DyingLoveGrape.

Complex Algebraic Curves, Part 1: Introduction to Homogeneous Polynomials.

Prereqs.

For this series, we're not going to require all that much; mainly just calculus or familiarity with partial derivatives. The first few posts will be somewhat general, focusing mostly on homogeneous polynomials — this can lead us in a number of different directions, but projective space is on the horizons. The book I will mainly be glancing through is Frances Kirwan's Complex Algebraic Curves, though I will deviate from it as I see it. I'm by no means proficient in this subject, so I will count on you, the loyal reader, to correct me in the comments below if I've made an error — though, I'll try my best not to make any particularly devistating mistakes.

Homogeneous Functions.

We're all familiar with functions; we see them all the time in calculus, algebra, and so forth. Let's recall that a polynomial in one variable is of the form $f(x) = a_{0} + a_{1}x + a_{2}x^{2} +\dots + a_{m}x^{m}$ where the $a_{i}$ are some numbers (real, complex, etc.), and $x$ stands for some independent variable which we can "plug in" the function and associate with some value, which we call $f(x)$; note that polynomials, by definition, have only a finite number of terms — that is, there can be only a finite number of $+$'s in our function. The degree of our polynomial $f$ is the highest power of $x$ that occurs in the function. For example, $f(x) = 3 + 2x + 5x^{2} + x^{3}$ has degree equal to 3; we usually will write $\deg(f) = 3$.

For a polynomial function of more than one variable, the same general idea applies: it's denoted by $f(x_{1}, x_{2}, \dots, x_{k})$ and equal to a bunch of terms which look like $a_{i}x_{1}^{m_{1}}x_{2}^{m_{2}}\cdots x_{k}^{m_{k}}$ where $a_{i}$ is a scalar, as usual, and the $m_{i}$ are all non-negative integers. For example, if we have a function of two variables, it might look something like this: $f(x,y) = 4 + 3x + 2y + 5xy + 6x^{2}y^{7} + 7x^{8}y$ and a three-variable function may look something like this $f(x,y,z) = 2 + x - y + 2z - x^{2}y + y^{2}z + xyz^{3}.$ Note that, for two and three variables we generally write our variables as $x,y,z$ instead of $x_{1},x_{2},x_{3}$, but the idea is the same.

The idea behind the degree of a multivariable polynomial is a bit stranger. Consider the polynomial $f(x,y) = 2 + x^{6} + y^{7} + x^{5}y^{6}.$ What should $\deg(f)$ be? The standard way to define the degree is to add the powers of each of the variables in a term, then take the largest value to be the degree. In this case, $\deg(f) = 11$, since $x^{5}y^{6}$ has a sum of 11 whereas the other two terms are of degree 0 (the constant), 6, and 7 respectively.

With that out of the way, let's talk a bit about the way that some functions can grow. Suppose that $f(x)$ models how much a worker gets paid per hour; that is, $f(x)$ is the amount he earns for working $x$ hours. This grows linearly; if he makes 8 dollars an hour, then he will make $8x$ dollars in $x$ hours. The way that we can write this is $f(x) = 8x.$ This function has an interesting property: for any real number $\alpha$ we have $f(\alpha x) = \alpha f(x)$. Essentially, we can "pull out" the $\alpha$ from inside the function notation.

Not every function is like this. Consider $g(x) = x^{2} + 1$. We have $g(2x) = 4x^{2} + 1 \neq 2x^{2} + 2 = 2g(x).$ Hence it's not the case that we can always "pull a number out of the function." Hmph. Well, let's try the function $f(x) = x^{3}$, just for kicks. $f(\alpha x) = \alpha^{3}x^{3} = \alpha^{3}f(x).$ In this case, we were able to pull out a few $\alpha$'s. It also (coincidentally?) happened to be equal to the degree of the function. Curious, indeed!.

