Which vector spaces are algebraic dual spaces?

Question

Let us say that a vector space $V$ is an algebraic dual space if there exist a vector space $U$ such that $V$ is isomorphic to $U^*$, the vector space of all linear maps from $U$ to the corresponding field of scalars.

It is known that if $V$ is finite-dimensional then $V$ and $V^*$ are isomorphic, hence $V$ is an algebraic dual space. On the other side, it is also known that the dimension of an algebraic dual space cannot be countably infinite, hence not all vector spaces are algebraic dual spaces.

Question: Is there any characterization of vector spaces that are algebraic dual spaces?

I am mostly interested in the case of vector spaces over reals.

Answer 1

Let $F$ be a field, $\kappa$ be a cardinal. Denote $F^\kappa$ the vector space of all functions $f\colon\kappa\to F$.

Let $V$ be a vector space over $F$. We shall use the following facts.

1. If $\kappa\ge\omega$ then $\dim F^\kappa=\left|F\right|^\kappa$.
2. If $\dim V=\kappa$ then $V$ is isomorphic to $\{v\in F^\kappa\!:\left|\mathrm{supp}(v)\right|<\omega\}$, where $\mathrm{supp}(v)=\{\alpha<\kappa\!:v(\alpha)\neq 0\}$.
3. If $\dim V=\kappa$ then $V^*$ is isomorphic to $F^\kappa$.
4. If $\dim V<\omega$ then $V^*$ is isomorphic to $V$.

For a proof of the first statement, see here. The second one follows from the fact that every element of $V$ can be uniquely represented by a finite linear combination of the elements of a basis of $V$. For the third statement, notice that every map from a basis of $V$ to $F$ can be uniquely extended to a linear map from $V$ to $F$. The last statement follows from the previous two.

Claim. Let $V$ be a vector space over a field $F$. Then $V$ is an algebraic dual space if and only if $\mathrm{dim}\ V<\omega$ or $\mathrm{dim}\ V=\left|F\right|^\kappa$ for some infinite cardinal $\kappa$.

Proof. Assume that $V=U^*$ and $\mathrm{dim}\ V\ge\omega$. Then $V$ is isomorphic to $F^\kappa$ where $\kappa=\mathrm{dim}\ U$. We have $\kappa\ge\omega$ since otherwise $V$ would be isomorphic to $U$ and thus $\mathrm{dim}\ V$ would be finite. By 1, $\dim V=\left|F\right|^\kappa$. This finishes one direction.

For the other direction, assume that $\dim V=\left|F\right|^\kappa$ for some $\kappa\ge\omega$. Then, by 1, $\dim V=\dim F^\kappa$, hence, by 2, $V$ and $F^\kappa$ are isomorphic, and, by 3, $V$ is an algebraic dual space. To complete the proof, notice that 4 implies that if $\dim V$ is finite then $V$ is an algebraic dual space. q.e.d.

Answer 2

(A note on notation: I will use simple vertical bars to denote the cardinality of a set. There isn't anything else involved to confuse it with.)

With just the algebraic structure, the only thing that matters is the cardinality of the basis. Let a basis of $U$ be $S$; the dual space $U^*$ can be realized as the space of functions from $S$ to the base field $F$. The cardinality there, $|F|^{|S|}$, comes out as whichever of $|F|$ and $2^{|S|}$ is larger.

Now, consider a vector space $V$ with an infinite basis $T$. This space can be realized as the space of finitely supported functions from $T$ to $F$. The cardinality here is the product of cardinalities $|T|\cdot |F|$, which is just whichever one is larger.

So then, if $|F|$ is small - say, if we're working with one of the prime fields $\mathbb{Z}/p$ or $\mathbb{Q}$, the cardinality of an infinite-dimensional vector space will be that of its basis, and the cardinality of a dual will be that of the basis' power set. The possible infinite cardinals that can be the dimension of a dual space are simply the power sets of infinite cardinals.

Now, what about larger fields? If we extend to a larger field while keeping the same basis, that's the same dimension for the vector space. What about its dual? Can we keep the same basis there?

Let $F$ be an extension of the prime field $K$, and let $B$ be a basis for $F$ over $K$. Suppose $U$ has basis $S$ over $K$, and $U'$ is the extension with basis $S$ over $F$. Then let $V=U^*$ and $V'=(U')^*$, with $T$ a basis for $V$ (over $K$).
Is $T$ linearly independent over $F$ in $V'$? Well, we can break down any linear dependence relation into components, one for each element of $B$ - and since $B$ is linearly independent over $K$, each component must give a linear dependence relation $\sum_i a_i (b_j t_i)$ over $K$. The elements of $T$ are linearly independent over $K$, so each of these linear dependence relations is trivial. Sum over $B$, and all the coefficients $\sum_j a_ib_j$ are trivial in $F$, so $T$ is linearly independent over $F$ as well. That's one direction - the dimension of $V'$ over $F$ is at least as large as that of $V$ over $K$, previously calculated as $2^{|S|}$.
Does $T$ span $V'$? Unfortunately, not necessarily. If we take a finite linear combination over $F$ of elements of $T$, the $K$-span of the values these functions take is contained in the $K$-span of the coefficients of the linear combination - which is finite-dimensional. If $F$ is large enough, there will be elements of $V'$ with function values that have an infinite-dimensional $K$-span.

Now, all of that isn't too much trouble in most cases. As long as $|F|\le 2^{|S|}$, we still have that upper bound to lock down the dimension of the dual space as $2^{|S|}$. This covers almost all examples, including all spaces over such fields as $\mathbb{R}$ or $\mathbb{C}$. Really big fields - I'll keep thinking.
[Added in edit] As noted in linked material, it is a known theorem that in those large field cases, the dimension is as large as possible - the full $|F|^{|S|}$. Such a number is always the cardinality of a power set, so no new possibilities for the dimension of a dual set are introduced this way.