1
$\begingroup$

Can you describe what is $S^1 \times$ [0,$\infty)$? Where "$\times$" stands for the product topology and the two factors are with the euclidean topology.

EDIT: What about $S^1 \times S^1$?

$\endgroup$
  • 5
    $\begingroup$ It's a half-infinite cylinder. $\endgroup$ – Parcly Taxel Jan 30 '18 at 16:32
  • 1
    $\begingroup$ Think of an infinite tube or an infinitely long hose. Now cut it in half (i.e., anywhere). $\endgroup$ – Michael Burr Jan 30 '18 at 16:34
  • $\begingroup$ What exactly are you looking for that the description "S^1 \times [0, \infty)$ doesn't already give you? $\endgroup$ – anomaly Jan 30 '18 at 16:35
  • $\begingroup$ It is the same as the closed unit disc minus the origin. $\endgroup$ – orole Jan 30 '18 at 16:39
  • $\begingroup$ @ParclyTaxel What about $S^1 \times S^1$? $\endgroup$ – qcc101 Jan 30 '18 at 16:47
6
$\begingroup$

One way to think of the cross product, is that at each point in the first factor, you are attaching a copy of the second space.

For example: $S^1 \times \{s\}$, where $s$ is a single point, is really just $S^1$, since at each point you are replacing it with a different point.

$S^1 \times \{s,t\}$ is a two copies of the circle. You can view this by taking a circle, and at each point, you are replacing it with two points, and looking at the full collection of these gives two circles.

Actually, $S^1 \times \{1, \dots, n\}$ is nothing but $n$ circles, and $S^1 \times \mathbb Z$ is a countable collection of circles. You can visualize them as stacked along some verticle axis, with a circle at each integer.

Going further, $S^1 \times \mathbb R$ is a circle, but whenever there was a point, you replace it with a line, so you get a circle of lines, or in other words, a cylinder.

$S^1 \times [a,\infty)$ is the same, but with a half open interval.

$S^1 \times S^1$ is a circle of circles, so at each point you attach a circle (for the sake of visualization, say you attach a circle with smaller radius), then you get a torus, with the traditional donut visual.

$\endgroup$
  • $\begingroup$ Addendum, this point of view lends itself well to one generalization of the cross product, known as a fiber bundle. Maybe just an idea to be aware of. One such example would be a mobius band, which looks kind of like $S^1 \times [0,1]$, but with a twist. $\endgroup$ – Andres Mejia Jan 30 '18 at 17:01

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.