The question is, when can we do this? For one-variable functions, the answer is fairly straightforward and the reader should attempt it. Hint: The function should have at most one term in it! But for a many-variable function, what can happen? Let's look at a few examples. $f(x,y) = x^{3}y^{2} + 3$ You can convince yourself that plugging in $\alpha x$ or $\alpha y$ (or both) won't give you anything nice; it's that darn constant that's throwing things off. What about $f(x,y) = x^{3}y^{2} + xy$ Note that $f(\alpha x, \alpha y) = \alpha^{5}x^{3}y^{2} + \alpha^{2}xy$ What does this equal in terms of $f(x,y)$? Well, we can pull out $\alpha^{2}$, but then we're left with $\alpha^{3}x^{3}y^{2} + xy$. This certainly isn't equal to $\alpha^{i}f(x,y)$ for any $i$ we put in. Sad. So we can't just pull out the $\alpha$'s from this one. What was the problem, though? We had a high power of $\alpha$ on the first term and a lower power of $\alpha$ on the second term, and we couldn't factor every single $\alpha$ out of the function. This comes from the fact that not every term had the same degree (sum of the powers of the variables); the first term has degree $3 + 2 = 5$, but the second term has degree $1 + 1 = 2$. Maybe this would work if all of the terms were of the same degree — let's try it. $f(x,y) = xy^{3} + 2x^{2}y^{2} - 3x^{3}y + x^{4}$ If we plug in $\alpha x$ for $x$ and $\alpha y$ for $y$, we obtain \begin{align*} f(\alpha x, \alpha y) &= \alpha^{4}xy^{3} + 2\alpha^{4}x^{2}y^{2} - 3\alpha x^{4}y + \alpha^{4}x^{4}\\ &= \alpha^{4}(xy^{3} + 2x^{2}y^{2} - 3x^{3}y + x^{4})\\ &= \alpha^{4}f(x,y).\end{align*} It looks like our guess was correct; it turns out, in fact, that a polynomial has this property if (and only if!) all of the terms are the same degree (which is also equal, therefore, to the degree of the polynomial). Formally,

Definition (Homogeneous Polynomial). Let $f(x_{1},x_{2},\dots, x_{n})$ be a polynomial in $n$ variables. If each term in $f$ has degree $m\in {\mathbb N}$ then we call $f$ a homogeneous polynomial of degree $m$.

Moreover,

Proposition. Given $f$ a homogeneous polynomial of degree $m$ in $n$ variables, for any $\alpha\in {\mathbb R}$ we have that $f(\alpha x_{1},\alpha x_{2}, \dots, \alpha x_{n}) = \alpha^{m}f(x_{1}, x_{2},\dots, x_{n}).$

This notion is surprisingly useful in some geometries. Specificaly, suppose we define some function $f(x,y)$ on two variables such that we know: \begin{align*} f(1,0) &= P\\ f(0,1) &= Q\\ \end{align*} If we knew nothing else about this function, it would be impossible for us to talk about any other points; if we happen to know that this is a homogeneous polynomial of degree $m$ then we note have the following things: \begin{align*} f(\alpha,0) &= \alpha^{m} f(1,0) = \alpha^{m} P\\ f(0,\alpha) &= \alpha^{m} f(0,1) = \alpha^{m} Q\\ \end{align*} This gives us a surprising amount of information: it gives us all of the points on the line $\ell_{1}$ which passes through the origin and $(1,0)$ and also all of the points on the line $\ell_{2}$ which passes through the origin and $(0,1)$. That's kind of neat. What if we want all of the values on the line that passes through the origin and, say, $(2,4)$? We note that $f(2,4) = 2^{m}f(1,2)$ which tells us if we want $(2,4)$ then it suffices to find $f(1,2)$. We could have also said $f(2,4) = 4^{m}f(\tfrac{1}{2},1)$ which is equivalent.

Notice something here: given any point $(a,b)$, we can always reduce it to the form $(1,\frac{b}{a})$ and, given a homogeneous polynomial of degree $m$, this gives us that $f(a,b) = a^{m}f(1,\frac{b}{a}).$ This might not seem like too big of a deal at first, but the following proposition makes apparent what is neat about this idea:

Proposition. Given a homogeneous function of degree $m$ in $n$ variables, $f(x_{1},x_{2},\dots, x_{n})$, if one knows the values of $\begin{array}{c} f(1,a_{2},a_{3},a_{4},\dots, a_{n-1},a_{n})\\ f(0,1,a_{3},a_{4},\dots, a_{n-1},a_{n})\\ f(0,0,1,a_{4}\dots, a_{n-1},a_{n})\\ \vdots\\ f(0,0,0,0,\dots, 0,1) \end{array}$ for each combination $a_{2},\dots, a_{n}\in {\mathbb R}$, then one knows the value of the function $f$ for each point $(a_{1},a_{2},\dots, a_{n})\in {\mathbb R}^{n}$.

