Understanding Separable Vector Spaces: The Basics Explained

[fog37]

Dear forum,

I am trying to understand what a separable vector space is. I know we can perform the tensor product of two or more vector space and obtain a new vector space. Is that vector space separable because it is the product of other vector spaces?

thanks

 

[Orodruin]

No. A separable space contains a countable dense subset. The product of two vector spaces will not be separable unless the two original vector spaces were.

 

[micromass]

No. A separable space contains a countable dense subset. The product of two vector spaces will not be separable unless the two original vector spaces were.

The notion of a vector space alone is not sufficient for separability to be well defined.

 

[fog37]

Thank you.

What would a countable dense subset be? I understand it contains a subset of vectors that has a particular property. Could you give me an example of what a countable dense subset means from ordinary finite dimensional linear algebra? Let's consider the vector space of vectors
$v=(v_{x} , v_{y}, v_{z})$...

 

[micromass]

Thank you.

What would a countable dense subset be? I understand it contains a subset of vectors that has a particular property. Could you give me an example of what a countable dense subset means from ordinary finite dimensional linear algebra? Let's consider the vector space of vectors
$v=(v_{x} , v_{y}, v_{z})$...

Dense makes no sense on a vector space only. You'll need further structure, like a norm, a distance or a topology.

For example, the vector space $\mathbb{R}^3$ can be equipped with the norm
$$\|v\| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$
A countable dense set can then be given by $$\{v~\vert~v_x, v_y,v_z\in \mathbb{Q}\}$$

But let's start with the basics, do you know what a norm is?

Also, where did you encounter the notion of separable? Seeing the context might help.

 

[Orodruin]

The notion of a vector space alone is not sufficient for separability to be well defined.

Obviously. This is implied by the "dense". Without some sort of topology, this is not well defined.

 

[micromass]

Obviously. This is implied by the "dense". Without some sort of topology, this is not well defined.

You know that. I know that. The point is that the OP might not.

 

[fog37]

Hi micromass,

The norm is, conceptually, the "length" of a vector. I run into the idea of separable vector space in introductory quantum mechanics where Hilbert vector space is said to be separable. this leads down to the discussion of such a space being a tensor product of other vector spaces...

 

[micromass]

Hi micromass,

The norm is, conceptually, the "length" of a vector. I run into the idea of separable vector space in introductory quantum mechanics where Hilbert vector space is said to be separable. this leads down to the discussion of such a space being a tensor product of other vector spaces...

OK, since it's in the context of Hilbert spaces, I can give a simpler definition of separable. All it means is that there is a countable orthonormal basis. That is: there is a subset of the Hilbert space $\{e_n~\vert~n\in I\}$ with $I$ countable, which intuitively means $I$ finite or $I=\mathbb{N}$ such that
1) $\|e_n\| = 1$ for all $n$
2) $<e_n, e_m> = 0$ for $n\neq m$
3) Every $x$ in the Hilbert space can be written as $x = \sum_{n\in I}\alpha_n e_n$ for some $\alpha_n\in \mathbb{C}$ (it can be shown that $\alpha_n =<x,e_n>$). Note this sum is an infinite sum (=series) when $I=\mathbb{N}$ and convergence comes into play, which makes it distinct from linear algebra where all sums are finite.

 

[micromass]

As an example in $\mathbb{C}^3$, take $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. These are orthonormal and form a basis.