Love and Math

by Frenkel, Edward

  2. More precisely, I proved that for each divisor d of n, the qth Betti number, where q = n(d − 2)/d, is equal to φ(d), and for each divisor d of n−1, the qth Betti number, where q = (n−1)(d−2)/d, is equal to φ(d). All other Betti numbers of B′n are equal to 0.

  3. In 1985, Mikhail Gorbachev came to power, and soon afterward he launched his policy of perestroika. As far as I know, systematic discrimination of Jewish applicants at the entrance exams to Mekh-Mat of the kind that I had experienced ended around 1990.

  4. S. Zdravkovska and P. Duren, Golden Years of Moscow Mathematics, American Mathematical Society, 1993, p. 221.

  5. Mathematician Yuly Ilyashenko argued that this event was a catalyst for the establishment of anti-Semitic policies at Mekh-Mat in the interview entitled The black 20 years at Mekh-Mat, published on the website Polit.ru on July 28, 2009: http://www.polit.ru/article/2009/07/28/ilyashenko2

  6. The question was to find in how many ways can one glue pairwise the sides of a regular polygon with 4n sides to obtain a Riemann surface of genus n. In Chapter 9, we will discuss a particular way of doing so when we identify the opposite sides of the polygon.

  7. Edward Frenkel, Cohomology of the commutator subgroup of the braid group, Functional Analysis and Applications, vol. 22, 1988, pp. 248–250.

  Chapter 7. The Grand Unified Theory

  1. Interview with Robert Langlands for the Mathematics Newsletter, University of British Columbia (2010), full version available at http://www.math.ubc.ca/Dept/Newsletters/Robert_Langlands_interview_2010.pdf

  2. Suppose that there exist natural numbers m and n such that . We can assume without loss of generality that the numbers m and n are relatively prime; that is, they are not simultaneously divisible by any natural number other than 1. Otherwise, we would have m = dm′ and n = dn′ and then . This process can be repeated, if needed, until we reach two numbers that are relatively prime.

  So let us assume that , where m and n are relatively prime. Squaring both sides of the formula , we obtain 2 = m2/n2. Multiplying both sides by n2, we have m2 = 2n2. This implies that m is even, for if it were odd, then m2 would also be odd, which would contradict this formula.

  If m is even, then m = 2p for some natural number p. Substituting this into the previous formula, we obtain 4p2 = 2n2, hence n2 = 2p2. But then n must also be even, by the same argument as the one we used to show that m is even. Thus, both m and n must be even, which contradicts our assumption that m and n are relatively prime. Hence, such m and n do not exist.

  This is a good example of a “proof by contradiction.” We start with the statement that is opposite to what we are trying to prove (in our case, we start with the statement that is a rational number, which is opposite to what we are trying to prove). If this implies a false statement (in our case, this implies that both m and n are even, even though we had assumed that they were relatively prime), then we conclude that the statement we started with is also false. Hence the statement we wanted to prove (that is not a rational number) is true. We will use this method again in Chapter 8: first, when we discuss the proof of Fermat’s Last Theorem, and then again, in endnote 6, when we give Euclid’s proof that there are infinitely many prime numbers.

  3. For example, let us multiply these two numbers: and 3 −. We simply open the brackets:

  But · = 2, so by collecting the terms, we obtain the following answer:

  It is a number of the same form, so it does belong to our new numerical system.

  4. We only consider the symmetries of our numerical system that are compatible with the operations of addition and multiplication, and such that 0 goes to 0, 1 goes to 1, additive inverse goes to additive inverse, and multiplicative inverse goes to multiplicative inverse. But if 1 goes to 1, then 2 = 1 + 1 must go to 1 + 1 = 2. Likewise, all natural numbers must be preserved, and then so are their negatives and multiplicative inverses. Hence, all rational numbers are preserved by such symmetries.

