Topological basis

  • #1
587
10
Why in ##l_2## space basis ##|1 \rangle=[1 \; 0 \; 0 ...]^{\mathsf{T}}##,
##|2 \rangle=[0 \; 1 \; 0 ...]^{\mathsf{T}}##, ##|3 \rangle=[0 \; 0 \; 1 \; 0...]^{\mathsf{T}}##... is called topological basis?
 

Answers and Replies

  • #2
martinbn
Science Advisor
1,914
603
Because, not every vector in ##l_2## is a finite linear combination of these basis vectors. Some are infnite combinations, convergence is needed, hence topological.
 
  • Like
Likes Math_QED
  • #3
13,563
10,681
Because, not every vector in ##l_2## is a finite linear combination of these basis vectors. Some are infnite combinations, convergence is needed, hence topological.
But they are not open, aren't they? And they aren't suited as a basis for closed sets. Your argument shows, that they cannot be a vector space basis since we needed finite linear combinations. Any open set is a union of ##|n\rangle ##, however, they are not part of the topology except for the discrete.
 
  • #4
martinbn
Science Advisor
1,914
603
But they are not open, aren't they? And they aren't suited as a basis for closed sets. Your argument shows, that they cannot be a vector space basis since we needed finite linear combinations. Any open set is a union of ##|n\rangle ##, however, they are not part of the topology except for the discrete.
What I meant is that if you take the span of those vectors you don't get the whole space you need to take the closer of the span.
 
  • #5
13,563
10,681
I think that this is an abuse of language. Those vectors do not constitute a topological basis, neither do they constitute a vector space basis. My assumption is therefore that the author uses topological basis in contrast to vector space basis, in order to get rid of the finiteness constriction of vector space bases, but they are neither.

They generate the direct product (not direct sum!) ##\ell_2=\mathbb{R}|1\rangle \times \mathbb{R}|2\rangle \times \mathbb{R}|3\rangle \times \ldots ##. To call it a topological basis is plain wrong in my opinion.
 
  • Like
Likes Math_QED
  • #6
martinbn
Science Advisor
1,914
603
I think that this is an abuse of language. Those vectors do not constitute a topological basis, neither do they constitute a vector space basis. My assumption is therefore that the author uses topological basis in contrast to vector space basis, in order to get rid of the finiteness constriction of vector space bases, but they are neither.

They generate the direct product (not direct sum!) ##\ell_2=\mathbb{R}|1\rangle \times \mathbb{R}|2\rangle \times \mathbb{R}|3\rangle \times \ldots ##. To call it a topological basis is plain wrong in my opinion.
Not sure what you mean! No one was talking about the direct sum nor the direct product. And, yes, it is just i name the author uses. I think it is quite standard.
 
  • #7
13,563
10,681
The basis of a topology is defined as subset of open sets. To distinguish it from a topological basis is hopefully not standard! And if, I will no longer use english and topology at the same time.
No one was talking about the direct sum nor the direct product.
Unfortunately. This is what it is about.
 
  • #8
martinbn
Science Advisor
1,914
603
The basis of a topology is defined as subset of open sets. To distinguish it from a topological basis is hopefully not standard! And if, I will no longer use english and topology at the same time.
No, it is not about bases of a topology, but about bases of topological vector spaces.
Unfortunately. This is what it is about.
I still don't understand the ##l_2## space is not the direct sum nor the direct product.
 
  • #9
13,563
10,681
No, it is not about bases of a topology, but about bases of topological vector spaces.
In this case, the wording is more than confusing.
I still don't understand the ##l_2## space is not the direct sum nor the direct product.
As ##\mathbb{R}##-module?
 
  • #10
martinbn
Science Advisor
1,914
603
In this case, the wording is more than confusing.
It can be, that's why the context is important.
As ##\mathbb{R}##-module?
The direct sum is the vector space of all finite sequences of real numbers. The direct product of all sequences. The ##l_2## of all square summable sequences.
 
  • #11
13,563
10,681
That's what I said. They generate the ##\mathbb{R}## module. No basis in either meaning, and not very related to one of the structures of ##\ell_2##. All it does is: We can write the elements of ##\ell_2## as ...

Calling it a topological basis when a basis of a topology is something else, is a very bad choice.
 
  • #12
Math_QED
Science Advisor
Homework Helper
2019 Award
1,703
724
I agree that topological basis is very bad use of language because you immediately think about a collection of opens in a topological space.

I also agree with @martinbn post #2. This is what the author had in mind.

Since ##l^2## is a Hilbert space, a more appropriate name would be "orthonormal basis".
 
  • Like
Likes fresh_42
  • #13
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,899
175
But it's still not a basis by the regular vector space definition, which the author thinks is more important to distinguish.
 
  • #14
Infrared
Science Advisor
Gold Member
789
412
I think this is often called a Schauder basis (https://en.wikipedia.org/wiki/Schauder_basis). I haven't seen the term "topological basis" used for this but it makes sense (although it's not a basis for a topology...)
 
  • #15
86
47
Here's a page from nLab, where the term "topological basis" (not in the sense "basis for the topology") is used, and it seems to be differentiated from both the Hamel and the Schauder basis (although I'm not exactly sure I understand the difference):
https://ncatlab.org/nlab/show/basis+in+functional+analysis
 
  • Like
Likes Infrared
  • #16
Infrared
Science Advisor
Gold Member
789
412
@Dragon27 I think the difference is something like this: Let ##V=C([0,1])##, with the sup norm. Then ##1,x,x^2,\ldots## is not a Schauder basis because not every continuous function is a power series, but it is a topological basis according to your link because every continuous function on a finite interval can be uniformly approximated by polynomials (and none of these monomials can be uniformly approximated by others).
 
  • #17
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,899
175
The link actually claims that you can uniformly approximate x on an interval with higher degree monomials at the bottom. But it does give a similar example using Fourier series.
 
  • Like
Likes Infrared
  • #18
Infrared
Science Advisor
Gold Member
789
412
Thanks, I should have read a bit further before posting...
 

Related Threads on Topological basis

Replies
18
Views
5K
  • Last Post
Replies
12
Views
3K
Replies
8
Views
1K
  • Last Post
Replies
11
Views
2K
Replies
9
Views
802
Replies
4
Views
2K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
10
Views
893
Top