A platonic solid is a regular, convex polyhedron. Informally, it is a two-dimensional, convex shape made by gluing together a number of copies of a regular polyhedron so that the same arrangements meet at every vertex. There are only five such solids, and the ancient world was fascinated with them. Plato himself associated them with earth, wind, air, fire, and the arrangement of the constellations.
Euclid gave simple proof classifying the Platonic solids, but there is also a nice way to predict the Platonic solids via Euler Characteristic. That is, via Euler’s famous formula , where are the number of vertices, edges, and faces of the polyhedra.
Let our polygons have -sides, with meeting at a vertex. We know . At the same time, , because every edge belongs to two faces, so that if we go through each face, count its edges, and add them all up, we will have double-counted edges. Moreover, we also know that , because to each vertex we can count the number of edges coming out of it, and adding this all together gives , but we will have double-counted edges because every edge belongs to two vertices. This gives
We can divide by to obtain
Since is positive, we see that
Moreover, we know that , we only get possibilities, . In fact, each of these gives a platonic solid, so we have found them all. The convenient short-hand for describing a regular, convex polyhedron we have given is known as a Schläfli symbol.
Can you try extending this technique to predict the regular, convex polytopes of higher dimensions?
In this post, we will use the Steinhaus Theorem to understand group homomorphisms from into , which will hopefully shed let on the uncountably many elements of . Recall
Proof: (i) We shall use the definition of coninuity. Let be in the image of , and let be some disk about . The preimage can be thought of as . That is, a copy of around each point in the preimage of . If we can show that for each copy of around some point there exist an open neighborhood of its center that maps within , we will have verified the definition of continuity.
Note that is a disk around the origin. Let us rename it for the sake of simplicity in notation, where is a disc about the origin of half the radius. Consider the countable collection . This collection covers , so that
If for all we had , then we would have
which, I assure you, is profoundly absurd. Thus we can find whose inverse has positive measure. Note, as before, that the inverse of is . Consider
By the Steinhaus theorem this must contain an open neighborhood of the origin. But this is just the set of points that map into . Since is a disc about the origin, we have . Thus, an open neighborhood about the origin maps to . Call this open neighborhood . Consider the collection that contains . Then this is the union of open sets, and hence open, and if the preimage is restricted to within this open set, then the image of is within , confirming continuity.
(ii) If is of the form described, then it is clearly continuous, and hence must be measurable. Conversely, suppose that is a measurable homomorphism, so continuous by the prior part of the exercise. Then let , where is the th standard basis vector for . To show linearity in all coordinates, it suffices to show linearity in each coordinate at a time, so without loss of generality it suffices to consider , the first coordinate of . We have
Let , we have
Lastly, if is an arbitrary real, let be a sequence of rationals converging to it. By continuity of ,
With this theorem in hand, we can now understand the automorphisms of (and indeed much more). Any such automorphism is also a homomorphism from to as vector spaces, and hence can be viewed as a map from , which is also a -dimensional real vector space. In general, an automorphism of as a field must send to and to . By the above theorem, if the automorphism is continuous it is linearly determined by where it sends . But , so that must be a square root of . If the image of is , we get the identity automorphism, and if the image of is , then we get the conjugation automorphism.
Thus, any other automorphism of cannot be continuous, and hence cannot be measurable. So, if you were looking for a concrete way to think of these automorphisms, you are out of luck! These maps are ridiculous and you might do well by stopping to think about them.
This post, and the one following following it, are based on a series of exercises in Tao’s “An Introduction to Measure Theory” that I found particularly interesting.
Our initial question is concerned with field automorphisms of . Certainly there are two obvious ones: the trivial automorphism, and the conjugation automorphism. Any more? Well, assuming choice, we have the following proposition.
Proof: Any automorphism of fixes . Let be a transcendence basis for (this exists by Zorn’s Lemma, which is equivalent to choice).
Proof of claim: We know that
Suppose that was countable. Then every element in would be the root of a polynomial over . However, since is countable, this gives countably many polynomials, each with finitely many roots, so that , which is impossible, proving the claim.
Now, we claim that there are uncountably many automorphisms of . This is because is isomorphic to the function field , and any permutation of the indices of the variables induces an automorphism of this function field.
