*To help keep me motivated and mathematically active, I will be blogging about "what I learned today" in my various projects. I will endeavor to post to this series daily, or as close to daily as I can. This is the first post in that series, and today we're talking about what I learned in universal algebra/hopf algebras.*

So I'm taking a break from my finite fields project and am focusing on a new one, purely in quantum groups. I "started" about three months ago, but between procrastination, my job, and other projects (exempli gratia, learning class field theory) I haven't actually learned any quantum group theory.

I had made some progress in learning quantum groups in July, but due to some personal issues that came up in August, I took a long break from mathematics and largely forgot everything I had learned. So I'm trying to learn it all again. Now, Majid's Quantum Groups Primer covers the material exactly needed for my project, but I'm having trouble with many of the algebraic subtleties. In fact, I'm pretty weak in algebra (or maths in general). I'm using Kassel's Quantum Groups as a companion text, and I spent an hour or two today going over some basic universal algebra. Here's what I learned: Let $A$ be an associative algebra over a field $k$. Then:

$$\text{Hom}_\text{Alg}(k[x_1,\ldots,x_n], A) \cong A^n$$You can see this because an algebra map from the polynomial algebra to $A$ is exactly determined by its action on the generators $x_1,\ldots,x_2$, hence a choice of an algebra map is a choice of a set map from $\{1,\ldots, n\}$ to $A$, id est, $A^n$.

This allows us to write the additive group structure of $A$ universally in terms of maps on $k[x]$ and $k[x', x'']$. That is, we will express addition on $A$ without reference to $A$ itself. The group addition on $A$ is a map $+: A^2 \to A$. Now, from above, $A^2 \cong \text{Hom}_\text{Alg}(k[x',x''],A)$. So if we let $f \in \text{Hom}_\text{Alg}(k[x',x''],A)$, we will reach our goal by expressing $+(f)$ in terms of the polynomial algebras.

$f$ is entirely determined by its action of $x'$ and $x''$, hence $f(x')$ and $f(x'')$ define $f$. Now let $\Delta: k[x] \to k[x', x'']$ by $x \mapsto x' + x''$. Then $f \circ \Delta (x) = f(x' + x'') = f(x') + f(x'') = +(f)$. Hence $\Delta$ corresponds to our group addition.

Similarly, the map $S:k[x] \to k[x]$ by $x \mapsto -x$ and $\epsilon: k[x] \to k$ by $x \mapsto 0$ correspond to the group inversion and the group identity, respectively, of $A$.

## No comments:

## Post a Comment