MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Dimension theory is quite a sophisticated topic (at least for me), it is fully settled in Shafarevich's book on the first 100 pages.

Shafarevich gives two nice applications of the theory. 1) A proof of Tzen's theorem 2) A proof of existence of a line on a cubic surface in $\mathbb P^3$.

Both applications are geometric statements. I would like to learn about some other applications that are easy to state. I am asking this partially because of an introductory course in algebraic geometry that I am teaching. I would like to give some motivation for developing the dimension theory.

share|cite|improve this question

It's hard to know where to start, since dimension theory is used everywhere. But here is another nice simple geometric application: Any smooth curve in projective space is isomorphic to a curve in $\mathbb{P}^3$. To prove this, start with a curve $C\subset\mathbb{P^n}$, with $n\ge 4$. Observe by computing dimensions that the set of secant and tangent lines won't sweep out of all of $\mathbb{P}^n$. So a general projection will map $C$ to an isomorphic curve in $\mathbb{P}^{n-1}$.

share|cite|improve this answer

Here is a cool idea which gets used a lot. Suppose that a connected algebraic group $G$ acts on a variety $X$. One can prove that the orbits are locally closed using Chevalley's theorem on constructible sets. Moreover, the boundary of an orbit must be a union of orbits. This implies that the orbits of minimal dimension are closed. The most important application of this is the following:

When a connected solvable group acts on a projective variety, there is at least one fixed point.

share|cite|improve this answer
Daniel, thank you for this answer. I am not quite sure though that I understand the phrase in grey. For example $\mathbb C^*$ is acting on an any elliptic curve over $\mathbb C$, and the action is without fixed points. – aglearner Feb 21 '14 at 20:46
@aglearner: no, $\mathbb C^*$ can't act on an elliptic curve, at least not through an algebraic action. Of course, identifying an elliptic curve $\mathbb C/(\mathbb Z+\tau\mathbb Z)$ with $\mathbb C^*/q^{\mathbb Z}$, for $q=e^{2\pi i\tau}$, one sees an obvious action of $\mathbb C^*$, but it is only holomorphic. – ACL Feb 21 '14 at 21:09
ACL, sure I had in mind the non-algebraic action. I can not yet grasp why dimension theory is doing the job in the algebraic case (while it fails in the analytic case)... – aglearner Feb 21 '14 at 21:14
More than dimension theory goes into the proof of the boxed result. The reason things break down for non algebraic actions is that stabilizers no longer need to be closed as illustrated by ACL – Daniel Barter Feb 21 '14 at 21:45
"Closed" meaning, in the Zariski topology. – Allen Knutson Feb 22 '14 at 17:27

This isn't a very direct answer, but one of the things I use all the time in algebraic geometry as opposed to non-algebraic is that if $X$ is dense in $Y$, then $\dim(Y\setminus X) < \dim X$. The topologist's sine curve is a counterexample to this in non-algebraic geometry.

share|cite|improve this answer

I've taught an introductory course recently and regarding students' reaction, Bertini was the best application ever.

share|cite|improve this answer

Most hypersurfaces in $\Bbb{P}^3$ of degree $d > 3$ have no lines! More precisely, there is an open dense subset of $\Bbb{P}^{\binom{d+3}{3} - 1}$ such that each point (or hypersurface) in here has no line. The proof is very cute: We set up an incidence correspondence $X \subseteq \Bbb{P}^{\binom{d+3}{3} - 1} \times \Bbb{G}(1,3) $ that makes precise the notion of a line in a hypersurface. One computes the dimension of $X$ and then shows for $d > 3$ the projection $X \to \Bbb{P}^{\binom{d+3}{3} - 1}$ cannot be dominant, proving the result! Of course in concluding the projection is not dominant we use the following result. Let $f : X \to Y$ be a dominant morphism of irreducible $k$-varieties. Then $\dim X \geq \dim Y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.