# Geometry of knots in real projective 3-space

Rama Mishra, Visakh Narayanan

###### Abstract.

This paper discusses some geometric ideas associated with knots in real
projective 3-space $\mathbb{R}P^{3}$. These ideas are borrowed from classical
knot theory. Since Knots in $\mathbb{R}P^{3}$ are classified into three
disjoint classes, - affine, class-$0$ non-affine and class-$1$ knots, it is
natural to wonder in which class a given knot belongs to. In this paper we
attempt to answer this question. We provide a structure theorem for these
knots which helps in describing their behaviour near the projective plane at
infinity. We propose a surgery operation on affine knots to produce several
examples of knots. We later show that this operation can be extended on an
arbitrary knot in $\mathbb{R}P^{3}$. We also define a notion of genus for
knots in $\mathbb{R}P^{3}$ and compare its properties with the Seifert genus
([10]) from classical knot theory. We prove that this genus detects
knottedness in $\mathbb{R}P^{3}$ and gives some criteria for a knot to be
affine and of class-$1$. We produce examples of class-$0$ non-affine knots
with genus $1$. And finally we introduce a notion of companionship of knots in
$\mathbb{R}P^{3}$ and using that we provide a geometric criteria for a knot to
be affine. Thus we highlight that, $\mathbb{R}P^{3}$ admits a knot theory with
a truly different flavour than that of $S^{3}$ or $\mathbb{R}^{3}$.

##  1\. Introduction

A knot, in a smooth manifold $M$ is an embedding $K:S^{1}\to M$ which is
recogonized upto ambient isotopy of $M$ ([6]). As usual, most of the times, we
would not distinguish between the embedding and its image. Classically, knot
theory was studied in $S^{3}$ or $\mathbb{R}^{3}$. Here we study the knot
theory of $\mathbb{R}P^{3}$. The canonical two sheeted covering map,
$\pi:S^{3}\to\mathbb{R}P^{3}$ is an essential tool for understanding the
geometry of $\mathbb{R}P^{3}$. The antipode map $a:S^{3}\to S^{3}$, is the
only non-trivial automorphism of the covering map. The fundamental group of
$\mathbb{R}P^{3}$ is $\frac{\mathbb{Z}}{2\mathbb{Z}}$. Thus, a knot in
$\mathbb{R}P^{3}$ can represent either $\bar{0}$ or $\bar{1}$ in the
fundamental group. We refer to knots in $\mathbb{R}P^{3}$ to be of class $0$
or of class $1$ accordingly. All class-$0$ knots, will lift to a link with two
components in $S^{3}$. The two components are antipodes of each other. The
class-$1$ knots lift to two paths which together form a knot in $S^{3}$. This
knot is the antipode of itself. Of late there has been a lot of interest in
studying knots in $\mathbb{R}P^{3}$ and their lifts in $S^{3}$ ([8]). Also
some of the knot invariants from the classical knot theory have been
generalized. For example, in [2] an analogue of the Jones polynomial for links
in $\mathbb{R}P^{3}$ and in [9] Twisted Alexander polynomial have been
developed. In this paper, we focus more on the geometric properties associated
with these knots.

Topology of $\mathbb{R}P^{3}$ can be understood in many ways. The simplest
description is to think of it is as a compactification of $\mathbb{R}^{3}$ by
adding a projective plane at infinity. Since, the complement of an open
$3$-ball in $\mathbb{R}^{3}$ is homeomorphic to $S^{2}\times[0,1)$, once we
add an $\mathbb{R}P^{2}$ at infinity the complement of a $3$-ball in
$\mathbb{R}P^{3}$ is homeomorphic to the space obtained by compactifying
$S^{2}\times[0,1)$ by adding an $\mathbb{R}P^{2}$, which is nothing but the
mapping cylinder ([3] ) of the two sheeted covering $S^{2}\to\mathbb{R}P^{2}$.
Thus $\mathbb{R}P^{3}$ can also be thought of as, joining a $3$-ball to this
mapping cylinder along their boundary sphere. We will use this decomposition
frequently.

All lines and planes in $\mathbb{R}^{3}$ get compactified as projective lines
and projective planes respectively in $\mathbb{R}P^{3}$. The complement of any
projective plane in $\mathbb{R}P^{3}$ is diffeomorphic to $\mathbb{R}^{3}$.
The rich geometry admitted by $\mathbb{R}P^{3}$, similar to the euclidean
space, can be exploited to undertand the knots in it. For example, a
diagrammatic theory for knots in $\mathbb{R}P^{3}$ ([2] and [4] ) has been
developed as follows:

Consider the disk model of $\mathbb{R}P^{3}$ presented as a quotient of a
$3-ball$, $D^{3}$. By choosing two diametrically opposite points as north and
south poles, we have a projection $\delta:D^{3}\to D^{2}$, onto a central disk
$D^{2}$ defined by circular arcs passing through the chosen poles. Refer to
Figure 1. The image of a tame knot in $\mathbb{R}P^{3}$, embedded in regular
position, under $\delta$, with the over-under crossing information provided at
each double point, is said to be a diagram of the knot. Just like the ball,
the diametrically opposite points of the disk on which the diagram is drawn
should be regarded as being identified. In a diagram of a knot in
$\mathbb{R}P^{3}$ on a disk $D^{2}$, any diameter of the disk is a projective
line in the quotient $\mathbb{R}P^{2}$. Notice that the pull back of such a
line under $\delta$, in $D^{3}$ is a $2-disk$ perpendicular to the $D^{2}$
which is the target of $\delta$. Refer to Figure 3. And in the quotient
$\mathbb{R}P^{3}$ this disk represents a projective plane. Thus projective
lines in a diagram can be pulled back to a projective plane.

Figure 1. The projection map used for making diagrams of knots. Figure 2. An
example of a knot diagram in $\mathbb{R}P^{3}$. Figure 3. Diameters of the
target $D^{2}$ are images of perpendicular disks

One of the first differences that one can observe between classical knot
theory and projective knot theory is, there are two different candidates which
deserve to be regarded as unknot in $\mathbb{R}P^{3}$.

