What Does Homogenisation Of An Equation Actually Mean?

For example, if we have a conic:

$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$

What does homogenising this equation with another line (say $ax + by + c = 0$ ) actually mean? As in, what are the graphical consequences of this? I've been told that it represents a pair of straight lines passing through the origin and the points of intersection (of the conic and the line), but how did this come about?

Update: I'm sorry for the lack of context, but I've just started algebraic geometry and I like understanding what things actually mean instead of mugging up a formula.

Solution 1:

A function $f:\mathbb{R}^n \to \mathbb{R}$ is homogeneous of degree $r$ when it has the following property:

$$f(\lambda x_1,\lambda x_2,\ldots,\lambda x_n) = \lambda^r f(x_1,x_2,\ldots,x_n)$$

Examples of such functions include:

• Linear functions, they are of degree 1. If you scale the graph of the function by a factor $\lambda$, you still get the same graph, except that all points have coordinates scaled up by the factor $\lambda$.

• The monomials in one variable: $f:\mathbb{R} \to \mathbb{R}: x \mapsto x^n$ is homogeneous of degree $n$, scale $x$ by a factor $\lambda$ and the function will scale by a factor $\lambda^n$.

• Any polynomial (function) in $n$ variables such that each term appearing in the polynomial is of degree $k$ is homogeneous of degree $k$. For instance $x^2 y + z^3$ is homogeneous of degree $3$.

The importance of homogeneity is the scale invariance of the functions. Which implies that the graphs of the functions will be scale invariant. Indeed, imagine a homogeneous function is used to define a geometrical object implicitly:

$$f(x_1,x_2,\ldots,x_n)=0$$

meaning all points with coordinates $(x_1,\ldots,x_n)$ that satisfy this equation will belong to the geometrical figure defined by $f$. If $f$ is homogeneous, it immediately follows that any multiple of these coordinates also satisfies the equations. In other words, any point that satisfies the equation immediately implies the entire ray going through that point and the origin of the space belong to the geometrical object.

Homogenizing an implicit polynomial equation means adding an extra variable $z$ and multiply any term by $z^k$ with $k$ such that the resulting polynomial is homogeneous. Of course, since any $z$-multiple of the polynomial will also be homogeneous, you choose the resulting homogeneous polynomial with smallest possible degree.

In your example, this would become

$$ax^2 + 2hxy + by^2 + 2gxz + 2fyz + cz^2 = 0$$

If you take the intersection of the geometrical figure defined by above equation with the plane $z=1$, you get back the original figure. If you choose another plane parallel with the $z=1$ plane, you get a scaled up or scaled down version of the figure.

Solution 2:

Disclaimer: I will try to keep my answer at a very elementary level, as you pointed out that you just started learning algebraic geometry. Also, I will be a bit imprecise at several points of the discussion, as I want to convey some intuition and not mathematical rigor.

If you're studying algebraic geometry, I think a good way to view homogenising an equation is to interpret it as "adding points at infinity" to the zero set of the polynomial.

In the two-dimensional case this works as follows: instead of the "normal" affine plane $\mathbb{A}^2$ (which consists just of 2-tuples coming from your ground field $k$) we consider the projective plane $\mathbb{P}^2$, which has some additional points: for each direction "out of the plane", we introduce a new point. This gives us a whole line of new points, which cannot be drawn in the usual picture of the affine plane, but are thought of as lying "at infinity". While this seems a bit odd at first it has many advantages. Consider for example two straight lines in the affine plane, then they will intersect in some point of the plane or they will be parallel (in which case they don't intersect). When we look at the same two lines in the projective plane, things get simpler: the lines will always intersect! Either the lines intersect in the "affine part" of our projective plane or they are parallel. In the second case they will both point "out of the plane" in the same direction, which means that they will intersect at the point of the projective plane that we've previously added for that particular direction. So, all straight lines will intersect in $\mathbb{P}^2$. Isn't that nice?

