I am trying to understand the conceptual role of the “point at infinity” in complex analysis and why it seems much more natural and useful there than in real analysis. In topology, one can always take the Alexandroff one-point compactification. So in principle both real and complex lines admit a one-point compactification. Is there a precise sense in which the compactification of $\mathbb{C}$ is canonical while the compactification of $\mathbb{R}^n$ is not? Are there frameworks in real analysis where a “point at infinity” plays a similarly natural role? I suspect that the presence of a holomorphic/conformal structure is part of the explanation, but I do not fully understand in what precise sense.
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4$\begingroup$ Possibly helpful: Why does $\mathbb{R}$ have two infinities instead of one? $\endgroup$Martin R– Martin R2026-02-28 18:31:42 +00:00Commented 10 hours ago
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$\begingroup$ Holomorphicity indeed relates the behavior at infinity to the behavior in the plane (often at $0$) in a way that smooth functions (and iirc even real analytic, to a certain degree) do not. As Martin mentioned, $\mathbb C$ also only has one end, while $\mathbb R$ has two. $\endgroup$Brevan Ellefsen– Brevan Ellefsen2026-02-28 18:35:55 +00:00Commented 10 hours ago
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2$\begingroup$ The Riemann sphere has a rich geometric structure to compliment the analytic structure on the complex plane and the point at infinity plays an important role there. I expect that's part of why it feels more natural. $\endgroup$CyclotomicField– CyclotomicField2026-02-28 18:59:15 +00:00Commented 10 hours ago
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$\begingroup$ To a knot theorist, the compactification of $\Bbb{R}^n$ to $\Bbb{S}^n$ is natural (because modifying the ambient space into a homogeneous one is a good idea). But knot theory in $\Bbb{R}^1$ is exceptionally simple, so this may not address the naturalness there. $\endgroup$Eric Towers– Eric Towers2026-02-28 22:11:01 +00:00Commented 6 hours ago
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$\begingroup$ One can also preserve some of the order structure of $\Bbb{R}$ vector subspaces of $\Bbb{C}$ by gluing a cross-cap onto the boundary of $\Bbb{C}$, obtaining a space equivalent to $\Bbb{RP}^2$. Here, parallel (or antiparallel) lines in $\Bbb{C}$ cross the cap along the same identification and non-parallel lines cross the cap along different identifications. This turns all lines into circles. $\endgroup$Eric Towers– Eric Towers2026-02-28 22:16:44 +00:00Commented 6 hours ago
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4 Answers
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You are comparing apples and oranges. Whether $K=\mathbb R$ or $K=\mathbb C$, the one-point compactification of $K$ is natural since it is naturally the projective line $KP^1$. Meanwhile, the one-point compactification of $\mathbb C^n$ for $n\geq2$ is, analytically speaking, just as unnatural as for $\mathbb R^n$. The natural compactification in each case is, of course, the projective space $KP^n$. It does not make much sense to compare compactifications of $\mathbb C$ and $\mathbb R^n$.
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I would say both are natural. But why is it that we consider the extended real line rather than one-point compactification of $\mathbb{R}$?
Well, how many finite amount of points can we add to $\mathbb{R}$, and how many to $\mathbb{C}$, in order to make it compact?
For $\mathbb{R}$ the answer is that we can add two points, and no more, and the space obtained is the extended real line. But for $\mathbb{C}$, there is only one-point compactification.
The article $N$-point compactifications by Magill is a nice discussion about compactifications by finite amount of points. Theorem 2.9 shows that in particular $\mathbb{R}^n$ for $n > 1$ does not have an $m$-point compactification for any $m > 1$. And so neither does $\mathbb{C}$ which is topologically the same as $\mathbb{R}^2$.
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1$\begingroup$ @BrianTung sorry, I sometimes write things that sound or appear similar to what I intend. I meant is $\endgroup$Jakobian– Jakobian2026-02-28 22:47:17 +00:00Commented 6 hours ago
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$\begingroup$ Just checking! :-) $\endgroup$Brian Tung– Brian Tung2026-02-28 22:49:52 +00:00Commented 6 hours ago
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A meromorphic function on $\mathbb{C}$ is a rational function, this is made possible because of how we define "holomorphic at $\infty$". In fact, working with a Riemann surface, more generally, requires us to use the one-point compactification. If you want to work with complex curves, and use techniques of complex analysis, to study those curves, then it is required to use the one-point compactification.
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3$\begingroup$ The question is about the importance of the one-point compactification of C relative to that of R. So a good answer should contrast the situation in complex analysis with the situation in real analysis. $\endgroup$MJD– MJD2026-02-28 22:42:22 +00:00Commented 6 hours ago
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$\begingroup$ This is not true. For many Riemann surfaces, there are multiple points at infinity. $\endgroup$Ted Shifrin– Ted Shifrin2026-02-28 23:04:12 +00:00Commented 6 hours ago
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$\begingroup$ @TedShifrin You are absolutely correct, however, we deal with maps $X\to S^2$, which are holomorphic, i.e. we compactify $\mathbb{C}^2$ with just one point at infinity. $\endgroup$Nicolas Bourbaki– Nicolas Bourbaki2026-03-01 01:42:47 +00:00Commented 3 hours ago
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Elliptic curves, even outside of real analysis (but still within the real domain), require a point at infinity.
Therefore, it is not only natural to have a point at infinity for them, but also required.
The assumption that the point at infinity is so natural for complex analysis is not correct.
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