Figure 4. Unknots in $\mathbb{R}P^{3}$

######  Definition 1.1.

A knot $K$ is said to be affine, if it is disjoint from some projective plane
in $\mathbb{R}P^{3}$.

######  Remark 1.1.

  1. (1)

Affine knots in $\mathbb{R}P^{3}$ are thus knots, contained in an open set
homeomorphic to $\mathbb{R}^{3}$. Thus any classical knot, which is a knot in
$\mathbb{R}^{3}$ by definition, has an affine knot in $\mathbb{R}P^{3}$
representing it.

  2. (2)

Since every knot in $\mathbb{R}^{3}$ is contractible, the affine knots in
$\mathbb{R}P^{3}$ are of class-$0$. Thus every class-$1$ knot is non-affine.

  3. (3)

The inverse image of an affine knot is a link in $S^{3}$ with two components.
The inverse image of any projective plane in $\mathbb{R}P^{3}$ is a great
2-sphere in $S^{3}$. Thus for an affine knot $K$, disjoint from a projective
plane $P$, the two components of its inverse image lie in the two hemispheres
of $S^{3}$ separated by the great 2-sphere lying over $P$. Hence, the linking
number of the inverse image of $K$ in $S^{3}$ must be zero.

Now suppose the lift of a class-$0$ knot $K$ is a link in $S^{3}$ with a non-
zero linking number then clearly $K$ is not an affine knot. The knot shown in
Figure 1 is an example where the inverse image has a non-zero linking number.
Thus we have three distinct sets of knots to study in $\mathbb{R}P^{3}$,
affine, class-0 non-affine and class-1 knots.

One simple criterion for detecting affineness from a diagram is the following.

######  Theorem 1.2.

If $K$ has a diagram on a disk which has a diameter disjoint from the diagram,
then $K$ is an affine knot.

Figure 5. Diameter disjoint from the diagram

Proof: The diameter (Figure 5) can clearly be extended to a projective line
disjoint from the diagram. Then this projective line pulls back to a properly
embedded projective plane disjoint from the knot in $\mathbb{R}P^{3}$. ∎

Many of the tools used to study classical knots are not available in
$\mathbb{R}P^{3}$. For instance, one of the techniques for constructing knots
in $S^{3}$ is the notion of connected sum for knots that gives rise to a
notion of primeness of a knot. Recall that the connected sum of knots $K_{1}$
and $K_{2}$ in $S^{3}$ is defined using a surgery procedure as : Think of
$K_{1}$ and $K_{2}$ to be contained in two distinct copies of $S^{3}$. Now
remove an open ball $B_{1}$ from the sphere containing $K_{1}$, which
intersects $K_{1}$ at a single arc with its two boundary points on
$\partial\bar{B_{1}}$. Similarly remove an open ball $B_{2}$ from the sphere
containing $K_{2}$, which intersect $K_{2}$ at a single arc. The complements
of $B_{1}$ and $B_{2}$ are closed 3-balls. By gluing their boundaries
together, with a diffeomorphism which also sends the boundary points of the
remaining arc of $K_{1}$ to the boundary points of the remaining arc of
$K_{2}$, we obtain a copy of $S^{3}$ containing a new knot. Note that the
ambiguity concerning the choice of points in $\partial(K_{1}\setminus B_{1})$
to be identified with specific points in $\partial(K_{2}\setminus B_{2})$ can
be taken care if we start with an orientation on both the knots $K_{1}$ and
$K_{2}$ [1]. This gives a well defined notion of $K_{1}\\#K_{2}$ as a new knot
in $S^{3}$. This procedure can be performed repeatedly as a sum of an
arbitrary number of knots and it is an associative binary operation which
defines a monoid structure on the set of knots in $S^{3}$.

If one introspects on this from a geometric perspective, the connected sum of
two knots in $S^{3}$ is a knot in $S^{3}$ only because $S^{3}\\#S^{3}\approx
S^{3}$. Thus we get a notion of addition of knots in $S^{3}$ as a knot in
$S^{3}$ itself. Now if we directly generalize this notion of connected sum to
the set of knots in $\mathbb{R}P^{3}$, then the connected sum of two knots in
$\mathbb{R}P^{3}$ would be a knot in $\mathbb{R}P^{3}\\#\mathbb{R}P^{3}$.
Furthermore, if we want to repeat this procedure, we would be defining the
connected sum of two knots in $\mathbb{R}P^{3}$ and a knot in
$\mathbb{R}P^{3}\\#\mathbb{R}P^{3}$ as a knot in
$\mathbb{R}P^{3}\\#\mathbb{R}P^{3}\\#\mathbb{R}P^{3}$. Thus this connected sum
defines a monoid structure on the collection of all knots in all closed
connected three manifolds and the set of classical knots, is a submonoid of
this large monoid. This clearly comes with the difficulty of dealing with so
many different three manifolds at the same time. In this paper, we develop a
surgery procedure which is a method of generating new knots in
$\mathbb{R}P^{3}$ from a given knot. In some sense it is described as a notion
of adding a projective line to the knot. This procedure can be repeated
arbitrarily many times and each time we perform it, the homology class of the
knot strictly changes. We refer to it as class changing surgery.

Another property which projective knots don’t have and classical knots have is
the existence of a Seifert surface. It is easy to see that if a knot in any
three manifold $M$, is the boundary of an orientable surface, then the knot
has to be homologically trivial in $M$. Hence for any three manifold which has
a non-trivial first homology, the knots in it cannot have a complete theory of
Seifert surfaces. For example the class-1 knots in $\mathbb{R}P^{3}$ cannot
have Seifert surfaces. Thus one cannot directly generalize the notion of
Seifert genus for projective knots. In this paper we define a notion of genus
for a knot in $\mathbb{R}P^{3}$. The philosophy we adopt here, is that, the
complexity of a knot should also reflect in the complexity of the surfaces on
which it can be embedded. The properties of this genus is very different from
that of the classical Seifert genus. Although there are some similarities too.
For instance this genus detects knottedness similar to the Seifert genus. It
also gives a criteria for a knot to be affine or to be of class-1. We also
construct several examples of class-0 non-affine knots of genus 1. But we do
not completely understand how the genus of this class behaves. And thereby it
reveals that the class-0 non-affine knots is a tough family of knots.

