Interactive Online Tutoring Services

December 13, 2022

The Knottedness and Chirality of the Trefoil

Filed under: knots,Mathematics,the knottedness and chirality of the trefoil — Rob burchett @ 3:15 am

Outline

Here is an outline in rough:

1. Start with an oriented trefoil in S^3.

2. Define two new spaces. A negative space and a mixed space.

3. Let the trefoil be in the mixed space.

4. If we wish, we can move the trefoil into positive space from the mixed space or move it into negative space from the mixed space. To move from positive to negative or from negative to positive we need to go through a transition space. (mixed space)

5. Make moving points (labels) a, b type to represent where the trefoil crosses over and under itself. These exist in mixed space and negative space but not in positive space. These moving points move as the diagram is moved, so that they are always above or below in a crossing. They can come out of the crossing and combine to form joinings (a(r) or (b(s) due to elimination via moves 1 and 2 type.

6. New labels s and r only exist in negative space and are counterparts to a and b. s and t may switch as the knot passes through itself as this is allowed in negative space.

7.Suppose I have a diagram with the same shadow as the usual representation of the trefoil. Then this is either an unknotted case with a trefoil shadow or a positive or negatively oriented trefoil.

8. Start with a positively oriented trefoil .Label the crossings in the transition space. Move the trefoil to the special format where I remove a, b and form ordered joining pairs (a(r) and (b(s) type.

9. After this, the trefoil may make any 1,2,3 moves recorded in r,s crossings. Eventually returning to have a trefoil shadow. These moves can all be shown to be mirrored as if we were in positive space. Then this represents the full knot.

10. At any time I can undo the diagram to see that the circle with (a(r), and (b(s) joining pairs is an invariant in the negative space.

11. I need to be able to reform a and b crossings by recombining the (a(r) and (b(s) joinings to get back into transition space before I can move to positive space. This must be either possible or not. The restriction on what type of crossings I can reform is given by the (a(r), (b(s) sequence. I can rotate the sequence to match with a copy of itself or overturn it. Since I am in the new space there are also other moves I can make.

12. Since I can’t form the unknot, the trefoil is knotted, since I can’t form the negative orientation, the trefoil is chiral.

December 10, 2022

Math Concept Sharing

Filed under: Concept Sharing,math concept sharing — Rob burchett @ 5:25 am

An interesting notion from considering something that happens in nature, the overlapping of two shadows from two different light sources.

For example, a cup of green tea could be placed on a table and lit from above from two light sources, one from each side. Two shadows are cast. The shadows overlap in the middle. See diagram below.

On the table where the two overlapping shadows are we can form this notion, this is because both shadows have zero height, but we have two together (there are two different shadows, a greater darkness can be observed).

So borrowing from this notion, since a point has position only, or the notion of no extent, with some origin, it might be possible to put two points together but not in coincidence.  We would just be unaware of two being there. While we can see that two shadows are present in the teacup diagram, when we are dealing with only points, the points would have to be different somehow from the usual idea of a point and also different from each other.

Concept Sharing of a Number:

We can start with the concept sharing of a natural number:

Consider a race where 2 people come in 2nd place. We could each give them the number 2 as to say they both came in second place. Or we could do something different, thinking about the teacup shadows, could there be 2, 2’s? Could we eliminate the number 2 and replace it with 2,2’’s. The 2’’s represent the second position, why can’t there be more than 1 number 2?

To explain further: Since a number is an exact concept it should be possible to have a mathematics where two different numbers, occupying the same position or being in coincidence (Two or more in a sense of combined two or more, rather than a separate two or more). Since numbers are exact these two ‘other numbers’ would simply be hidden from us. We could specify from the outset how many we would like there to be in a specific mathematics. Usually then we only have one level.

Since a number is an exact position it should be possible to somehow have 2 number 2’’s, if only we could replace the number 2. It’s possible if together the 2, 2’’s don’t combine to form one 2’(so 2’ is different from 2). Also the 2’’s have to be different in some other way from each other. There must be then a deeper concept of “number of numbers”. Now we have two or more numbers brought into existence because of the necessity of being able to remove the number 2 (since we could have 2, 2”s as long as the 2’s are somehow different).

Also think about the world of duplicates. Most of what we are surrounded by isn’t unique. There are many copies of things which are necessary for ease of manufacture and for people to function. Each of these copies can be considered to have numbers associated with them, such as dimensions, if there are printed words with them, the letters are all representing positions on an alphabet…etc.

