{"id":276,"date":"2026-10-19T02:11:05","date_gmt":"2026-10-19T00:11:05","guid":{"rendered":"https:\/\/blogs.ethz.ch\/sirawit\/?p=276"},"modified":"2026-10-19T02:11:33","modified_gmt":"2026-10-19T00:11:33","slug":"computer-vision-and-algebraic-geometry","status":"publish","type":"post","link":"https:\/\/blogs.ethz.ch\/sirawit\/computer-vision-and-algebraic-geometry\/","title":{"rendered":"Computer Vision and Algebraic Geometry"},"content":{"rendered":"\n

It might be weird for me to register for the computer vision course alongside my Master’s thesis1<\/a><\/sup>, but as my friends know, I’m pretty much originally a gamer before I’m even started to be interested in mathematics (and then computer science). After playing some video games, I started to form another dream of building my own robot! Interesting!<\/em><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

So I asked Mr. Dae (i.e. Assoc. Prof. Nattee Niparnan, Computer Engineering, Chulalangkorn University<\/a>) to teach me robotics, and I sent him the current course offerings at ETH. He suggest me to begin with Computer Vision<\/a>, so here I am.<\/p>\n\n\n\n

(Also, if at some point, I turn out to be too busy doing these than spending adequate time with the thesis, I might consider withdrawing from the courses later on… For now let’s just see how it goes…)<\/p>\n\n\n\n

Now, as a chuunibyou and edgy person as I am right now, I’m often attracted to recasting things into an algebro-geometric language.<\/p>\n\n\n\n

Note:<\/em> I wrote this post myself, but I used ChatGPT to catch some of the errors. In particular, let me put a big disclaimer here. Disclaimer:<\/strong> Some statements are false! I try to fix them as much as I could, and the main idea should still work, but there may be bugs and I don’t have enough time and energy to polish to the cleanest point, so that’s it.<\/p>\n\n\n\n

The Basics of Geometry<\/h2>\n\n\n\n

Let us fix a base field \"\mathbb{C}\" and recall the basic settings. The affine \"n\"-space, denoted by \"\mathbb{A}^n\", is simply the set \"\mathbb{C}^n\" without the additive structure of a vector space. We equip this with Zariski topology and the sheaf of regular functions to see that this is just the scheme \"\operatorname{Spec} \mathbb{C}[z_1, \dots, z_n]\" over \"\operatorname{Spec}\mathbb{C}\".<\/p>\n\n\n\n

Now we go to the projective case, which is the main protagonist of the relevant geometry here. We define the projective space \"\mathbb{P}^n\" as \"\mathbb{C}^{n+1}/{\sim}\", where \"(z_0, z_1, \dots, z_n) \sim (z'_0, z'_1, \dots, z'_n)\" whenever there exists some \"\lambda \in \mathbb{C}^\times\" such that \"z'_i = \lambda z_i\" for all \"i \in \{0, \dots, n\}\". Equipping this with Zariski topology and the sheaf of regular functions turns this into the scheme \"\operatorname{Proj} \mathbb{C}[z_0, \dots, z_n]_\bullet\" over \"\operatorname{Spec}\mathbb{C}\".<\/p>\n\n\n\n

Now, there is another important construction, which many books don’t appreciate but I think we should introduce it here: the dual projective space. We define \"\mathbb{P}^{n*}\" to be the projectivization of the vector space \"(\mathbb{C}^{n+1})^*\" of \"\mathbb{C}\"-linear forms of \"\mathbb{C}^{n+1}\". We can think of an element of \"\mathbb{P}^{n*}\" as a hyperplane, because the zero locus of a projectivized linear form is really just a hyperplane in \"\mathbb{P}^n\".<\/p>\n\n\n\n

There is now an obvious map<\/strong> \"\mathbb{P}^n \to \mathbb{P}^{n*}\". How? Let the point \"[w_0 : \dots : w_n]\" be sent to the linear form \"(z_0, \dots, z_n) \mapsto \sum_{i=0}^n w_i z_i\", and then collapsed through the projectivization into its class in \"\mathbb{P}^{n*}\". In this blog post, my decision is to denote this map by the symbol \"\intercal\", and write \"x \mapsto x^\intercal\" for such operation (for \"x \in \mathbb{P}^n\"). You may appreciate (or hate) this notation later on.<\/p>\n\n\n\n