We also generalize the notion of companionship of knots from classical knot
theory to $\mathbb{R}P^{3}$. It is known that the classical unknot is a
companion of every knot in $S^{3}$. As a parallel to this, we prove that the
projective unknot is a companion to every knot in $\mathbb{R}P^{3}$. And
finally we provide a geometric characterization for affineness of a knot which
we call as projective inessentiality.

Organization of the paper: In Section 2, the terms residual tangles and the
class changing surgery have been introduced using which a structure theorem
for projective knots has been proved. In Section 3, the genus for projective
knots is defined and some of its properties are discussed. Section 4 defines
the concept of companionship for knots in $\mathbb{R}P^{3}$ and discusses its
comparison with the existing notion of companionship for classical knots. At
the end of Section 4 the notion of projective inessentiality is defined and it
has been shown that it provides a complete criterion for a given knot in
$\mathbb{R}P^{3}$ to be affine.

##  2\. A structure theorem for knots in $\mathbb{R}P^{3}$

Consider a smooth 2-sphere $S$ which bounds a ball in $\mathbb{R}P^{3}$.
Removing a tubular neighbourhood of $S$ produces two connected components, a
$3-ball$ $B$ and a mapping cylinder, $C$, of the two sheeted covering map from
$S^{2}$ to $\mathbb{R}P^{2}$. Since the boundaries of $B$ and $C$ are
naturally identified to $S$, for the sake of simplicity, most of the times we
will refer to both $\partial B$ and $\partial C$ by $S$. As mentioned above,
if we join two knots in $\mathbb{R}P^{3}$ by removing an unknotted arc
contained in a ball and identifying the boundary points as in the classical
case, then the resulting knot is in $M:=\mathbb{R}P^{3}\\#\mathbb{R}P^{3}$.

The above mentioned process seems to be the only technique to define a
connected sum operation for all knots in $\mathbb{R}P^{3}$. And it comes with
the hassle of dealing with knots in an arbitrary $3$ manifold $M$ and
$\mathbb{R}P^{3}$ at the same time. But for certain special families of knots
in $\mathbb{R}P^{3}$ we can have some operation very similar to the connected
sum. Below is a procedure, that may be regarded as a connected sum of an
affine knot with the projective unknot.

###  2.1. Projectivization of affine knots

Consider an affine knot, $K$, in $\mathbb{R}P^{3}$ which is embedded inside a
ball, $B$ with a boundary sphere, $S$. Then we can push a small arc of $K$ out
of $B$ in such a way that $K$ now intersects $S$ at exactly two points and the
complement $C$ of $B$ intersects $K$ at an unknotted arc disjoint from a
projective planes in $C$. Then clearly this arc is a trivial cycle in
$H_{1}(C,S)$. Now consider a standardly embedded projective unknot, $J$, in
another copy of $\mathbb{R}P^{3}$. Removing a small ball, $B^{\prime}$, with a
boundary $S^{\prime}$, containing a unknotted arc of $J$, produces a copy of
the mapping cylinder, $C^{\prime}$. And $J\cap C^{\prime}$ represents a non-
trivial cycle in $H_{1}(C^{\prime},S^{\prime})$. Refer to Figure 8. Now by
gluing $B$ and $C^{\prime}$ along their boundary spheres in such a way that
$K\cap S$ gets identified with $J\cap S^{\prime}$ produces a knot in
$\mathbb{R}P^{3}$. Refer to Figure 6. Notice that the initial knot $K$ is of
$class-0$ while the new one is a $class-1$ knot. Thus this is a process of
producing a $class-1$ knot from a $class-0$ knot, which we will call
projectivization. A knot in $\mathbb{R}P^{3}$ which intersects a projective
plane exactly at one point can be thought of as produced from an affine knot
via the above surgery. Hence we wish to make the following definition.

Figure 6. Projectivization of an affine trefoil

######  Definition 2.1.

A knot, $K$ in $\mathbb{R}P^{3}$ is called a projectivized affine knot, if
there is a seperating sphere $S$ intersecting $K$ at exactly two points such
that the corresponding $C\cap K$ is isotopic to a standardly embedded non-
trivial cycle in $H_{1}(C,S)$.

###  2.2. Tangles in the mapping cylinder

Any knot in $\mathbb{R}P^{3}$ can intersect the mapping cylinders in a
collection of curves. And if a knot, $K$, is not entirely contained in a
mapping cylinder $C$, then $K\cap C$ is a tangle consisting of only intervals
meeting the boundary sphere at exactly two points. They also can be linked
within $C$. Refer to Figure 7. Thus the problem we are dealing with is the
theory of tangles in the mapping cylinder. This tangle is in no way unique to
the knot or to the separating sphere. In what follows we wish to study the
tangles created by a knot in the mapping cylinders in $\mathbb{R}P^{3}$. If
needed one can indeed push out a small arc from the knot outside the ball and
add an extra component for the tangle. Hence it is in no way unique to the
knot.

Figure 7. A tangle in the mapping cylinder

Notice that $C$ is a 3-manifold with a boundary sphere and an embedded
projective plane. And tangles, induced by knots, are composed of intervals
with there boundary points on the sphere and interiors embedded in the
interior of $C$. The arcs may or may not intersect the projective plane. The
patterns of their intersections are definitely interesting to study. Such
studies can give topological information on the knot. For example the $mod-2$
intersection number of the tangle and the projective plane is the homology
class of the knot in the integral homology of $\mathbb{R}P^{3}$. Thus the
$mod-2$ intersection number of the tangle induced by $K$ in $C$ with the
projective plane is independent of the tangle and depends only on the knot.
Thus the projectivized affine knots, defined above, are knots with a simple
tangle which consists of just one arc intersecting the projective plane at
exactly one point.

