Linear Functionals, Dual Spaces & Linear Transformations Between Them

[sponsoredwalk]

I have a question about mappings that go from a vector space to the dual space, the
notation is quite strange.

A linear functional is just a linear map f : V → F.
The dual space of V is the vector space L(V,F) = (V)*, i.e. the space
of linear functionals, i.e. maps from V to F.
L(V,F)= { f : V → F | [f(x + y) = f(x) + f(y)] ⋀ [ f(λx) = λf(x) ]}.
That' supposed to read "The set of all functions from V to F such that f is linear" (i.e. a
linear functional), if you have a better, clearer way to express that please post it!

When you're dealing with a linear transformation from a vector space to the dual space
I am really worried about the form it takes: T : V → L(V,F), i.e. T : V → (V)*.
Reading the start of Spivak's Calculus on Manifolds he mentions this explicitly and more
as below. This is what I find so weird, you're mapping an element of a vector space
to a map, not just some element but to an element of a vector space which
itself is a map, mapping from the vector space to the field...

If (ℝⁿ)* is the dual space to ℝⁿ, with x ∈ ℝⁿ,
define φx ∈ (ℝⁿ)* by φx(y) = <x,y>,
define T : ℝⁿ → (ℝⁿ)* by T(x) = φx. This comes from Spivak's CoM, He's
clearly written that φx ∈ (ℝⁿ)*, but I get the feeling that sometimes
you're working with the scalar value <x,y> also supposed to be in (ℝⁿ)*
& other times not. For example in Susskind's quantum mechanics lectures he explicitly
says that this is all just a way to play with inner products & get complex numbers out,
but again I want to get this rigorous because it's troubling me with regard to differential
forms in particular. But to put it all into my notation of:

T : ℝⁿ → (ℝⁿ)* | x ↦ T(x) = φx.

the element φx is itself a map from ℝⁿ to ℝ so you pick a vector from the
dual space and map it onto the inner product. To spell it all out:

T : ℝⁿ → L(ℝⁿ,ℝ) | x ↦ T(x) = φx : ℝⁿ → ℝ | y ↦ φx(y) = <x,y>.

Here x ∈ ℝⁿ is mapped onto T(x) = φx ∈ (ℝⁿ)* = L(ℝⁿ,ℝ) but this
itself is mapping y ∈ ℝⁿ to <x,y> ∈ ℝ

I'm going to say that this is very weird. For example, does this:

T : ℝⁿ → (ℝⁿ)* | (λx + y) ↦ T(λx + y) = λφx + φy

then

λφx : ℝⁿ → ℝ | z ↦ λφx(z) = λ<x,z>
φy : ℝⁿ → ℝ | z ↦ φy(z) = <y,z>,

or more explicitly,

T : ℝⁿ → (ℝⁿ)* | (λx + y) ↦ T(λx + y) = λφx + φy = [λφx : ℝⁿ → ℝ | z ↦ λφx(z) = λ<x,z>] + [φy : ℝⁿ → ℝ | z ↦ φy(z) = <y,z>] = λ<x,z> + <y,z>


make sense? Clearly here you've got this big chain of map to the value of the map which
itself is a map to the real numbers & I point this out because I see no reason to think it's
wrong. Any thoughts? The notation is the most important thing here but if you would be
able to include some thoughts on how this notation works in quantum mechanics &
differential forms that would be great.

 

[Fredrik]

It gets weird towards the end, where you dropped the subscript on the φ and then confused y with z and so on, but you seem to have understood these maps pretty well. So I'm not sure what you need help with.

In QM, we're dealing with a complex Hilbert space \mathcal H. The inner product is antilinear in the first variable and linear in the second. (That's the physicist's convention. Mathematicians do it the other way round). For each x\in\mathcal H, we can define \phi_x\in\mathcal H^* by \phi_x(y)=\langle x,y\rangle for all y\in\mathcal H. Actually, I prefer to just use the notation \langle x,\cdot\rangle instead of introducing the new symbol \phi_x. It can be shown that the map x\mapsto\langle x,\cdot\rangle is an antilinear isometric bijection from \mathcal H onto \mathcal H^*. (Details in posts 13-14 here. There are a few typos in there, and a broken LaTeX formula, but I don't think that will cause too much confusion). The fact that this map is surjective ensures that for every \phi\in\mathcal H^*, there's an x\in\mathcal H such that \phi=\langle x,\cdot\rangle.

 

[sponsoredwalk]

Wow, some of those mistakes of mine were inexcusable! Thanks, they're fixed now.
So you consider the example I gave, where you pick z out from ℝⁿ in L(ℝⁿ,ℝ)
& dot it with the maps λφx + φy to be correct?
I'm just not used to working with something like this so I needed clarification on that part.

Thanks for the link, that part makes sense but the whole concept itself is quite roundabout
and weird at first, has caused me untold confusion & stress to get to be able to write it all
out the way it is in my OP. I would prefer the <x,.> notation as well, it's just clearer.
I was watching the Susskind quantum mechanics lectures and this issue came up, it's basically
a longstanding issue from something I'm trying to understand as regards differential forms & only
this minute can I proceed due to this post so thanks.

 

[micromass]

Hi sponsoredwalk!

Everything you did in your OP looks correct to me. Except:

T : ℝⁿ → (ℝⁿ)* | (λx + y) ↦ T(λx + y) = λφx + φy = [λφx : ℝⁿ → ℝ | z ↦ λφx(z) = λ<x,z>] + [φy : ℝⁿ → ℝ | z ↦ φy(z) = <y,z>] = λ<x,z> + <y,z>

the last equation isn't really true. You have a function and on the right-hand side you have a number. You seem to be saying that

\phi_x=&lt;x,z&gt;

which isn't true. It is true however that

\phi_x(z)=&lt;x,z&gt;.

