
\enddemo

\proclaim{Corollary 10} If $f:M^n \to \br^n$ is an immersion, then $\deg \hn =
\chi(M)$.
\endproclaim

\subheading{5. Lifting vector fields}

In this section we shall draw some consequences of Theorem 5 and Proposition 8.
First we obtain a formula for $\Ind(f^*V)$ in terms of local conditions. Then
we compare it with theorem 5, which is a global formula.

Consider as usual a smooth map $f:M^n \to \br^n$. Suppose that $V$ is a vector
field on $\br^n$ with isolated zeroes at $x_i$. For each zero $x_i$, let $n_i$
stand for the number of noncritical points in $f^{-1}(x_i)$. Now consider the
zeroes of $f^*V$. They fall into two classes. There are the {\it regular}
zeroes, that is, those zeroes which are non singular points of $f$; and the
{\it latent} zeroes, those zeroes which occur at the singular points of $f$.
The image of a regular zero is a zero of $V$, but the image of a latent zero
need not be a zero of $V$. Now if $m_j$ is a latent zero, we will denote its
local index by $r_j$. If $m$ is a regular zero, then the local index of $m$
equals $v_i$, the local index of $x_i = f(m)$. This is seen since $f$
restricted to a small neighborhood of $m$ is a diffeomorphism, hence locally by
Proposition 8, $\Ind f^*V = \Ind V^* = \Ind V$. The second equality follows
from (7). It is the fact that index is invariant under diffeomorphism.

\proclaim{Proposition 11} $\Ind f^*V = \Sigma n_iv_i + \Sigma r_j$ when the
$n_i$ are finite. Then in the case where the singular set $\Sigma$ does not
intersect $\partial M$, we combine Proposition 11 and Theorem 5 and Proposition
8, if $V^*$ exists, to get the equation
$$
\Ind V^* = \Ind f^*V = \Sigma \omega_iv_i + \chi(M) - \deg \hn = \Sigma n_iv_i
+ \Sigma r_j\tag24
$$
\endproclaim

Now we draw some consequences of equation (24). We introduce the following
notation. Given $f:M^n \to \br^n$ with a vector field $V$ defined on $\br^n$,
we let $Z \subset \br^n$ denote the set of zeroes of $V$, and $\Sigma$ denote
the singular points of $f$ and $\Delta = f(\Sigma)$. We call $\Delta$ the {\it
discriminant} of $f$. In this terminology, the conditions for equation (24) to
hold are $\partial M \cap \Sigma = \phi$ and $f(\partial M) \cap Z = \phi$
and $f^{-1}(Z) - \Sigma$ is finite.

\noindent Then applying (24) we see

\noindent (25) $\Ind f^*V = \chi(M) - \deg \hn = \Sigma r_j$ if $Z = \phi$.

\noindent (26) $\Ind f^*V = 0 = \Sigma r_j$ if $Z = \phi$ and $n$ odd.

\noindent (27) $\Ind V^* = \Sigma \omega_iv_i + \chi(M) - \deg \hn = \Sigma
n_iv_i$ if $Z \cap \Delta = \phi$.

\noindent (28) $\Ind V^* = \Sigma \omega_iv_i = \Sigma n_iv_i$ if $Z \cap
\Delta = \phi$ and $n$ odd.

The condition that the vector fields involved have isolated singularities is
not meant seriously. In fact, we can define the local index of isolated
connected components of $Z$, by using equation (7). So for each connected
component $Z_i$ of $Z$, we define the local index $v_i$. Since $Z \cap
f(\partial M) = \phi$, we see that $\omega_i$ is constant at every point of
$Z_i$. So the first two equations of (24) are true in this sense.

The last equation of (24) makes sense if we add the conditions that $f(\partial
M) \cap \Delta = \phi$ and that $Z_i \subset \Delta$ if $Z_i$ intersects
$\Delta$ and $f^{-1}(Z)$ consists of a finite number of path components. Then
the $r_j$ are the indices of those components of the zeroes of $f^*V$, or of
$V^*$, which are contained in $\Sigma$. And $v_i$ is the index of the sets
$Z_i$ in $\br^n$. The $n_i$ are constant for each point in $Z_i$ since
$f(\partial M) \cap \Delta = \phi$.

