6
$\begingroup$

Let $E$ be a locally convex topological vector space over $\mathbb{R}$. The projectivization $PE$ is the quotient of $E\backslash\{0_{E}\}$ with respect to the equivalence relation $e\sim f$ if $e=\lambda f$.

Is $PE$ a Tychonoff (i.e. completely regular Hausdorff) space?

As far as I can tell, the theorems about the quotient uniform spaces do not apply. On the other hand, it is plausible to expect that this is a known fact.

I can show that $PE$ is completely Hausdorff, i.e. any two points can be separated by a real-valued continuous function. Indeed, if $e\not\sim f$, take $\mu,\nu\in E^{*}$ such that $\left<\mu,e\right>=1, \left<\nu,e\right>=0, \left<\mu,f\right>=0, \left<\nu,f\right>=1$, and consider the map $\mu\oplus \nu:E\to \mathbb{R}^2$. By the definition of the quotient, this map induces a map $\varphi: PE\to P\mathbb{R}^2=S^1$. Since the latter is Tychonoff, we can separate the images of the classes of $e$ and $f$ by a continuous function.

$\endgroup$
3
  • $\begingroup$ I am afraid that your proof of the complete Hausdorffness can contain a gap because for the locally convex space $\mathbb R^\omega$ the projective space is not Urysohn: for any sequence of non-empty open sets $U_1,\dots,U_n$ in $P\mathbb R^\omega$ the intersection $\overline U_1\cap\dots\cap\overline U_n$ is not empty. The fact that the projective space of $\mathbb R^\omega$ is not Urysohn (and hence not completely Hausdorff) has been first noticed by Gelfand and Fuks, the reference to their paper can be found in: doi.org/10.1016/j.topol.2021.107909 $\endgroup$ Nov 19, 2021 at 21:04
  • $\begingroup$ @TarasBanakh thank you, you are right, and here is the gap: the map $\mu\oplus\nu$ does not work because it has a non-trivial kernel. Perhaps you could post your comment as an answer? Also, are you sure that the closures are needed? take the pre-images of $U_i$'s, they contain "shifted" subspaces of finite co-dimension, which necessarily intersect $\endgroup$
    – erz
    Nov 20, 2021 at 6:53
  • $\begingroup$ The closures are needed because the projective spaces are Hausdorff. $\endgroup$ Nov 20, 2021 at 7:24

1 Answer 1

5
$\begingroup$

The projective space $PE$ of a topological vector space $E$ is Hausdorff but in general is not Tychonoff, not functionally Hausdorff and even not Urysohn (let us recall that a topological space is Urysohn if any distinct points have disjoint closed neighborhoods).

As a suitable counterexample, consider the countable product of lines $E=\mathbb R^\omega$. The projective space $P\mathbb R^\omega$ is superconnected in the sense that for any non-empty open sets $U_1,\dots,U_n$ in $P\mathbb R^\omega$ the intersection of their closures $\overline U_1\cap\dots\cap\overline U_n$ is not empty. This pathological property of the projective space $P\mathbb R^\omega$ was first noticed by Gelfand and Fuks in 1967.

A countable counterpart of the projective space $P\mathbb R^\omega$ is the projective space $\mathbb QP^\infty$, whose topology has been characterized in this paper.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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