But what, exactly, is a blue moon? We think back to my New Year's Eve calendar post. Clearly "blue moon" has something to do with the moon. In that post, I mentioned how the solar year cannot be divided evenly into lunar months. And recall that a "lunar month" is the period of time from one new moon to the next -- or equivalently, one full moon to the next, since blue moons have something to do with full moons.
In particular, the solar year is not exactly twelve lunar months. We see what happens in the Islamic Calendar, where the years consist of exactly 12 months -- each year ends up being slightly shorter than a solar year. In order to obtain solar years, some of the years must have 13 full moons -- which means that one month must contain two moons. And so we define the blue moon to be the second full moon during the calendar month.
We notice that a lunar month is about 29 1/2 days. This means that, if a month were to contain two full moons, the first one would occur on the first or second day of the month, and the second would occur on the 30th or 31st day of the month. And this is exactly what happens in July 2015. A full moon occurred on July 1st, and the second full moon, the blue moon, occurs tonight -- officially after midnight in most time zones, so the blue moon appears on July 31st on calendars.
How often do blue moons occur? Recall that I mentioned the ancient Greek astronomer Meton, who discovered that we must add seven months to a lunar calendar every 19 years in order to obtain a lunisolar calendar. These extra seven months can be interpreted as the blue moons, so we conclude that a blue moon occurs once every 19/7 years -- about once every three years. (A better estimate is that it occurs once every e years.) So this is what "once in a blue moon" actually means. Notice that blue moons occur somewhat more often than leap years in the Gregorian calendar.
There is another definition of blue moon that sometimes appears -- this one is based on dividing the year into seasons rather than months. A solar year, with its four seasons, normally contains 12 full moons, so that divides naturally into three full moons per season. But if there are 13 full moons in the year, one season would have to contain four full moons. Historically, it is the third of these four full moons that would be the blue moon.
Notice that since the seasons don't line up with the months, a blue moon using the monthly definition can never be a blue moon using the seasonal definition. In particular, tonight's full moon is the second of three full moons during Summer 2015, so it can't be a seasonal blue moon.
When is the next seasonal blue moon? One way to determine this is to look at future Easter dates, since Easter is based on the full moon after the spring equinox (a seasonal marker). We notice that Easter falls on March 27th, 2016, and then on April 16th, 2017. Whenever there is an early Easter followed by a late Easter, we know that the late Easter must occur 13 full moons after the early Easter, and so a seasonal blue moon must occur between the two Easters.
If we consult a 2016 calendar, we see that both a full moon and the summer solstice occur on June 20th, 2016, with the full moon about 12 hours before the solstice. This means that the June 20th full moon is actually the fourth full moon of spring, and so the third full moon of spring -- May 21st, 2016 -- must be the seasonal blue moon.
We notice that the seasonal blue moon occurs about one year after the monthly blue moon. There is a reason for this -- we start by noting that the reason for a monthly blue moon in July 2015 is the fact that the first full moon occurs on the first of the month. If we count 12 lunar months (i.e., one Islamic year, 354 days) from July 1st, 2015, we obtain June 20th, 2016 -- the summer solstice -- and it's the full moon on that day that causes May 21st, 2016 to be a seasonal blue moon. In other words, seasonal blue moons generally occur about a year after monthly blue moons due to the coincidence that the difference between 12 lunar months and one solar year (i.e., 11 or 12 days) is nearly equal to the difference between any equinox/solstice and the first day of the following month (in this case, the difference between June 20th and July 1st).
Here's one final difference between monthly and seasonal full moons -- there can never be a monthly full moon in February, since that month is shorter than a lunar month (even if February contains a Leap Day, it is still about 12 hours shorter than a lunar month). On the other hand, February is one of only four possible months for a seasonal blue moon, since seasonal blue moons must occur about one month before an equinox or solstice, and February is one month before the spring equinox (just as May 21st, 2016 is about one month before the summer solstice).
