Why is $S^1 \times S^1=$ Torus and $S^2 \ne S^1 \times S^1$

Question

I understand that topologically a Torus, $S^1 \times S^1$ is very different from a 2-Sphere, $S^2$.

My question is more specifically aimed at understand the meaning of $\times$ that is use $S^1 \times S^1$.

1. What is this $\times$ called in mathematics and what are its properties?

I am familiar with Cartesian Product as in $\mathbb{R}^2=\mathbb{R}\times\mathbb{R}$, which is just a tuple in coordinate representation (which is what I understand). But the above encounters seems to suggest that my understanding is incorrect/ partial.

2. So, what exactly am I missing?

Answer 1

Abstractly, the product $\times$ for topological spaces is defined as follows:

Given two topological spaces $X,Y$ the product $X\times Y$ is the unique space (up to unique homeomorphism) such that

1. There are two continuous functions $p_1:X\times Y\to X$ and $p_2:X\times Y\to Y$
2. For any space $Z$ with continuous functions $f_1:Z\to X$ and $f_2:Z\to Y$, there is a unique continuous function $f:Z\to X\times Y$ such that $f_i=p_i\circ f$ for each $i$

This definition carries over to other categories too, like sets (swap "continuous function" with "function"), groups (swap "continuous function" with "homomorphism"), vector spaces (swap "continuous function" with "linear transformation"), and so on. That's the beauty of abstract definitions. The downside is that it is, well, abstract. That kind of definition takes some real work to get used to.

However, we can also construct the product more concretely. The topological space $X\times Y$ has the set of all pairs $\{(x,y)\mid x\in X, y\in Y\}$ as its points. The topology is given by a basis of "rectangles"; any set of the form $U\times V=\{(u,v)\mid u\in U, v\in V\}$ where $U\subseteq X$ and $V\subseteq Y$ are open (we can also restrict ourselves to $U,V$ being elements of some basis for the topologies of $X$ and $Y$ rather than general open sets) (these sets are in fact actual rectangles in the case of $\Bbb R\times\Bbb R$ when $U,V$ are open intervals).

This is a construction you have to use a few times to get used to, but it's nowhere as tough as the abstract definition. Many other categories, like the ones mentioned above, use the same base set for their product. Then in a similar fashion to how we constructed the topology on the product, they build the relevant structure on top of this set. The category of sets is the simplest example, where there is no additional structure, you just have the set of pairs, and that's it.

For later, it could be worth noting that the abstract definition (suitably generalized) is a big contributor to why a product of infinitely many spaces doesn't have the topology you might intuitively expect the first time you encounter it.

Answer 2

The cartesian product is a set-theoretic operation. You can take two sets and form their cartesian product. However, when we talk about the cartesian product of some richer structures, like topological spaces, we mean that the set-theoretic cartesian product has been endowed with a natural structure. For example, when we have two topological spaces, one can define the product topology on their cartesian product which forms the product space. The writing $S^1\times S^1$ refers to this operation.

To see that $S^1\times S^1$ is not homeomorphic to $S^2$, the easiest thing to do is find a topological property of the one that is not enjoyed by the other. I can't think of something very elementary besides the fact that $S^2$ is simply connected and $S^1\times S^1$ is not: every loop (i.e. continuous path with the same endpoints) on $S^2$ can be continuously deformed to a single point. This is not true for $S^1\times S^1$.

Answer 3

We can view $S^1$ as subset of $\Bbb R^2$ in a standard way: $$S^1=\{\,(x_1,x_2)\in\Bbb R^2\mid x_1^2+x_2^2=1\,\}. $$ This allows us to find $S^1\times S^1$ also in a standard way inside $\Bbb R^4$ (or more precisely: $\Bbb R^2\times \Bbb R^2$, but that's "the same") as $$S^1\times S^1=\{\,(x_1,x_2,x_3,x_4)\in\Bbb R^4\mid x_1^2+x_2^2=1\land x_3^2+x_4^2=1\,\}. $$ This is very different from $$S^2=\{\,(x_1,x_2,x_3)\in\Bbb R^3\mid x_1^2+x_2^2+x_3^2=1\,\} $$ (as well as from $S^3=\{\,(x_1,x_2,x_3,x_4)\in\Bbb R^3\mid x_1^2+x_2^2+x_3^2+x_4^2=1\,\} $ ).