Anyway, I also prefer the notation &lt;x,\cdot&gt;, actually...

 

[sponsoredwalk]

I think you're right micromass but I got this idea from the notation used for differential forms.
I'm now trying to work out the consequences for differential forms of this new approach to
the notation, I think I'll have to post some definitions & ask you guys how it all makes
sense because it's not so clear. We'll see how I get on.

 

[Fredrik]

So you consider the example I gave, where you pick z out from ℝⁿ in L(ℝⁿ,ℝ)
& dot it with the maps λφx + φy to be correct?

I wouldn't describe it as dotting a member of the vector space with a map, but your calculation looks OK except for the for the problem that micromass pointed out. I would just write T(\lambda x+y)(z)=(\lambda Tx+Ty)(z)=(\lambda\phi_x+\phi_y)(z)=\lambda\phi_x(z)+\phi_y(z)=\lambda\langle x,z\rangle+\langle y,z\rangle. Note that the on the left, T is acting on λx+y, and the result of that is acting on z.

 

[sponsoredwalk]

Yeah bad language on my part, apologies. So for T : ℝⁿ → (ℝⁿ)* could I write:

T : ℝⁿ → (ℝⁿ)* | (λx + y) ↦ T(λx + y)(z) = (λφx + φy)(z) = λφx(z) + φy)(z) = λ<x,z> + <y,z>.

Is that flawed? Do you need to stop once you've got (λφx + φy)
and then construct a new transform from z to (λφx + φy)(z)?

It seems that the T : ℝⁿ → (ℝⁿ)* | (λx + y) ↦ T(λx + y)(z)
stuff is the correct way to do this but is this not falling into the problem micromass
pointed out? Does T : ℝⁿ → (ℝⁿ)* not map to an element of (ℝⁿ)*, some function,
that you then need to manipulate to get out a scalar value? You see the notation
used in differential forms makes it appear, at least to me, that T : ℝⁿ → (ℝⁿ)* is
spitting out a scalar value at the end.

Let me give a definition for differential forms to show you what I mean, it looks like they
are doing it the way I am with T : ℝⁿ → (ℝⁿ)* | (λx + y) ↦ T(λx + y)(z)
but I'm just not be sure.

Definition: Let Ω be an open set of ℝⁿ.

A differential form ω in Ω is a map ω : Ω → L(ℝⁿ,ℝ) that associates to every x ∈ Ω a linear map ω(x) : ℝⁿ → ℝ.

where:

ω(x) = ∑ᵢⁿ ωᵢ(x) dxᵢ,

ωᵢ(x) := < ω(x), eᵢ >, x ∈ Ω,

http://books.google.ie/books?id=do6...resnum=1&ved=0CBYQ6AEwAA#v=onepage&q&f=false"

So I think you can see I'm not sure what the story is on that issue.

In any case I would love to understand differential 1-forms via this definition in the way
we're doing it here.

I think we can write something like:

ω : Ω → L(ℝⁿ,ℝ) | x ↦ ω(x)(y) = ωx(y) = ∑ᵢⁿ ωᵢ(x) dxᵢ

That's a bit off I'm sure but I think it's close. Could you help me clean that up?

There are two issues, the question of whether you construct a new transform midway
& the question of understanding that definition I gave in terms of what we've written in
here.

 

[micromass]

Yeah bad language on my part, apologies. So for T : ℝⁿ → (ℝⁿ)* could I write:

T : ℝⁿ → (ℝⁿ)* | (λx + y) ↦ T(λx + y)(z) = (λφx + φy)(z) = λφx(z) + φy)(z) = λ<x,z> + <y,z>.

No, I don't like that either. It should be

T:\mathbb{R}^n\rightarrow (\mathbb{R}^n)^*:(\lambda x+y)\rightarrow T(\lambda x+y)

where T(\lambda x+y) is a function:

T(\lambda x+y):\mathbb{R}^n\rightarrow \mathbb{R}:z\rightarrow T(\lambda x+y)(z)=\lambda &lt;x,z&gt;+&lt;y,z&gt;

 

[sponsoredwalk]

Cool, alright I'll keep that distinction in mind. So it would mean that:

ω : ℝⁿ → (ℝⁿ)* | x ↦ ω(x)

where

ω(x) : ℝⁿ → ℝ | y ↦ ω(x)(y) = < ω(x), y>

is a differential form. Basically some standard expression for a differential form
ω = Pdx + Qdy + Rdz could be translated into:

ω : ℝ³ → (ℝ³)* | (x,y,z) ↦ ω(x,y,z) = (P(x,y,z),Q(x,y,z),R(x,y,z))

where:

ω(x,y,z) : ℝ³ → ℝ | (dx,dy,dz) ↦ ω(x,y,z)(dx,dy,dz) = <(P(x,y,z),Q(x,y,z),R(x,y,z)),(dx,dy,dz)> = P(x,y,z)dx+ Q(x,y,z)dy + R(x,y,z)dz,

or so it would seem from page 609 of this link & pages 3 to 5 of this pdf.

Now if you go to the example of Lang's on page 610 he has the equivalent of me writing
ω(x,y,z) = P(x,y,z)dx+ Q(x,y,z)dy + R(x,y,z)dz. My guess is that he is being lazy in the
way he described he would be on page 609, furthermore I would guess there is a heirarchy
of laziness:ω = Pdx+ Qdy + Rdz
ω(x,y,z) = P(x,y,z)dx+ Q(x,y,z)dy + R(x,y,z)dz
ω(x,y,z)(v) = ω(x,y,z)(dx,dy,dz) = P(x,y,z)dx+ Q(x,y,z)dy + R(x,y,z)dz.
Where v = (dx,dy,dz)[/size]
What do you think?