Figure 8. Diagram of a class-1 arc in the mapping cylinder

Since the affine knots are homologically trivial, one can see that the process
of projectivization also changes a class-0 knot into a class-1 knot. And
indeed this is a way of defining connected sums of a knot with the projective
unknot which has a clear variant of connected sum with the affine unknot. But
the procedure described above is limited to affine knots. We would, here, like
to define a general version of this process. And thereby obtain a classifying
theme for all knots in $\mathbb{R}P^{3}$.

###  2.3. Residual tangles and space bending surgery

Let $L^{n}$ represent the projective closure of $n$ disjoint non-parallel
lines in $\mathbb{R}^{3}$.

######  Definition 2.2.

The tangle defined by $L^{n}$ in the complement mapping cylinder of some open
ball centered at origin intersecting all lines in $L^{n}$ will be called the
$n^{th}$ residual tangle denoted by $T^{n}$.

Figure 9. Residual tangles

Notice that $T^{n}$ is composed only of $class-1$ arcs in a mapping cylinder
$C^{\prime}$ intersecting a projective plane at exactly one point. Each arc
has two boundary points on $S^{\prime}:=\partial C$. Hence the boundary of
$T^{n}$ is a collection of $2n$ points on $S^{\prime}$. It is easy to see that
for every $n$ there is a unique such $T^{n}$ whose boundary has $2n$ points.
Consider an affine knot $K$ in $\mathbb{R}P^{3}$ with a non-empty intersection
with a disk $B$. Now the boundary of $B$ is a seperating sphere, $S$, and the
complement of $B$ in $\mathbb{R}P^{3}$ is a mapping cylinder $C$. Notice that
$C\cap K$ is a finite collection of intervals with boundary points on $S$.
Thus $B\cap K$ is a tangle with some even number, say $2m$, boundary points on
$S$.
Let $f:S\to S^{\prime}$ be a diffeomorphism sending the boundary of $K\cap B$
to $\partial T^{m}$. Now gluing $B$ and $C^{\prime}$ along $f$ produces a copy
of $\mathbb{R}P^{3}$, with a new knot which is formed by gluing $K\cap B$ and
$T^{m}$ along their boundary. We wish to denote the knot obtained by the above
surgery as $\Sigma(K,S,f)$. This procedure will be refered to as space bending
surgery.

######  Remark 2.1.

The philosophy behind the name is that, an affine knot is basically a
classical knot in $\mathbb{R}^{3}$ with a new identity once we add a
projcetive plane at infinity. Thus, through this surgery, we are connecting
this knot with the projective plane at infinity, and we are making it aware of
the change in the ambient space.

If $K$ is an affine knot and $S$ intersects $K$ at exactly two points then any
$\Sigma(K,S,f)$ is a projectivization of $K$ as defined above. Thus we can
clearly see that projectivization for a given affine knot is not unique. It
depends on the choice of intersection of $K$ with $S$ and the diffeomorphism
$f$. The following theorem says, every knot in $\mathbb{R}P^{3}$ is isotopic
to a knot of the form $\Sigma(K,S,f)$.

######  Theorem 2.2.

(The structure theorem): Any knot in $\mathbb{R}P^{3}$ can be obtained from an
affine knot by a space bending surgery.

Proof: Let $J$ be a knot in $\mathbb{R}P^{3}$. Consider a diagram of $J$ on
some $D^{2}$ as below. Each smoothly embedded circle in the interior of
$D^{2}$ will seperate the diagram into a smaller disk and an annulus (Figure
10). One of the boundary circles of the annulus is the boundary of the initial
$D^{2}$. Since $J$ is a tame knot, assumed to be in regular position, the set
of double points in the diagram will be finite. For each such diagram, we can
choose a separating circle $\sigma$ in $D^{2}$ such that all the crossings
(double points) are contained only in the disk bounded by $\sigma$.

Figure 10.

And the corresponding annulus contains only a finite set of non-intersecting
intervals. There are three kinds of intervals according to the way their
boundaries are placed in the annulus.
Case 1: Suppose all the intervals have their one boundary point on $\sigma$
and the other on $\partial D^{2}$. Then the number of intervals is even and
corresponding to an interval with one boundary point $p\in\partial D^{2}$
there is an interval with one boundary point at the antipodal point to $p$.
Case 2: Suppose both boundary points of an interval in the annulus is on
$\sigma$. Since there are no double points on the annulus, we can push this
curve inside the disk (Figure 11) bounded by $\sigma$ and thereby remove it
from the annulus.

Figure 11. Removing a class-0 arc from the annulus.

Case 3: Suppose both the boundary points of an interval $\tau$ in the annulus
is on $\partial D^{2}$. Since there are no double points on the annulus, one
can enlarge $\tau$ without creating any new double point, inorder to meet
$\sigma$ transversally at exactly two points. Refer to Figure 12. This removes
$\tau$ from the annulus and adds two more intervals in the annulus both
sharing one of their boundary points with $\tau$ and another boundary point on
$\sigma$ and one interval in the disk bounded by $\sigma$ connecting the
boundary points of the new intervals lying on $\sigma$.