So then we can give the same duplicate numbers to each copy of an item.

So here’s what we can do differently: instead of considering the racers or the duplicates as having the same number, concept share the number itself. So for the racers for example we can have two number 2s. 2’(1) and 2’(2), and give one concept shared number to each racer or duplicate.

If I have 2 objects (first level 2) I can give each object a symbol (2’(1)and 2’(2)). Each object is part of the total. We also have (2’(1)(2’(2)). Indicating that the sum doesn’t resolve into 2’(concept sharing 2’- together they do not equal 2). We have a number of numbers, in this case, 2 (first level 2). Also we require 2’(1) and 2’(2) to be different in some other way given by their definition in the level of the number of numbers.

The notation (() introduced to indicate objects placed together. At the far right I have an object in complete parenthesis, indicating an object we can concept share and to the left I am revealing what is hidden with the object in the complete parenthesis.(as if I expanded the shared concept to the left)

Concept sharing is different from coincidence in that the original concept is removed, given the necessity of a concept of concept space, and replaced by two or more of the original concepts which are different in that they do not combine to form the original concept. They are different in some other way.

The concept of two or more is granted by the addition of another number level, a number of numbers level.

There would have to exist another level of a number of numbers. Normally this would be 1, but we can look at other cases where this could equal 2, 3, etc.

Rather than the number of numbers being 1, the number of numbers is now 2. For this to work, we have to be able to remove the number 2 which is already present and replace it with 2 number 2’s. These other number 2s we can notate 2’. 

The number 2 indicates the 2nd position. Having 2 of them there doesn’t change what’s already laid down. 2’(1) and 2’(2) are both in the second position. There can be two there and we could be unaware of it (due to the exactness of a number).

2 is an exact position , as such 2, 2s could exist there under the right circumstances. (the creation of a level of number of numbers and another difference between the two 2’’s).

This does occur in nature too. If one thinks of DNA. The two strands separate and then they rebuild to form two copies after eliminating the original.

The case of two points:

Then returning to the case of two(2) (referring to the second 2, 2  2’s concept sharing a second level of 2)points. The two(2) points would be different, in the following way: Since we are to have two points together, we must remove the situation where the two points form a single point (coincidence). So it must be possible to take this point out. Meaning that for a plane there is another level of place, a place of places. (think of taking apart a jigsaw puzzle).

These are coexisting with the places, but are not the same. Then the two points can be different. I can move one in this plane, place (2), while keeping the other one still. The difference can be that one point is fixed, while the other is mobile. 

For the motion of the other point, I need another type of displacement. Since I am moving away from zero distance and this is not positive displacement, let it be negative. Then if I take away some negative distance I get closer to zero. So I am going below zero. Let there be a distance below zero to the left and to the right. At the middle there is some bottom place.

Then the numbers associated with the points can be 1’(1) and 1’(2), 2’(1) and 2’(2)… as has been described.  When the two are together, they are indistinguishable, but one is still fixed and the other one is mobile.

Knots:

Working on knots I started with the concept of two points existing together, not forming a single point but coexisting as two points “concept sharing” a “deeper” concept of place. A place of places. Could there be another form of distance where I might take places out into?

It is not obvious to our senses but follows logically from the notion of a point having no extent. As Descartes has shown us, we cannot rely on our senses or even on our imagination. It is hard to imagine two points existing together, but logically, it follows.

So this started with the concept of a point and then this point sharing this extended concept at another level of concept. That is, we allow concepts to have deeper dimensions. Two points together are not usually considered but this can be if we have another level of place and some difference between the points. The difference can be that we keep one point fixed and allow the other to move. When the two are together, they are indistinguishable. ex. (1’(1’).

That is a more basic level of place is needed so that two usual level points can share this new point. There could be any natural number of points at one place of place (place(2)) but we can limit it to a certain geometry, or say that we can arbitrarily have as many points(2) here as we require for some finite geometry.

Also, since there are only two solutions to the distance equation (positive and negative) it might be that there are only two levels here. That is two places of places, place(2)’s, could share at another level, place(3). But I don’t know how to separate them as there are no further solutions to the distance formula.

Since two points can exist together and still be two points, (under the right circumstances) we need to remove the point that is already there and this leads to the necessity and existence of the more basic level. 

The new level isn’t actually lower or higher than the usual level, just different. Places and place(2)’s coexist. Similar to taking a jigsaw puzzle apart, places can have places(2)(places of places).

Then this underlying geometry was shown to lead to a better understanding of knots and space itself.

