Cartesian product of $S^1$ and $S^2$ Can anyone help me imagine $S^1 \times S^2$? I understand that $T^2=S^1 \times S^1$, but I don't know what to do with the spheres. I am not even sure if it's in 3D. Thank you!
 A: This is a good example of a 3-dimensional manifold that cannot be embedded into the Euclidean space $\mathbb R^3$.
The best way you can think of it is as a circle such that it has a sphere associated to each and every point of it.
I bet you already know how to parameterise both the circle $S^1$ (call the parameter $\rho$, for instance) and the sphere $S^2$ (say the parameters are $\varphi$ and $\theta$). Then, you can just put your parameterisations together and get 
$$\Phi:(\rho, \varphi,\theta)\mapsto \Phi(\rho, \varphi,\theta)\in S^1 \times S^2.
$$
If you are concerned with regards to its embedding into an Euclidean space $\mathbb R^n$, the best I can offer you is $n=5$. That would come with the $\Phi$ from above, which embeds $S^1$ into $\mathbb R^2$ and $S^2$ into $\mathbb R^3$.
A: Here is something you could try to visualize to help: When you do $S^1 \times S^1$ to get the torus, one way you could have imagined that was by taking a $S^1$ and gluing it to another $S^1$, and keep gluing more $S^1$ until you get a tube. Then you glue the first $S^1$ to the last $S^1$ to close the ends of the tube, and you get your torus.
Now to do $S^2 \times S^1$. You can hold a $S^2$ in your hand, and wrap it inside another $S^2$. Then you wrap another one over that, and keep wrapping more $S^2$ until you have an Earth-sized ball, only with a hollow core. Finally, you glue the outside of the Earth's surface to the inside of the hollow center.
If this is hard to imagine, you could also go back to the torus construction, but instead of making a "tube" by gluing $S^1$s, grab a pencil and paper and draw a circle, then a slightly bigger circle around that, and keep going until you've drawn a disc with a hole in the middle. This is topologically equivalent to the tube and you make the torus by gluing the hole in the annulus to the outside rim of the annulus. 
Now for the higher dimension, instead of a big 2-ball with a little 2-ball missing from the center, you have a big 3-ball with a little 3-ball missing from the center.
Happy visualizing
A: It does not embed in $\mathbb{R}^3$, but can be embedded in $\mathbb{R}^4$ (with the $S^1$ being a tiny circle in radial times the fourth dimension).  Of course, if you can think in $\mathbb{R}^5$ then it is just the product of $S^1\subset\mathbb{R}^2$ with $S^2\subset\mathbb{R}^3$.