Meanwhile, I finally finished Eugenia Cheng's How to Bake Pi. In the last two chapters, Cheng explains that categories have initial and terminal objects. For example, in the category that I mentioned in my last post:
Parallelogram --> Rectangle
Rhombus --> Square
the object "parallelogram" is an initial object, while "square" is a terminal object.
One of Cheng's favorite examples is clock arithmetic. We know that even though 11 + 2 = 13, on the clock, two hours after 11:00 is 1:00 -- that is, 11 + 2 = 1 on the 12-hour clock. Here is a YouTube video from Square One TV that describes clock arithmetic in more detail:
As Cheng points out, there can be any number of hours on the clock, not just 12. For each natural number n, there is a group with n elements, called the cyclic group of integers modulo n. I mentioned earlier on the blog that we can think of the chromatic scale in music as being isomorphic to the group of integers mod 12 (Z/12Z) -- but we could also invent a scale isomorphic to the integers mod 19 (that is, Z/19Z) instead.
Cheng writes that in category theory, if two groups are objects, an example of an arrow between them can be a function that "treats the group operation sensibly." Since I took group theory at UCLA, I know exactly what word she's trying to avoid using here -- homomorphism.
An example of a homomorphism is from Z/24Z to Z/12Z -- that is, it maps the 24-hour clock to the 12-hour clock in the obvious way. We perform this homomorphism every time we try to convert from military time to civilian time. Another homomorphism maps Z/12Z to Z/60Z -- and we perform this homomorphism every time we convert the number the minute hand is pointing at to actual minutes. I point out that this is a homomorphism because it respects the group operation of addition -- we can either add first (1 + 2 = 3) and convert this to minutes (minute hand on 3 = 15 minutes), or we can convert to minutes first (5 minutes and 10 minutes) and then add (5 + 10 = 15 minutes), and we obtain the same answer.
I can see why Cheng would want to avoid using the word homomorphism -- that word can be confused easily with homeomorphism, which is a map between two topologically equivalent figures. Notice that both homomorphisms and homeomorphisms are morphisms -- the arrows in category theory.
Cheng's book was certainly a delight to read, and I highly recommend it to anyone who wants a brief introduction to category theory -- with some dessert recipes thrown in as a bonus!
Of course, I can't say that I'm an expert in category theory yet. I often had to reread some of what Cheng wrote in order for me to understand it -- and I feel that I'm at an advantage only because I've taken a class in group theory. How much more difficult, then, would it be for, say, one of you who hasn't taken group theory to understand Cheng's book? How much even more difficult would it be for one of our students to understand her book?
This is one reason that I like to read math books that challenge me, as Cheng's book did. Just as I find category theory a challenge, my students find High School Geometry a challenge. It's too easy for me to think back to my own days as a student and finding most of Geometry easy (though I admit that even I found some of final chapters, such as trig and especially circles, a bit difficult). But category theory -- now that's something that I find as difficult as my students find Geometry. I can understand better why my students are having trouble with geometrical similarity when I compare it to my own struggle to understand category theory.
In fact, sometimes I wonder whether I, if I ever get my own classroom, should teach myself some difficult subject on the side, such as category theory. Or it doesn't even have to be math -- it could be something like a foreign language. As a Californian, I have many Spanish speakers in my classes, so if I try to understand what they are saying in Spanish, I am doing something that's as hard for me as Geometry is for them. In that way, I can sympathize with my students better.
This is another thing that I've learned about myself in the year that I've started this blog. I've been told that one of my strengths is my ability to teach English learners. When I speak, I repeat myself -- often it's because I can't think of anything to say. Ironically, this can be a strength when I am in a classroom of English learners, because they need that repetition. This is why many of the worksheets that I post here on the blog begin with a vocabulary section.
On the other hand, my repetition is a weakness when I am in a class of English natives and stronger math students. I wonder whether my problem is that as a student growing up, I've been way ahead of most of my fellow students in math, so I subconsciously think of most of my students as having low levels of math and gear my instruction towards them.