Convention:<\/strong> Here I try to adapt to both the culture of algebraic geometry and the culture of computer vision, so, for a point \"p \in \mathbb{P}^n\" (with a normal font for \"p\"), this is just a point, and the homogeneous coordinates are written \"p_0, p_1, \dots, p_n\". For a single point \"p\" there may be multiple choices for its coordinate, i.e., if \"p_0, p_1, \dots, p_n\" is a coordinate, then one can just pick \"\lambda \in \mathbb{C}^\times\" and then \"\lambda p_0, \lambda p_1, \dots, \lambda p_n\" would also be another coordinate. The convention is that if I write \"p = [p_0 : \dots : p_n]\" then this means “let \"p_0, \dots, p_n\" be any possible form of the homogeneous coordinates for \"p\"; it doesn’t matter what we pick”<\/em>. The “it doesn’t matter what we pick” is an assertion, and technically has to be check every time I write this and claim something, but we will implicitly keep this in the back of our minds and just work with it (and stop ourselves from doing any operation that isn’t invariant under the choice of such coordinate). Now, when I write \"\mathbf{p}\" (in boldface), it is meant to be understood as the vector \"\begin{pmatrix}p_0\\p_1\\\vdots\\p_n\end{pmatrix} \in \mathbb{C}^{n+1}\" or also the corresponding matrix in \"\operatorname{Mat}_{(n+1) \times 1}(\mathbb{C})\". That is, boldfaces represent viewing things in a linear algebraic manner<\/strong>, while normal font represents viewing things abstractly (and also in the algebro-geometric sense). For points, we will often use \"\mathbf{x}\" to denote the vector of coordinates corresponding to a point in \"\mathbb{P}^2\" and use \"\mathbf{X}\" to for a point in \"\mathbb{P}^3\".<\/p>\n\n\n\n

Points and Lines<\/h2>\n\n\n\n

There is a computational trick that I didn’t know before. If \"p, q \in \mathbb{P}^2\" are 2D projective points, then what would be the line passing through them? Before that, what is a line in \"\mathbb{P}^2\"? Algebraically we may think of a line as a projective subvariety defined as the zero locus of a linear polynomial. And this can be interpreted computationally easily: let \"a, b, c \in \mathbb{C}\" be not all zero, then the projective variety \"V_{\mathbb{P}^2}(az_0 + bz_1 + cz_2)\" is said to be a line. And we can then interpret this linear-algebraically: a coordinate \"(z_0, z_1, z_2)\" is on the line whenever \"az_0 + bz_1 + cz_2 = 0\" (and then projectivize this). What I want to say is that the projectivization of the linear form \"(z_0, z_1, z_2) \mapsto az_0 + bz_1 + cz_2\" is an element of \"\mathbb{P}^{2*}\" corresponding to this line! Therefore, a line in \"\mathbb{P}^2\" is just an element of \"\mathbb{P}^{2*}\" (one has to check the reverse correspondence, but this is not hard).<\/p>\n\n\n\n

An element of \"\mathbb{P}^{2*}\" can therefore be represented by homogeneous coordinates \"[a : b : c]\", where its line locus is \"\{z \in \mathbb{P}^2 \colon az_0 + bz_1 + cz_2 = 0\}\". This means, for a line to pass through \"p\" and \"q\", such line, parametrized by \"[a : b : c]\", must satisfy<\/p>\n\n\n\n

  <\/span>   <\/span>\"\[\begin{cases}ap_0 + bp_1 + cp_2 = 0\\aq_0 + bq_1 + cq_2 = 0.\end{cases}\]\"<\/p><\/p>\n\n\n\n

What are the answers to this linear system of equations? Surprisingly, a clean answer here is \"\begin{pmatrix}a\\b\\c\end{pmatrix} = \mathbf{p} \times \mathbf{q}\" (cross product). We can check this explicitly. Recall that<\/p>\n\n\n\n

  <\/span>   <\/span>\"\[\mathbf{p} \times \mathbf{q} = \begin{pmatrix}p_1q_2 - p_2q_1\\p_2q_0 - p_0q_2\\p_0q_1 - p_1q_0\end{pmatrix}.\]\"<\/p><\/p>\n\n\n\n

