Is the category of vector spaces with row-finite linear maps an abelian self-dual category?

Question

Fix a field $K$. Given a vector space $V$ with a basis $B$, a vector space $W$ with a basis $C$ and a linear map $f: V \to W$, let $\{f_{c, b}\}_{c, b}$ be the representing matrix of $f$, meaning that $$ f\left(\sum\limits_{b \in B} \lambda_b b \right) = \sum\limits_{c \in C} \left( \sum\limits_{b \in B} f_{c, b} \lambda_b \right) c $$ (where only finitely many $\lambda_b$ are nonzero). It should be noted that $\{f_{c, b}\}_{c, b}$ is necessarily column-finite, meaning that $\forall b \in B: |\{c \in C: f_{c, b} \neq 0\}| < \infty$ (in particular, this condition gives us that the above sum is well-defined). For more information on representing matrices on infinite-dimensional spaces and column-finiteness, see https://math.stackexchange.com/a/4486/491450.

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Let $\mathcal{C}$ be the following category:

• The objects are pairs $(V, B)$ where $B$ is a basis of $V$

• Take objects $(V, B)$ and $(W, C)$ of $\mathcal{C}$. Then the morphisms $(V, B) \to (W, C)$ are linear maps $f: V \to W$ such that $\{f_{c, b}\}_{c, b}$ is row-finite, meaning that $\forall c \in C: |\{b \in B: f_{c, b} \neq 0\}| < \infty$.

Since $\operatorname{id}_{b_1, b_2} = \delta_{b_1, b_2}$, the identity is row-finite (and column-finite). Given linear row-finite maps $f: V \to W$ and $g: W \to X$, we have that $(g f)_{d, b} = \sum\limits_{c \in C} g_{d, c} f_{c, b}$. Hence, we have that for each $d \in D$, $$ \{b \in B \mid (g f)_{d, b} \neq 0\} \subseteq \bigcup\limits_{\substack{c \in C \\ g_{d, c} \neq 0}}\{b \in B \mid f_{c, b} \neq 0\} $$ As a finite union of finite sets, this is finite, so $g f$ is also row-finite (analogously, $g f$ is also column-finite). This shows that $\mathcal{C}$ actually is a category.

Now, consider the functor $F: \mathcal{C} \to \mathcal{C}^{\operatorname{op}}$ that is the identity on objects and that transposes maps. This is well-defined since all maps are assumed to be row-finite, hence their transpose is column-finite and hence actually defines a linear map. Considering also the functor $G: \mathcal{C}^{\operatorname{op}} \to \mathcal{C}$ which again is the identity on objects and that transposes maps, we see that $G \circ F = \operatorname{id}_{\mathcal{C}}$ and $F \circ G = \operatorname{id}_{\mathcal{C}^{\operatorname{op}}}$ and hence $F$ is an equivalence of categories.

My questions are:

1. Is $\mathcal{C}$ abelian? I suspect that the answer is yes.

2. Is my argument that $\mathcal{C}$ is self-dual correct?

3. Can we somehow describe $\mathcal{C}$ without using bases? I suspect that the answer is no, since the requirement of row finiteness is not a part of the "vector space information" associated to $f$.

Answer 1

$\mathcal{C}$ is not an abelian category.

In an abelian category, a morphism that is both epic and monic is an isomorphism. I'll give an example of a morphism in $\mathcal{C}$ that is epic and monic but not invertible.

Let $V$ be a countable dimensional vector space, with a basis $B$ indexed by the natural numbers.

Let $X=(x_{ij})$ be the (row-finite) $\mathbb{N}\times\mathbb{N}$ matrix with $$x_{ij}=\begin{cases} 1&\text{if } i=j\\ -1&\text{if } i+1=j\\ 0&\text{otherwise.} \end{cases}$$

So $X$ describes a morphism $\varphi:(V,B)\to (V,B)$ in $\mathcal{C}$.

$X$ has an inverse $Y=(y_{ij})$ with $$y_{ij}=\begin{cases} 1&\text{if } i\leq j\\0&\text{otherwise.} \end{cases}$$

Note that $Y$ is not row-finite, and so represents an inverse to $\varphi$ in the ordinary category of vector spaces, but not in $\mathcal{C}$.

Since $\varphi$ is epic and monic in the ordinary category of vector spaces, it is monic and epic in $\mathcal{C}$. But it is not invertible in $\mathcal{C}$.

Answer 2

Some thoughts on the third question. A vector space $V$ equipped with a basis $B$ is also equipped with a nondegenerate symmetric bilinear form (call this an "inner product" for short) $\langle -, - \rangle$ given by the usual "dot product" with respect to $B$-coordinates, i.e. by linearly extending

$$\langle b_i, b_j \rangle = \delta_{ij}$$

where $b_i \in B$ are distinct. Given two such vector spaces $(V, B), (W, C)$ and a linear map $f : V \to W$ we can ask for an adjoint $f^{\dagger} : W \to V$, namely a map, if it exists, satisfying

$$\langle f(v), w \rangle = \langle v, f^{\dagger}(w) \rangle.$$

If $f^{\dagger}$ exists we might say that $f$ is adjointable. This corresponds exactly to row-finiteness. So, we might say that the category consists of vector spaces equipped with inner products which are... let's call them "Euclidean," in the sense of having orthonormal bases as above, with morphisms the adjointable linear maps. Then the adjoint exhibits the self-duality, as you've said; this is true more generally in any dagger category, which this is an example of.

This definition of "Euclidean" is a bit unsatisfying since it still depends on the existence of a basis, but at least the basis plays no role in the definition of the morphisms. It would be nice to characterize the Euclidean inner products by some property, but as far as I know the classification of inner products over a field $K$ is quite sensitive to arithmetic details about $K$, e.g. details about which elements have square roots, that sort of thing.