This also explains why I can't fully agree with the traditionalists who want to push everyone into Calculus classes and set up all math classes from eighth grade on to prepare them for Calculus. I know that the types of students that I teach would get D's, if not F's, in every single math class from eighth grade on if they were forced through the Calculus door.
Today is a spherical geometry day -- and here we go again! Look at how long this post is, and I've yet to say a single word about spherical geometry.
In the news today there is a story about the remains of a plane found on Reunion Island. I thought this was interesting because Reunion Island may be the closest human inhabited island to the antipodes of my California home. Recall that almost the entire continental United States -- including all of California -- has its antipodes in the Indian Ocean. If I decided that I wanted to go on a vacation and travel as far away from California as I possibly can, Reunion Island would be a great choice.
Here is a link to Reunion Island on the Degree Confluence page:
Notice that much of this page is in French, as Reunion is a French possession. The author writes that there is no actual confluence on the island -- all seven confluences are in the ocean, but all are close enough to the island for it to be visible on the horizon (which is a requirement).
The coordinates of this confluence is 21S, 55E. The coordinates of my actual antipodal point would be 34S, 62E, which is nearly a thousand miles away. The website also states that there is another nearby island, Mauritius, that is visible from the confluence 21S, 57E. This may actually be closer to my antipodal point than Reunion, since 57E is closer to 62E. If I wanted to go on my near-antipodean vacation, Mauritius may be a better choice than Reunion, especially since Mauritius is a former British colony (now it's independent), and so its inhabitants are more likely to speak English. (What did I just say in this post about learning languages again? Even though I took French in high school, I doubt that I know it enough to have a conversation with someone on Reunion Island, and so I can avoid the foreign language by going to Mauritius instead.)
Reunion and Mauritius have about a million inhabitants each. It is now summer in California, and so it is currently winter in Reunion and Mauritius. Both islands would normally be in a time zone 12 hours ahead of mine, But in reality, both islands are only 11 hours ahead of California because we observe Daylight Saving Time here. (Recall that these islands aren't true antipodes but are actually much closer to the Equator than California. Tropical areas almost never observe DST because the length of the day doesn't change enough to justify DST.)
Speaking of French, I know that these spherical geometry posts are supposed to be all about the French mathematician Legendre and his Elements of Geometry. But I often like to discuss the geometry of the earth in my spherical geometry posts -- including things like degrees, angles, and especially antipodes as such antipodal points are critical in spherical geometry. But now let's get back to Legendre's book. For those who don't remember the link to the Legendre text, here it is again:
We are currently on pages 165-166 of Legendre. Notice that the first thing that appears on page 165 is a demonstration (proof) of the last proposition on page 164, and so this has already been covered. As it turns out, there are only four propositions left on these two pages, Propositions 477 through 480 --but these are some of the most important theorems of spherical geometry. So we definitely want to spend some extra time on these.
Legendre's Proposition 477 is a "scholium" (follow-up) to his 476:
477. Scholium. It may be remarked that, beside the triangle DEF, three others may be formed by the intersection of the three arcs DE, EF, DF. But the proposition only applies to the central triangle, which is distinguished from the three others by this, that the two angles A, D are situated on the same side of BC, the two B, E on the same side of AC, and the two C, F, on the same side of AB.
Different names shall be given to the triangles ABC, DEF; we shall call them polar triangles.
So Legendre tells us that in his 476 and 477, he is defining something called "polar triangles." In particular, given any triangle ABC, there exists a unique triangle DEF such that ABC and DEF are polar triangles.
In 476, Legendre tells us how to find, given triangle ABC, a triangle DEF that is polar to ABC. The name "polar" implies poles -- recall that every circle (including great circles, since the sides of a spherical triangle are arcs of great circles) has exactly two poles. We then let D be the pole of BC, E the pole of AC, and F the pole of AB.
This may be easier to visualize with a specific triangle, Let A and C both lie on the Equator, and point B lie somewhere due north of C. We can easily find the location of point E -- this point must be the pole of AC, and since AC is the Equator, E must be the North Pole. Of course, every line actually has two poles, so how do we know that E is the North Pole and not the South Pole? This is what the "scholium" in 477 tells us -- B and E must lie on the same side of AC (the Equator), and since B is north of C, E must be the North Pole.