Figure 12. Removing a class-0 arc, by creating two class-1 arcs

Clearly corresponding to $\sigma$ there is a unique standardly embedded
2-sphere $S$ in $D^{3}$ whose intersection with the chosen $D^{2}$ is
$\sigma$. By the methods suggested in Case 2 and Case 3 one can see that we
can always isotopically move $J$ in $D^{3}$ so that the diagram we obtain is
always in Case 1. In the splitting induced by $S$, the mapping cylinder $C$
contains a residual tangle isotopic to $T^{n}$ where $n$ is half of the number
of points in the intersection of $J$ with $S$. In this residual tangle, each
$class-1$ arc is represented in the diagram by two intervals with a pair of
antipodally opposite points in their boundary as described in Case 1. Let the
corresponding ball in the seperation be denoted by $B$. Then $J\cap B$ is a
tangle consisting of only intervals whose endpoints lie on $\partial B$. By
connecting the endpoints by $class-0$ arcs in $C$ in such a way that the union
of the new arcs with the tangle in $B$ is connected, we can obtain an affine
knot $K$. By cutting $\mathbb{R}P^{3}$ along a small enough tubular
neighbourhood of $S$ produces $B^{\prime}\approx B$ and $C^{\prime}\approx C$
both whose boundaries are naturally identified by a diffeomorphism $f:\partial
B^{\prime}\to\partial C^{\prime}$ sending $K\cap\partial B^{\prime}$ onto
$J\cap\partial C^{\prime}$. Then it is obvious that $J$ is isotopic to
$\Sigma(K,\partial B^{\prime},f)$.∎
The following diagram shows an example which demonstrates the dependence on
the $f$ in the surgery.

Figure 13. Surgery on affine unknot producing affine unknot Figure 14. Surgery
on affine unknot producing a non-affine knot

The knots on the left are both obtained from the affine unknot by space
bending surgery. The separating spheres are both the same. But by using two
different $f$’s we can obtain two distinct knots. The one above is again an
affine unknot and the one below is a non-affine class-0 knot.

######  Corollary 2.3.

For every knot in $\mathbb{R}P^{3}$, there exist a separating sphere such that
in the corresponding separation, the tangle in the mapping cylinder is a
residual tangle.

######  Definition 2.3.

Let $K$ be a knot in $\mathbb{R}P^{3}$. A separating sphere which induces a
splitting such that the part of $K$ in the mapping cylinder is a residual
tangle, is said to be a residual sphere for $K$.

###  2.4. Class changing surgery

Space bending surgeries are reversible, and can be altered to produce many
knots from a given knot. For example, let $S$ be a residual sphere for a knot
$K$. Then in the induced splitting, $\mathbb{R}P^{3}:=B\cup C$, the tangle in
$C$ is a residual tangle. Now by pushing an arc of $K$ from $B$ to $C$, such
that the boundary points are diametrically opposite in $S$. Now remove this
class-0 arc in $C$ and replace it by a class-1 arc in $C$. Thus we obtain a
new knot (Figure 15). This can be repeated arbitrary number of times. This
procedure is just a small variation to the space bending surgery. Each time it
is performed, it changes the homology class of the knot. Thus we would refer
to this procedure as class changing surgeries. The following picture
demonstrate such a surgery on a non-affine knot.

Figure 15. Surgery on a projective trefoil producing an affine trefoil

Notice that the class-1 knot in the example above is a projcetivized affine
trefoil. We may refer to it as a projective trefoil. It is easy to see that
the structure theorem facilitates the surgery to be performed on arbitrary
knots, rather than on just affine knots. This is because, when we are pushing
out arcs of $K$ out of a ball $B$, if boundary of $B$ is a residual sphere,
then it can be done in such a way that the arc of $K$ or the class-1 arc which
is replacing it, would not be interlinked with the tangle outside.

##  3\. Genus for knots in $\mathbb{R}P^{3}$

Let $S$ be a closed connected surface in $\mathbb{R}P^{3}$ containing a knot
$K$. We know that $\widetilde{S}:=\pi^{-1}(S)$ is a closed surface in $S^{3}$
containing the link, $\pi^{-1}(K=)\widetilde{K}$. By removing any point in the
complement of $\widetilde{S}$, we can obtain an embedding of $\widetilde{S}$
in $\mathbb{R}^{3}$. Every closed surface embedded in $\mathbb{R}^{3}$ is
orientable, and hence, $\widetilde{S}$ is orientable. If $S$ is contained in
some open 3-ball in $\mathbb{R}P^{3}$, that is, it is contained in the affine
part of $\mathbb{R}P^{3}$, then $\widetilde{S}$ is disconnected.

######  Definition 3.1.

A closed surface $S$ containing $K$ in $\mathbb{R}P^{3}$ is said to be a good
surface for $K$ if the inverse image $\widetilde{S}:=\pi^{-1}(S)$ is
connected.

Notice that, if $S$ is a good surface for $K$, it may or may not be
orientable. As an example, consider the affine unknot, $K_{0}$. Clearly, a
projective plane $P$ and a torus $T$ intersecting $P$ at only $K_{0}$, are
both good surfaces for $K_{0}$. But $T$ is orientable, while $P$ is non-
orientable.

######  Theorem 3.1.

Every knot in $\mathbb{R}P^{3}$ has a good surface.

Proof:  Let $K$ be a knot in $\mathbb{R}P^{3}$. Consider a diagram of $K$ on
some projective plane $P$. If $K$ is an unknot, (affine or projective) there
is a diagram of $K$ with no crossing. The inverse image of $P$ is a great
$2-sphere$ in $S^{3}$ and thus $P$ itself is a good surface for $K$. Suppose
$K$ is knotted. Then for each crossing of $K$ we attach a handle on $P$
containing the crossing as shown in Figure 16. Thus, we will obtain a surface,
$S$ containing $K$. To show that this is a good surface for $K$, we note that
the inverse image $\widetilde{S}$ is a surface obtained by attaching handles
on the great $2-sphere$, $\pi^{-1}(P)$. Here for each handle attached on $P$,
there would be two handles on $\pi^{-1}(P)$ covering it. Thus $\widetilde{S}$
is connected and we are done. ∎

Figure 16. A good surface containing projective trefoil

######  Remark 3.2.

The good surfaces obtained by the above construction are all non-orientable
since they are obtained by attaching handles on a projective plane. However,
there may be orientable good surfaces of $K$ depending on $K$.

######  Definition 3.2.

The minimum seifert genus of an inverse image surface varied over all the good
surfaces for $K$ will be called the genus of $K$.

