Polynomial Invariants Dylan Wilson October 9, 2014 (1) Today we will be interested in the following Question 1.1. What are all the possible polynomials in two variables f(x, y) such that f(x, y) = f(y, x)? Of course, this question is vague. It s unclear what sort of answer I m looking for. To explain, let s start with an example. Example 1.2. A natural condition on a polynomial in one variable, p(x) is that it be even, i.e. p(x) = p( x) Let s try to describe the set of all even polynomials. To start with, let s consider the case of quadratics p(x) = ax 2 + bx + c. The condition that this be even says that we have an equality ax 2 + bx + c = a( x) 2 + b( x) + c Subtracting the right from the left we get the condition = ax 2 bx + c 2bx = 0 which can only happen if b = 0. (Here and for the rest of the talk, when I write down p(x) = 0 I mean that the polynomial is zero as a function, not that some value of x makes it zero.) Thus, quadratic polynomials are even if and only if they are of the form ax 2 + c. More generally, if p(x) = a n x n + a n 1 x n 1 + + a 0 then p(x) = p( x) means a n x n + a n 1 x n 1 + + a 0 = a n ( x) n + a n 1 ( x) n 1 + + a 0 Two polynomials are equal if and only if their coefficients are equal (take this either as a definition or an exercise), so this says that a k = ( 1) k a k If k is even this doesn t say anything, just a k = a k. If k is odd, it says a k = a k which can only happen if a k = 0. Thus we have shown that a polynomial is even if and only if it can be written a 2n x 2n +a 2n 2 x 2n 2 + + a 2 x 2 + a 0. In other words, a polynomial is even if and only if it is a polynomial in the variable x 2. We write this suggestively by saying that the collection of even polynomials is R[x 2 ] (just like the collection of all polynomials was R[x]). Ok, so much for even polynomials. But now maybe it s clearer what I mean by the question above. Let s make it even clearer in the following set of exercise: 1
Exercise 1.3. Show that the set of even polynomials is a vector subspace of R[x] directly from the definition of even (i.e. not using the result we proved above describing the set of even polynomials another way.) Exercise 1.4. Recall that the degree of a monomial ax k, where a 0, is k, and the degree of a polynomial (which is a sum of monomials) is the highest degree nonzero monomial that appears. Prove that polynomials of degree at most n form a vector subspace of R[x]. We will write this subspace as R[x] n. Prove that the even polynomials of degree at most n form a vector subspace, we ll denote this by Even n. (Hint: Something you proved during your quiz could be helpful...) Exercise 1.5. Prove that any even polynomial of degree 6 can be written uniquely in the form ax 6 + bx 4 + cx 2 + d. (Hint: To prove something like this, you show that any two ways of writing the polynomial in that form are equal. That is, start by writing down two expressions as above and setting them equal to each other. Then prove they were the same expression.) Exercise 1.6. Show that the function f : R 4 Even 6 given by f(a, b, c, d) = ax 6 + bx 4 + cx 2 + d is (i) linear, (ii) injective (look it up), (iii) surjective (look it up). Such a map is called an isomorphism. Show, more generally, that we have an isomorphism R n Even 2n 2 for n 1. You ll talk more about dimension in class, but whatever we mean by dimension it should be the case that R n has dimension n. So here is a sort of cheap definition of dimension: Definition 1.7. A vector space V has dimension n if there is a linear isomorphism R n V. Warning 1.8. We haven t proved anything about this concept yet! As of now, it could be the case that a vector space can have different dimensions at the same time... For example, how do we know there is no linear isomorphism R R 2? Why isn t R 2 2-dimensional and 1-dimensional? So what have we accomplished? We ve cooked up some basic polynomials, {1, x 2, x 4,...}, such that any even polynomial is a unique linear combination of these ones. A linear combination of vectors v i in a vector space V is just any vector of the form a 1 v 1 + a 2 v 2 + + a n v n, a i R A basic observation is the following: Proposition 1.9. A vector space V has dimension n if and only if there are vectors v 1,..., v n V such that every element of V is a unique linear combination of the vectors {v 1,..., v n }. Proof. Given such vectors v 1,..., v n define a map f : R n V as follows: This is linear because f(a 1,..., a n ) = a 1 v 1 + + a n v n f(λa 1 + µb 1,..., λa n + µb n ) = (λa i + µb i )v i = (λa i v i + µb i v i ) ( ) ( ) = λai v i + µbi v i = λf(a 1,..., a n ) + µf(b 1,..., b n ) It is surjective since any vector v can be written as v = a 1 v 1 + + a n v n = f(a 1,..., a n ), by assumption, and it is injective because the choice of a i is unique, again by assumption. On the other hand, if we start off with a linear isomorphism f : R n V, then define v i = f(e i ) where e i is the vector with a 1 in the ith spot and zeros elsewhere. Now pick v V. Since isomorphisms are surjective, there is some vector (a 1,..., a n ) R n such that f(a 1,..., a n ) = v. By linearity, v = f(a 1,..., a n ) = f(a 1 e 1 + + a n e n ) = a 1 f(e 1 ) + + a n f(e n ) = a 1 v 1 + + a n v n 2
