# Is there a categorical construction of the general linear group?

This question is related to the answer of Qiaochu in this one. Since the object $X=\mathbb{F}_2^2$ generates the category of vector spaces of dimension $2^n$ over $\mathbb{F}_2$, and since we know $\text{End}(X)$ (which is isomorphic to $S_3$), is it possible to $\it construct$ the general linear group for any vector space of dimension $2^n$ from the knowledge of the structure of this category ?

(As was discussed in the answer of that other question, if one is considering only the morphisms of $X$ and the symmetric operation, one ends up with wreath products. But the actual general linear group is bigger than that, so in other terms, where are the other morphisms coming from ?)

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Short answer: No. Recall the definition of a generator, and why it should be better called separator: The terminology is quite unfortunate. – Martin Brandenburg Feb 15 '13 at 13:23
@MartinBrandenburg: could you elaborate on this notion of separator ? – AlexPof Feb 17 '13 at 16:51

What Martin says is correct, but one should note that $\mathbb{F}_2$ and $\mathbb{F}_2^2$ together do generate (in the strongest possible sense) the category of vector spaces over $\mathbb{F}_2$. Actually, all this goes through for an arbitrary (not necessarily commutative) ring $A$, so I will work in that generality.
Let $\mathbf{\Theta}$ be the full subcategory of $\textbf{Mod}_A$ spanned by $A$ and $A \oplus A$. There is an evident functor $N : \textbf{Mod}_A \to [\mathbf{\Theta}^\textrm{op}, \textbf{Set}]$ sending a right $A$-module $M$ to the restricted hom functor $\textrm{Hom}_A (-, M)$. This is certainly faithful, since an element of $M$ is essentially the same thing as a homomorphism $A \to M$, and in fact it is even full. Indeed, for each element $a$ of $A$, the map $x \mapsto a x$ defines a right $A$-module homomorphism $A \to A$, and thus a map $a^* : \textrm{Hom}_A (A, M) \to \textrm{Hom}_A (A, M)$; if we identify $\textrm{Hom}_A (A, M)$ with $M$ in the natural way, $a^*$ turns out to be the map $m \mapsto m a$. On the other hand, the homomorphism $p : A \to A \oplus A$ defined by $a \mapsto (a, a)$ induces a map $p^* : \textrm{Hom}_A (A \oplus A, M) \to \textrm{Hom}_A (A, M)$ which, if we identify $\textrm{Hom}_A (A \oplus A, M)$ with $M \times M$ and $\textrm{Hom}_A (A, M)$ with $M$, is easily seen to be the map $(m_1, m_2) \mapsto m_1 + m_2$. Thus, any natural transformation $N M' \Rightarrow N M$ necessarily comes from a homomorphism $M' \to M$.
Next, observe that $N : \textbf{Mod}_A \to [\mathbf{\Theta}^\textrm{op}, \textbf{Set}]$ preserves all limits. Thus, we can identify $N (A^n)$ with $(N A)^n$, and because $N$ is fully faithful, we have the following monoid isomorphism: $$\textrm{Hom}_A (A^n, A^n) \cong \textrm{Nat}((N A)^n, (N A)^n)$$ Now, $N A : \mathbf{\Theta}^\textrm{op} \to \textbf{Set}$ is a representable functor with respect to $\mathbf{\Theta}$, so the right hand side of the above isomorphism can be constructed using categorical nonsense from $\mathbf{\Theta}$ alone. Thus, to construct $\textrm{Mat}_n (A)$ as a multiplicative monoid, it is enough to know about $a \times b$ matrices for $a$ and $b$ in $\{ 1, 2 \}$, and it is easy to recover $\mathrm{GL}_n (A)$ from the multiplicative structure of $\textrm{Mat}_n (A)$.
What is $\textrm{Mat}_n (A)$ in your last paragraph ? – AlexPof Mar 6 '13 at 12:06
The ring of $n \times n$ matrices over $A$, of course. – Zhen Lin Mar 6 '13 at 12:52