5
$\begingroup$

I would greatly appreciate if I could get some help in clarifying my understanding. (This is a special topic I am studying as a 2nd year University student - I haven't taken topology yet - so please keep this in mind if answering the question, I probably won't understand the answer if it requires a lot of background in topology)

I'm reading The Knot Book by Colin C. Adams, and from my understanding of the introductory chapters, when two knots are 'composed' together - they can form different knots depending on where the arc is removed, but we must consider the orientation of the knots.

There are two cases:

(1) If the orientations of both knots match up - then no matter where we remove the arc to make the composition, we could slide one of the knots to any other projection of the same knot. Which seems to make intuitive sense.

(2) If the orientations of the knots clash, then composing the two knots from different arc openings can yield different knots.

(Q1) For the 2nd case, Is this because the orientations clash and you can't 'slide' the knots around?

(Q2) The book does say that if at least one of the knots are invertible - then it can always be deformed to reverse the orientation so that it matches the other knot. But when would this occur? Surely it would be before the two knots are composed. So would that mean - reversing the orientation yields a distinct knot, and not reversing the orientation potentially yields another distinct knot?

Would that also mean if we disregard orientation - that all compositions of two knots would yield a single distinct knot?

Many thanks in advance.

Extra Clarification Added:

Just to further clarify, my confusion lies in the order of the operations:

Say I have two distinct knots, I can compose it on one side, or I can say, flip one of the knots upside down and compose it from the other side.

The fundamental question is this: I want to know whether I can turn one of these projections to the other.

The book says that if one of the knots is invertible then yes one projection can be made into the other. Does that mean the knot can be made to go in the other direction after it is connected (from the other side)?

Because if it requires it to be inverted before it is connected, then isn't that not answering my question? (I'm asking if doing a particular thing (flipping a knot and putting it on the other side) will yield the same result as the first thing, and I am being told if you do something else (inverting a knot) it will give the same result - but that's not my question?)

Does that make sense? Or have I got the idea entirely wrong?

$\endgroup$
  • 4
    $\begingroup$ Regarding Q1, yes. Q2 also yes. Regarding your final (unnumbered) question, no. If $K_1$ and $K_2$ are both non-invertible, and if $\iota K_2$ denotes the inverse/reverse of $K_2$, then $K_1 \# K_2$ and $K_1 \# \iota K_2$ are not isotopic knots. $\endgroup$ – Ryan Budney Mar 13 '14 at 7:38
  • $\begingroup$ So is the inverse/reverse of a Knot considered to be the same as the original? $\endgroup$ – JackReacher Mar 13 '14 at 7:53
  • 2
    $\begingroup$ @mathstudent No. The knot $9_32$ is chiral, which means it is not the same as its reverse or its mirror image (and therefore the reverse of the mirror image also). Many knots are the same as their reverse though. $\endgroup$ – N. Owad Mar 13 '14 at 18:44
  • $\begingroup$ Oops. I meant $9_{32}$. $\endgroup$ – N. Owad Mar 14 '14 at 15:57
  • 1
    $\begingroup$ @mathstudent If a knot is equivalent to its reverse, you can just say that you switch the orientation, because being equivalent to its reverse means there is an ambient isotopy which between them. There is a physical way to visual it, but that is usually not as important as just knowing it is possible. $\endgroup$ – N. Owad Mar 17 '14 at 17:06
1
$\begingroup$

This question has been answered in comments.

To try and summarize:

Take two oriented knots $K$ and $K'$. We wish to form a connected sum of $K$ and $K'$. One option we have is to join the two so that the orientations match. Call this result $K_1$. There is an obvious choice of orientation on $K_1$, given by the orientations on $K$ and $K'$. The knot $K_1$ does not depend on where we join the two knots, as by shrinking one enough we can slide it around the other to any other location.

If we were in a strange mood, we might instead choose the other orientation. Let's call this $K_2$. Thus $K_2$ is the inverse of $K_1$.

Our other option is to join $K$ and $K'$ so that the orientations disagree. Here again we have two choices of orientation on the result (one agreeing with $K$, and one agreeing with $K'$); there is no obvious preferred choice though. Let's call the former $K_3$ and the latter $K_4$. Again, $K_4$ is the inverse of $K_3$. Note that, as before, we can change the point of joining $K$ and $K'$ by sliding one around the other. However, this does not in general enable us to move between $K_3$ and $K_4$.

Now we can ask which, if any, of $K_1$, $K_2$, $K_3$, $K_4$ are the same knot, given properties of $K$ and $K'$.

Suppose, for example, $K$ is invertible, so we have an isotopy taking it back to itself with orientation reversed. Form $K_1$, say, with $K'$ very small. If we make $K'$ small enough, we can do `the same' isotopy of $K$ without $K'$ interfering. So we get a new isotopy, this time from one connected sum to another. The question is, which connected sums did we show are isotopic? The first connected sum we had was $K$ and $K'$ summed so their orientations agree, and taking the natural orientation. The second connected sum is the inverse of $K$ summed with $K'$ (because we didn't do anything to $K'$ during the isotopy). Moreover, the isotopy has not changed the way the orientations go, so this is again the connected sum with the orientations agreeing and the natural orientation chosen.

Like this it's possible to work through all the combinations to see under what circumstances two connected sums could be the same. Showing that two are the same can essentially be done intuitively, provided you take enough care. I think that whenever intuitively two should not be the same they are indeed not the same, but actually proving that is more difficult, and requires at least a bit of knowledge of topology.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.