\demo{Remark} The reader may interpret $\omega_i$ as a local degree for the map
$f$. This is not quite accurate. We will give an example which should eliminate
some misconceptions. Let $M$ be the Mobius band. Hence $M$ is not orientable!
Let $f:M \to \br^2$ be a map so that $\Sigma$ is the middle circle of the
Mobius band. Suppose that $\Delta = f(\Sigma)$ is the unit circles in $\br^2$.
Suppose that the rest of $M$ is mapped outside of the unit circle and that
$\partial M = S^1$ is mapped with winding number $2$ about the origin. Now let
$V$ be the vector field given by an instantaneous rotation of $\br^2$ about the
origin. Then the origin is the only zero of $V$ and it has index $v = 1$. Note
that this zero is not in the image of $f$! Now $V$ lifts to a vector field
$V^*$ which has no zeroes. Hence applying theorem 5
$$
0 = \Ind V^* = 2 \times 1 + 0 - 2
$$
which verifies that the equation is true in this situation.

\subheading{6. The Euler--Poincar\'e number of 
${\bolditalic{f^{-1}(f(\Sigma))}}$}

Here we obtain a formula for $\chi(f^{-1}(f(\Sigma))$ where $f:M^n \to \br^n$
is a map satisfying mild conditions by lifting vector fields. We define a
vector field $V$ on $\br^n$ which can be lifted, and using (24) we determine
the index of the lifting $V^*$ in two ways. Combining them yields the formula.

\proclaim{Theorem 12} Let $f:M^n \to \br^n$ be a smooth map such that
$f(\Sigma) \cap f(\partial M) = \phi$. Suppose there exists a compact
submanifold $N^n \subset \br^n$ so that $f(\Sigma)$ is a deformation retract of
$N$ and $f^{-1}(f(\Sigma))$ is a deformation retract of $f^{-1}(N)$.

Then
$$
\chi(f^{-1}(f(\Sigma))) - \sum_i \omega_i\chi(\Sigma_i) + (-1)^n \sum_j
(\omega_j-n_j)\chi(D_j) + \chi(M) - \deg\hn.\tag29
$$
\endproclaim
Here $\Sigma_i$ are the connected components of $f(\Sigma)$. The $D_j$ are the
bounded components of $\br^n-f(\Sigma)$. For each $D_j$ we choose a point $x_j
\in D_j$. Then $n_j$ is the cardinality of $f^{-1}(x_j)$ and $\omega_i$ is the
winding number about $\Sigma_i$.

\demo{Proof} We construct a vector field $V$ on $\br^n$ and a lifting $V^*$ as
follows. First we let $V$ be zero on $f(\Sigma)$ and $V^*$ be zero on
$f^{-1}(f(\Sigma))$. Now any continuous extension of $V$ will automatically
give a unique lifting $V^*$ on $M$ extending $V^*$ on $f^{-1}(f(\Sigma))$ since
outside of $\Sigma$ the map $f$ is an immersion.

Now extend $V$ over $N$ so that $V$ points outside of $\partial N$ at every
point of $\partial N$. Then $V^*$ will point outside of $f^{-1}(N)$ at the
boundary $f^{-1}(\partial N) = \partial f^{-1}(N)$. Now the local index for
$N_i$, which is the component of $N$ containing $\Sigma_i$, is $v_i =
\chi(\Sigma_i)$. The index of $V^*$ on $f^{-1}(N)$ is equal to
$\chi(f^{-1}(N)) = \chi(f^{-1}(f(\Sigma))$.

Now extend $V$ to all of $\br^n$. On the unbounded components of $\br^n - N$
this can be done without introducing a zero. On a bounded component $D_j$, we
may extend $V$ so that there is only one zero at some arbitrary point $x_j$. We
choose $x_j \notin f(\partial M)$ so that $V^*$ is non zero on $\partial M$.

Now we apply (24) to $V^*$. First we note that the local index $v_j$ about
$x_j$ is equal to $(-1)^n\chi(D_j)$. The $(-1)^n$ comes in since the vector
field $V$ is pointing inside at the boundary of $\br^n - \overset\circ\to{N}$.

Then (24) gives
$$
\Sigma \omega_i(\Sigma_i) + (-1)^n \Sigma \omega_j\chi(D_j) + \chi(M) - \deg
\hn = (-1)^n \Sigma n_j\chi(D_j) + \chi\left(f^{-1}(f(\Sigma))\right).
$$
Solving for $\chi\left(f^{-1}(f(\Sigma))\right)$ gives the result.

