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The categories of vector spaces and finite dimensional vector spaces are pretty much as nice as can be, I think.

I was wondering what portions of basic linear algebra (first couple of courses) fall out by saying "big"(er) words, and also what standard facts admit a clarifying categorical phrasing.

What are some interesting examples of facts about vector spaces and linear maps that admit a nice categorical formulation?

Edit. I'm not looking for (completely) elementary things like definitions of universal constructions by universal properties instead of concrete ad hoc realizations, though I can't think of any "nonelementary" things. If you write an elementary property of the category of vectors spaces, e.g a property of any abelian category, please give some nice examples of where it lends its power.

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  • $\begingroup$ The first that comes to my mind is biproducts, followed by the other abelian category axioms. Experience shows those are some pretty powerful facts. $\endgroup$ – Todd Trimble Jul 2 '16 at 13:36
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    $\begingroup$ It gives a sense in which finite-dimensional vector spaces are "naturally isomorphic" to their double duals (this example was in Eilenberg and Mac Lane's first paper on category theory) and it clears up what certain constructions like tensor products or exterior powers are all about by making them solutions to a suitable universal mapping problem. $\endgroup$ – KConrad Jul 2 '16 at 13:37
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    $\begingroup$ Probably, the Freyd-Mitchell embedding theorem can be of some help? Maybe take an abelian category that has all finite limits and colimits and somehow restrict it to be "over-a-field-like". If we succeed in this, literally all facts concerning vector spaces will become category-theoretic. $\endgroup$ – lisyarus Jul 2 '16 at 13:39
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    $\begingroup$ @lisyarus that sounds good except I don't know what "over-a-field-like" would be except maybe by working in the dual setting and looking at terminal spectrums... $\endgroup$ – Arrow Jul 2 '16 at 13:50
  • $\begingroup$ Related: mathoverflow.net/questions/118246/… $\endgroup$ – Dag Oskar Madsen Jul 3 '16 at 8:42
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To my mind there are two classes of interesting categorical facts here, loosely speaking "additive" facts and "multiplicative" facts. Some additive facts:

  1. Finite-dimensional vector spaces over $k$ has biproducts, and every object is a finite biproduct of copies of a single object, namely $k$. The categories with this property are precisely the categories of finite rank free modules over a semiring $R$ (the endomorphisms of the single object). These biproduct decompositions encapsulate both the idea that vector spaces have bases and that a choice of bases can be used to write linear maps as matrices.

  2. The single object $k$ above is simple, and so every object is a finite biproduct of simple objects. The categories with this property, in addition to 1, are precisely the categories of finite rank free modules over a division semiring $R$.

Note that additive facts can't see anything about fields being commutative. The multiplicative facts can:

  1. Finite-dimensional vector spaces over $k$ is symmetric monoidal with respect to tensor product, and is also closed monoidal and has duals (sometimes called compact closed). This observation encapsulates the yoga surrounding tensors of various types (e.g. endomorphisms $V \to V$ correspond to elements of $V \otimes V^{\ast}$), as well as the existence and basic properties (e.g. cyclic symmetry) of the trace.

  2. The single object $k$ above is the tensor unit, and so every object is a finite biproduct of copies of the unit. Also, the monoidal structure is additive in each variable. I believe, but haven't carefully checked, that the (symmetric monoidal) categories with this property, in addition to 1 and 2 above, are precisely the (symmetric monoidal) categories of finite rank free modules over a commutative division semiring ("semifield") $R$. This encapsulates the concrete description of tensor products as a functor in terms of Kronecker products.

There's surprisingly little to say as far as fields being rings as opposed to semirings, though. This mostly becomes relevant when we reduce computing (co)equalizers to computing (co)kernels by subtracting morphisms, as in any abelian category.

