# In what capacity is $S^{3}$ locally the same as $S^{2}\times S^{1}$?

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:

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

And of course there is the isomorphism:

$$S^{1}=U(1)$$

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?

• (principle?) $S^1$-bundles on $S^2$? Jun 1, 2021 at 5:32
• @user10354138 so they're related through the Hopf fibration? Jun 1, 2021 at 5:33
• @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 $(S^2\times S^1,\pi_1,S^2)$. Jun 1, 2021 at 6:15

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 $$SU(2)$$

• +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, Jun 1, 2021 at 7:49
• There is no Lie group homeomorphic to $S^2\times S^1$. Jun 1, 2021 at 8:50
• @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. Jun 1, 2021 at 9:46
• @R.Rankin: The frame bundle over any oriented 3-manifold is trivial. Jun 1, 2021 at 13:12
• @MoisheKohan please see another way of asking what I was thinking: math.stackexchange.com/questions/4158377/… Jun 1, 2021 at 23:25