For odd dimension $n$, equation (29) simplifies to
$$
\chi\left(f^{-1}(f(\Sigma))\right) = \Sigma \omega_i\chi(\Sigma_i) +
(-1)^n\Sigma (\omega_j-n_j)\chi(D_j).\tag30
$$
\enddemo

\subheading{7.~Fixed point theory for bodies in space}

Suppose $M^n \subset \br^n$ is a compact manifold. We shall call it a body.
Suppose that $f:M^n \to \br^n$ is a continuous map. Then we define a vector
field $V_f$ on $M^n$ by
$$
V_f(m) = \left(m-f(m)\right)_m.\tag31
$$
The zeroes of $V_f$ are precisely the fixed points of $f$. The local index of a
zero of $V_f$ is precisely the fixed point index for the fixed point. Now
equation (7) and Lemma 3 will combine to give us various formulas analogous to
the Lefschetz equation equating the Lefschetz number $\Lambda_f$ and the sum of
the local fixed point indices.

First we consider the example where $f(M) \subset M$, so that $f$ can be
regarded as a self map. If $f$ has no fixed points on $\partial M$, then we
obtain equation (3), \cite{1,2,3}
$$
\Lambda_f + \Lambda_{-\hv_f,\hn} = \chi(M).
$$
Next we suppose that $f:M^n \to \br^n$ is {\it virtually transverse} on
$\partial M$. That is we suppose the function
$\partial M \to \{-1,+1\}$ given by 
$$
m \mapsto \left((m-f(m)) \bullet N(m)\right)/|(m-f(m)) \bullet N(m)|
$$ 
is well defined and continuous. That is
the same as saying that $m-f(m)$ never points both normally inside and outside
on the same connected component of $\partial M$. If the vector $m-f(m)$ were
never tangent to $\partial M$. this is a map {\it transverse} to $\partial M$
and an example of one virtually transverse to $\partial M$.

If $f$ is virtually transverse to $\partial M$, then
$$
\Ind(\partial_-V_f) = \Sigma \chi(\partial M_i)\tag32
$$
where $\partial M_i$ are components of $\partial M$ so that $V_f$ does not
point normally outside. For even dimensional $M$, we see that
$\Ind(\partial_-V) = 0$. This gives us the following theorem.

\proclaim{Theorem 13} If $f:M^n \to \br^n$ is virtually transverse to $\partial
M$, then $f$ has a fixed point if $\chi(M^n) - \Sigma\chi(\partial M_i) \not=
0$.
\endproclaim
\noindent Hence\qquad\qquad a)\ If $n$ is even, $f$ has a fixed point if
$\chi(M) \not= 0.$\newline
\noindent\phantom{Hence\qquad\qquad }b) If $n$ is odd, $f$ has a fixed point if 
$\chi(M) \not\equiv 0 (\mod 2)$.\newline
\noindent As an example we get Dold's generalization of the Brouwer fixed point 
theorem, \cite{3}. Namely, if $f:B^n \to \br^n$ is a virtually transverse 
map of the unit ball, then $f$ has fixed point. Because $\chi(B^n) = 1$.

\subheading{8.~The Euler--Poincar\'e number and curvature}

Our third example of equation (7) applied to bodies in space relates curvature
of $\partial M$ and the Euler--Poincar\'e numbers of the body $M$ and $M \cap
\Pi^k$ where $\Pi^k$ is a $k$--dimensional hyperplane. Given $\Pi^k$ we
consider the vector field $V$ given by $V(m) = p(m) - m$ where $p:\br^n \to
\Pi^k$ is the orthogonal projection onto $\Pi^k$. Consider $V$ restricted to a
body $M$. To use equation (7) we must calculate $\Ind V$ and $\Ind
\partial_-V$. First we deal with $\Ind \partial_-V$.

Let $m \in \partial M$. We define the {\it convexity number} $C(m,\Pi^k)$ as
follows. Consider the line normal to $\partial M$ at $m$. If this line is not
normal to $\Pi^k$, we let $C(m,\Pi^k) = 0$. If the line is normal to $\Pi^k$,
we observe that the vector field $\partial V$ must have a zero at $m$. Then
$C(m,\Pi^k)$ is defined to be equal to the index of $\partial V$ at $m$. If $m$
is not isolated, we define it to be the index of the connected component of the
zeroes of $\partial V$ containing $m$.