Starting with this idea, one wants to make all of this precise and more rigorous. The formal way to define curves in the projective plane is to only consider zero sets of homogeneous polynomials, that is polynomials in which each monomial has the same total degree. In order to get the right thing, we also need to introduce an extra variable and change the underlying set of our space (it no longer consists 2-tuples but of 3-tuples, and we mod out by some equivalence relation). For the details of the construction you could read the corresponding Wikipedia article on Projective space. I will also try to give a short summary in the following: define the projective plane $\mathbb{P}^2$ over your ground field $k$ as follows: the underlying set is $k^3 \setminus \{(0,0,0)\}/\sim$, where $\sim$ denotes the following equivalence relation: $(a,b,c) \sim (d,e,f)$ iff there is a $\lambda \in k \setminus \{0\}$ such that $(\lambda a, \lambda b, \lambda c) = (d,e,f)$ (in other words, if the two points lie on the same line through the origin in $k^3$). We denote the equivalence class of a point $(a,b,c)$ by $(a:b:c)$, as the equivalence relation we introduced tells us that we only care about the ratios of the coordinates. We also endow this set with a certain topology (the Zariski topology), but I do not want to elablorate on the details here. But also without knowledge of the topology, you can see that, as a set, we have created a space that contains the underlying set of the affine plane: If you look at the subset $\{(a:b:c) \subset \mathbb{P}^2| c \neq 0\}$, then you see that every equivalence class in that set has a unique representation as $(a':b':1)$, as multiplying $(a:b:c)$ by $\frac{1}{c}$ will not change the equivalence class of $(a:b:c)$. Now, we can just forget about the 1 in $(a':b':1)$ and only look at $(a',b')$. Now, I intentionally separate $a'$ and $b'$ by a comma, because the representation $(a':b':1)$ was unique, which means that we don't need to consider equivalence classes any further. If you think about it, this means that we have identified $\{(a:b:c) \subset \mathbb{P}^2| c \neq 0\}$ with the subset $k \times k$, which was the underlying subset of our affine plane $\mathbb{A}^2$. Thus $\mathbb{P}^2 = \mathbb{A}^2 \cup \{(a:b:c) \subset \mathbb{P}^2| c = 0\}$. You can check for yourself that $\{(a:b:c) \subset \mathbb{P}^2| c = 0\}$ is a copy of $k$, i.e. a line (plus an additional point, to be precise). This is why we can say that $\mathbb{P}^2$ is the affine plane plus a line at infinity.

Now we also want to define curves in $\mathbb{P}^2$ by zero sets of polynomials. This is more complicated than previously: our points in $\mathbb{P}^2$ have three coordinates , therefore we need an additional variable. Also, we need to cope with the difficulty that we're defining a function on an equivalence class of points. In the end, we solve these problems by only considering homogeneous polynomials in the variables $x,y,z$: for a homogeneous polynomial f(x,y,z), we have that $f(\lambda x, \lambda y, \lambda z) = 0 \Leftrightarrow f(x, y, z) = 0$ for all $\lambda \in k \setminus \{0\}$. Thus, if $f$ is zero on one representative of an equivalence class $(a:b:c)$, it will be on all of them, which is just what we need. The set of equivalence classes $(a:b:c)$, where $f$ is zero will be the curve defined by $f$ in $\mathbb{P}^2$. If you rememeber how we "extracted" $\mathbb{A}^2$ from $\mathbb{P}^2$, you can also check now, what $f$ looks like on the $\mathbb{A}^2$-part of $\mathbb{P}^2$: just set the $z$-variable of $f$ equal to 1 and look at the zero set of $f(x,y,1)$ in the affine plane, as you did before. This process is called de-homogenising.

Now you can understand what you're doing when you're homogenising a polynomial in two variables: you reverse the process of de-homogenesing, i.e. you "recover" a curve that is defined on the projective plane, by the zero set of the homogeneous polynomial. On the "affine part" of your projective plane, your curve will just look like the zero set of the polynomial that you started with.

Example: look at the line in $\mathbb{A}^2$ defined by $ax +by + c =0$, with $a,b,c \neq 0$. we can homogenize this to $ax +by + cz =0$. This is the equation of a line in $\mathbb{P}^2$, and on the affine part $z=1$, it looks like the line $ax +by + c =0$. On the "infinte part" part $z = 0$, it has additional points: the equation then reads $ax +by =0$. By linear algebra, you know that this has a one-dimensional solution space, i.e. all solutions are of the form $(\lambda d, \lambda e)$ for some fixed $d,e \in k$ and all $\lambda \in k$. Thus on the "infinite part" of $\mathbb{P}^2$ the solutions look like $(\lambda d: \lambda e : 0)$, which is just a single point in $\mathbb{P}^2$ by the defining equivalence relation. Thus, homogenizing the equation $ax +by + c=0$ resulted in obtaining one additional point of the line at infinity.

Solution 3:

This might help (a post by Qiaochu Yuan). Suppose you have some curve $C$ defined by $x^2+y^2 = 1$ on the plane $z=1$. We want each side of the equation to have the same degree. So we multiply each term on the LHS by $z^0$ and the term on RHS by $z^2$.

This is also a useful reference.