Now let's find point D. We see that D is the pole of BC, and B lies north of C. This means that BC is a meridian. We've mentioned that the pole of any meridian lies on the Equator, with longitude exactly 90 degrees away from that meridian. So we know that D lies somewhere on the Equator.
Point F is unfortunately difficult to locate, since it's the pole of AB and this line is neither the Equator nor a meridian. But we already know a few things about triangle DEF. We notice that ABC might be a very small triangle compared to the size of the globe -- A and C could be just a few miles apart on the equator, and B just a few miles north of C. But its polar triangle DEF must be quite large, because one of its vertices D lies on the Equator and another vertex E lies at the North Pole! So ABC and DEF are in general not congruent triangles -- nor did Legendre intend them to be congruent. Indeed, making ABC smaller results in making DEF larger, not smaller.
Now suppose point B isn't merely any point north of C, but it is in fact the North Pole itself. This is a very special case, but it's the only case where it's easy to calculate the poles of all three sides (since we only know how to find the poles of the Equator and the meridians). In particular, we let C be the intersection of the Equator and the Prime Meridian. We want A to lie on the Equator as well, so why don't we place it on my home meridian -- 118W. Then this makes the angle at B measure 118 degrees, while the angles at A and C are both right angles. (That's right -- the angles of the triangle clearly don't add up to 180 degrees, but we're still waiting for Proposition 489 to tell us what the sum of the angles acually is.)
Now we see that E is still at the North Pole, since it's a pole of the Equator AC. (Notice that the points B and E must be on the same side of the Equator AC, but they are almost never actually the same point except in this special case -- which is why I saved it for last.) Point D is the pole of the Prime Meridian BC, so it must lie on the Equator at longitude 90 degrees -- west, since it must lie on the same side of the Prime Meridian as point A. Point F is the pole of the meridian 118W, so it must lie on the Equator 90 degrees away from 118W in the direction of the Prime Meridian, so it must lie at longitude 28W.
Far from being congruent to ABC, DEF actually lies completely inside ABC. Now Legendre's 476 tells us that the angles of DEF are not necessarily congruent to the corresponding angles of ABC -- instead they are supplementary. Indeed, this is the case as angles D and F are still right angles, while the measure of angle E at the North Pole is now 62 degrees as it lies between 90W and 28W -- and 62 degrees is indeed supplementary to the original 118-degree angle.
Notice that the polar triangle of the polar triangle is the original triangle. We see that the polar triangle of DEF still has a vertex at the North Pole, and the other two vertices are on the Equator -- the side 90W moves 90 degrees east back to the Prime Meridian and the 28W moves 90 degrees west back to the 118W meridian.
Despite this, notice that polar triangles are not a transformation in the ordinary sense since we transform entire triangles, not individual points. Indeed, the location of point B tells us nothing about where point E is -- except that B and E must be in the same hemisphere defined by AC. We saw that point E would have been at the North Pole no matter where B was as long as it was in the Northern Hemisphere -- and if B were in the Southern Hemisphere, E would have been at the South Pole, all because A and C are on the Equator.
Let's move on to Legendre's Propositions 478 and 479. These two go together, since 479 is simply a "scholium" of 478:
478. The triangle ABC being given, if, from the pole A and with the distance AC, an arc of a small circle DEC be described, if, also, from the pole B, and with the distance BC, the arc DFC be described, and from the point D where the arcs DEC, DFC cut each other, the arcs of the great circles AD, DB be drawn; we say that, of the triangle ADB thus formed, the parts will be equal to those of the triangle ACB.
Demonstration. For, by the construction, the side AD = AC, DB = BC, and AB is common; therefore the two triangles will have the sides equal, each to each. We say, moreover, that the angles opposite to the equal sides are equal.