With the above definition of $C(m,\Pi^k)$, our next theorem will be correct in
every situation. However in the non degenerate smooth situations, we can give a
very geometric interpretation of $C(m,\Pi^k)$ which depends on the curvature of
$\partial M$ at $m$. We let $\ell_m$ denote the ray which begins at $m$ and
leaves $\partial M$ in a normal direction. The ray intersects $\Pi^k$
orthogonally at a point, call it $y$. Let $\Pi^\perp$ denote the $(n-k)$
dimensional hyperplane orthogonal to $\Pi^k$ and passing through $y$. Now we
consider the $(n-k-1)$ dimensional surface $\Pi^\perp \cap \partial M$ in the
$(n-k)$ space $\Pi^\perp$. This surface contains $m$ and it should have
$(n-k-1)$ principal curvatures at $m$. For each principal curvature there is a
center of curvature which lies on the line containing $\ell_m$. Suppose that
there are $r$ centers of curvature on $\ell_m$ between $y$ and $\infty$. In
computing $r$, if two centers of curvature coincide, we count them twice. Then
$$
C(m,\Pi^k) = (-1)^r (\sign(m,y))\tag33
$$
where $\sign(m,y)$ denotes the sign of the curvature of $\partial M$ at $m$
calculated on the side facing $y$. For even dimensional $\partial M$ it does
not matter which side of $\partial M$ we take to compute the curvature's sign.
We are taking the curvature to be the product of the curvatures of the $n-1$
principal curves. The following figure shows these concepts for $\br^2$.
$$
\alignedat4
& &\hskip1truein &\ \Pi^1 &\hskip.80truein & &\hskip.75truein &\\
\\
\\
&_{}\phantom{\bullet}\ell_m &\hskip1truein &{}_\bullet y&\hskip.80truein
&{}_\bullet &\hskip.75truein &{}_\bullet m\\
& &\hskip1truein& &\hskip.80truein\text{cent}&\text{er of}
&\hskip.75truein &\\
& &\hskip1truein & &\hskip.80truein\text{curv}&\text{ature} &\hskip.75truein 
&\partial M
\endalignedat
$$
In this case $C(m,\Pi^1) = +1$. If we translate $\Pi^1$ parallel to itself to
the other side of $\partial M$ then we would get $C(m,\Pi^1) = -1$. Now let
$\Pi^\circ = y$. In the picture as drawn $C(m,y) = +1$ since $n = 0$. As we 
move $y$ along the line $\ell_m$ towards the right, the sign $C(m,y)$ changes
after we pass the center of curvature and remains negative even after we pass
through $m$ since the fact that $r$ changes from $1$ to $0$ is compensated by
the fact that the curvature changes sign as we change sides.

Now we can state our result.

\proclaim{Theorem 14} Suppose that $M^n$ is a smooth body and $\Pi^k$ is a
$k$--dimensional hyperplane which is nowhere tangent to $\partial M$. Then
$$
\chi(M^n) = (-1)^{n-k}\chi\left(M^n \cap \Pi^k\right) + \sum_m C(m,\Pi^k)
$$
where the sum is taken over those $m$ for which $\ell_m$ points initially
inward.
\endproclaim

\demo{Proof} We want to apply (7) to $V$. The set of zeroes of $V$ is the
submanifold $M \cap \Pi^k$. Unfortunately, there are zeroes on $\partial M$ if
$\partial M \cap \Pi^k \not= 0$. We will change $V$ slightly to $V'$ such that
$V'$ has no zeroes on $\partial M$ and so that $\partial V = \partial V'$.
Since $\Pi^k$ is nowhere tangent to $\partial M$, there is an $\varepsilon > 0$
so that $V(m) + \frac\varepsilon2 N(m)$ is never zero when $m \in \partial M$
is of distance less than $\varepsilon$ to $\Pi^k$. We choose a small collar
neighborhood $C$ of $\partial M$ so that $C \cap \Pi^k$ is a collar
neighborhood of $\partial M \cap \Pi^k$. We can extend $\frac\varepsilon2 N$ to
a vector field $W$ on $M$ so that $W$ is zero on $M - C$ and $V + W$ is never
zero on $C$. Let $V' = V + W$. Then $V'$ has no zeroes on $\partial M$ and
$\partial V' = \partial V$.

The zeroes of $\partial V'$ coming from $\partial M \cap \Pi^k$ are now in
$\partial_+V'$, hence $\Ind \partial_-V' = \Ind \partial_-V = \Sigma
C(m,\Pi^k)$ where the sum is taken over the inward pointing rays. What is $\Ind
V'$? Consider a small compact neighborhood of $M \cap \Pi^k$ of the form $(M
\cap \Pi^k) \times I^{n-k}$. Now $V'$ restricted to this neighborhood can be
homotopied, without changing the zeroes, to a vector field of othe form $A
\times B$, where $A$ is a vector field on $M \cap \Pi^k$ which points outside
on $\partial M \cap \Pi^k$, and $B$ is the vector field on $I^{n-k}$ given by
$B(m) = - m$. Then $\Ind A = \chi(M \cap \Pi^k)$ and $\Ind B = (-1)^{n-k}$.
Since the index of a product of two vector fields is the product of the index,
we get
$$
\Ind  V' = (-1)^{n-k}\chi(M \cap \Pi^k).\tag34
$$
Combining (34) and (7) gives the theorem.