######  Theorem 3.3.

Both the affine and projective unknots have genus $0.$

Proof:  For the affine unknot $K_{0}$, consider a disk containing it in
$\mathbb{R}P^{3}$. Being affine, we can think of this as a flat disk in
$\mathbb{R}^{3}$. Then there exist a flat plane in $\mathbb{R}^{3}$ containing
it. The closure of this plane in $\mathbb{R}P^{3}$ would be a projective plane
$P$ containing the knot. The inverse image of $P$ is a great 2-sphere
containing $\widetilde{K_{0}}$. Thus it has a good surface, $P$, whose inverse
image has seifert genus $0.$ Thus the affine unknot has genus 0.
Let $K$ be the projective unknot. Then it is represented by a projective line
in $\mathbb{R}P^{3}$. Any projective line in $\mathbb{R}P^{3}$ is contained in
a projective plane, $P$. And as above, $P$ is a good surface for $K$ and the
genus of $K$ is also $0$. Hence we are done. ∎

######  Theorem 3.4.

A knot in $\mathbb{R}P^{3}$ has genus 0, only if it is an unknot.

Proof: Case 1: Let $K$ be a class-$0$ knot. Then $\widetilde{K}$ is a link
with two components. If $K$ has genus $0$, then there exists a 2-sphere $S$,
containing $\widetilde{K}$ in $S^{3}$. Then both the components of
$\widetilde{K}$ are circles on a sphere, thus they both bound disks in $S$ and
these disks are antipodes of each other, since $S$ is the antipode of itself.
Thus the push-down of any one of these disk, is a disk, $D$, in
$\mathbb{R}P^{3}$ bounded by $K$. Choose a small enough circle, $K^{\prime}$
in the interior of $D$ disjoint from some projective plane. Then there is an
isotopy shrinking the boundary of $D$, that is $K$, to $K^{\prime}$. Thus $K$
is isotopic to the affine unknot.
Case 2:  Let $K$ be a class-1 knot. Then $\widetilde{K}$ is a knot in $S^{3}$.
Let $S$ be a sphere containing $\widetilde{K}$. Being contained in a sphere,
$\widetilde{K}$ bounds a disk and hence is unknotted. $S$ is the inverse image
of surface $G$ in $\mathbb{R}P^{3}$. Being the inverse image under a covering
map, $S$ is closed under the deck transformations of $\pi$. Thus $S$ is a
2-sphere in $S^{3}$ closed under the antipode map of $S^{3}$. And similarly
$\widetilde{K}$ is a circle in $S^{3}$ closed under the antipode map. Thus $G$
is obtained by identifying antipodally opposite points of a sphere, and thus
is a projective plane and $K$ is a projective line in $G$.
Hence we are done. ∎

######  Corollary 3.5.

A knot in $\mathbb{R}P^{3}$ is an unknot if and only if it has genus 0.

######  Theorem 3.6.

All non-trivial class-$1$ knots have genus 1.

Proof:  Let $K$ be a class-$1$ knot in $\mathbb{R}P^{3}$. Let $U$ be a tubular
nbd of $K$. The boundary of $\bar{U}$ is a torus. By choosing an arbitrary
non-vanishing vector field, $X$ on $K$ and pushing $K$ into
$T:=\partial\bar{U}$ along $X$ we can obtain a parallel of $K$ isotopic to $K$
inside $T$. Now the inverse image of $U$ is a tubualar nbd of $\widetilde{K}$
and hence is a solid torus in $S^{3}$. The boundary of this tubular nbd is a
torus containing the inverse image of the parallel of $K$ contained in $T$.
Thus $T$ is a good surface for $K$ (or its parallel). And since $K$ is non-
trivial, the genus of $K$ is $1.$ Hence we are done. ∎

Now we will study the properties of genus for affine knots. Notice that any
affine knot, has two lifts in $S^{3}$ which are unlinked. Any of these lifts,
being a knot in $S^{3}$ has a Seifert genus. We would like to compare, for an
affine knot, the Seifert genus of a lift in $S^{3}$ and the genus the knot in
$\mathbb{R}P^{3}$.
Consider an orientable surface, $S$, contained in an affine region of
$\mathbb{R}P^{3}$. Let $P$ be a projective plane disjoint from it. By removing
a small disk from both $S$ and $P$, and connecting them by a tube, we can
always obtain a surface in $\mathbb{R}P^{3}$ which is homeomorphic to the
connected sum, $P\\#S$. The inverse image of this surface would be a great
sphere (the inverse image of $P$) attached with two copies of $S$ with tubes,
one inside and one outside. This is a connected surface. Thus for any knot $K$
contained in a surface $S$ in an affine region, the connected sum of $S$ with
a projective plane is always a good surface for $K$.

######  Lemma 3.7.

Every non-trivial affine knot has genus strictly greater than $1$.

Proof: Let $K$ be a non-trivial affine knot. If $K$ had genus 0, then it is an
unknot by Corollary 4.4, hence cannot be a non-trivial knot. Thus $K$ cannot
be of genus 0. Now suppose $K$ had genus $1$. Then it has a good surface $G$,
whose inverse image is a torus in $S^{3}$, which is a two sheeted cover for
$G$. Then, $G$ is Klein’s bottle or a torus. Since the Klein’s bottle cannot
be embedded in $\mathbb{R}P^{3}$ [5], $G$ is a torus.
We know that, since $K$ is an affine knot, there exist a projective plane,
$P$, disjoint from it. We may assume the intersection of $P$ with $G$ is
transversal and thus is a collection of circles. If any of these circles are
contractible in $G$, then such intersections may be removed by modifying $P$
after cutting along such circles. Thus we may assume all circles in $P\cap G$
are essential. Hence $G\setminus P$ is a collection of cylinders. Since $K$ is
a knot, it is contained in one cylinder, say $C$, in $G\setminus P$. Being
disjoint from a $P$, $C$ is contained in an affine part of $\mathbb{R}P^{3}$.
We may consider $C$ to be a closed cylinder with two boundary circles, by
removing a open tubular neighbourhood of $P$ from $G$. Since $K$ has no self
intersections, $K$ with some orientation is a generator for
$H_{1}(C,\mathbb{Z})$. But we know that, another representative for a
generator of $H_{1}(C,\mathbb{Z})$ is a boundary circle of $C$, which is an
affine unknot, $\alpha$. Then, $K$ is homologous to $\alpha$ in $C$ and
together they bound a cylinder inside $C$. By shrinking this cylinder we can
construct an isotopy of $K$ to $\alpha$. Thus $K$ is isotopic to the affine
unknot. Which is a contradiction. Thus the genus of $K$ cannot be $1$, and
hence it has to be strictly greater than $1$. ∎

