3
$\begingroup$

Consider a real vector space $V$ with dimension $n$ (say $V=\mathbb{R}^n$). The construction I'm describing is similar to the construction of the projective space. Instead of the space of lines, it is the set of (affine) halfspaces, or equivalently the space of "oriented" (affine) hyperplanes.

Define $A$ the space of non-constant affine functions from $V$ to $\mathbb{R}$. For each $f\in A$ define $S(f)$ as the set of solutions of the equation $f(x)\geq 0$:

$$S(f)=\{x\in V /f(x)\geq 0\}$$

Now give to $S(A)$ the final topology with respect to $S$: open sets of $S(A)$ are those whose preimage by $S$ are open. If you prefer, $S(A)$ is the quotient space of $A$ by the equivalence relation $f\sim g : S(f)=S(g)$.

Question : describe $S(A)$ topologically (find a "simple" topological space homeomorphic to $S(A)$).

I'm very interested in special cases esp. $n=2$.

$\endgroup$
  • 5
    $\begingroup$ Isn't this just $\mathbb{S}^n$, the $n$ dimensional sphere? $\endgroup$ – WhatsUp Jul 6 '17 at 12:06
  • $\begingroup$ If you replace affine by linear, it's $\mathbb{S}^{n-1}$. But with affine I don't see a sphere. With $n=1$ it is two separated lines : left halves and right halves. $\endgroup$ – Benoit Sanchez Jul 6 '17 at 16:04
6
$\begingroup$

The space is just $\mathbb{R} \times S^{n-1}$: each oriented hyperplane $H$ is identified by the normal unit vector $v$ together with the inner product of $v$ with an arbitrary $w \in H$.

$\endgroup$
  • 2
    $\begingroup$ I agree with this. If you add the constant non-zero affine functions, (two points : empty and full sets), then I think we get $S^n$. $\endgroup$ – Benoit Sanchez Jul 7 '17 at 7:48
4
$\begingroup$

Each hyperplane not passing through the origin is uniquely determined by its closest point to the origin, plus sign (the normal vector points towards or away from the origin), so the set of such hyperplanes is simply $(\mathbb{R}^n \backslash \{0\})\times \{-1, 1\}.$ The two connected components are glued together along $\mathbb{S}^{n-1}.$

$\endgroup$
  • $\begingroup$ So two copies of $\mathbb{R} \times S^{n-1}$ glued together along $S^{n-1}$? $\endgroup$ – Geoffrey Irving Jul 7 '17 at 6:02
2
$\begingroup$

I get $\mathbb R^n$ with the origin removed, as follows. Choose a continuous order preserving bijection $\delta:\mathbb R^{>0}\to\mathbb R$, say, $\delta(d)=\log(d)$; to a nonzero vector $v$ of length $|v|$ assign the hyperplane orthogonal to $v$ at the distance $\delta(|v|)$ from the origin in the direction of $v$, and the orientation according to the direction of $v$.

$\endgroup$
  • 2
    $\begingroup$ This is homeomorphic to $\mathbb R\times S^{n-1}$ in Adam P. Goucher's answer by mapping $v\mapsto(e^{|v|},v/|v|)$. $\endgroup$ – Sebastian Goette Jul 7 '17 at 7:18
  • 2
    $\begingroup$ @SebastianGoette Well all correct answers must give homeomorphic results :D It is just an alternative way to see a homeomorphism, hopefully handy at least for some cases... $\endgroup$ – მამუკა ჯიბლაძე Jul 7 '17 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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