Every finite-dimensional vector space is isomorphic to its dual. However for an infinite-dimensional vector space $E$ over a field $K$ this is always false since its dual $E^\ast$ is a vector space of strictly larger dimension: $dim_KE \lt dim_K E^\ast $ (dimensions are cardinals of course). This is a non-trivial statement for which our friend Andrea Ferretti has given an astonishingly unexpected proof here. This implies for example that a vector space of countably infinite dimension over a field $K$, like the polynomial ring $K[X]$, cannot be the dual of any $K$-vector space whatsoever.

So an infinite-dimensional vector space is not isomorphic to its dual but it could be isomorphic to the dual of another vector space and my question is: which vector spaces are isomorphic to the dual of some other vector space and which are not?

In order to make the question a little more precise, let me remind you of an amazing theorem, ascribed by Jacobson (page 246) to Kaplanski and Erdős:

The Kaplanski-Erdős theorem : Let $K$ be a field and $E$ an infinite-dimensional $K$-vector space . Then for the dual $E^\ast$ of $E$ the formula $dim_K (E^\ast) = card (E^\ast)$ obtains.

So now I can ask

A precise question : Is there a converse to the Kaplanski-Erdős condition i.e. if a $K$- vector space $V$ (automatically infinite dimensional !) satisfies $dim_K (V) = card (V)$ , is it the dual of some other vector space : $V \simeq E^\ast$? For example, is $\mathbb R ^{(\mathbb R)}$ - which satisfies the Kaplanski-Erdős condition ( cf. the "useful formula" below) - a dual?

A vague request : Could you please give "concrete" examples of duals and non-duals among infinite-dimensional vector spaces?

A useful formula : In this context we have the pleasant formula for the cardinality of an infinite-dimensional $K$-vector space $V$ ( for which you can find a proof by another of our friends, Todd Trimble, here )

$$ card \: V= (card \: K) \; . (dim_K V) $$

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    $\begingroup$ This is really set theory rather than linear algebra, no? $\endgroup$ – Mariano Suárez-Álvarez Feb 3 '11 at 8:47
  • $\begingroup$ Dear Mariano, we'll see in the answers what this really is (Kaplanski-Erdős involves hard linear algebra), but you are definitely right that I should add the tag "set-theory". I have just done it: thanks for your remark. $\endgroup$ – Georges Elencwajg Feb 3 '11 at 9:04
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    $\begingroup$ The precise question seems indeed to reduce to set theory (see my answer below). $\endgroup$ – François Brunault Feb 3 '11 at 11:12
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    $\begingroup$ Note that if $\mathrm{dim}_K(V) = \lambda \geq \omega$, then a $K$-v.s. $W$ is (isomorphic to) $V^{\ast}$ iff $|W| = |K|^{\lambda}$ iff $\mathrm{dim}_K(W) = |K|^{\lambda}$. So a $K$-v.s. is isomorphic to another infinite dimensional $K$-v.s. iff it's cardinality, or equivalently, its dimension over $K$, is an infinite power of $|K|$. Now you may ask, which cardinals are infinite powers of $|K|$? The short answer is, there's a few basic things we can say about cardinal exponentiation, but aside from those basic restrictions, any other possibility is consistent relative to ZFC. $\endgroup$ – Amit Kumar Gupta Feb 3 '11 at 17:15
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    $\begingroup$ For some basics of cardinal arithmetic, see chapter 5 in Jech's "Set Theory." Easton showed that for $\kappa$ regular, $2^{\kappa}$ can be anything of cofinality greater than $\kappa$. Chapter 15 in that same book deals with applications of forcing and covers Easton forcing. The function $\gimel (\kappa) = \kappa ^{\mathrm{cf}(\kappa)}$ plays an important role in cardinal arithmetic, and in light of Easton's theorem, it's of most interest in the case of singular $\kappa$. Abraham and Magidor's article "Cardinal Arithmetic" in "The Handbook of Set Theory" is a good reference. $\endgroup$ – Amit Kumar Gupta Feb 3 '11 at 19:08

The $\mathbf{R}$-vector space $\mathbf{R}^{(\mathbf{R})}$ has dimension $\operatorname{card} \mathbf{R}$ by definition, so it is isomorphic to $\mathbf{R}^{\mathbf{N}}$ (by the Erdős-Kaplansky theorem and because $\operatorname{card}(\mathbf{R}^{\mathbf{N}}) = \operatorname{card} \mathbf{R}$). So $\mathbf{R}^{(\mathbf{R})}$ is isomorphic to the dual of $\mathbf{R}[X]$.

In general, your precise question is equivalent to the following purely set-theoretical question (which seems difficult). By the useful formula, the identity $\operatorname{card} V = \operatorname{dim}_K V$ is equivalent to $\operatorname{card} K \leq \operatorname{dim}_K V$. Let $\kappa = \operatorname{card} K$ and $\lambda = \dim_K V$, and assume $\kappa \leq \lambda$. Does there always exist an infinite cardinal $\alpha$ such that $\kappa^\alpha = \lambda$? (here $\alpha$ is meant to be the dimension of the vector space whose dual is $V$). In general, Stephen's answer shows that there are counterexamples.

