1
$\begingroup$

A topological group is a group $G$ whose underlying set is equipped with a topology such that:

$(1)$ the multiplication map $\mu: G \times G \rightarrow G,$ given by $(x,y) \rightarrow xy$ is continuous if $G \times G$ has the prouct topology.

$(2)$ the inversion map $I : G \rightarrow G$, given by $x \rightarrow x^{-1}$ is continuous.

I don't understand how a function between the two groups $G \times G$ to $G$ can be continuous. How do you define a continuous map between two groups?

$\endgroup$
  • 2
    $\begingroup$ "whose underlying set is equipped with a topology" $\endgroup$ – Michael Albanese May 23 '17 at 15:26
  • $\begingroup$ We are considering continuous functions between topological spaces. $\endgroup$ – Dietrich Burde May 23 '17 at 15:28
  • $\begingroup$ But how do you define $xy$ in a topological space? Isn't this an operation? $\endgroup$ – Oliver G May 23 '17 at 15:29
  • 2
    $\begingroup$ It's true that $\mu$ defines a group operation here (by hypothesis). But that's not important for the question, which is whether the map $\mu : G \times G \to G$ is continuous with respect to the topologies on $G$ and $G \times G$. $\endgroup$ – Travis Willse May 23 '17 at 15:34
  • $\begingroup$ The definition gives a map between two groups, but it says that the map between the two groups is continuous. How do you define a continuous map between two groups? Is it referring to the topological spaces themselves instead of the groups? If so, how do you shows that a function $(x,y) \rightarrow xy$ is continuous? I don't understand how the multiplication operation from the two groups $xy$ can be an element of $G$. Is it just take $(x,y)$ from $G \times G$, transform it to $xy = z \in G$, and then treat that element $z$ as an element in $G$ as a topological space? $\endgroup$ – Oliver G May 23 '17 at 15:42
3
$\begingroup$

If you look at your statement, it says "is equipped with a topology . . ."

So, the way you think of a continuous map $G\times G \longrightarrow G$ is the same as you think of any continuous map, that is, the preimage of any open set is open. You may want to consider that in this case, you are looking at the product topology on $G\times G,$ and then you can use what you know about the definition of the product topology to help you out.

$\endgroup$
  • $\begingroup$ So the elements of $G$ from this map are of the form $xy$? How do you define this product operation in a topological space? $\endgroup$ – Oliver G May 23 '17 at 15:33
  • $\begingroup$ A topological group is a set that you equip with a topology as well as a group structure (satisfying certain axioms). For example, consider the set $\mathbb{R}$ with its usual topology. Then $+$ gives us a group structure on $\mathbb R$. Further, the maps $(x,y)\mapsto x+y$ and $x\mapsto -x$ are continuous, so $(\mathbb{R},+)$ forms a topological group. $\endgroup$ – florence May 23 '17 at 15:39
  • $\begingroup$ I can see how $x \rightarrow -x$ is continuous since it's just the negative of $x$, but how would you show that $(x,y) \rightarrow x + y$ is continuous? How do you define inverse images of elements of the form $x+y$? $\endgroup$ – Oliver G May 23 '17 at 15:49
  • $\begingroup$ We're most interested in the inverse images of the open intervals (because these form a basis for $\mathbb{R}$). If $\mu$ is the binary operation, then we'd be looking to see if $\mu^{-1}\left( (a,b)\right)$ is open for all $a,b\in \mathbb{R}$. We're trying to find $\{x,y\in \mathbb{R}| (x+y)\in(a,b)\}$ which can be represented as a compound inequality in two variables. $\endgroup$ – rnrstopstraffic May 23 '17 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.