6
$\begingroup$

Given an $n$ dimensional vector space $V$, we know it is isomorphic to $\mathbb{R}^n$. Can we not use this isomorphism to define a topology on $V$?... If you want to know if a set of elements in $V$ is closed or open, just look at the image of these elements under the isomorphism; if that set is open/closed, so is its preimage.

I guess you'd have to pick a basis for $V$ first, but still.

Is this right?

In general, say we have a topological algebra $T$ and an algebra $A$, and an isomorphism between them. Can't we use the same argument above to use the topology on $T$ to define a topology on $A$? Thanks!!

$\endgroup$
1
  • 4
    $\begingroup$ Note that the first sentence tacitly assumes that the field of scalars is $\mathbb{R}$. This isn't relevant to the actual question, but it's worth keeping in mind that that needn't be so. $\endgroup$ Sep 10, 2020 at 5:28

2 Answers 2

13
$\begingroup$

That's right, but a priori the topology could end up depending on which choice of basis you pick. That it doesn't is basically due to the fact that linear transformations $\mathbb{R}^n \to \mathbb{R}^n$ are continuous.

In other words, finite-dimensional vector spaces over $\mathbb{R}$ have a canonical topology on them; one way of characterizing this topology is that it's the unique Hausdorff topology making addition and scalar multiplication continuous (exercise). This result is very false in infinite dimensions which is why we need to study all sorts of different topological vector spaces.

$\endgroup$
10
$\begingroup$

Yes, and in fact the algebra is a red herring: we can always transport structure along arbitrary bijections.

For example:

  • If $f:A\rightarrow B$ is a bijection and $*$ is a binary operation on $A$ such that $(A,*)$ is a group, then the binary operation $\star$ on $B$ defined by $$x\star y=f(f^{-1}(x)*f^{-1}(y))$$ has the property that $(B,\star)$ is a group isomorphic to $(A,*)$ (and indeed $f$ itself provides an isomorphism between them).

  • If $f:A\rightarrow B$ is a bijection and $\tau$ is a topology on $A$, then the set $$\sigma=\{\{x\in B: f^{-1}(x)\in U\}: U\in\tau\}$$ is a topology on $B$ such that $(A,\tau)\cong(B,\sigma)$ - with, again, $f$ itself providing a homeomorphism.

And so on.

$\endgroup$
3
  • 1
    $\begingroup$ 100% agree with this answer. There is the opportunity for surprise when there are multiple bijections between $A$ and $B$ that need not give the same topology. (Homeomorphisms are bijections, but bijections need not be homeomorphisms.) $\endgroup$ Sep 10, 2020 at 14:40
  • 1
    $\begingroup$ As an example of what @EricTowers mentions, not too far from the question, you can find a bijection from any finite-dimensional vector space $V$ over $\mathbb{R}$ to any other finite-dimensional vector space $W$ over $\mathbb{R}$ . (Because these are equipotent as sets.) But that does not mean that any choice of bijection will give the "correct" topology (the topology that makes the vector space operations continuous) if you "transport" the natural topology from one space to the other. If the dimensions are different, you cannot even choose a bijection that respects both natural topologies. $\endgroup$ Sep 10, 2020 at 16:35
  • $\begingroup$ (continues) That is a consequence of the Invariance of domain. $\endgroup$ Sep 10, 2020 at 16:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .