Skip to main content

L&S Cat Theory: Attempted solutions to "Conceptual Mathematics" by Lawvere and Schanuel

Ryan Ruff

Section 3.7 Session 7: Isomorphisms and coordinates

It turns out that last week’s lack of exercises wasn’t as much of an anomaly as I thought it was. After realizing that this week’s session didn’t have any exercises either, I scanned ahead and noticed that there are several sessions like this before eventually reaching a "Quiz" and "Test". I’m not sure how I plan to handle those yet, but I think I’m going to keep with my chill pace for now. This session did include several side questions throughout that I can treat as I would the exercises.

Example 3.7.1. Aside 1.

What, approximately, is \(coordinate(x)\) for the point \(x\) in the picture?
Solution.
I took a modern approach and used technology to provide a more accurate estimation.
  1. I began by holding the page as flat as I could while taking a photo with my phone and uploaded to the cloud.
  2. I downloaded this photo from my PC and opened it in GIMP
     1 
    gimp.org
    .
  3. I used the cursor to identify the following coordinates in pixels: \(r = plot(-4.3) = (897,2904)\text{,}\) \(x = (1370,2926)\text{,}\) \(p = plot(0) = (1689,2937)\text{,}\) \(q = plot(3.5) = (2453,2970)\text{.}\)
  4. Next I calculated the distances from \(r\) to \(q\) and from \(p\) to \(x\text{.}\) Specifically, \(||q-r|| \approx 1557.4\) and \(||p-x|| \approx 319.19\text{.}\)
  5. With this information in hand, I can set up a ratio to determine \(x\text{.}\) Solving \(\frac{3.5-(-4.3)}{0-x} \approx -\frac{1557.4}{319.19}\) results in a solution of \(x \approx -1.5\text{.}\)
This seems like a reasonable estimate based on the diagram.

Example 3.7.2. Aside 2.

What are the choice of origin, positive direction, and unit of ’distance’ involved in specifying this isomorphism?...
Solution.
Oooh this is a fun one! Let’s presume we meet some alien species and are trying to sync up our timelines. We’re going to need both space and time coordinates if we really want to do this right. One possibility is for both parties to use "the big bang" as a point of origin and "the light cone" for a direction and magnitude. This choice makes converting easy, but potentially imprecise because of the sheer size of those values.
A much more practical approach would be to use the meeting time and location as a point of origin, use physical constants to communicate SI units, then work backwards to calculate the local time for each party using a Lorentz transformation. We’d just need to where they came from and how fast they traveled to convert their home civilization’s time to their current time.
I think the point that the authors were trying to communicate was that our choice of units and magnitude for "time" are very specific to our origin on planet Earth. If we’re talking about units of "years" as being "one orbit around the Sun" and the origin as a socially defined "common era", then an alien race from another plant would likely have totally different system of dating then us. However, most SI units are based on physical properties of our universe. Even if this species chose a different set of base units than us, we could match up measurements of certain elements to develop an isomorphism between our two systems.

Example 3.7.3. Aside 3.

...rational explanation for the curious bracketing...
Solution.
The first thing that jumps out at me about the bracket is that the sums of the adjacent ranks are constant:
Figure 3.7.4. 8 Player Bracket from L&S Session 7
In other words, the best player is alway paired with the worst player, the next best with the next worst, and so on until all pairs have been assigned. However, the fun doesn’t stop there.
The bracket also appears to balance "skill gap" between the top half and the bottom half. The winner of the match-up with the highest rank differential goes on to face the winner of the match with the lowest differential in the next round. Those differentials sum to a constant much like the ranks did.
Figure 3.7.5. 8 player bracket with rank differentials
Between those two conditions, that should be enough to determine the 4 and 16 player brackets. The four player one is easy, since it’s essentially embeeded in the original.
Figure 3.7.6. 4 player bracket
To make the 16 player bracket, I’m going to start by pairing off players based on rank: The player in rank 1 should be paired with the player in rank 16, rank 2 with rank 15, rank 3 with rank 14, and so on until all 8 pairs have been assigned. Since these pairings go from largest score differential to smallest, we can assign the pairs to each of the "ranks" in the 8 player bracket to extend it further.
Figure 3.7.7. 16 player bracket
This works out nicely because if the top 8 players all win their match-ups in the first round of the 16-player bracket, they’d wind up in a bracket that is basically identical to their placement in the 8-player bracket. Likewise, if the top four players continue to win then we end up in a situation identical to the 4-player bracket. It has a nice recursively defined structure.
PrevTopNext