  5. It is easy to check that this symmetry is indeed compatible with addition, subtraction, multiplication, and division. Let’s do this for the operation of addition. Consider two different numbers in our new numerical system:

  where x, y, x′, y′ are rational numbers. Let’s add them:

  We can apply our symmetry to each of them. We then get:

  Now let’s add these:

  We see that the number we get is equal to the number obtained by applying our symmetry to the original sum

  In other words, we can apply the symmetry to each of the two numbers individually and then add them up. Or we can first add them up, and then apply the symmetry. The result will be the same. This is what we mean by saying that our symmetry is compatible with the operation of addition. Likewise, we can check that our symmetry is compatible with the operations of subtraction, multiplication, and division.

  6. For example, in the case of the number field obtained by adjoining to the rational numbers, the Galois group consists of two symmetries: the identity and the symmetry exchanging and . Denote the identity by I and the symmetry exchanging and by S. Let’s write down explicitly what the compositions of these symmetries are:

  and the most interesting one:

  Indeed, if we exchange and and then do this again, the net result will be the identity:

  We have now completely described the Galois group of this number field: it consists of two elements, I and S, and their compositions are given by the above formulas.

  7. A few years earlier, Niels Henrik Abel showed that there was a quintic equation that could not be solved in radicals (Joseph-Louis Lagrange and Paolo Ruffini also made important contributions). However, Galois’ proof was more general and more conceptual. For more on the Galois groups and the rich history of solving polynomial equations, see Mario Livio, The Equation That Couldn’t Be Solved, Simon & Schuster, 2005.

  8. More generally, consider the quadratic equation ax2 + bx + c = 0 with rational coefficients a,b,c. Its solutions x1 and x2 are given by the formulas

  If the discriminant b2 − 4ac is not the square of a rational number, then these solutions are not rational numbers. Hence, if we adjoin x1 and x2 to the rational numbers, we obtain a new number field. The group of symmetries of this number field also consists of two elements: the identity and the symmetry exchanging the two solutions, x1 and x2. In other words, this symmetry exchanges and .

  But we don’t need to write down explicit formulas for the solutions to describe this Galois group. Indeed, because the degree of the polynomial is two, we know that there are two solutions, so let’s just denote them by x1 and x2. Then we have

  Opening the brackets, we find that , so that . We also have because x1 is a solution of the above equation. Therefore, if the discriminant is not the square of a rational number, then the number field obtained by adjoining x1 and x2 to the rational numbers consists of all numbers of the form α + βx1, where α and β are two rational numbers. Under the symmetry exchanging x1 and x2, the number α + βx1 goes to

  This symmetry is compatible with the operations of addition, etc., because both x1 and x2 solve the same equation with rational coefficients. We obtain that the Galois group of this number field consists of the identity and the symmetry exchanging x1 and x2. I stress again that we did not use any knowledge whatsoever about how to express x1 and x2 in terms of a,b,c.

  9. To illustrate this point, consider, for example, the equation x3 = 2. One of its solutions is the cubic root of 2, . There are two more solutions, which are complex numbers: and , where

  (see the discussion of complex numbers in Chapter 9). The smallest number field containing these three solutions should also contain their squares: and , as well as their ratios: ω and ω2. So it looks like to construct this number field, we have to adjoin eight numbers to the rationals. However, we have a relation:

  which allows us to express ω2 in terms of 1 and ω:

  Hence we also have

  Therefore to obtain our number field, we only need to adjoin five numbers to the rationals
: , and . Hence a general element of this number field, called the splitting field of the equation x3 = 2, will be a combination of six terms: a rational number plus a rational number times ω plus a rational number times , and so on. Compare this with the splitting field of the equation x2 = 2, whose elements have two terms: a rational number plus a rational number times .

  We have seen above that the elements of the Galois group of the splitting field of the equation x2 = 2 permute the two solutions of this equation, and . There are two such permutations: the one switching these two solutions and the identity.