Note that theorem 14 gives an inductive definition of the Euler--Poincar\'e
number in terms of sections of lower dimensions. As such it extends and
illuminates work of \cite{7}.

There is another theorem similar in spirit. Let $\Pi^{n-2}$ be a hyperplane in
$\br^n$ and let $M^n$ be a body. Let $V$ be the velocity vector field of a
rotation about $\Pi^{n-2}$. Thus in $\br^2,\ V$ is the velocity field rotating
about a point and in $\br^3,\ V$ is the velocity field rotating about an axis.
Where are the zeroes of $\partial V$? They occur at points $m \in \partial M$
where there is a tangent $(n-1)$--plane which contains $\Pi^{n-2}$. We define
$d(m,\Pi^{n-2})$ to be the index of $\partial V$ at $m$. When $m$ is an
isolated zero in the non degenerate smooth case, then
$$
d(m,\Pi^{n-2}) = \sign (m)\tag35
$$
where $\sign(m) = +1$ if the curvature is positive when calculated on the side
of $\partial M$ to which $m$ is moving under the velocity $m$. If the curvature
is negative, then $\sign(m) = -1$ and if the curvature is zero, then $\sign(m)
= 0$.

Now to apply (7) to $V$ we must eliminate the zeroes of $V$ at $\partial M \cap
\Pi^{n-2}$. If $\Pi^{n-2}$ is nowhere tangent to $\partial M$, we can find a
vector field $V'$ as in the proof of theorem 14 so that $\partial V = \partial
V'$ and $\Ind \partial_-V' = \Ind \partial_-V'$. This time the homotopy of $V'$
in a compact neighborhood $M \cap \Pi^{n-2} \times I^2$ results in a vector
field $A \times B$ so that $A$ on $M \cap \Pi^{n-2}$ has index equal to $\chi(M
\cap \Pi^{n-2})$ and $B$ on $I^2$ has index equal to $1$. Thus we get $\Ind V'
= \chi(M^n)$ and hence from (7),
\enddemo

\proclaim{Theorem 15}\qquad\qquad$\chi(M^n) = \chi(M \cap \Pi^{n-2}) + \Sigma
d_i(m)$\newline
\noindent where $\Pi^{n-2}$ is nowhere tangent to $\partial M$ and the sum is 
taken over those points $m$ which are being rotated into $M$.
\endproclaim

As an example of the use of the above theorem we consider the case of a solid 
torus in three dimensional Euclidean space. Suppose that $\Pi^{1}$ is a line 
which intersects the torus in two segments. Then the theorem results in the 
equation $ \Sigma d_i(m) = -2 $. From this we conclude that there are two points
of negative curvature on the boundary of the solid torus whose tangent planes
contain the line $\Pi^{1}$. If we reverse the direction of rotation in the 
theorem we get two other points with the same property. Thus there are four 
points of negative curvature whose tangent planes contain $\Pi^{1}$.
 
Now let us play with theorem 14. Suppose we are in a space ship in space and 
there
is a bounded medium $M$. By shooting out laser beams we can tell which
directions are normal to the boundary $\partial M$ and we can determine the
convexity numbers by using the reflection at the boundary of the laser beam.
Suppose the different amounts of reflection tell us whether or not the normal
beam is passing from inside to outside. What can we tell about $M$?

Answer:~We can determine $\chi(M)$ and whether or not we are inside $M$. The
sum of all the convexity numbers gives the $\chi(\partial M)$ and hence
$\chi(M) = \frac12 \chi(\partial M)$. Then using theorem 14 we subtract the sum
of the inward pointing convexity numbers from $\chi(M)$ to find $\chi(\Pi^\circ
\cap M)$. If it is one, we are inside, if it is zero, we are outside.

We should reflect how lucky we are to live inside an odd dimensional universe.
If we were in an even dimension universe we could not use the sum of the
convexity numbers to determine $\chi(M)$ and hence we would not know
if we were inside or outside $M$. If we add the capability of measuring the
distances to the normal points, then we would know whether we were inside $M$,
since we look at the closest normal point and determine if the beam is exiting
or entering $M$. Then we would know $\chi(\Pi^\circ \cap M)$ and we could
determine $\chi(M)$ in even dimensional universes also.

If there were no reflections of beams going from inside to outside and we did
not know if we were inside or outside $M$, then the sums of the convexity
numbers we could see would give us either $\chi(M)$ or $\chi(M)+1$.

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\enddocument