######  Theorem 3.8.

The genus of any non-trivial affine knot in $\mathbb{R}P^{3}$ is $2.$

Proof:  Let $K$ be an affine knot, contained in the affine part of
$\mathbb{R}P^{3}$. Consider a torus (possibly knotted) containing it in the
affine part and take a connected sum with a projective plane (Figure 17). Thus
we can get a good surface for $K$ in $\mathbb{R}P^{3}$. Then the genus of the
inverse image of this good surface is $2$. That is the genus of any affine
knot, is less than or equal to $2$. Then by Lemma 3.6, the genus of $K$ is
exactly $2.$ ∎

Figure 17. A minimal good surface for affine trefoil

We see that the genus of an affine knot behaves differently from the Seifert
genus of its lift. And that for most affine knots, the genus takes much
smaller values than the Seifert genus of its lift. For example, if $T(p,q)$ is
a torus knot, then its Seifert genus, is $(p-1)(q-1)/2$. The theorem above,
shows that the actual value of the genus is not dependent on $p$ and $q$ and
for almost all affine torus knots in $\mathbb{R}P^{3}$, the genus is much
smaller than the Seifert genus of its lift.

######  Corollary 3.9.

All knots of genus $1$ are non-affine.

Just like knots, any subset of $\mathbb{R}P^{3}$ has a notion of being affine
or non-affine.

######  Definition 3.3.

We will call a torus $T^{2}$ in $\mathbb{R}P^{3}$ to be class-$1$ if it is the
boundary of a tubular nbd of a class-$1$knot.

Clealy the inverse image of such a torus under $\pi$ is a connected torus in
$S^{3}$. Thus for a knot embedded on a class-$1$ torus, it is a good surface.
Since every knot of genus $0$ is an unknot, a non-trivial knot embedded on a
class-$1$ torus has genus $1.$

Figure 18. A class-0 non-affine knot of genus 1 on a class-1 torus

The homology of $T^{2}$ is
$H_{1}(T^{2},\mathbb{Z})\approx\mathbb{Z}\oplus\mathbb{Z}$. Any knot on the
torus $T^{2}$ has a homology class of the form $(p,q)$ where $p$ and $q$ are
coprime. A $(p,q)$ knot on a Heegaard torus as above, will intersect the
projective plane at exactly $p$ points. Thus it is of class-0 exactly when $p$
is even. Since all these knots have genus 1, they must be non-affine. Thus
there are infinitely many class-$0$ non-affine knots with genus $1.$

Figure 19. A class-$0$ non-affine knot on a knotted class-$1$ torus

The knot shown in Figure 19, is non-affine, since the inverse image in $S^{3}$
is a two component link with linking number 2 between its components. Clearly
this knot, is boundary of a band in $\mathbb{R}P^{3}$. This band can be
thickened to be a class-$1$ torus on which the knot represents the $(2,5)$
circle. Thus it has genus $1.$ The core of the solid torus bounded by this
torus is class-$1$ knot intersecting the projective plane at infinity at 3
points.

######  Conjecture 3.10.

There exists class-$0$ non-affine knots, with genus not equal to $1.$

As a support for the conjecture, in Figure 20, we present a diagram of a knot,
which we suspect to be of genus $3.$ The surface on whichithe knot is lying is
good and it is covered by a genus-$3$ surface in $S^{3}$.

Figure 20. A non-affine knot on a good surface

##  4\. Companionship and Projective inessentiality

The following is a definition of companionship for knots in $\mathbb{R}P^{3}$
directly generalized from that of $S^{3}$. A subset of a solid torus is said
to be geometrically essential, if every meridinal disk intersects it.

######  Definition 4.1.

A knot $K_{1}$ in $\mathbb{R}P^{3}$ is called a companion for a knot $K_{2}$
if there exists a tubular neighbourhood, $V$ of $K_{1}$ containing $K_{2}$
such that $K_{2}$ is geometrically essential in $V$.

######  Theorem 4.1.

The projective unknot is a companion of every knot $K$, in $\mathbb{R}P^{3}$.

Proof: Consider a diagram of $K$ on a disk $D$. Choose a separating circle
$\sigma$ which separates the diagram into the diagram of a residual tangle in
the annulus $A$ and the diagram of the complementary tangle in the disk
$D\setminus A$.
Case 1: Suppose $K$ is a non-affine knot. Then it has a non-empty residual
tangle in $A$. Now, as shown in Figure 21, choose a pair of half disks
intersecting the boundary circle of $D$.

Figure 21. The green shaded region is a 2-disk in the quotient
$\mathbb{R}P^{2}$

The two half disks represents a $2$-disk in the quotient projective plane,
which is the target of the projection. The inverse image of this disk together
with the north pole, the source of projection, added is diffeomorphic to a
solid torus pinched at a point. Refer to Figure 22.