If there were some other way of writing v as such a linear combination, i.e. v = b 1 v 1 + + b n v n then, again by linearity, v = f(b 1,..., b n ). But f is injective, so this implies that (b 1,..., b n ) = (a 1,..., a n ), i.e. a i = b i for all i. In other words, there is a unique choice of the a i that give v. This completes the proof. The v i above serve as basic building blocks for the vector space V. They are called a basis. Our goal will be to find similar basic building blocks for symmetric polynomials. (2) First we should lay the groundwork. Remember, we re interested in polynomials in two variables x and y, like 5x 2 y + 3x + 2, 3y, (x + 2) 2 (where the last expression really means x 2 + 4x + 4). We define the degree of a monomial ax i y j to be i + j if a is nonzero. The degree of a polynomial is the degree of its largest nonzero monomial. Exercise 2.1. Show that the set of symmetric polynomials, Symm, is a subspace of R[x, y]. Show that the set of polynomials of degree at most n, R[x, y] n, and the set of symmetric polynomials of degree at most n, Symm n, are also subspaces. Definition 2.2. A homogeneous polynomial of degree d is a sum of monomials of degree exactly d. convention, 0 is considered homogeneous of every degree. By Write down an example of a non- Exercise 2.3. Write down examples of homogeneous polynomials. homogeneous polynomial. Exercise 2.4. Show that the set of homogeneous polynomials of degree d, R[x, y] n, is a subspace of R[x, y] and of R[x, y] n. Show that the set of homogeneous, symmetric polynomials of degree d is a subspace. Exercise 2.5. Show that every nonzero polynomial can be written uniquely as a sum of nonzero homogeneous polynomials. Exercise 2.6. Show that a polynomial is symmetric if and only if it is a sum of homogeneous, symmetric polynomials. All told, then, we will have answered our original question if we can answer: Question 2.7. How can we describe the space of homogeneous, symmetric polynomials of degree d, for arbitrary d? Example 2.8. (d = 0). The only degree 0 polynomials are constants. They are all symmetric. So there is a one-dimensional space of such things. Example 2.9. (d = 1). Degree 1 homogeneous polynomials are called linear and look like this: 2x + 3y, 3x, 5x 5y, ax + by The symmetry condition says ax + by = bx + ay, i.e. a = b. In other words, every degree 1, homogeneous polynomial is a multiple of the polynomial x + y. So, again, we get a one-dimensional space. Example 2.10. (d = 2) These look like ax 2 + bxy + cy 2 Being symmetric means that a = c. So they look like a(x 2 + y 2 ) + bxy. You can check that this space is 2-dimensional. 3
Example 2.11. (d = 3) These are: ax 3 + bx 2 y + cxy 2 + dy 3 where a = d and b = c. In other words, they look like a(x 3 + y 3 ) + b(x 2 y + xy 2 ). As we do these examples, we start to collect our building blocks. They look a little more complicated than what we saw in the case of even polynomials in one variable. We ve picked out the polynomials: 1, x + y, x 2 + y 2, xy, x 3 + y 3, x 2 y + xy 2 A bit too early for a pattern, but we at least notice the polynomials x n + y n appear. Indeed, these are obviously homogeneous and symmetric. Now for an observation, again in the form of an exercise: Exercise 2.12. Show that if f and g are homogeneous, symmetric polynomials of degrees d and e, respectively, then f g (the product) is a homogeneous, symmetric polynomial of degree d + e. In particular, (x + y) 2 is homogeneous and symmetric. That means we can write it in terms of the building blocks x 2 + y 2 and xy: (x + y) 2 = (x 2 + y 2 ) + 2xy Let s turn this on its head: x 2 + y 2 = (x + y) 2 2xy In other words: the building block x 2 + y 2 can be written in terms of xy and (x + y) 2. It follows that every degree 2, symmetric, homogeneous polynomial is a unique linear combination of xy and (x + y) 2. (Why?) So our new list of building blocks looks like: 1, x + y, xy, (x + y) 2, x 3 + y 3, x 2 y + xy 2 and, actually, let s notice that and x 2 y + xy 2 = xy(x + y) thus So we write the basic building blocks as: At this point we might guess: (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 3 = (x 3 + y 3 ) + 3xy(x + y) x 3 + y 3 = (x + y) 3 3xy(x + y) 1, x + y, xy, (x + y) 2, (x + y) 3, xy(x + y) Proposition 2.13. Every symmetric, homoegenous polynomial of degree d can be written uniquely as a linear combination of the polynomials (x + y) d, (x + y) d 2 xy, (x + y) d 4 (xy) 2,... In other words, these polynomials form a basis for the degree d homogeneous, symmetric polynomials. In particular, this space has dimension d/2 + 1 if d is even and (d 1)/2 + 1 if d is odd. The proof is a little involved, so let me not give it at the moment. (Later in the course I ll be able to give a fairly quick proof, after we learn some tricks for checking when some set of vectors is a basis.) Something better is true, actually: 4
Theorem 2.14. Let f(x, y) be a symmetric polynomial (not necessarily homogeneous). unique polynomial h(w, z) in two variable such that Then there is a f(x, y) = h(x + y, xy) In other words, we can think of symmetric polynomials as polynomials in the elementary symmetric polynomials x + y and xy. It follows from what we ve done so far that we can find such a polynomial. The uniqueness part is a little tricky, but also follows from what we ve done so far. At this point I should probably stop, but let me point out some obvious questions, which will be in the form of challenge problems: Problem 2.15. Write x n + y n in terms of the basis we ve given. Problem 2.16. Run the same analysis for polynomials f(x, y) such that f(x, y) = f( x, y) = f(x, y). Prove that these are built from x 2 and y 2 just as symmetric polynomials are built from x + y and xy. Compute the dimensions of the spaces of such polynomials which are homogeneous of degree d. Problem 2.17. Do the same for polynomials f(x, y) such that f(x, y) = f( x, y). The dimension part is a little trickier because not all degrees will behave the same way... 5