In knots we can place knots in the new extended space. Introduce labels (necessary if we want to freeze the flowing points(2) in one configuration), and create a concept sharing knot from a closed non-intersecting curve.

We can let one part of the knot move out into negative space and keep the other still. The one that is still moving can move to create new negative labels as well as move along concept shared positive labels.

 We can create flow, using moving points(2). Points(2) can pass through other points(2). Two knots can be compared(we take “half” of one and compare it to “half” of the other to seek a congruence) in this way if we want to see if they are equivalent

Reidemeister moves can be made once the knot is moving in negative space and the labels so made will be different from that which has already been labeled- Since they are created in negative space. These new labels can be combined with the labels made in positive space.

Then upon reforming the knot in positive space the negative labels must be eliminated and we end up with a different diagram using the same positive labels. All this would be normally hidden. 

The labels are rearranged. It is possible to change the writhe of the diagram. One thing we can do is start with a minimal diagram and find all other minimal diagrams.

Space:

Space itself can now be looked at differently as having a negative as well as a positive distance associated with it. I can move one point away from its shared counterpart in place(2) space. if I create negative distance co-existing with positive distance. These can exist together, and we can choose to either obey one or the other, as there can be two solutions to the Cartesian distance formula. The positive distance has the same notion but now has two levels.

This can be so for any point in space and distance between two points can be positive as well as negative. This shows the nature of space to be different from what was previously thought. However, this is not inconsistent or alien from Cartesian space as I am still basing it on the notion of a point having no extent, with an origin of the fixed places(2) space. It could lead to a simpler approach to questions such as the Poincare conjecture, since in space without “holes” everywhere, I am simply connected.

So I have an underlying place of places (fixed) and a set of fixed places and a set of moving places.

The Physical World:

Since “point” is such a basic concept in geometry and topology it makes sense that an improvement here can also lead to better understanding of the physical world as nature and mathematics are related through the use of idealizing the very small as point-like.

The Creation of a Plane of Numbers:


We can build another set of numbers and combine this with the new plane which was previously created. In this way create a number plane rather than a number line.

Start with place(2) space, free of any places, and an arbitrary number of points(2)(these are the points which can move) placed at some origin. We can have two number axes perpendicular to each other at this origin.

Now we can build a new number line by moving two points(2) out to some unit length of 1 unit. Do this for both axes. So that in each direction we have a distance of 1 unit marked, now also I have 1 point(2) at 1 and mark it with 1’. This is a location (0,0) moved from the origin, so represents some negative distance. 1’ is identical to 1, except that if I want to I can move these numbers in the place(2)a axis as they are associated with points(2) I can have operations like addition and multiplication on the 1’, 2’, 3’…etc. as well mirroring the usual operations on ordinary numbers. 

In the first quadrant we can make a number (1’(1’) by moving a part at 1’ on one axis and the other part at 1’ on the other axis to combine at the place(2)  (1,1). But actually move 1’(1) and1’(2) and give it  the number(2), (1′(1′). To do this we have to move two parts out to each axis, but this is okay because we can have an arbitrary number of parts there.

Now we can make (2′(2′) by the same process. Move two parts from the origin out 2 units and move one part up 2. units from the x axis and one part from the y axis to come together at the number (2′(2′).

We can go on making a midline of new numbers (1′(1′), (2′(2′), (3′(3′),….

Right now we are only interested in the numbers in this number plane that makes sense to us. That is, the numbers on the midline. 1’ and 3’ could also share but there would be no higher number here at the first level.

Similar to negative numbers, these could have some meaning, but I won’t explore this here.

We can have operations on these numbers such as addition (1′(1′) +(1′(1′)= (2′(2′) = 2’(1′(1′)- if we combine the new number system with the system of numbers point(2) numbers-and multiplication (1′(1′)* (1′(1′)= (1′(1′) or (1′(1′) *(2′(2′) = (2′(2′)= 2’*(1′(1′) as well.

But here is something interesting, since we are combining the two point(2) number systems (1′(1′) =sqrt(2’), if I move the lower axis to be congruent to the midline. Now we have to make this space different from that of which we are already aware. Let any distance be negative as has been described. Here we do not have a positive distance.

Then since (1′(1′)^n=(1′(1′) then sqrt(2’) =sqrt(2’)*sqrt(2’) =sqrt(2’)*sqrt(2’)*sqrt(2’)… (this is for the moving sqrt(2’) not the fixed sqrt(2)-not seen in this geometry). And we have an equivalence of these numbers. This is also seen intuitively as I am moving parts out to the different place(2)’s and these parts are equivalent for a certain set of numbers in this space.

