Do Finite-Dimensional Vector Spaces Have Limits?

Title: Do Finite-Dimensional Vector Spaces Have Limits? Category: mathematics

Here, “limits” means categorical limits, not limits of sequences or functions. Let $\mathrm{FinVect}_k$ be the category whose objects are finite-dimensional vector spaces over a field $k$ and whose morphisms are linear maps.

This category has every finite limit. It does not have every small limit: limits of infinite diagrams can require infinite-dimensional vector spaces. The distinction is whether the diagram has finitely many objects and arrows, not whether its vector spaces are finite-dimensional (they are, by definition, in this category).

The basic finite limits

Two constructions give the main ingredients.

Finite products. The product of two finite-dimensional spaces $V$ and $W$ is their direct sum $V\oplus W$. In the category of vector spaces, a finite direct sum is also a product: a linear map into $V\oplus W$ is uniquely determined by its two component maps. Since $\dim(V\oplus W)=\dim(V)+\dim(W)$, it remains finite-dimensional. The product of no objects, called the terminal object, is the zero vector space: there is exactly one linear map from any space to $0$.

Equalizers. Given linear maps $f,g:V\to W$, their equalizer is the inclusion of the subspace

$$\ker(f-g)=\{v\in V:f(v)=g(v)\}$$

into $V$. This subspace is finite-dimensional because it is a subspace of a finite-dimensional space.

Finite products and equalizers can be used to construct any finite limit. More directly, a finite diagram’s limit is the subspace of compatible tuples described below.

Constructing the limit of a finite diagram

Suppose a finite diagram has a vector space $D(j)$ for each object $j$ of its indexing category, and a linear map $D(a):D(j)\to D(\ell)$ for each arrow $a:j\to\ell$. Form the finite direct sum of all its spaces, then take the subspace

$$L=\left\{(x_j)_j\in\bigoplus_j D(j): D(a)(x_j)=x_\ell\text{ for every arrow }a:j\to\ell\right\}.$$

The equations say that the tuple agrees with every map in the diagram. Because there are finitely many spaces, the direct sum is finite-dimensional, and its subspace $L$ is finite-dimensional too. The coordinate projections from $L$ to the $D(j)$ form a cone. Any other cone gives a unique linear map into $L$ by collecting its component maps into a tuple, so $L$ satisfies the limit’s universal property.

Worked example: a pullback

A pullback is the limit of a pair of maps with a common target. Let $f:V\to Z$ and $g:W\to Z$. Its pullback is

$$P=\{(v,w)\in V\oplus W:f(v)=g(w)\}.$$

For a concrete case, take $V=W=Z=k^2$, $f(x,y)=(x,0)$, and $g(u,v)=(0,v)$. The pullback condition is

$$(x,0)=(0,v),$$

so $x=0$ and $v=0$. Thus

$$P=\{((0,y),(u,0)):y,u\in k\},$$

a two-dimensional subspace of $V\oplus W$. The two coordinate projections give maps from $P$ to $V$ and $W$, and their composites into $Z$ agree, as required.

Why infinite limits can fail

If $I$ is an infinite index set, the product of $I$ copies of $k$ in the category of all vector spaces is $k^I$, the space of all families of scalars indexed by $I$. It is infinite-dimensional: the vectors that are $1$ at one chosen coordinate and $0$ elsewhere form an infinite linearly independent set.

There cannot be a finite-dimensional product object in $\mathrm{FinVect}_k$ instead. If such a product $P$ existed, its universal property applied to the test space $k$ would give

$$P\cong \operatorname{Hom}_k(k,P)\cong \prod_{i\in I}\operatorname{Hom}_k(k,k)\cong k^I.$$

That would make a finite-dimensional space isomorphic to an infinite-dimensional one, which is impossible.

The same issue occurs for some infinite diagrams even when every space in the diagram is finite-dimensional. Consider the inverse sequence $k^1\leftarrow k^2\leftarrow k^3\leftarrow\cdots$, where each map drops the last coordinate. A compatible tuple consists of an arbitrary infinite sequence of scalars, so its inverse limit in the category of all vector spaces is $k^{\mathbb{N}}$. It is not finite-dimensional. More precisely, if a limit object $P$ existed in $\mathrm{FinVect}_k$, its universal property applied to the test space $k$ would identify $\operatorname{Hom}_k(k,P)\cong P$ with the vector space of compatible tuples in the $k^n$, which is $k^{\mathbb{N}}$. This is impossible for finite-dimensional $P$.

Common mistakes

• Confusing finite-dimensional objects with finite diagrams: A diagram can contain only finite-dimensional spaces and still have an infinite limit.
• Treating an infinite product like a direct sum: The product $k^I$ contains all scalar families; the direct sum contains only families with finitely many nonzero entries. They differ when $I$ is infinite.
• Saying the category has no limits: It has all finite limits. It is not complete, because completeness requires limits for all small diagrams, including infinite ones.