Edit: I haven't mentioned the determinant yet. This mixes additive and multiplicative: abstractly the point is that we have a natural graded isomorphism

$$\wedge^{\bullet}(V \oplus W) \cong \wedge^{\bullet}(V) \otimes \wedge^{\bullet}(W)$$

where $\wedge^{\bullet}$ denotes the exterior algebra (which we need the symmetric monoidal structure, together with the existence of certain colimits, to describe). It follows that if $L_i$ are objects which have the property that $\wedge^k(L_i) = 0$ for $k \ge 2$ then

$$\wedge^n(L_1 \oplus \dots \oplus L_n) = L_1 \otimes \dots \otimes L_n.$$

Combined with the facts above this gives the existence and basic properties of the determinant, more or less. Note that the exterior algebra can be defined by a universal property, but to verify that the standard construction has this universal property we need the symmetric monoidal structure to distribute over finite colimits. Fortunately this is implied by compact closure.

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Qiaochu's answer captures very nicely the ordinary categorical aspects of the category $\mathbf{Vect}_k$ of finite-dimensional vector spaces over $k$; I'll just add some homotopy-theoretic comments that I find interesting.

I am fond of the fact that we can think of vector bundles in the following way: for a topological field $k$, the category $\mathbf{Vect}_k$ is closed monoidal, and thus also enriched over $\mathbf{Top}$, the category of topological spaces. We can thus consider $\mathbf{Vect}_k$ as an $(\infty,1)$-category, and vector bundles over a space $X$ correspond to functors $X\to\mathbf{Vect}_k$; this is exactly analogous to how Grothendieck opfibrations $E\to B$ correspond to pseudofunctors $B\to \mathbf{Cat}$ (and coCartesian fibrations of quasicategories generalize both these examples).

To put things in perspective, this is really what's going on when we consider the classifying spaces $BO(n)$ or $BU(n)$ (recall that homotopy classes of maps $X\to BO(n)$ classify $n$-dimensional real vector bundles over $X$ and similarly with $BU(n)$ for complex vector bundles). Recall that spaces are $\infty$-groupoids. If $X$ is connected, then it's an $\infty$-groupoid with only one object (up to equivalence), so a functor $X\to\mathbf{Vect}_k$ is defined on objects just by choosing a single object $V\in\mathbf{Vect}_k$, and we can consider functors $X\to\mathcal{V}$, where $\mathcal{V}$ is the full sub-$\mathbf{Top}$-category of $\mathbf{Vect}_k$ with $V$ as its only object. And because $X$ is an $\infty$-groupoid, any such functor factors through $\mathrm{Core}(\mathcal{V})$, where $\mathrm{Core}(\mathcal{V})$ is the subcategory of $\mathcal{V}$ of automorphisms of $V$. When we go back to thinking of the $\infty$-groupoid $\mathrm{Core}(\mathcal{V})$ as a space, we wind up with $BO(n)$ or $BU(n)$ (for real and complex vector spaces of dimension $n$ respectively).

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My favorite example: Construct the functor $\Lambda^n : \mathsf{Vect}_k \to \mathsf{Vect}_k$. Show that $\Lambda^n(V)$ is of dimension $\binom{d}{n}$ when $d$ is the dimension of $V$ (this was roughly explained by Qiaochu Yuan). In particular, $\Lambda^d(V)$ is $1$-dimensional. Conclude that $\Lambda^d$ induces a multiplicative map $\mathrm{End}(V) \to \mathrm{End}(\Lambda^d(V)) = k$ and call it the determinant ... In this approach, the multiplicativity of the determinant follows from the functoriality of $\Lambda^n$, and the latter is trivial. Also, the proof that $\Lambda^d(V)$ is $1$-dimensional can be used to derive Leibniz's formula for the determinant of a matrix.

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  • $\begingroup$ I know that this is just one example of the "basic properties of the determinant" mentioned Qiaochu's answer, but I thought that it would be worth to make this point more explicit. $\endgroup$ – HeinrichD Sep 23 '16 at 14:03

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