  Likewise, for any other equation with rational coefficients, we define its splitting field as the field obtained by adjoining all of its solutions to the rational numbers. By the same argument as in endnote 4 above, any symmetry of this number field compatible with the operations of addition and multiplication preserves the rational numbers. Therefore, under such a symmetry, any solution of this equation must go to another solution. Hence, we obtain a permutation of these solutions. In the case of the equation x3 = 2, there are three solutions listed above. Under each permutation, the first, , goes to any one of the three solutions; the second, , goes to one of the remaining two solutions; and then the third, , must go to one remaining solution (a permutation must be one-to-one in order to have an inverse). Therefore, there are 3 · 2 = 6 possible permutations of these three solutions. These permutations form a group, and it turns out that this group is in a one-to-one correspondence with the Galois group of the splitting field of the equation x3 = 2. Thus, we obtain an explicit description of the Galois group in terms of permutation of the solutions.

  In the above calculation, we used explicit formulas for the solutions of the equation. But a similar argument can be made for an arbitrary cubic equation with rational coefficients, and we don’t need a formula for its solutions in terms of the coefficients. The result is the following: let us denote the solutions of the equation by x1, x2, and x3. Assume that all of them are irrational. However, it is easy to see that the discriminant of the equation, defined as

  is always a rational number. It turns out that if its square root is not a rational number, then the Galois group of the splitting field of this equation is the group of all permutations of these solutions (it then consists of six elements). If the the square root of the discriminant is a rational number, then the Galois group consists of three permutations: the identity, the cyclic permutation , and its inverse.

  10. For example, it is not difficult to show that for a typical quintic equation (that is, one with n = 5), for which we have five solutions, the Galois group is the group of all permutations of these five numbers. A permutation is a one-to-one reshuffling of these numbers, such as the one shown on the picture below.

  Under such a permutation, the solution x1 goes to any of the five (possibly, to itself), so we have five choices, then x2 has to go to one of the remaining four solutions, x3 to one of the remaining three, and so on. Hence, altogether, there are 5 · 4 · 3 · 2 · 1 = 120 permutations, and so the Galois group consists of 120 elements.

  (The group of permutations of a set of n elements, also known as the symmetric group on n letters, consists of n! = n · (n−1) ·...2 · 1 elements.) Unlike the Galois groups of the quadratic, cubic, and quartic equations, it is not a solvable group. Therefore, according to Galois’ argument, we cannot express solutions of the general quintic equation in terms of radicals.

  11. It is now available on the website of the Institute for Advanced Study in Princeton, http://publications.ias.edu/sites/default/files/weil1.pdf

  12. Quoted from the image available at the Institute for Advanced Study Digital Collections, http://cdm.itg.ias.edu/cdm/compoundobject/collection/coll12/id/1682/rec/1

  Chapter 8. Magic Numbers

  1. Robert Langlands, Is there beauty in mathematical theories?, in The Many Faces of Beauty, ed. Vittorio Hösle, University of Notre Dame Press, 2013, available online at http://publications.ias.edu/sites/default/files/ND.pdf

  2. For more on conjectures, see this insightful article: Barry Mazur, Conjecture, Synthèse, vol. 111, 1997, pp. 197–210.

  3. For more on the history of Fermat’s Last Theorem, see Simon Singh, Fermat’s Enigma: The Epic Quest to Solve the World’s Greatest Mathematical Problem, Anchor, 1998.

  4. See Andrew Wiles, Modular elliptic curves and Fermat’s last theorem, Annals of Mathematics, vol. 141, 1995, pp. 443–551;

  Richard Taylor and Andrew Wiles, Ring-theoretic properties of certain Hecke algebras, Annals of Mathematics, vol. 141, 1995, 553–572.

  They proved the Shimura–Taniyama–Weil conjecture in the most typical (the so-called semi-stable) case, which turned out to be sufficient to settle Fermat’s Last Theorem. A few years later, the remaining cases of the conjecture were proved by C. Breuil, B. Conrad, F. Diamond, and R. Taylor.

  Because it is now proved, it would be more proper to refer to the Shimura–Taniyama–Weil conjecture as a theorem. And in fact, many mathematicians now refer to it as the “modularity theorem.” But old habits die hard, and some, like me, still use its old name. Ironically, Fermat’s Last Theorem has always been referred to as a theorem, even though it was in fact a conjecture. No doubt, initially this was done out of respect for Fermat’s claim that he had found a proof.