Indeed, if the center of the sphere be supposed in O, we can suppose a solid angle formed by the three plane angles AOB, AOC, BOC, we can suppose, likewise, a second solid angle formed by the three plane angles AOB, AOD, BOD. And since the sides of the triangle ABC are equal to those of the triangle ADB, it follows that the plane angles, which form one of the solid angles, are equal to the plane angles, are equal to the plane angles which form the other solid angle, each to each. But the planes of any two angles in the one solid have the same inclination to each other as the planes of the homologous angles in the other (359); consequently, the angles of the spherical triangle DAB are equal to those of the triangle CAB, namely DAB = BAC, DBA = CBA, and ADB = ACB; therefore the sides and angles of the triangle ADB are equal to the sides and angles of the triangle ACB.
479. Scholium. The equality of these triangles does not depend on an absolute equality, or equality by superposition, for it would be impossible to make them coincide by applying the one to the other, at least except they should happen to be isosceles. The equality, then, under consideration, is that which we have already called equality by symmetry, and, for this reason, we shall call the triangles ACB, ADB, symmetrical triangles.
Okay, Legendre is doing several things with these two propositions. Just as in 476, here we are starting with a triangle ABC and forming a second triangle. But unlike 476, the new triangle formed in 478 actually is congruent to ABC.
Notice that Legendre refers to poles of small circles, but we can easily imagine simply placing a large compass with the tip at A and the pencil C, drawing an arc on the other side of AB, and then placing the tip at B and the pencil at D, drawing another arc on the other side of AB. Then the point at which the two compass arcs intersect is the new point D.
In this proof, we have AD = AC, DB = BC, and AB = AB, so it would appear that we already have the two triangles ABC and ADC congruent by SSS. The problem is that we haven't actually proved that SSS works for spherical triangles yet! So we must show that the angles are congruent as well. We see this occur in traditionalist Euclidean geometry texts as well -- when students first learn about triangle congruence, about to learn about SSS and the other postulates in the next section. The students are given two triangles and they must show that all of the corresponding parts are congruent. The final step would be that the two triangles are congruent by "definition of congruent triangles" (i.e., that all the corresponding parts are congruent). Of course, as soon as students learn about SSS and the other postulates, they'll never use the sufficient condition part of "definition of congruent triangles" in a proof again.
But Legendre now uses the traditionalist "definition of congruent triangles" in his proof. We've already proved that the sides are congruent, so now we work on the angles. Notice that he goes back to solid angles, and uses our trick that since the sides (arcs of great circles) are congruent, so must the plane angles (the central angles of those arcs). Again, I point out that if the sphere has radius 1, the sides of the spherical triangles equals the radian measures of the plane angles.
Legendre mentions that he uses Proposition 359 in his proof. This theorem essentially states that if two solid angles consist of three or corresponding congruent plane angles, then the two solid angles are themselves congruent, and so are the dihedral angles (the "inclinations") between the planes. This matters because Legendre defines angles of a spherical triangle in terms of these dihedral angles. And so the two triangles ACB and ADB must be congruent.
Now in the "scholium," Legendre states that these two triangles are equal "by symmetry." If we think about it, we notice that a simple reflection actually maps Triangle ACB to Triangle ADB -- the reflection over the (plane containing the) great circle AB. And so we see our first spherical isometry.
The concept of "superposition" goes back to Euclid. If we consider Euclid's proof of SAS, he uses superposition by translating and rotating one of the triangles until it coincides with the other. We see that Euclid omits reflections here -- so strictly speaking, he didn't actually prove SAS. Legendre corrects this error by counting only translations and rotations (isometries that preserve orientation) as "superposition" -- two figures that are mirror images of each other are equal "by symmetry" instead of equal "by superposition," except in the special case that the triangles are isosceles (since we could then rotate one triangle 180 degrees with the pole at the midpoint of AB to obtain the other).
I've mentioned earlier that in Euclidean geometry, there are four isometries (rotation, translation, reflection, and glide reflection), while in hyperbolic geometry, there are five isometries (rotation, translation, horolation, reflection, and glide reflection). So now we wonder, how many isometries are there in spherical geometry?
