Are algebroids "just matrices"?

Question

$\DeclareMathOperator\Vect{Vect}\DeclareMathOperator\Mat{Mat}$This question was originally asked on MSE but may be better here.

Algebroids are particularly interesting structures: they are basically categories enriched over $\Vect_K$ for some field $K$. This means the hom-sets are all vector spaces, and composition is "bilinear" in a certain sense. (Let's just focus on when $K$ is a field for now rather than an arbitrary ring.)

Most examples I can think of are basically equivalent to some subset of matrices with the usual addition and multiplication rules, as long as we are willing to be creative and allow "infinite matrices" of various cardinalities to exist. In general, for any $n$ and $m$, the set of $n \times m$ matrices forms an algebroid, as long as we also allow the $n \times n$ and $m \times m$ identity matrices to exist. The union of any such sets also forms an algebroid, as long as matrix compositions exist when expected (meaning if we have $5 \times 4$ and $4 \times 3$ matrices, we must also have $5 \times 3$ matrices, as well as the relevant identity matrices). This is also true if $n$ and $m$ are arbitrary infinite cardinals, with the caveat that only finitely many elements of each column of the matrix can be nonzero.

So, we can ask if this is basically "what algebroids are." It is a little bit difficult to even figure out how to formalize the question, but we can think about it in the following sense:

Let's say that the category $\Mat^+_K$ is basically an extension of the usual $\Mat_K$, but with the objects as all possible cardinals rather than only natural numbers. For objects $\kappa$, $\lambda$, the morphisms $\kappa \to \lambda$ are (possibly infinite) matrices of size $\kappa \times \lambda$ (treating these cardinals as initial ordinals), with finitely many nonzero coefficients in each column, taking values in $K$.

As a sort of first pass, we would then like to ask the following "pre-question":

Naive Pre-Question: is every possible $K$-algebroid equivalent to a subcategory of $\Mat_K^+$?

The main problem with this is that we can take a disjoint union of two $\Mat_K^+$'s. This will be a $K$-algebroid, but as we haven't drawn explicit isomorphism arrows between the two copies of each object, I don't think the two categories will be equivalent.

But, we can salvage the spirit of the question by noting that $\Mat_K^+$ is a skeleton category of $\Vect_K$, which has infinitely many copies of the things in $\Mat_K^+$ with all possible linear transformations (and thus isomorphisms, when they exist) between them. So, we can thus ask our real question:

Real Question: is every possible $K$-algebroid equivalent to a subcategory of $\Vect_K$?

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EDIT: It has been pointed out in the comment that this can fail due to size issues, e.g. you can have an algebroid so large that it isn't concretizable. I guess that technically answers the way I formulated the "real question" above, as I hadn't thought of that snag, but still looking to get clarity on the algebraic relationship of these things beyond that sort of thing, so would be happy to see it addressed for small algebroids.

Answer 1

If we take the large but locally small category $\mathcal{C}$ described by Isbell in Example 2.4 of

• Two set-theoretical theorems in categories, Fundamenta Mathematicae 53 Issue 1 (1964) pp 43-49, (EuDML),

namely, with class of objects $(\{1,2\}\times\mathrm{ORD}) \sqcup \{X,Y\}$, and generated by maps $f_{\alpha,X}\colon (1,\alpha) \to X$, $f_{\alpha,Y}\colon (1,\alpha) \to Y$, $g_{X,\alpha}\colon X\to (2,\alpha)$, $g_{Y,\alpha}\colon Y\to (2,\alpha)$, for each ordinal $\alpha \in \mathrm{ORD}$, with the relation that $\forall \alpha$ forces $g_{X,\alpha}\circ f_{\alpha,X} = g_{Y,\alpha}\circ f_{\alpha,Y}$, and such that we also keep the two composites $g_{X,\beta}\circ f_{\alpha,X}$ and $g_{Y,\beta}\circ f_{\alpha,Y}$ distinct for $\alpha\neq \beta$.

The free $K$-algebroid $K[\mathcal{C}]$ on this category, namely take the free vector space on the hom-sets over any field $K$, and extend composition bilinearly, cannot admit a faithful functor to $K\text{-}\mathrm{Vect}$, since otherwise there would be a faithful functor $\mathcal{C} \to K[\mathcal{C}] \to K\text{-}\mathrm{Vect} \to \mathrm{Set}$, which Isbell proved cannot happen.

EDIT: For small algebroids, there is no issue: for every small algebroid $\mathcal{A}$ we can send an object $a$ to the vector space that is the direct sum over all $\mathcal{A}(x,a)$ as $x$ varies over all objects. This gives a faithful functor to vector spaces.

It is only for large algebroids where the issues come in. There are large algebroids where the construction in the previous paragraph still works and gives a functor to the category of vector spaces. For example, a large algebroid $\mathcal{A}$ such that for each object $a$, only a set's worth of hom-spaces $\mathcal{A}(x,a)$ are nontrivial vector spaces. For example the free $K$-algebroid on the opposite of $\mathrm{ORD}$ (arrows point to smaller ordinals), so that hom-spaces are either $K$ or $0$, and only set-many are $K$.