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I was trying to visualize the 3-sphere, $S^3$.

One way is with the help of hopf fibration, as there is a fibre bundle with total space being $S^3$, $$S^1 \hookrightarrow S^3 \rightarrow S^2$$ and this can be thought of as a $S^1$ fiber over $S^2$. The way I visualize a fibre is like drawing the fiber at each point of the base space, so if we have the base space as $S^1$, a circle, and an $S^1$ fiber is thought of as attaching each point of the circle another circle. If this is done in such a way as is usually represented in lectures, what we get is a torus, $T^2$. And also, we know that a 2-torus is nothing but the cartesian product of two circles, i.e. $S^1\times S^1$.

I tend to extend this way of thinking and tend to construct a 3-sphere as the cartesian product of a 2-sphere and a 1-sphere.

My doubt is that whether this way of thinking is right or not. Or is $S^2\times S^1$ is homeomorphic to $S^3$

Eden Zane
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    They are not homeomorphic, since they have different homologies. – simo210 Feb 08 '23 at 09:35
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    The point with fibrations is that they are only locally like a product. For example, if $I$ is the standard interval, then the Moebius strip $M$ is a fibration $I \hookrightarrow M \to S^1$, but this doesn't mean $M = S^1 \times I$. – Jo Mo Feb 08 '23 at 09:37
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    They are not homemorphic, since the surgery knot of $S^2 \times S^1$ in $S^3$ is the unknot, but this is not Kirby equivalent to the surgery knot of $S^3$, which is empty. – Jo Mo Feb 08 '23 at 09:39
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    The probability that someone who cannot distinguish these two spaces can use their descriptions as the result of surgeries to do that is -0.73 – Mariano Suárez-Álvarez Feb 08 '23 at 09:45
  • @JoMo I got it, loosely speaking there will be twist when looking globally in the case of Mobius, otherwise it is a cylinder, and a cylider is obviously not homeomorphic to a mobius. – Eden Zane Feb 08 '23 at 10:09
  • I have to look for what a surgery knot is.. do you have any reference to suggest for that? – Eden Zane Feb 08 '23 at 10:10

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No, these are not homeomorphic. You can see this by comparing their fundamental groups. The fundamental group of $ S^{3} $ is trivial, so $\pi_1(S^3)=0 $ because $ S^{3} $ is simply connected. Meanwhile $\pi_1(S^2 \times S^1) = \pi_1(S^2) \times \pi_1(S^1) = 0 \times \mathbb{Z} = \mathbb{Z} $

PSWadder
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  • Yeah, this calculation makes it clear. But how is connectedness related to the fundamental group explicitly? I mean apart from an indicator for simply connectedness, is there a general relation – Eden Zane Feb 08 '23 at 10:17
  • So generally speaking the choice of a "base point" matters when working with fundamental group, however, in path connected spaces the fundamental group is isomorphic for all choices of base point. Path + trivial fundamental group is one definition of simply connected. – PSWadder Feb 09 '23 at 18:40
  • what is the defintion for doubly connected or in general multiply connected.. and how is it different from "n-connected", I mean how are 2-connected and doubly connected different? – Eden Zane Feb 10 '23 at 08:39