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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 $\dim(_DV^{\ast})> \dim(V_D)$. 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.

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.

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 $\dim(_DV^{\ast})> \dim(V_D)$. 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.

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 $\dim(_DV^{\ast})> \dim(V_D)$. 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.

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 $\dim(_DV^{\ast})> \dim(V_D)$. However, perhaps less well know 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.

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 $\dim(_DV^{\ast})> \dim(V_D)$. 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.