Vector space is not isomorphic to its dual

Conversation started Aug 15, 2017 at 8:49.

Leaky Nun

Aug 15, 2017 08:49

How difficult is it to prove that the dual space of an infinite-dimensional vector space is never isomorphic to itself? I tried using cardinalities but it wouldn't work

@BalarkaSen

Tobias Kildetoft

@LeakyNun Why didn't it work?

Balarka Sen

@Daminark Well, let's say image of $f$ is $[a, b]$ and the critical points are $a, x, y, b$.

Consider $f^{-1}[a, x)$. That's a disk in the torus of height smaller than the first saddle.

Leaky Nun

@TobiasKildetoft because R and R^N have the same cardinality

Tobias Kildetoft

@LeakyNun Right, and the dual space has strictly larger cardinality

Leaky Nun

@TobiasKildetoft I don't think so

consider the real sequences with finite support. Its cardinality is c. So is its dual space's.

Daminark

Aug 15, 2017 08:53

@BalarkaSen Would it be a disk? From the picture it seems open

LLL

math.stackexchange.com/questions/2393463/…. I have again edited the question.

Tobias Kildetoft

@LeakyNun Hmm

Balarka Sen

@Daminark Let's move to DC, it's too crowded here

Tobias Kildetoft

@LeakyNun Right, so instead we need to just find sufficiently many linearly independent maps that we exceed the dimension of the original space

Leaky Nun

@TobiasKildetoft I had some progress when restricting the coefficients to 0 and 1, but not much.

Tobias Kildetoft

Aug 15, 2017 08:55

Consider a basis and the corresponding duals. Now imagine adding infinitely many of these in any way you want. Make this precise to get a larger set of independent maps

Martin Sleziak

I'll also add a link to this: Why are vector spaces not isomorphic to their duals? Other posts which are linked there might be of interest, too.

And this MO question is linked in comments: Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional

Secret

A weird extended question, must a mathematical object A be isomorphic to itself because A = A is a tautology?

Leaky Nun

@MartinSleziak how long is that?

@Secret I suggest that you look up the terms you are trying to use before asking questions about them...

in this case, "isomorphic".

Martin Sleziak

@LeakyNun I'm not sure I understand the question...?

Tobias Kildetoft

@Secret There are "places" where this does not hold, but we don't go there :)

Leaky Nun

Aug 15, 2017 09:07

@MartinSleziak how long is the answer... so long

Martin Sleziak

So it was a rhetorical question, right? (And as such, I should not have answered it.)

Tobias Kildetoft

More precisely, those places are called semicategories

Leaky Nun

@MartinSleziak you can say so

Martin Sleziak

Is this a rhetorical question (Y/N)? :-)

Leaky Nun

@MartinSleziak :p

Martin Sleziak

Aug 15, 2017 09:09

A joke stolen from QI. (At least that's where I heard it.)

Tobias Kildetoft

I actually spent some time trying to find the correct notion of a fiat $2$-semicategory, but that turned out to be more trouble than it was worth

Leaky Nun

> Now let $V^*$ be the dual of $V$. Since $V^* = \mathcal{L}(V,F)$ (where $\mathcal{L}(V,W)$ is the vector space of all $F$-linear maps from $V$ to $W$), and $V=\mathop{\oplus}\limits_{i\in\kappa}F$, then again from abstract nonsense we know that
$$V^*\cong \prod_{i\in\kappa}\mathcal{L}(F,F) \cong \prod_{i\in\kappa}F.$$
Therefore, $|V^*| = |F|^{\kappa}$.

I know that "abstract nonsense" is a reference to category theory, but could somebody explain?

@MartinSleziak

Tobias Kildetoft

@LeakyNun Hom sends direct sums (on the left) to direct products.

Martin Sleziak

To add a bit of context for others, thet above quote is from Arturo Magidin's answer.

Leaky Nun

@TobiasKildetoft can it be proved?

Tobias Kildetoft

Aug 15, 2017 09:19

@LeakyNun Sure, it is just a matter of writing up the definitions (and making sure you are working in the correct categories for each object, which is not a problem here)

Leaky Nun

@TobiasKildetoft without categories?

Tobias Kildetoft

@LeakyNun Yes, but it takes a bit more work, writing up what everything actually does

Leaky Nun

@TobiasKildetoft in other words can I prove that an element in $\operatorname{Hom}(V,F)$ is uniquely determined by its value in the standard basis of $V$?

Tobias Kildetoft

@LeakyNun That is a standard linear algebra thing

linear maps are uniquely determined by their value on a basis

Martin Sleziak

Basically, what you're saying Tobias is this? $$\mathcal L (\prod\limits_{i\in I} V_i,W) \cong \prod\limits_{i\in I} \mathcal L (V_i,W)$$

Leaky Nun

Aug 15, 2017 09:21

@MartinSleziak no, sum

$\bigoplus$

$\displaystyle \mathcal L \left( \bigoplus_{i\in I} V_i,W \right) \cong \prod_{i\in I} \mathcal L (V_i,W)$

Tobias Kildetoft

Because $\operatorname{Hom}(-,W)$ is contravariant, so it turns direct sums into direct products (well, given that it turns them into something)

Leaky Nun

I like to think of $\Bbb N$ as a vector space, over $F_2$

with addition being $\oplus$

7+9 = 14

Secret

From a very brief skim read of the first few lines in nlab, semicategories kinda reminds of semirings and semigroups without identies a bit. Interesting. However will save that for later after the foundation is solid

Tobias Kildetoft

@Secret semicategories are terrible things where nothing makes sense any more.

Secret

I like the weirdness and so far I am not afraid of them yet. But to explore the weird, one needs to be well prepared

Tobias Kildetoft

Aug 15, 2017 09:27

Certainly not much has been written about them that I could find. Though it is interesting to note that the forgetful functor from categories to semicategories has both a left- and a right- adjoint, where one is the "naive" additions of identities, and the other is the Karoubi envelope (which I had never considered before in this context)

Leaky Nun

Aug 15, 2017 09:54

Let $\{v_a : a \in A\}$ be vectors indexed by infinite set $A$. What does it mean that they are linearly independent?

@AkivaWeinberger buenos dias

Akiva Weinberger

Buenos días

Tobias Kildetoft

@LeakyNun Same as if they were indexed by a finite set

Leaky Nun

@TobiasKildetoft but infinite linear combination may not be well-defined

Tobias Kildetoft

i.e. no linear combination gives $0$ except the one

Right, linear combinations are still finite

Leaky Nun

alright, thanks

@TobiasKildetoft @AkivaWeinberger consider the vector space $\displaystyle \bigoplus_{r \in \Bbb R} F_2$ over the field $F_2$...

Akiva Weinberger

Aug 15, 2017 09:57

@LeakyNun Similarly, the span of $\{1,x,x^2,\dots,x^n,\dots\}$ is the set of finite sums, i.e. it's the polynomials rather than the formal series

Leaky Nun

@AkivaWeinberger right

Akiva Weinberger

The idea is that it's the smallest ring that contains all of them @LeakyNun

And the polynomials is smaller than the the formal series

Leaky Nun

And then consider the span of $\left\{\dfrac1{1-x},1,x,x^2,\cdots,x^n,\cdots\right\}$ lol

Akiva Weinberger

That would be linearly independent :)

And still not the entirety of formal series, I think that requires an uncountable number of things to span it

Leaky Nun

I know

Akiva Weinberger

Aug 15, 2017 10:02

and you probably need choice to get a minimal spanning set? A basis?

Leaky Nun

@AkivaWeinberger I think so

Conversation ended Aug 15, 2017 at 10:03.

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