## Dual space rambling

In this post I’m going to describe a (probably ridiculous) idea I came up with, which seems to only be interesting for infinite-dimensional vector spaces.

Recall that given a vector space $\mathsf{V}$ over a field $\mathbb{F}$, the dual space $\mathsf{V}^*$ is defined as follows:

$\displaystyle \mathsf{V}^* := \{ \mathsf{L} : \mathsf{V} \to \mathbb{F} : \mathsf{L} \text{ is linear} \}$.

It therefore makes sense to talk about the dual of any vector space. Even $\mathsf{V}^*$ has a dual, which we denote $\mathsf{V}^{**}$. It’s defined exactly the way $\mathsf{V}^*$ is defined, except of course we replace all the occurrences of $\mathsf{V}$ with $\mathsf{V}^*$. So why do we call this set a dual space? Well, we can turn it into a vector space over $\mathbb{F}$ by making some no-brainer definitions:

• For all $f, g \in \mathsf{V}^*$, we define the function $f+g$ by $(f+g)(v) := f(v) + g(v)$ for all $v \in \mathsf{V}$.
• For all $f \in \mathsf{V}^*$ and $\lambda \in \mathbb{F}$, we define the function $\lambda f$ by $(\lambda f)(v) := \lambda f(v)$ for all $v \in \mathsf{V}$.

I’ll leave it to you to check the axioms 🙂 but it does indeed work. Anyways, my idea takes an interest in the relationship between $\mathsf{V}$ and its double-dual $\mathsf{V}^{**}$, because there is a natural way we can send the elements of $\mathsf{V}$ into $\mathsf{V}^{**}$. How, you ask? Well, we proceed as follows. For each vector $v \in \mathsf{V}$, define the evaluator $\mathcal{E}_v : \mathsf{V}^* \to \mathbb{F}$ of $v$ by $\mathcal{E}_v(f) := f(v)$ for all $f \in \mathsf{V}^*$. That is, the evaluator of $v$ takes in an element of $\mathsf{V}^*$ (a linear functional on $\mathsf{V}$) and spits out the result of evaluating that functional at $v$. Pretty straightforward, I think. Verify yourself that $\mathcal{E}_v \in \mathsf{V}^{**}$.

Now, we want to find a map from $\mathsf{V}$ into $\mathsf{V}^{**}$. What better way to do this than to send each vector $v \in \mathsf{V}$ to its evaluator $\mathcal{E}_v \in \mathsf{V}^{**}$? Therefore, define $\Phi : \mathsf{V} \to \mathsf{V}^{**}$ by putting $\Phi(v) := \mathcal{E}_v$ for all $v \in \mathsf{V}$. It’s completely natural — depends on no choice of basis at all! This is wonderful!

It turns out that this map $\Phi$ is linear, and it’s always an injection. Here’s why my idea only involves infinite-dimensional vector spaces: for finite-dimensional vector spaces, the map $\Phi$ turns out to be a vector space isomorphism. In other words, $\mathsf{V}$ and $\mathsf{V}^{**}$ turn out be structurally identical. This is boring.

The basic idea I’ll describe in words: I want to examine the vector spaces $\mathsf{V}^{**}/\mathsf{V}$, $\mathsf{V}^{***}/\mathsf{V}^*$, and so on. Of course, $\mathsf{V}$ is not a subspace of $\mathsf{V}^{**}$, heck, it’s not even a subset — but we have this map $\Phi$ which we can use to “embed” vector spaces into their double-duals. So really, I mean $\mathsf{V}^{**}/\Phi(\mathsf{V})$, $\mathsf{V}^{***}/\Phi'(\mathsf{V}^*)$, and so on.

In a more “proper” (aka terse) way of speaking, define $D_1 := \mathsf{V}$ and recursively put $D_{n+1} := {D_n}^*$ for $n \geq 1$. Now for each $n \geq 1$, define the map $\Phi_n : D_n \to D_{n+2}$ by putting $\Phi_n(v) := \mathcal{E}_v$ for all $v \in D_n$. Put $H_n := D_{n+2}/\Phi_n(D_n)$. What can be said about the sequence $(H_n)$?