EDIT : in order to explain why the question is difficult, consider a field $K$ such that $\operatorname{card} K = \aleph_1$ (for example, one could take $K=\mathbf{Q}((T_i)_{i \in I})$ with $\operatorname{card} I =\aleph_1$). Take a $K$-vector space $V$ of dimension $\aleph_1$. Then $V$ is a dual if and only if there exists $\alpha \geq \aleph_0$ such that $\aleph_1 = (\aleph_1)^\alpha$, which amounts to say that $\aleph_1 = 2^{\aleph_0}$. In other words, $V$ is a dual if and only if the continuum hypothesis holds.

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    $\begingroup$ François, let me point out that (regardless of the continuum hypothesis or any such assumption) there are cardinals that are not of the form $\kappa^\alpha$ for any infinite $\kappa,\alpha$. For example, take $\beth_\omega$, the supremum of the sequence ${}|{\mathbb N}|,2^{|{\mathbb N}|},2^{2^{|{\mathbb N}|}},\dots$ This is because of a result of König that asserts that $\kappa^{{\rm cf}(\kappa)}>\kappa$ for any infinite $\kappa$. Here, ${\rm cf}$ is the cofinality function, en.wikipedia.org/wiki/Cofinality $\endgroup$ – Andrés E. Caicedo Feb 4 '11 at 7:41
  • $\begingroup$ Very nice, François. Merci! $\endgroup$ – Georges Elencwajg Feb 5 '11 at 2:54
  • $\begingroup$ I hate to resurrect an ancient thread, but something seems very wrong here. How can we have $\mathbb{R}^\mathbb R \cong \mathbb{R}^\mathbb N$ when, for example, $|\mathbb{R}^\mathbb N | = |\mathbb R| = \mathfrak c$ while $|\mathbb{R}^\mathbb R| = |2^\mathbb R| > \mathfrak c$? $\endgroup$ – Reinstate Monica Jan 11 '16 at 6:42
  • $\begingroup$ @silvascientist. It is $\mathbf{R}^{(\mathbf{R})}$, not $\mathbf{R}^{\mathbf{R}}$, which is isomorphic to $\mathbf{R}^{\mathbf{N}}$. The vector space $\mathbf{R}^{\mathbf{R}}$ has dimension $2^\mathfrak{c}$ by Erdős-Kaplansky. $\endgroup$ – François Brunault Jan 12 '16 at 9:19
  • $\begingroup$ I see, I guess I'm just not familiar with the parenthetical notation for direct sum. Please forgive my naïveté. $\endgroup$ – Reinstate Monica Jan 12 '16 at 17:10

If $V$ is a vector space of infinite dimension over a field $K$, then the dimension of its dual is given by $\dim(V^*)=|K|^{\dim(V)}$. This is essentially just a restatement of what you have called the Kaplanski-Erdős theorem, because elements of $V^*$ correspond to functions $B\to K$, where $B$ is a basis of $V$, so $\dim(V^*)=|V^*|=|K^B|=|K|^{\dim(V)}$.

To answer your "precise question": Let $V$ be a vector space of dimension $\aleph_0$ over $\mathbb{Q}$. Then $|V|=\dim(V)$, yet $V$ is not isomorphic to the dual of any vector space, as its dimension is not of the form $|\mathbb{Q}|^\lambda$ for any infinite cardinal $\lambda$. Moreover, the same is true if $\aleph_0$ is replaced by any other strong limit cardinal, such as $\beth_\omega$.

Note that $\mathbb{R}^{(\mathbb{R})}$ is isomorphic to the dual of $\mathbb{R}^{(\mathbb{N})}$.

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    $\begingroup$ For more details on $\beth_\omega$ see en.wikipedia.org/wiki/Beth_number $\endgroup$ – François Brunault Feb 3 '11 at 10:36
  • $\begingroup$ Stephen: Your "the same is true if..." is not quite correct. For example, $\beth_{\omega_1}^{\aleph_0}=\beth_{\omega_1}$. Here, $\omega_1$ is the first uncountable ordinal. (Your remark is true with $\beth_\omega$ and even with $\beth_\alpha$ whenever $\alpha$ has cofinality $\omega$, for example.) $\endgroup$ – Andrés E. Caicedo Feb 4 '11 at 7:37
  • $\begingroup$ @Andres Caicedo - I'm not claiming that $\beth_{\omega_1}^{\aleph_0}>\beth_{\omega_1}$. I only need that $\beth_{\omega_1}\cdot{\aleph_0}=\beth_{\omega_1}$ (to get $|V|=\dim(V)$) and that there is no cardinal $\lambda$ such that $\aleph_0^\lambda=\beth_{\omega_1}$ - and these statements are true, and don't involve $\beth_{\omega_1}^{\aleph_0}$. $\endgroup$ – Stephen S Feb 4 '11 at 9:19

Let $K$ be a field of some size $\kappa$ and let $V$ be a $K$-vector space of dimension $\lambda$. If $\kappa\leq\lambda$, then the dimension of the dual of $V$ is simply $2^\lambda$. This is implies that vector spaces of large dimensions are duals iff their dimension (and hence their size) is a power of 2. I don't have the time right now to elaborate what happens if the dimension of $V$ is small relatively to the size of $K$.

  • $\begingroup$ I hope you'll find the time to come back to your answer in the future, Stefan.Anyway, thank you for your answer. $\endgroup$ – Georges Elencwajg Feb 5 '11 at 2:53

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