This might not seem like a huge deal, but what it does for us is essentially cuts the amount of work we have to do down significantly: instead of having to find values every possible combination $f(a_{1},\dots, a_{n})$, we need only work with, at worst, $f(1,a_{2},\dots, a_{n})$. You may want to try to prove this claim yourself: the idea is, if the first coordinate is not zero then factor it out to make the first coordinate equal to 1 (that is, divide all the other coordinates by the first coordinate's value). If it's 0, then go to the next coordinate. Notice that we don't need to ever worry about $f(0,0,0,\dots,0)$, since its value is always $0$.

Corollary. Given a homogeneous function of degree $m$ in one variable, $f(x)$, it is enough to know the value of $f(1)$ to completely determine all values of the function.

Note that this is the case since $f(\alpha) = \alpha^{m}f(1)$ gives us the value for any real number $\alpha$. Note also that this forces $f(0) = 0$; in fact, this is true for any homogeneous function (why?). Also,

Corollary. Given a homogeneous function of degree $m$ in two variables, $f(,yx)$, it is enough to know the values of $f(1,y)$ and $f(0,y)$ to completely determine all values of the function.

This isn't that bad, all things considered. The "picture" of this is also somewhat nice:

The idea behind this picture is, given any point on the plane (let's say in the first quadrant; if it's negative, we do the same thing but we need to multiply it by $(-1)^{m}$) just draw a line from the point to the origin; it will go through exactly one point on the dashed blue line which represents $(1,y)$. That's one of the kinds of points that we're talking about in the corollary above. Similarly, if the point is on the $y$-axis, then it will be of the form $(0,y)$, which is the other kind of point from the corollary above. There are similar nice pictures for other dimensions, but we'll get to that soon —

Properties of Homogeneous Polynomials: Zeros.

Homogeneous polynomials are pretty neat. But, being functions, we can do a bunch of different things with them. What's the first thing we usually think of when we talk about functions? Zeros are a big deal, so let's see what happens with a homogeneous polynomial.

One zero of the function is given by $(0,0,\dots, 0)$, certainly (why?). Suppose $f$ is homogeneous of degree $m$ and suppose that another zero is given by $f(a_{1},a_{2},\dots, a_{n}) = 0$ for some $a_{1}, \dots, a_{n}\in {\mathbb R}$. Then any multiple will also be equal to 0. Note that since for $\alpha>0$ we have $f(\alpha a_{1},\dots, \alpha a_{n}) = \alpha^{m}f(a_{1},\dots,a_{n}) = \alpha^{m}\cdot 0 = 0$ we also have that any multiple of the coordinates will be a zero as well. That is,

Proposition.For $f$ homogeneous of degree $m$, if $(a_{1},\dots, a_{n})$ is a zero of $f$, then so is $(\alpha a_{1},\dots, \alpha a_{n})$ for any $\alpha\in {\mathbb R}$.

Interestingly, this means, in particular, that we need only look for zeros for $f$ on the unit circle (or, in higher dimensions, on the unit $n$-sphere):

And, further, we only need half of the unit circle (or unit $n$-sphere) since each of these lines goes through exactly two points on the circle. On ${\mathbb R}^{2}$, for example, we may take the $x>0$ part of the circle along with the point $(0,1)$ as the places we need to investigate for potential zeros for $f$. Note that $(0,1)$ comes from the vertical line through the origin; we could choose either that one or $(0,-1)$ since both of them lie on the circle, but let's stick with $(0,1)$. Let's do an example to illustrate the process of finding a zero in a simple case.

Example. Let $f(x,y) = x^2 + y^{2} + 2xy$. Check that this is homogeneous of order $2$. There are two ways of doing this: the first is to check the unit circle with $x > 0$ and the point $(0,1)$ for a solution by noting that, in the former case, $x = \sqrt{1 - y^{2}}$; the second is to note that $f(x,y) = (x + y)^{2}$ and hence $0 = (x + y)^{2}$ implies $y = -x$. The two points on the circle associated to this would be $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. Note though that $f(\alpha, -\alpha) = \alpha^{2}f(1,-1)$ is a zero, as we expected.

Sometimes it's nice to use algebraic manipulation to find the zeros of homogeneous polynomials, but other times it may require a lot of effort. The nice thing about using the unit $n$-sphere is that we may eliminate one or more variables, thus making the calculation potentially easier.