  5. If N is not a prime, then we can write N = xy for two natural numbers x and y between 1 and N − 1. Then x does not have a multiplicative inverse modulo N. In other words, there does not exist a natural number z between 1 and N − 1 such that

  Indeed, if this equality were satisfied, we would multiply both sides by y, and we would obtain

  But xy = N, so the left-hand side is Nz, which means that y is divisible by N. But then y cannot be between 1 and N−1.

  6. The proof attributed to Euclid goes as follows. We apply the method of “proof by contradiction,” which we have already used in this chapter when we discussed the proof of Fermat’s Last Theorem.

  Suppose that there are only finitely many primes: p1, p2,...,pN. Consider the number A obtained by taking their product and adding 1; that is, set A = p1p2...pN + 1. I claim that it is a prime number. We prove this by contradiction: if it is not a prime number, then it is divisible by a natural number other than 1 and itself. Hence A has to be divisible by one of the prime numbers – let’s say, by pi. But if A is divisible by pi, then A = 0 modulo pi, whereas it follows from the definition of A that A = 1 modulo pi.

  We have arrived at a contradiction. This means that A is not divisible by any natural number other than 1 and itself. Hence A is itself a prime number.

  But since A is clearly larger than any of the numbers p1,p2,...,pN, this contradicts our assumption that p1,p2,...,pN were the only prime numbers. Therefore our initial statement that there are only finitely many primes is false. Hence, there are infinitely many prime numbers.

  7. Let’s spell this out: within a particular numerical system, a multiplicative inverse of a number a is a number b such that a · b = 1. So, for example, within the numerical system of rational numbers, the multiplicative inverse of the rational number is . Within the numerical system we are considering now, the inverse of a natural number a between 1 and p − 1 is another natural number b in the same range such that

  No matter which numerical system we consider, the number 0, the additive identity, never has a multiplicative inverse. That’s why we exclude it.

  8. Here is the proof. Let us pick a natural number a between 1 and p − 1, where p is a prime number. Let’s multiply a by all other numbers b in this range and take the result modulo p. We will compile a table consisting of two columns: in the first column will be the number b, and in the second column will be the number a · b modulo p.

  For instance, if p = 5 and a = 2, this table looks as follows:

  We see right away that each of the numbers 1,2,3,4 appears in the right column exactly once. What happens when we multiply by 2 is that we obtain the same set of numbers, but they are permuted in a certain way. In particular, num
ber 1 appears in the third line. This means that when we multiply 3 by 2, we get 1 modulo 5. In other words, 3 is the inverse of 2 if we do arithmetic modulo 5.

  The same phenomenon holds in general: if we compile a table like the one above for any prime p and any number a from the list 1,2,..., p−1, then each of the numbers 1,2,..., p−1 will appear in the right column exactly once.

  Let us prove this, again employing the trick of proof by contradiction: suppose that this is not the case. Then one of the numbers from the set 1,2,..., p−1, call it n, has to appear in the right column at least two times. This means that there are two numbers from the set 1,2,..., p−1, call them c1 and c2 (suppose that c1 > c2), such that

  But then we have

  The last formula means that a · (c1 − c2) is divisible by p. But this is impossible because p is a prime and both a and c1 − c2 are from the set {1,2,..., p−1}.

  We conclude that in the right column of our table each of the numbers 1,2,..., p−1 appears no more than once. But because there are exactly p−1 of those numbers and we have the same number of lines in our table, p−1, the only possibility for this to happen is for each number to appear exactly once. But then number 1 has to appear somewhere in the right column and exactly once. Let b be the corresponding number in the left column. But then we have

  This completes the proof.

  9. For example, we can divide 4 by 3 in the finite field of 5 elements:

  (here we use the fact that 2 is the multiplicative inverse of 3 modulo 5).

  10. Let us observe that for any number a whose absolute value is less than 1 we have