Question

I am curious. Is there a "slick" way of showing that given an arbitrary algebra $A$ with generating set $X$, an algebra endomorphism $\alpha : A\to A$ and a function (satisfying some conditions) $d : X\to A$, that $d$ extends uniquely to an $\alpha$-derivation $D : A\to A$?

Answer 1

Suppose $V$ is a vector space, let $TV=\bigoplus_{n\geq0}V^{\otimes n}$ be the tensor algebra on $V$ and let $I$ be an ideal of $TV$. Let $A=TV/I$ be the quotient algebra, let $p:TV\to A$ be the canonical projection and let $\alpha:A\to A$ be an endomorphism of algebras. Let, moreover, $\delta:V\to A$ be any linear map.

There is a (in fact unique) linear map $\delta_1:TV\to A$ such that

• the restriction of $\delta_1$ to $V\subseteq TV$ is $\delta$, and
• $\delta_1(xy)=\alpha(p(x))\delta(y)+\delta(x)p(y)$ for all $x$, $y\in TV$.

Indeed, these two conditions show that its restriction to $V^{\otimes n}$ must be given by $$\delta_1(x_1\otimes\cdots\otimes x_n)=\sum_{i=1}^n\alpha(p(x_1\cdots x_{i-1}))\delta(x_i)p(x_{i+1}\cdots x_n),$$ and if we use this formula to define $\delta_1$, a boring verification will show that we get a map that actually works.

Let $I$ is generated by elements ${r_j}_{j\in J}\subseteq TV$ and let us suppose that

$$\text{$\delta_1(r_j)=0$ for all $j\in J$.} \tag{$\star$}$$

It is then easy to see that $\delta_1(I)=0$, using the fact that the ideal $I$ is the linear span of all elements of the form $xr_jy$ with $x$, $y\in TV$ and $j\in J$. As a consequence, $\delta_1$ passes down to the quotient to give a linear map $\delta_2:A=TV/I\to A$ which by design is an $\alpha$-derivation.

We conclude that $(\star)$ is a sufficient condition for the existence of an $\alpha$-derivation extending $d:V\to A$, and a little reflection will show that it is also necessary. As Martin has shown earlier that we have uniqueness, we are happy.

The condition is one that one can check in concrete examples with little trouble.

An example. Let $q$ be a scalar and let $A$ be the free algebra generated by $K$, $L$, and $F$ subject to the relations \begin{gather} KL=1=LK, \\ FK=q^2KF. \end{gather} The first two relations tell us that $L=K^{-1}$, and using the second one one can see without much pain that ${F^aK^b:a\in\mathbb N,b\in\mathbb Z}$ is a basis of $A$. A little extra work will show that there is an automorphism $\alpha:A\to A$ such that $$\alpha(F^aK^b)=q^{-2b}F^aK^b.$$ We want to construct an $\alpha$-derivation $d$ of $A$ such that $$d(F)=\frac{K-K^{-1}}{q-q^{-1}}$$ and $$d(K)=d(L)=0.$$

We let $V$ be the vector space with basis ${K,L,F}$, define $\delta:V\to A$ putting $\delta(K)=\delta(L)=0$ and $\delta(F)=(q-q^{-1})^{-1}(K-K^{-1})$ and use the technology developed above. We have the map $\delta_1:TV\to A$ amd we have to check that it vanishes on the elements $KL-1$, $LK-1$ and $FK-q^2KF$. We have, for example, $$\delta_1(KL-1)=\alpha(K)\delta(L)-\delta(L)K=0$$ simply because $\delta$ kills $K$ and $L$, and similarly for the second relator. The third one is more interesting: \begin{align} \delta_1(FK-q^2KF)&=\alpha(F)\delta(K)+\delta(F)K-q^2\alpha(K)\delta(F)-q^2\delta(K)F\\ &=\delta(F)K-q^2\alpha(K)\delta(F)\\ &=\frac{K-K^{-1}}{q-q^{-1}}K-q^2(q^{-2}K)\frac{K-K^{-1}}{q-q^{-1}}\\ &=0. \end{align} We have checked our condition $(\star)$, so there is an $\alpha$-derivation $d:A\to A$ which does what we wanted.

This example is one of the steps required in showing that $U_q(\mathfrak{sl}_2)$ is an iterated Ore extension —indeed, $U_q(\mathfrak{sl}_2)=A[E;\alpha, d]$.

Answer 2

The key is to observe that the definition of a derivation $d$ immediately implies that $\ker(d)$ is a subalgebra of $A$. Now if $d,d'$ are derivations, then $d-d'$ is a derivation. Thus, if $d$ and $d'$ coincide on algebra generators, they coincide everywhere.