Lastly, we need to show that we can extend these uncountably many elements of to elements of . However, is algebraic, so this follows from standard results in Galois theory. .
This result is quite interesting. Not only are there more automorphisms than the trivial automorphism and conjugation, there are uncountably many automorphisms! If so, what exactly do these field automorphisms look like? How should we best go about picturing them? To answer that question, we take a detour into the theory of Lebesgue measure and integration.
Recall the construction of the convolution: Let be Lebesgue measurable functions such that is absolutely integrable and is essentially bounded. Then the convolution is defined by
Proof: Let be essentially bounded by . Note that
so that the convolution is both well defined and bounded. To see that it is continuous, note that
Since the Lebesgue integrable is translation invariant, we can translate the first integral by $y \to y + h$. Then we have
But if we let , then since translation is continuous in (a proof of which can be found in Tao’s text on Measure Theory, Prop. 1.6.13)
With the machinery of convolutions, we now prove the Steinhaus theorem.
Proof: Without loss of generality, is bounded, else we can replace it with some that has positive measure (such an must exist, by the upward monotone convergence theorem for measurable sets). Then , so if the former contains an open set about the origin, so must the latter.
Consider the convolution
This is well defined, since the functions in question are both and essentially bounded (by the boundedness of ).Note that
For an arbitrary , there exists for which iff and , so that . So if is not in , then . Consider the set
that is, the set of for which the integral defining the convolution is positive. We have shown that and that . Since the convolution is continuous, and is an open set, is an open set containing the origin inside of .
It may seem that our initial question in field theory, and our latter discussion of real analysis, are completely unrelated. This will be resolved in the following post, where we will use the Steinhaus theorem to better understand the nature of these bizarre field automorphisms of produced by the axiom of choice.
I gave a talk this Wednesday on Dehn Surgery and Decayed Knots, summarizing some of the reading I’ve been doing as part of an REU with Prof. Liam Watson. It was also my 22nd birthday.
For my the final report of Math 191, I wrote the following report on Distinguishing Numbers. Check it out!
The following is a problem from Enderton’s “A Mathematical Introduction to Logic”:
10. Say that a set of well-formed formulas (wffs) is equivalent to a set of wffs if for any wff , we have iff . A set is independent iff no member of is tautologically implied by the remaining members of .
Parts (a) and (b) of the problem ask us to show that any finite set of wffs has an independent equivalent subset, but that this might fail for infinite sets of wffs. Both of these are very easy, but the interesting part is (c).
(c) Let be any infinite sequence on wffs. Show that some independent equivalent set exists.
Solution: WLOG, is consistent, otherwise we can replace it with the independent, equivalent set consisting of a single false statement. We are going to construct a set as follows. Firstly, enumerate all the well-formed formulas, and go through the list one element at a time. At each element , make the following considerations:
1. Does ? If not, skip .
2. Does adding to affect the independence of ? If not, add to . Otherwise, proceed to step 3.
3. Why does adding to interfere with the independence of ? If it is because we can already prove from , then is of no use to us, so do not add it to . Otherwise, it is because we can prove some from the other elements of . In a sense, we might say that is strictly stronger than some of the statements in . In this case, proceed to the last step.
4. Suppose that at this point in the construction
Instead of adding to , add
We claim that adding to does not affect its independence. This is easily seen because:
(a) The other elements of cannot prove , as that would imply that they could prove , contrary to hypothesis.
(b) We cannot use to help prove some other , as if we do not assume every to begin with, is meaningless. To put it precisely, if we can use to prove , then we can get from only and . Since
and since is presumed independent from the other elements of , we can only get from and if we already knew , contrary to hypothesis.
Now, we complete this construction to obtain . By construction, and by Compactness, is independent, as it is independent at every finite step. Moreover, it is equivalent to , because if , then either:
1. is already in .
2. can be proven from elements that were already in .
3. isn’t in , but a statement of the form
is, with the , so that .
