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Let $k$ be a field and $\text{Mat}_n(k)$ be $n \times n$ matrices over $k$. Let's consider $\text{Mat}_n(k)$ as an associative algebra and denote $gl_n(k)$ be the same $k$-linear space as $\text{Mat}_n(k)$ but considered as a Lie algebra. We can form the universal enveloping algebra $U(gl_n(k))$ of $gl_n(k)$, which satisfies the universal property, $\forall$ unital associative $k$-algebra $A$, we have $$ Hom_{k-alg}(U(gl_n(k)),A)\cong Hom_{Lie}(gl_n(k),F(A)), $$ where $F$ is the forgetful functor from $k$-alg to Lie.

Now we know that the associative algebra $\text{Mat}_n(k)$ is not the universal enveloping algebra $U(gl_n(k))$. However, by the universal property, we have $$ Hom_{k-alg}(U(gl_n(k)),\text{Mat}_n(k))\cong Hom_{Lie}(gl_n(k),gl_n(k)), $$ which makes $\text{Mat}_n(k)$ "special" in some sense.

$\textbf{My question}$ is: Could we make the above observation more precise? Does $\text{Mat}_n(k)$ have some universal properties similar to $U(gl_n(k))$?

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    $\begingroup$ It seems odd to think about matrices as an object having a universal property (in the same way that it seems odd to think about $S_n$ as an object having a universal property). I would rather say that it's the endomorphism monoid of an object in a category I care about (namely vector spaces). $\endgroup$ – Qiaochu Yuan Sep 17 '13 at 21:26
  • $\begingroup$ @QiaochuYuan, yes, you are right, but/and this could be revised-away from the question, just as the coordinates could be removed from $GL_n(\mathbb R)$ and such... $\endgroup$ – paul garrett Sep 17 '13 at 22:31
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Of course, what's a universal property is open to interpretation, but $\mathrm{Mat}_n(k)$ corepresents something very nice: $\mathrm{Hom}_{k-alg}(A,\mathrm{Mat}_n(k))$ is the space of representations of $A$ on $k^n$ (not up to any kind of equivalence, of course). This can also be interpreted as the $k$ points of an affine algebraic $k$-variety $\mathrm{Rep}_n(A)$, as long as $A$ is finitely generated: if $a_1,\dots, a_m$ are generators of $A$, and $r_1(\mathbf{a}), \dots, r_\ell(\mathbf{a})$ are the non-commutative polynomials giving relations, then $\mathrm{Rep}_n(A)$ is defined by taking $n\times n$ matrices $M_1,\dots, M_m$ with each matrix coefficient a variable, and letting the relations be $r_k(\mathbf{M})=0$.

In fancier language, we can define its points over any $k$-algebra $R$ by $$\mathrm{Rep}_n(A,R)=\mathrm{Hom}_{R-alg}(A\otimes_k R,\mathrm{Mat}_n(R)).$$

You can interpret this as saying that the functor from commutative to non-commutive $k$-algebras given by $\mathrm{Mat}_n$ is the right adjoint to the functor sending $A$ to $k[\mathrm{Rep}_n(A)]$, much like the universal enveloping algebra is left adjoint to the forgetful functor from algebras to Lie algebras that just remembers the bracket.

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    $\begingroup$ I think this is "the right answer". :) $\endgroup$ – paul garrett Sep 17 '13 at 22:29
  • $\begingroup$ Based on the OP's earlier interest in maps out of the universal enveloping algebra I would have guessed that the OP was interested in maps out of the matrix algebra, not maps in. $\endgroup$ – Qiaochu Yuan Sep 18 '13 at 1:19
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    $\begingroup$ In the words of Mick Jagger: you can't always get what you want. $\endgroup$ – Ben Webster♦ Sep 18 '13 at 13:13

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