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Let $V_F$ be an infinite dimensional right $F$-vector space (over a field $F$, or even over a division ring). The dual space $V^{\ast}={\rm Hom}(V,F)$ is naturally a left $F$-vector space (coming from the left $F$-module structure on the second coordinate $F$).

It is well-known that $\dim(_FV^{\ast})> \dim(V_F)$. However, perhaps less well known is that $V$ and $V^{\ast}$ can have the same cardinality and the same structure as abelian groups.

Now, suppose that $V$ is in fact an $F$-$F$-bimodule. Then $V^{\ast}$ is also an $F$-$F$-bimodule, with the right $F$-module structure induced from the left structure on $V$.

Can the dimension $V$ and $V^{\ast}$ be the same, on the respective sides?

As a "pie-in-the-sky" hope, it would be nice to find an example with $V\cong V^{\ast}$ as bimodules, as this would answer Faith's Conjecture on quasi-Frobenius ring in the negative. So I'm only trying to understand the more modest question above.

Gro-Tsen in the comments asks for an example where the dimensions are the same on at least one side. Here is one such example, where the dimensions on the left are the same. (I don't know how to make the dimensions on the right the same, given the set-up above.)

Let $F$ be a field with continuum cardinality and with a field injection $\varphi:F\to F$ such that $F$ is continuum-dimensional over $\varphi(F)$. (For instance we can take $F=\mathbb{C}$, or more simply $F=\mathbb{Q}(\{x_i\}_{i\in I})$ where the $x_i$ are independent transcendentals and $I$ is a set of continuum cardinality.)

Now let $V'=tF$ be a $1$-dimensional right $F$-space. Define left multiplication from $F$ by $\alpha t = t\varphi(\alpha)$. This makes $V'$ a bimodule, and taking $V$ to be a countable direct sum of copies of $V'$ will suffice.

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  • $\begingroup$ Is this the sort of example of the "perhaps less well known" fact? If $F$ is the field of rational functions over $\mathbb{Q}$ in $2^{\aleph_0}$ indeterminates, which has dimension $2^{\aleph_0}$ over $\mathbb{Q}$ (I think), and if $V$ has dimension $\aleph_0$ over $F$ then $V^*$ has dimension $2^{\aleph_0}$ over $F$, so they both have dimension $2^{\aleph_0}$ over $\mathbb{Q}$ (so they have the same cardinal and abelian group structure). Right? $\endgroup$
    – Gro-Tsen
    Commented Feb 3, 2016 at 22:01
  • $\begingroup$ Also: I suspect one can — and I suspect you know how to — manufacture an example where $V$ and $V^*$ have the same dimension on one side. If so, you should maybe state this, and give an example. (And if not, you should maybe add this as an easier or preliminary question.) $\endgroup$
    – Gro-Tsen
    Commented Feb 3, 2016 at 22:09
  • $\begingroup$ @Gro-Tsen: Regarding your first comment, the answer is yes. As long as the cardinalities of $V$ and $V^{\ast}$ are equal, and bigger than the cardinality of the prime subfield of $F$, you should get isomorphic abelian groups. Regarding your second comment, it isn't too difficult to get a bimodule $V$ where on the right the dimension is countable and on the left it is continuum. I'll post such an example above. $\endgroup$
    – Pace Nielsen
    Commented Feb 3, 2016 at 22:30
  • $\begingroup$ Do you know examples of bimodules $_EV_F$ for fields $E$ and $F$ where the $E$-dimension of $\operatorname{Hom}_F(V,F)$ is strictly Iess than the $E$-dimension of $V$? By the way, I think that in the second paragraph $D$ should be $F$? $\endgroup$
    – Jeremy Rickard
    Commented Feb 4, 2016 at 14:15
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    $\begingroup$ Just to expand on a previous comment. If $E$ and $F$ are fields and $V$ is an $E$-$F$-bimodule, infinite-dimensional over $F$, then $V^\ast=\operatorname{Hom}_F(V,F)$ is an $F$-$E$-bimodule whose $F$-dimension is strictly greater than that of $V$. So if $E=F$ and $V$ answers the question, we'd need the $E$-dimension of $V^\ast$ to be strictly less than that of $V$. Even without $E=F$ I don't know if this is possible, although it is possible for the $E$-dimensions of $V$ and $V^\ast$ to be equal (and infinite). $\endgroup$
    – Jeremy Rickard
    Commented Feb 17, 2016 at 11:19

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