We recall the U of Chicago text and its definition of rotation -- it is the composite of two reflections in intersecting lines. Meanwhile, the definition of translation is that it is the composite of two reflections in parallel lines. But in spherical geometry, there are no parallel lines. Therefore, in spherical geometry there are no translations!
Notice that any transformation that appears to be a translation is actually a rotation, since any apparent "translation" is actually motion along either a great or small circle. The center of the rotation is simply the pole of this circle. So a "translation" east or west is a rotation centered at the North (or South) Poles, while a "translation" north or south is a rotation centered at a point on the Equator that is 90 degrees away from the current longitude. In either case, the magnitude (degree) of the rotation is equal to the change in longitude or latitude of the point.
This leaves us with the glide reflection. We notice that a glide reflection in Euclidean geometry is the composite of a reflection and a translation, but in spherical geometry there are no translations. Still, in spherical there exist transformations that are the composite of three reflections, yet are not themselves simple reflections. One may still call these "glide reflections" even though there are no translations -- these glide reflections are also the composite of a reflection and a rotation.
One key glide reflection is the antipodal map -- the function mapping each point to its antipodes. This map is the composite of a reflection in any line (great circle) and a rotation of 180 degrees centered at the pole of that great circle.
We can also think of the isometries of the spherical geometry in terms of the (three-dimensional) symmetries of the sphere itself. In this case, line (great circle) reflections correspond to plane reflections, point rotations correspond to axis rotations, and glide reflections correspond to what we call "roto-reflections," with the antipodal map corresponding to point inversion about the center of the original sphere.
If we think in terms of elliptic rather than spherical geometry, then all isometries are rotations! This is because we identity a point with its antipodes. So to perform a reflection on a figure, we simply take the (reversed) copy of that figure at the antipodes and rotate it all the way around the globe 180 degrees until it arrives at the place where we want the reflection to be. So in elliptic geometry, all isometries are "superpositions."
Let's get to our final proposition of the day -- but it's an important one:
480. Two triangles situated on the same sphere, or on equal spheres, are equal in all their parts, when two sides and the included angle of one are equal to two sides and the included angle of the other, each to each.
Demonstration. Let the side AB = EF, the side AC = EG, and the angle BAC = FEG, the triangle EFG can be placed upon the triangle ABC, or upon the triangle symmetrical with it ABD, in the same manner as two plane triangles are applied, when they have two sides and the included angle of one respectively equal to two sides and the included angle of the other (36). Therefore all parts of the triangle EFG will be equal to those of the triangle ABC; that is, besides the three parts which were supposed equal, we shall have the side BC = FG, the angle ABC = EFG, and the angle ACB = EGF.
Here, we see that Legendre is proving SAS Congruence for spherical triangles. He compares this to his Proposition 36, which is SAS Congruence for ordinary plane triangles.
Legendre proves SAS first by superposition -- EF can be rotated so that it lands on AB. Now there are only two possible places where G can end up -- C and D, where D is the vertex of ADB, the triangle that is "equal by symmetry" (i.e., the reflection image) to ACB.
The proof of SAS Congruence in the U of Chicago text is similar to this. The text transforms the original triangle ABC until its image lies on EF. Then the Isosceles Triangle Theorem is used to show that since CDG is isosceles, the reflection image of C must be G. Notice that as we traditionally use SAS to prove the Isosceles Triangle Theorem (and Legendre will do exactly this), we can avoid circularity by using the Ruler and Protractor Postulates to demonstrate why C reflected over AB must be the point G.
In both the Euclidean and spherical cases, it may be instructive to find the actual transformation that maps AB onto EF. On the sphere, we notice that if AB and EF are on the same great circle, the correct rotation is obvious. If AB and EF aren't on the same great circle, then we notice that their great circles must intersect since all great circles intersect. The points of intersection can be taken as the poles of a rotation which maps the great circle of AB to that of EF. Now this reduces us to the first case, so a simple rotation maps the image of AB to EF.
Thus concludes this post.