Figure 22. The pull back of a disk under the projection representing a solid
torus pinched at a point

It can be smoothed out to be an unknotted solid torus, $V_{2}$, in
$\mathbb{R}P^{3}$ disjoint from $K$. The complement,
$V_{1}:=\mathbb{R}P^{3}\setminus{\kern 0.0ptV_{2}}^{\mathrm{o}}$, is a solid
torus containing $K$. Then,$\mathbb{R}P^{3}=V_{1}\bigcup V_{2}$ is a Dehn
surgery description of $\mathbb{R}P^{3}$ as obtained from $S^{3}$ by a $(2,1)$
surgery on the unknot. Then any non-singular loop in $V_{1}$ representing a
generator of the $\pi_{1}(V_{1})$ is a projective line in $\mathbb{R}P^{3}$.
Now since two 1-dimensional submanifolds in $V_{1}$ can never intersect
transversally, we can always choose such a projective line $L$, disjoint from
$K$ such that $V_{1}$ is a tubular neighbourhood for $L$. Now any meridinal
disk in $V_{1}$ shares its boundary with a Mobius strip in $V_{2}$ and
together they form a projective plane. Thus if $K$ was disjoint from a
meridinal disk then it is disjoint from a projective plane. This contradicts
the initial assumption that $K$ is not affine. Thus $L$ is a companion of $K$.
Case 2: Suppose $K$ is an affine knot. Let $\sigma$ be a separating circle as
before. Then the diagram of the residual tangle inside $A$ may be empty. In
which case, by performing a standard move of pushing an arc of $K$ out of the
projective plane representing $\partial D$, we may create a non-empty residual
tangle for $K$. Thus we can always choose a diagram of $K$ with a non-empty
residual tangle, $T^{n}$ in $A$. Let $E:=D\setminus{\kern
0.0ptA}^{\mathrm{o}}$ be the disk cut out by $\sigma$. The part of the diagram
inside $E$, is a diagram of a tangle inside a $3-ball$, which is a collection
of paths in $E$ connecting a point of $\partial T^{n}$ to another. We will
refer to this part of the diagram as $K\setminus T^{n}$. Let $\delta$ be a
semicircular arc as shown in red in Figure 23. Each of these arcs are images
of projective planes. The part of these arcs outside the green half disks are
the projections of meridinal disks of $V$. Since $K$ is contained in $V_{1}$
and the images of every meridinal disk intersect the image of $K$, it is
geometrically essential in $V_{1}$.

Figure 23. Geometric essentiality of $K$ in $V_{1}$

Notice that the inverse image of any diameter of $\partial D$ with the north
pole added is a projective plane. Now consider two half-disks as above, whose
centers are placed at the boundary points of $\delta$. The image of any
meridinal disk will be a path connecting a point in the boundary of one half
disk to a point in the boundary of the other half disk. It is easy to see that
such a path has to intersect either $\alpha$ or the arcs of the residual
tangle which extends $\alpha$. Thus $K$ intersects every meridinal disk of
$V_{1}$. And the core of $V_{1}$ is a projective line $L$, companion to $K$. ∎

######  Definition 4.2.

A Mobius strip, $M$, neatly embedded in a solid torus $V$, such that $\partial
M$ represents the $(2,1)$ class in $H_{1}(\partial V,\mathbb{Z})$ will be
called an untwisted Mobius strip.

Notice that $\mathbb{R}P^{3}$ is diffeomorphic with $L(2,1)$, a lens space. A
tubular neighbourhood of a projective unknot, is a solid torus whose
complement is also a solid torus. Thus the boundary of a tubular nbd of a
projective unknot induces a Heegaard splitting for $\mathbb{R}P^{3}$. The
meridian of a Heegaard torus is glued to a $(2,1)$ curve on the boundary of
the other one in order to construct $L(2,1)$. The Heegaard surface for this
splitting is a torus. The characteristic curve on the Heegaard surface, bounds
a 2-disk in one solid torus and a Mobius strip on the other. They together
glue to form an $\mathbb{R}P^{2}$ in $L(2,1)$.

######  Definition 4.3.

Let $K$ be a knot in $\mathbb{R}P^{3}$. $K$ will be called projectively
inessential if it is contained in a solid torus $V$ from a Heegaard splitting
such that there exists an untwisted Mobius strip in $V$ disjoint to $K$.

######  Theorem 4.2.

A knot in $\mathbb{R}P^{3}$ is affine if and only if it is projectively
inessential.

Proof: (If:)Let $K$ be a projectively inessential knot. By definition, there
exist a solid torus $V$ containing $K$ whose complement
$U:=\mathbb{R}P^{3}\setminus{\kern 0.0ptV}^{\mathrm{o}}$ is also solid torus.
$\mathbb{R}P^{3}=V\cup U$ is a Heegaard splitting. And there is an untwisted
Mobius strip $M$ in $V$ disjoint from $K$. The circle $\partial M$ is a
characteristic curve for the splitting and hence is a meridian for $U$ which
bounds a disk $D$ in $U$. Then $D\cup M$ is a projective plane and it is
disjoint from $K$. Thus $K$ is an affine knot.
(Only if:) Suppose $K$ is an affine knot in $\mathbb{R}P^{3}$. Choose a
projective plane $P$ disjoint from $K$. There exists a ball model for
$\mathbb{R}P^{3}$ where $P$ is the closure of an equitorial disk, $D$. Choose
a disk $D^{\prime}$ perpendicular to $D$ and consider a diagram of $K$ on the
closure projective plane, $P^{\prime}$ of $D^{\prime}$. The image of $P$ under
the projection is a projective line $L$, disjoint from the image of $K$.
Choose a residual sphere for $K$ which will intersect $P^{\prime}$ on a
circle. Choose a 2-disk on $P^{\prime}$ disjoint from the diagram of $K$ and
$L$. Refer to Figure A above. Then the inverse image of this disk can be
modified in to a solid torus $V$, whose complement solid torus $U$ contains
both $K$ and $L$. Since $K$ is tame, $V$ can be chosen in a way that, it is
disjoint from $K$ and $\partial V$ is a transversal to $P$. Clearly $V\cap P$
is a 2-disk, hence $U\cap P$ is an untwisted Mobius strip. Thus $K$ is
projectively inessential. And we are done. ∎

This theorem provides a purely geometric criterion for a knot to be affine. We
also wish here to mention a theorem announced by J.Viro and O. Viro in [4]

######  Theorem 4.3 (J.Viro-O.Viro).

A knot $K$ is affine if and only if $\pi_{1}(\mathbb{R}P^{3}\setminus K)$ has
an element of order 2.

This is a purely algebraic criterion for a knot to be affine. Thus we have
many techniques to detect whether a given knot is affine.

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