You can plug this into the equations and see that they are satisfied. Now, I think the uniqueness question isn’t hard. By linear algebra, the solution space \"(a, b, c) \in \mathbb{C}^3\" cut out by two linear equations is a one-dimensional subspace (if \"p \ne q\"), or a two-dimensional subspace (if \"p = q\"). For the case of \"p \ne q\", by projectivization, such one-dimensional subspace reduces to a single object in \"\mathbb{P}^{2*}\". This is the line passing through \"p\" and \"q\". For the \"p = q\" case there are still infinitely many lines passing through them, and so there is not much we can do to try and extract one correct line. We have:<\/p>\n\n\n\n

Proposition 1.<\/strong> For a pair \"(p, q) \in (\mathbb{P}^2)^2\" of distinct points, there exists a unique line passing through them. If \"p \ne q\" then \"(\mathbf{p} \times \mathbf{q})^\intercal \in \mathbb{P}^{2*}\" is that line. \"\square\"<\/p>\n\n\n\n

Now, in computer vision, we often represent things with arrays (and hence vectors and matrices), so instead of keeping things in \"\mathbb{P}^{2*}\", we often write down \"\mathbf{l} \in \operatorname{Mat}_{3 \times 1}(\mathbb{C})\", representing an element \"l \in \mathbb{P}^2\" such that \"l^\intercal \in \mathbb{P}^{2*}\" is its corresponding line. In practice, we’d use the following characterization: a point \"x \in \mathbb{P}^2\" lies on the line \"l\" if and only if \"\mathbf{l}^\intercal\mathbf{x} = 0\" (if and only if \"l^\intercal(x) = 0\" as well).<\/p>\n\n\n\n

Onto the next computational trick, suppose \"l, l' \in \mathbb{P}^2\" are “lines” (i.e. \"l^\intercal, l'^\intercal \in \mathbb{P}^{2*}\" are the corresponding lines). What is the intersection of these two lines? The surprising computational trick is that we can write the lines’ data in vectors \"\mathbf{l}, \mathbf{l'} \in \mathbb{C}^3\", and then consider the cross product \"\mathbf{x} := \mathbf{l} \times \mathbf{l'}\". Then \"x \in \mathbb{P}^2\" is the intersection point (with its homogeneous coordinates being \"\mathbf{x}\") of the two lines! This works whenever \"l \ne l'\". Let us show this. Let \"l = [l_0 : l_1 : l_2]\" and \"l' = [l'_0 : l'_1 : l'_2]\". Then \"\mathbf{x} = \mathbf{l} \times \mathbf{l'} = \begin{pmatrix}l_1l'_2 - l_2l'_1\\l_2l'_0-l_0l'_2\\l_0l'_1-l_1l'_0\end{pmatrix}\". If \"\mathbf{x} = 0\" then \"l_1l'_2 = l_2l'_1\", \"l_2l'_0 = l_0l'_2\", and \"l_0l'_1 = l_1l'_0\", i.e. \"l = l'\", a contradiction to our assumption that \"l \ne l'\". So \"\mathbf{x} \ne 0\", and so \"x\" is well-defined as an element of \"\mathbb{P}^2\". Now let us check that \"x\" lies on \"l\" and \"l'\". Indeed, \"\mathbf{l}^\intercal\mathbf{x} = l_0(l_1l'_2-l_2l'_1) + l_1(l_2l'_0-l_0l'_2) + l_2(l_0l'_1-l_1l'_0) = 0\", so \"x\" lies on \"l\". Similarly one can check that \"\mathbf{l'}^\intercal\mathbf{x} = 0\", and so \"x\" lies on \"l'\" as well. Observe that maybe \"l\" and \"l'\" can be parallel in the standard affine patch, but then that means they intersect at the point at infinity, and that’s fine. This gives the following dual to the Proposition 1:<\/p>\n\n\n\n

Proposition 2.<\/strong> For a pair \"(l, l') \in (\mathbb{P}^2)^2\" of distinct “lines”, there exists a unique point \"x \in \mathbb{P}^2\" lying on the intersection of \"l^\intercal\" and \"l'^\intercal\", and a homogeneous coordinates of \"x\" can be taken to be \"\mathbf{x} := \mathbf{l} \times \mathbf{l'}\". \"\square\"<\/p>\n\n\n\n

(Uniqueness can be proven similarly to Proposition 1.)<\/p>\n\n\n\n

Conics<\/h2>\n\n\n\n
\"\"
JensVyff, CC BY-SA 4.0 https:\/\/creativecommons.org\/licenses\/by-sa\/4.0<\/a>, via Wikimedia Commons<\/figcaption><\/figure>\n\n\n\n

Depending on where you’re from, you might define a conic as:<\/p>\n\n\n\n