Fermat’s Last Theorem:

Now I want to take a look at the general statement of Fermat’s Last Theorem, seeing if I can find a way to show its truth. It is said that it is unlikely that Fermat had a proof but my ideas could have been known at his time( since in a sense they are basic), so maybe with this new understanding he worked out what I did.

Many people have tried to prove the theorem in the past with elementary methods that would have been available at Fermat’s time. But here I am introducing something new and powerful which didn’t exist before, but might have been known and then subsequently lost.

Now consider that I might have three numbers a, b,c with c related to a and b. I am thinking specifically of a-b=c or a^2-b^2=c^2 or a^3-b^3=c^3…..

Let these be extended to the second form of numbers so that I have a’^n(1′(1′), b’^n(1′(1′), c’^n(1′(1′)=a’^n(1’(1’)-b’^n(1′(1′), n is some natural number (first level natural) all on the midline.

Now look at a’(1′(1′), b’(1′(1′) and c’(1′(1′). C’ might or might not be a natural number so we call it (c’^n)^(1/n) in order that I might operate with c’^n a natural number, being a’^n-b’^n the difference of two natural numbers.

(c’^3)(1/3)(1’(1’) is between b’(1’(1’) and a’(1’(1’). A’ is the largest number. If we focus for a while on the cubic case.

We can ask if a and b are lattice points(2). Then is (c’^3)(1/3) a lattice point(2)? I can move the midline so it is congruent with the horizontal axis.

Since (1′(1′)^n=(1′(1′) and that means sqrt(2’)=sqrt(2’)*sqrt(2’)…(this is for the mobile sqrt(2’) and not the still one-not seen in this geometry)

Does (c’^3)(1/3)*sqrt(2’)^n=m’*sqrt(2’) for m’ a natural number ? (n is an ordinary natural number) Since these are the only equivalent parts in the space.

Then geometrically we can move and expand the line b’*sqrt(2’)——–(c’^3)^(1/3)*sqrt(2’)———a’*sqrt(2’) by factors of sqrt(2’) and look to see if the point(2) (c’^3)^(1/3)*sqrt(2’)^n falls on any lattice point(2), m’*sqrt(2’). A’ and b’ and n can vary.

So an equation we can come up with is m’=(c’^3)^(1/3)*sqrt(2’)^(n-1).

But m’ (c’^3) and 2’ are not points but points(2) which can move. Let (c’^3) and 2’ all vary on the places(2) axis. These are fixed at their values in place(1) so the equation still holds. Places(2) numbers taking on the exponents from the points(2), in this way taking over from the points(2).

We can ask if there exists place(2)’s, p, on the axis which I can take all points(2) to and the equation still holds? Move all points(2) to a single place(2) which could be any place(2) thus creating the generality of the equation holding in place(2).  For then the equation will hold in place(2) as well as place(1).

The equation has to hold here for both points(2) and places(2) like it does originally for both points(2) and places(2) otherwise I can’t form a map back and forth in the extended space. This shows the allowable values of n, a natural number through the exponent of p, a general place(2).

Then let c’^3 and 2’ and m’ move together on the horizontal place(2) axis to m so that at some place(2),p: p’ = (p’^)(⅓)*(p’)^(½)(n-1).  I can reverse the motion of the points(2), 2, c^3 and m’ and the equation would work for these values of n. At the beginning where I have the points(2) similar to being fixed, the equation works for both points(2) and places(2). Where I bring all the points(2) together at the same place(2) this must also be so since this is a one-to-one correspondence, it will only work for these values of n.

Looking at the exponents then, we can ask is there any n for which 1=(1/n)+(½)(n-1)? In the cubic case, i.e. let n=3. For which the answer is no. For the squared case 1=(½)+(½)*1 and the linear case 1=1+0. Any case past cubic is also not possible. So for the cases which work, I can work backwards to the equation I started with.

So for these numbers (1′(1′), (2′(2′), (3′(3′),… Fermat’s Last theorem seems to hold. If we project these to the number axis on either the left or below we will get the same result for regular numbers as all numbers(2) have their counterparts in places(2).

Further work:

One may also consider concept sharing as it applies to the concept of a set or a group. It may be that concept sharing is a unifying idea in math.

Additionally, concept sharing works mathematically because such concepts are exact. But could concepts in other fields be made exact so concept sharing could work beyond mathematics?

Contact us today to get started