Note the reason it’s “boring” in the finite-dimensional case is because all the spaces $H_n$ end up being trivial… lol. But in the infinite-dimensional case it seems like there are many questions we can ask. It’s odd because using this construction, we come up with a quotient space that is naturally associated with $\mathsf{V}$

just another guy trying to make the diagrams commute.
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### 16 Responses to Dual space rambling

1. William says:

Wait a second, what exactly do you mean by $\mathsf{V^{**}}/\Phi(\mathsf{V})$ and $\mathsf{V^{***}}/\Phi'(\mathsf{V^{*}})$? Isn’t $\Phi(\mathsf{V})$ some potentially abtract field whereas $\mathsf{V^{**}}$ a set of linear functions that take in linear functions as its arguments?

Unless I’m confusing notation, or you somehow ordered $\mathsf{V^{**}}$ such that it forms some finite group under $\Phi(\mathsf{V})$, I’m not too sure.

e.g. I’m thinking of $\mathbb{Z}/n\mathbb{Z}$

2. dx7hymcxnpq says:

Err. Groups? Fields? I don’t see how any of that relates to what I’m talking about. $\Phi(\mathsf{V})$ is merely the image of a vector space under a linear transformation — namely, a subspace of its codomain. I am forming quotient spaces.

3. dx7hymcxnpq says:

He defined what a “quotient space” is in class. I think you’re thinking of quotient groups. (Also, you don’t need an order to have a group…)

4. William says:

… Back to reviewing then. I’ll think about it and get to you tomorrow.

5. Luis Gabriel Marques Reginato says:

Hello, Michael! I’m Luis Gabriel from Brazil, and sorry for my poor English. I graduated in Statistic in 2006, but someday I’ll finish the Pure Math course as well.
Have you ever heard of Synchronicity? I’ve always found the “Dual Spaces” theme very intriguing, but never had googled for it, until today. Then I watched a couple of your videos tonight and for my surprise later I realized that you posted them very recently!
Anyway, thanks a lot for the excellent explanations! Good luck for the rest of your course!

• dx7hymcxnpq says:

Thanks for the comment, Luis! 😀 Glad to hear my videos helped. (Your English is pretty good!)

6. William says:

… My geometric intuition has been fried trying to imagine what you’re quotient space would even look, feel, or seem like.

To start, could you please explain why the $H_n$ spaces end up being trivial? The double duals of a finite dimensional space V should be of same dimension since the evaluator map is an isomorphism.

7. dx7hymcxnpq says:

Suppose $\mathsf{V}$ is some vector space and $\mathsf{U} \subseteq \mathsf{V}$ is a subspace. To construct the quotient space $\mathsf{V} / \mathsf{U}$ you define an equivalence relation $\sim$ by saying that $v \sim u$ if and only if $v-u \in \mathsf{U}$. At this point $\mathsf{V} / \mathsf{U} = \{ [v] : v \in \mathsf{V} \}$ that is, you look at the set of equivalence classes, and define $[v] + [w] = [v+w]$, as well as $\lambda [v] = [\lambda v]$ (you can check that these are well-defined).

OK, so with this in hand, note that $\Phi_n$ is a v.s. isomorphism as you pointed out, which means it’s a bijective linear map from $D_n$ onto $D_{n+2}$. Look at the quotient space $H_n = D_{n+2} / \Phi_n(D_n)$. This is defined to be the set of all those equivalence classes. However, since $\Phi_n$ is a v.s. isomorphism it follows that $\Phi_n(D_n) = D_{n+2}$, so that $H_n = D_{n+2} / D_{n+2}$. You’re quotienting out by the entire space, which means (if you look at the definition) there will only be one equivalence class of the equivalence relation $\sim$. That’s why the spaces $H_n$ become trivial when $\Phi_n$ are v.s. isomorphisms.

• William says:

@ 2nd paragraph: If you quotient out a space with itself, doesn’t that just leave you with the singleton set of the original space? If so, how exactly is this a space, since the original space can be non-trivial.

@ “a quotient space that is naturally associated with V” : I’m not too sure about this. Can we be certain that dim V = dim [ V**/Phi(V)] ? Also, can we construct a natural isomorphism between the two such as between V and V**?

… Wait, how exactly do we find dim [ V**/Phi(V)] as usually in the finite case this is equal to Nullity(V**).