Properties of Homogeneous Polynomials: Derivatives.

Usually, given some function, we want to find a few things besides zeros: maximums, minimums, and so forth. Most of this is done via a derivative, so it's not all that surprising that we'd want to see if there are any neat properties which derivatives of homogeneous polynomials have.

Let's try to do this for a few special cases. If we have a single-variable homogeneous polynomial of order $m$, say $f(x)$, then we note that $f(\lambda x) = \lambda^{m}f(x)$ which, differentiating with respect to $x$ on both sides (also, why is this function differentiable?), we obtain $\lambda f'(\lambda x) = \lambda^{m}f'(x).$ We don't need to worry about $\lambda = 0$ since $f(0) = 0$ for any homogeneous polynomial, so dividing out we obtain $f'(\lambda x) = \lambda^{m-1}f'(x).$ What do we obtain? A homogeneous polynomial of order $m-1$. This makes sense if you think about what homogeneous functions of one variable look like.

Let's try for a two-variable homogeneous polynomial of order $m$; call it $f(x,y)$. We note that every such (complex) polynomial is of the form $f(x,y) = \sum_{i+j = m} a_{i,j}x^{i}y^{j}$ where $a_{i,j}\in{\mathbb C}$ and the notation under the sum means that we sum over all such $i$'s and $j$'s such that $i+j = m$. Let's differentiate: $\frac{d}{dx}f = \sum_{i+j=m} a_{i,j}ix^{i-1}y^{j}$ $\frac{d}{dy}f = \sum_{i+j=m} a_{i,j}jx^{i}y^{j-1}$ Note that in these last cases, these become homogeneous polynomials of degree $m-1$. We'd expect this from what we did previously with the one-variable homogeneous polynomials. But now there's a nice relation: look at how similar both of these sums look! They're just slightly different at each term: there's an $i$ where a $j$ is, and vice versa. But we know the relation $i + j = m$, so maybe we ought to try to sum these things up... $\frac{d}{dx}f + \frac{d}{dy}f = \sum_{i+j = m} a_{i,j}(ix^{i-1}y^{j} + j x^{i}y^{j-1}).$ Well, this is nice looking, but it's kind of irritating that we have $x^{i-1}y^{j}$ and $x^{i}y^{j-1}$ instead of just $x^{i}y^{j}$. But, if we multiply $\frac{d}{dx}f$ by $x$ and $\frac{d}{dy}$ by $y$, then this becomes... \begin{align*}x\frac{d}{dx}f + y\frac{d}{dy}f &= \sum_{i+j = m} a_{i,j}(ix^{i}y^{j} + j x^{i}y^{j})\\ &=\sum_{i+j=m} a_{i,j}x^{i}y^{j}(i+j) \\ &=\sum_{i+j=m} a_{i,j}x^{i}y^{j}m \\ &=m\sum_{i+j=m} a_{i,j}x^{i}y^{j} \\ &=mf(x,y) \end{align*} Where we've used that $i+j = m$. Look at this! We've reconstructed the function, multiplied by its degree, by taking derivatives and multiplying by their respective variables! Sort of neat!

Proposition. Given a homogeneous complex polynomial $f(x,y)$ of degree $m$, we have the equality $x\frac{d}{dx}f(x,y) + y\frac{d}{dy}f(x,y) = mf(x,y).$

Note that the reader should try to prove something similar about the one-variable case. The reader should also do this for the $n$-variable case, since it's virtually the same deal. We'll give it its own box, because it is a pretty nice relation. This is sometimes called Euler's Relation for Homogeneous Polynomials.

Proposition. Given a homogeneous complex polynomial $f(x_{1},\dots, x_{n})$ of degree $m$, we have the equality $\sum_{i=1}^{n}x_{i}\frac{d}{dx_{i}}f(x_{1},\dots, x_{n}) = mf(x_{1},\dots, x_{n}).$

Besides writing this out "formally" as we did above, there is another slick proof which is similar to how we looked at the one-dimensional case — see if you can figure it out!

Next Time...(Interpreting Homogeneous Polynomials.)

At this point, we know a bunch about homogeneous polynomials but the reader may have that who cares! feeling about them. We ought to talk about them as if they were geometric objects but, as we've seen, if we plot them in the plane they are a bit dull. It turns out that these kinds of polynomials are perfect to represent functions in something called the projective plane. Next time, we'll talk about what the projective plane is and what these functions represent there.