Clearly they're not the same topologically as the former is simply connected and the latter not so.

One can consider the three-sphere as diffeomorphic to $SU(2)$

$$S^{3}\simeq SU(2)$$

Now, we also know that there is a relationship for the two-sphere:


And of course there is the isomorphism:


So we can write:

$$S^{2}\times S^{1}\simeq\frac{SU(2)}{U(1)}\times U(1)$$

I'm guessing I can't just cancel out these factors to say that $S^{2}\times S^{1}\simeq SU(2)$, (As these $U(1)$ groups may be different ones in fact) but I'm thinking that maybe locally they're somehow equivalent, perhaps a lie bracket of vector fields on both adhere to the $su(2)$ lie algebra??? Since both manifolds are parallelizable this appears feasible. Could someone please enlighten me?

  • 2
    $\begingroup$ (principle?) $S^1$-bundles on $S^2$? $\endgroup$ Jun 1, 2021 at 5:32
  • $\begingroup$ @user10354138 so they're related through the Hopf fibration? $\endgroup$
    – R. Rankin
    Jun 1, 2021 at 5:33
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    $\begingroup$ @R.Rankin More that the Hopf fibration $h:S^3\to S^2$ and the projection onto the first factor $\pi_1:S^2\times S^1$ are two quotients over the same space with the same fiber, both of which can be given principal bundle structures. In this way the bundle $(S^3,h,S^2)$ can be thought of as a "twisted" version of $(S^2\times S^1,\pi_1,S^2)$. $\endgroup$
    – Kajelad
    Jun 1, 2021 at 6:15

1 Answer 1


There are several levels to this questions. On one hand, they are both three-dimensional manifolds, so technically they are trivially locally the same: they are both locally the same as $\mathbb R^3$. But I guess this is not what you are really looking for.

What you really noticed is the following: for both $S^3$ and $S^2\times S^1$ there is a surjective map to $S^2$ such that the fibers (i.e. the pre-images of each point of $S^2$ through this map) are some copies of $S^1$.

Algebraically, for $S^3$ this map is the group projection to the quotient. For $S^2\times S^1$ this is just the projection to the first factor.

This also means the following: both $S^3$ and $S^2\times S^1$ look like a union of a family of $S^1$'s, and this family is parameterized by a $S^2$. For $S^2\times S^1$ this is obvious: the circles all have the form $\{x\}\times S^1$, where the parameter $x$ ranges throught the points of $S^2$. The projection map we were talking about before is the map that "identifies" each thiat circle to a point. You can visualize this, like trying to "shrink" these $S^1$ till they become points.

For $S^3$, yes, this is the Hopf fibration. This describes the 3-sphere as a union of 1-circles parameterized by a 2-sphere. And the corresponding projection map $S^3\to S^2$ can be seen exactly as the projection map to the quotient $SU(2)/U(1)$.

All that can be summarized as follows: both are $S^1$-bundles on $S^2$, as user10354138 observed.

Topologically, this is similar to what happens with the Moebius strip and with the cylinder: both are "unions of lines, that move around a circle", or $\mathbb R^1$-bundles on $S^1$. The fibers just move around in a different way: in the Moebius strip and in the Hopf fibration, they rotate in a notrivial way. In the cylinder or in $S^2\times S^1$ they just move around in the simplest straightforward way.

I think it would be interesting to rephrase all this from the point of view of Lie groups and Lie algebras, or from vector fields, but I warn against some of the most naive attempts, by reminding that there is no Lie group structure on $S^2$, by the hairy ball theorem. In particular we cannot just write a short exact sequence

$$ 1\to U(1) \to G \to SU(2)/U(1) \to 1 $$

of groups, because the last quotient is not a group quotient, is it? I have not looked properly into this, but maybe we can simply look at which Lie groups are $S^2\times S^1$ and see how much do they look like or unlike $SU(2)$

  • $\begingroup$ +1 Great and thank you! though I'll admit I'm coming from a physics background, I'm using these manifolds as spacelike surfaces connected by Lorentz cobordisms in spacetime. I ultimately to get results have to formulate them in terms of vector fields, $\endgroup$
    – R. Rankin
    Jun 1, 2021 at 7:49
  • $\begingroup$ There is no Lie group homeomorphic to $S^2\times S^1$. $\endgroup$ Jun 1, 2021 at 8:50
  • $\begingroup$ @MoisheKohan I now realize I'm interested in the principle frame bundle over $S^{2}\times S^{1}$ and what it's total space would be as compared to the principle bundle over $S^{3}$ which is trivial. $\endgroup$
    – R. Rankin
    Jun 1, 2021 at 9:46
  • $\begingroup$ @R.Rankin: The frame bundle over any oriented 3-manifold is trivial. $\endgroup$ Jun 1, 2021 at 13:12
  • $\begingroup$ @MoisheKohan please see another way of asking what I was thinking: math.stackexchange.com/questions/4158377/… $\endgroup$
    – R. Rankin
    Jun 1, 2021 at 23:25

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