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I'm looking at motivating the standard deformation of $U(\mathfrak{sl}(2))$. As an algebra $U(\mathfrak{sl}(2))$ is generated by $X,Y$ and $H$ and subject to the relations $[X,Y] = H$, $[H,X] = 2X$ and $[H,Y] = -2Y$. From $q$-analysis I have that an integer $n$ is deformed according to $[n] = \frac{q^{n}-q^{-n}}{q-q^{-1}}$.

This seems to motivate, at least partially, the means of deforming $U(\mathfrak{sl}(2))$. That is, set $[X,Y] = [H] = \frac{q^{H}-q^{-H}}{q-q^{-1}}$ and then set $K = q^{H}$ and $K^{-1} = q^{-H}$. But what motivates the relations $KX = q^{2}XK$ and $KY = q^{-2}YK$?

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Hi Ryan,

You can prove that if $a,b$ are some elements in an algebra such that $[a,b]=\lambda b$ for $\lambda$ a scalar, then (in a context where this expression makes sense) $q^a b q^{-a}=q^{\lambda} b$: rewrite the relation as $$ab=b(a+\lambda)$$ then $$a^nb=b(a+\lambda)^n$$ therefore $$q^ab=\sum \frac{\log(q)^na^n}{n!}b=b\sum \frac{\log(q)^n(a+\lambda)^n}{n!}= bq^{(a+\lambda)}=q^{\lambda}bq^a$$ since $\lambda$ commutes with everything.

Therefore it's in fact not a deformation but the same relation with $q^H$ instead of $H$.

In my opinion, what is more miraculous is the existence of a non-trivial deformation of the Hopf structure. Although it does not leads to explicit formulas, the conceptual explanation for that comes from the Kohno-Drinfeld theorem, or more generally from the Etingof-Kazhdan quantization functor.

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Thanks for the reply. What's the best way to prove this? –  Ryan Mar 24 '12 at 1:09
    
It's a direct computation, I edited my answer. –  Adrien Mar 24 '12 at 3:01
    
Thank you Adrien for clarifying. If I may, I'm trying to fill in some background as I go. Where does the the assignment $q^{a} = \sum\frac{log(q)^{n}a^{n}}{n!}$ come from? –  Ryan Mar 24 '12 at 5:39
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$q^a$ is by definition $\exp(\log(q) a)=\sum (\log(q)a)^n/n!$. –  Adrien Mar 24 '12 at 10:37
    
Where does this definition come from? I've looked in several texts and none of them mention this definition for $q^{a}$. –  Ryan Mar 24 '12 at 16:37
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Allow me the excuse to mention a cute way of constructing Uq(sl2) as a q-analogue. Presumably this is ahistorical, since I suspect that initially Uq(sl2) was constructed first over the formal disc, then later it was realised that it was definable over $\mathbb{Q}(q)$.

Anyhow, let S be a finite set. Consider a vector space with basis $\{[I]\mid I\subset S\}$. Define operators e,f,h on this vector space by $$e[I]=\sum_{i\notin I}[I\cup\{i\}]$$ $$f[I]=\sum_{i\in I}[I\setminus \{i\}]$$ $$h[I]=(2|I|-|S|)[I].$$

You can check that this is a representation of sl2 (it's just a tensor power of the standard represenation in disguise).

Now q-analogise! How? Well we should replace the idea of a subset of a set with that of a subspace of a vector space (of course working over the finite field with q elements).

Let V be a finite dimensional vector space over $\mathbb{F}_q$. We should consider the space of all complex-valued functions on the Grassmannian Gr(V) of all subspaces of V. Well actually we'd really like something of dimension 2^n, so we should consider only those functions invariant under the action of a Borel subgroup.

Let F be the space of pairs (W,W') with $W\subset W'$ two subspaces of V satifying dim(W')=dim(W)+1. There are two obvious maps $p_1,p_2:F\to Gr(V)$.

Now we can define two operators on our space of functions on Gr(V), namely $E=(p_2)_!(p_1)^!$ and $F=(p_1)_!(p_2)^!$. Here p! is pullback of functions and p! means sum the function over the fibre. These are almost the generators of the quantum group we're after, to make the formulae match up nicely, we need to modify the definition of p! by multiplying it by a factor of q-d/2 where d is the dimension of the fibre summed over (a Tate twist).

Now you have two operators E and F, and when you compute their commutator, you naturally find another operator K such that $$EF-FE=\frac{K-K^{-1}}{q^{1/2}-q^{-1/2}}.$$ Computing KE=qEK and qKF=FK is now easy.

But even better, this approach gives more than just the product structure on the quantum group. You can also get the coproduct. I will be brief since I haven't checked all powers that q that appear. I haven't thought about obtaining the antipode.

Consider a short exact sequence of vector spaces 0-->V'-->V-->V''-->0. Given functions f' on Gr(V') and f'' on Gr(V''), one can define a function on Gr(V) by $$(f'\otimes f'')(X)=f'(X\cap V')f''(X/(X\cap V')).$$

There is a reasonable chance this needs correction by a power of q. The point however is that this defines an isomorphism between Borel-equivariant functions on Gr(V) and the tensor product of Borel-equivariant functions on Gr(V') with Borel-equivariant functions on Gr(V''). (If you drop Borel-equivariant, then you're still getting a map, you're just not deforming a representation of sl2).

I leave it as an exercise to the interested reader to use this to deduce the coproduct formulae in Uq(sl2).

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I've got shreik pushforwards and pullbacks in this answer not because they're the correct thing, but putting in stars required dealing with the bugginess that markdown introduces. Feel free to edit appropriately. Also: Exercise: work sheafily! –  Peter McNamara Mar 25 '12 at 4:35
    
Thanks for the neat insights! –  Ryan Mar 26 '12 at 5:03
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Since $K=q^H$, the relation $KX=q^2XK$ is really $$q^HX=q^2Xq^H.$$ Now $\frac{d}{dq}q^H=Hq^{-1}q^H$ so differentiating with respect to $q$ the previous equality we get $$Hq^{-1}q^HX=2qXq^H+q^2XHq^{-1}q^H.$$ If we now let $q=1$, this becomes $HX=2X+XH$, which becomes, after rearranging terms, $$[H,X]=2X$$ which is the classical commutation relation.

We thus see that the relations defining the exponential are a deformation which infinitessimally, at $q=1$, gives the usual enveloping algebra.

(All of this can be put in a context where it actually makes sense...)

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Thanks for the reply. That makes sense, and it seems easy if one starts with the result and then recovers the appropriate relation, but what what would be the process of arriving at $KX = q^{2}XK$ just starting from $[H,X] = 2X$? –  Ryan Mar 24 '12 at 1:17
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