• dx7hymcxnpq says:

1. If $\mathsf{V}$ is a vector space then $\mathsf{V} / \mathsf{V} = \{ [v] : v \in \mathsf{V} \}$ under the equivalence relation “$u \sim v$ if and only if $u-v \in \mathsf{V}$“. But this equivalence relation $\sim$ is trivial: every vector in $\mathsf{V}$ is congruent modulo $\sim$ to every other vector in $\mathsf{V}$. That is, the relation $\sim$ is equal to the entire Cartesian product $\mathsf{V} \times \mathsf{V}$… so you get only one equivalence class (which every vector in $\mathsf{V}$ lies in). Since $\mathsf{V} / \mathsf{V}$ is nothing more than a space of these equivalence classes, there’s literally just $[0]$ in there, so $\mathsf{V} / \mathsf{V}$ is the zero vector space (boring).

On the other hand, if you consider the quotient $\mathsf{V} / \{ 0 \}$, you end up with something naturally isomorphic to $\mathsf{V}$ itself. Quotient something small out, you get something big. Quotient something big out, you get something small.

2. I said “associated with”, not “identified with”. I’m not claiming any similarity between $\mathsf{V}$ and $\mathsf{V}^{**} / \Phi(\mathsf{V})$, merely saying that constructing the latter from the former requires no arbitrary/ugly choices.

3. I don’t know what you’re talking about here. $\mathsf{V}^{**}$ is a vector space. What do you mean by the nullity of a vector space?

• William says:

1. Ah, okay. I was going by the definition from our notes.

2. Okay, that makes sense. Have you derived any further conclusions upon examining V**/Phi(V) ?

3. Nullity of a vector space V is just the dimension of the null space of V. In the finite dimensional case, the nullity of V/U for U a subset of U is usually the nullity of U.

• dx7hymcxnpq says:

Sorry that I replied to a higher post — I don’t like the way the horizontal real estate progressively gets smushed at higher levels of indentation…

1. Actually as far as I know, the definition in my course notes of a quotient space is done in terms of the cosets $x + \mathsf{V}$. It’s equivalent to what I said.

2. I haven’t looked any further into what conclusions we can draw about $(H_n)$ just yet.

3. I think you’re confusing vector spaces and linear transformations. There’s no such thing as the kernel of a vector space — only of a linear map. (It is the set of elements in the domain which get sent to zero).

• William says:

@ 3. Whoops, confused notation again. I probably meant the orthogonal complement of U for V/U where U is a subspace of V.

• dx7hymcxnpq says:

To restate what I said earlier in the notation of the course notes, essentially the reason $\mathsf{V} / \mathsf{V}$ is trivial is because regardless of which $w \in \mathsf{V}$ you choose, the coset $w + \mathsf{V}$ turns out to be $\mathsf{V}$ anyways, since $u + \mathsf{V} := \{ u + v : v \in \mathsf{V} \}$ by definition.

8. dx7hymcxnpq says:

I guess I can give an example of how you could visualize a quotient space, in the finite-dimensional case. Suppose $\mathsf{V}$ is nothing more than $\mathbb{R}^2$ considered as a v.s. over $\mathbb{R}$. Let $\mathsf{U}$ be the span of $(1,1)$. Let’s think about what $\mathsf{V} / \mathsf{U}$ might look like. Well, any two vectors in $\mathsf{V}$ which differ only by something in $\mathsf{U}$ are thrown into the same equivalence class. But such a pair of vectors is merely a pair of vectors who are both sitting on one line parallel to $\mathsf{U}$ (think about direction vectors, etc.)

So the equivalence classes you obtain can be thought of as “lines” parallel to $\mathsf{U}$. This means $\mathsf{V} / \mathsf{U}$ is one-dimensional. But in general, the quotient space is a new vector space, so you shouldn’t attempt to visualize it as something “sitting inside” of $\mathsf{V}$

9. dx7hymcxnpq says:

Also, just a cosmetic issue but if I had to choose a name for $\Phi$, “natural embedding” seems like the best choice in this situation. The evaluators are maps sitting in $\mathsf{V}^{**}$, but the natural embedding $\Phi$ is what actually transforms vectors $v \in \mathsf{V}$ to their corresponding evaluators $\mathcal{E}_v \in \mathsf{V}^{**}$.