**Knot
theory**

**
IV. Knot invariants: Classical theory**

In this lesson, we define some classical knot invariants.

**Section1. Minimum
number of crossing points**

A regular diagram of a knot **K** has at most a finite
number of crossing points. However, this number** c(D)** is NOT a knot
invariant. For example, the trivial knot has two regular diagrams **D** and
**E**, which have a different number of crossing points, fig. 32.

**fig. 32. the number of crossing points of D and E are
respectively, c(D)=0, but c(E)=1**

Consider, instead, all the regular diagrams of a knot **K**, let **c(K)
***be the minimum number of crossing points of ALL regular diagrams*. *
This ***c(K)*** is a knot invariant*. Mathematically speaking, we use
the following equation to stand for the formula for such an expression, where S
is the set of all regular diagrams of K :

**Theorem 4.1** **c(K)** is a
knot invariant.

Note: The above quantity is called the * minimum number of crossing
points *of

Proof: Suppose that **D1** is the minimum regular
diagram of **K1**. Let **K2** be a knot that is equivalent to **K1**,
and suppose that **D2** is its minimum regular diagram. Since we can think of
**D2** as a regular diagram of **K1** (since **K1** and **K2** are
equivalent), from the definition we have that

However, since **D2** is a regular diagram of **K2**, it again follows
from the definition that

Hence, combining these two inequalities, we obtain

That is, **c(D1)** *is the minimum number of crossing points for all
knots equivalent to ***K**. Consequently, it is a knot invariant.

**Exercise 4.1**
Show that for **c(D)=0, 1, 2**, the trivial knot is the only knot that
possesses a regular diagram **D** with one of these values.
( Hint: It is pretty clear for c(D)=0. For the other two values, use elementary
knot moves/Reidemeister moves to get back the trivial knot )

**Exercise 4.2 ** Show that the trefoil
knot, **K**, has **c(K)=3**. Further, show that among all knots, the
trefoil knot is the only one with **c(K) =3. **(For the latter part, again,
suppose you have such a knot with three crossing points, you should be able to
deduce that trefoil knot is the only available choice, by using elementary knot
moves. )

In general, there is no known method to determine **c(K)**. It has only
been computed for some specific types of knots. The difficult hinges on the fact
that when the knot is very complicatedly knotted, then we are forced to deal
with a gigantic number of regular diagrams which are equivalent to the original
one. So the set **S**, in the definition of the minimum regular crossing
points of knot, is just huge and extremely difficult to handle.

**Section2. The
bridge number**

At each crossing point of a regular diagram **D** of a
knot **K** (or link), let us remove from **D** a fairly small segment **
AB** that passes over the crossing point. The result of removing these
segments is a collection of disconnected (without any crossing points) polygonal
curves, see fig. 33. We may think of the original regular diagram, **D**, as
the resulting diagram that occurs when we attach the segments **AB**, that
pass over to the endpoints of these disconnected polygonal curves on the plane.

**fig. 33. The crossed lines in the first and the third
pictures stand for the bridges, which pass over the diagram. To put it in
the other way, we can say that the first and the third pictures are the regular
diagrams of their respective original knots after attaching the crossed segments
(which pass over) to the endpoints of the disconnected polygonal curves in the
second and the fourth pictures, respectively.**

Since these segments **AB** pass over the segments on
the place, these segment **AB** are called ** bridges**. For a given

**Example ** The minimum bridge
number of fig. 34 is 2. It is called the **hopf link**

**fig. 34**

**Exercise 4.3**

**(A)** Show that
**br(K1)=3, br(K2)=2**

**(B) **By manipulations using your hands, show that **
K1** is equivalent to **K2**

**fig. 35**

By exercise 4.3, we know that bridge number is NOT a knot invariant. However,
as in the previous section, if we consider all the regular diagrams for a given
knot, then the ** minimum bridge number** over all these regular
diagrams of a given knot

**Theorem 4.2** ** br(K)** is a knot
invariant.

**Exercise 4.4** By considering
the proof of theorem 4.1, prove theorem 4.2 by similar method.

**Exercise 4.5** Show that if **
br(K)=1**, then **K** is the trivial knot, and the trivial knot is the only
knot with bridge number equal to 1

**Exercise 4.6** Show that if **L**
is a n-component link (that is, consisting of n knots) then

In the specific case if **br(K)=2**, there are many knots with this bridge
number, including the trefoil knot (look at the two representations in fig. 35
again, we knot that the minimum bridge number of the trefoil knot is at most 2.
But by theorem 4.5, we know that the minimum bridge number must be 2) and the
hopf link in fig.34. These knots, called for the obvious reasons *2-bridge
knots*, have been extensively studied, to the point that they have been
completely classified. In general, however, no method has yet been found to
allow us to determine the minimum bridge number for a random knot **K**. The
reason is again, that the set **S** consisting of all regular diagrams of a
knot, is huge when the knot is highly knotted in a complicatedly way.

To end this section, let us state the following theorem, which is proved in
Schubert [Sc]. First, recall the meaning of a connected sum, **K1*K2** of two
knots **K1**, and **K2** from lesson 2.

**Theorem 4.3** Suppose K1, K2 are
two arbitrary, knots (or links). Then we have the following formula:

**Section3. The
unknotting number**

At one of the crossing points of a regular diagram, **D**,
of a knot (or link) **K** exchange locally the over-crossing and
under-crossing segments. Since this type of alteration is NOT an elementary knot
move, in general what we obtain is a regular diagram of some other knots, see
the following example, fig. 36.

**fig. 36. If we exchange the under-crossing and the
over-crossing segments within the small dotted circle in the first picture, the
subsequent regular diagram can readily be seen as the trivial knot.**

The following theorem is stated without a proof. For a proof, see K. Murasugi [Mu].

**Theorem 4.4** We can change a
regular diagram D, of an arbitrary knot (or link) to the regular diagram of the
trivial knot (or link) by exchanging the over-crossing and the under-crossings
segments at several crossing points of D. Notice that it may also be necessary
to use Reidemeister moves.

We define the ** unknotting number of D** as the

**Exercise 4.7**

**(A) ** By manipulations, show that the
two knots in fig. 37 are equivalent.

**(B)** Show that the first
diagram has unknotting number 1, that is, it only requires 1 unknotting
operation to change it into a trivial knot.

**(C)** Show that the second
diagram has unknotting number 2, that is, it requires 2 unknotting operations to
change it into a trivial knot.

**fig. 37**

As in the previous two sections in this lesson, *if we consider all the
regular diagrams for a given knot, then the minimum number of unknotting
operations from all the regular diagrams is a knot invariant. *
Mathematically, we define * minimum unknotting number* as the minimum
of unknotting numbers of all possible regular diagrams of a given knot

**Theorem 4.5** The minimum
knotting number is a knot invariant.

**Exercise 4.8 **By considering
the proof in theorem 4.1, prove theorem 4.5

If we exclude the trivial knot, then for all the other knots **K**, we
have:

However, to actually compute **u(K) **for a given knot is very difficult.
Even for very specific types of knots, there is virtually no method available
(so you can say that the computation of this minimum unknotting number is every
more difficult than the minimum bridge number and the minimum number of crossing
points.). To finish this section, let's try the following exercises on
your own.

**Exercise 4.9 ** Show that the knot in
fig. 38 has minimum unknotting number at most 3.

**fig. 38**

**Exercise 4.10 **Show that the knot in fig. 39 has
minimum unknotting number 1.

**fig. 39**