Theorem 1: Euler Product Formula
Proof: We know that . Let . Then
We now write
Multiplying out these parenthesis, we see by the Fundamental Theorem of Arithmetic that we get
So what is the probability that any two integers are coprime? Let and be integers. For a fixed prime the probability that any number is divisible by is , as every th number is a multiple of . The probability that two numbers are both divisible by is . The probability that at least one is not divisible by is . For any two primes, the events that are both divisible by either prime are mutually independent. Thus, to find out the probability that any prime divides both and , we need take the product
By Theorem 1, this is just . Pick your favorite proof that
Then the probability that any two numbers are coprime is .
From this, it is easy to see that the probability that any numbers are coprime is .
The following excerpt is taken directly from Carl E. Linderholm’s, “Mathematics Made Difficult”.
Mathematically, it is indeed inconclusive that nobody has ever succeeded in finding any numbers that divide exactly, other than the numbers and which obviously do this; this does not at all show that no such numbers exist. On the other hand, it may be quite easy to show that there are no such numbers. In mathematics, if one approach does not work, one must always try another; according to the principle catus multifariam deglubitur.
I answer that the number is prime. It is easily seen that the only numbers between and , including but excluding , are or . Thus the remainder left by any number on division by is either or . Hence the quotient ring
where is the ideal in generated by , has only the elements and , where these are the images of and under the canonical quotient map. Since must be a unit of this ring, every element of this ring except is a unit, and the ring is a field. As such, it has no zero-divisors other than . But looking back now at , this shows that if , then one of is an element , i.e., is an even number. In other words, we have either or ; say the former. By cancellation, , so that both and are units. (Or one may argue that since is a field, the ideal is maximal in , and hence prime, which implies that the generator is prime.)
The fact that the number is prime is useful in many ways. It prevents us, for instance, from indulging in the time-wasting habit, to which the ignorant are so prone, of attempting to discover a proper factor for the number . It shows that it would be very unwise to program a computer to keep on trying numbers one after the other until the computer should have found a divisor of distinct from and . It saves the trouble of looking for a rectangle with integral sides and area …
This post is about classifying primary ideals in a PID. Nothing too exciting, to be honest. You can take this problem in a few different directions, such as noticing that a PID is a UFD, in which case the primary ideals are generated by powers of primes. In that light, I’m going to offer a different solution, which I think is rather pleasant to read.
Claim: Let be a PID. Then is primary if and only if is maximal.
First Direction: Assume is primary. Then is prime.
Proof: Let . Then . Thus either or is in , so that either or is in .
Moreover, prime ideals in a PID are maximal.
Proof: Let be a prime ideal in , a PID. Suppose that . Then we can write . This implies that . Since is prime, either or is in . If is the latter, then . If it is the former, then we can write . This gives us
Since is a domain, . Then is a unit. This implies that .
Second Direction: Conversely, suppose that is not primary. Thus there exists such that , but , . Then .
Observe that is a proper subset of the ideal . Moreover, , for suppose that . Then there exists such that
Multiplying by ,
However, the RHS is the sum of two elements in , so is in . Yet by hypothesis.
Furthermore, is a proper subset of the ideal , as only the latter contains . However, , for suppose that . Then , which is impossible. We conclude that is not maximal.
Last time, we showed that was -categorial for uncountable cardinalities . In this post, we make steps towards proving the completeness and decidability of .
Recall Tarski’s Test : Let be -structures. Then if and only if, for every -formula for and a single variable, and every we have the following: if then there is such that .
Lemma: Suppose the theory has infinite models. Then for any , has a model of size .
Proof: Let be infinite. By inflating the model with extra elements if necessary, we can assume . Now, take with . Set , the model generated by . Note that , as extending from the set to a structure requires completion under functions and interpretation symbols in , which cannot increase the cardinality, for .
Now, add a witness to for every -formula and , such that . What we are trying to do is to fulfill the requirements of Tarski’s test, to get a model of size elementarily equivalent with . Call the resulting set (after adding witnesses), , which may not the be universe of a substructure. However, we can generate a substructure from it, i.e. . Again, . However, we are not done here, as we only took witnesses for parameters in , and now we need to consider paramaters in . To avoid this problem, we interate this process countably many times.
Note that . Now, put . Then is the universe of a substructure of , and by construction, satisfies the hypothesis of Tarski’s Test, so that . As a result, , and we have obtained the model we were looking for.