Introduction to Weyl Algebra

 

Basic introduction to Weyl Algebra

The motivation we study Wely Algebra is for learning $\mathcal{D}-$$\mathcal D-$module, whcih is a way that use algebra method to solve linear differential equations. The basis example of Weyl Algebra is $A_1(\mathbb{C})=\mathbb{C}[z,D]$$A_1(\mathbb C)=\mathbb C[z,D]$, here $D=\frac{d}{dx}$$D=\frac{d}{dx}$, whcih correspond the ode with polynomial coefficient. In general, nth-Well Algebra over a charatestic $0$$0$ field is defined as:

$$\begin{array}{}\text{(1)}& A_n(\mathbb{K})=\mathbb{K}[z_1,...,z_n,{\partial }_1,...,{\partial }_n]\end{array}$$

We need a simple lemma to understand what happen next.

Remark. Some readers may oberseve the $A_1(\mathbb{C})$$A_1(\mathbb C)$ looks like $\mathbb{C}[e^{\lambda x},e^{-\lambda x},D]$$\mathbb C[e^{\lambda x},e^{-\lambda x},D]$ appear at my work (PDF) ODE: An Algebraic Approach (researchgate.net). We see that $e^{\lambda x}\cdot D\cdot e^{-\lambda t}=D-\lambda$$e^{\lambda x}\cdot D\cdot e^{-\lambda t}=D-\lambda$, which is really helpful for solving linear constant coefficient ode.

Lemma. Any ring $R$$R$ is a subring of ${\mathrm{End}}_{\mathrm{Ab}}(R)$$\mathrm{End}_{\mathrm{Ab}}(R)$. That is, there exists a embedding $\iota :R\to {\mathrm{End}}_{\mathrm{Ab}}(R)$$\iota:R\to\mathrm{End}_{\mathrm{Ab}}(R)$.

Proof. $\mathrm{\forall }a,b\in R,\iota (a)b:=ab$$\forall a,b\in R,\iota(a)b:=ab$. $\iota (a)(b+b^{\prime }):=ab+a{b}^{\prime }=\iota (a)(b)+\iota (a)(b^{\prime })$$\iota(a)(b+b'):=ab+ab'=\iota(a)(b)+\iota(a)(b')$. $◻$$\square$​

So, we should $A_n(\mathbb{K})$$A_n(\mathbb K)$) as a subring of ${\mathrm{End}}_{\mathbb{K}-\mathrm{vect}}(\mathbb{K}[z_1,...,z_n])$$\mathrm{End}_{\mathbb K-\mathrm{vect}}(\mathbb K[z_1,...,z_n])$. The product in Weyl Algebra is the composition of $\mathbb{K}-$$\mathbb K-$linear map, denote as $\cdot$$\cdot$​.

Definition. Lie bracket or commutator over a ring $R$$R$ is defined as $[X,Y]=XY-YX$$[X,Y]=XY-YX$ for any $X,Y\in R$$X,Y\in R$. It is a no-assotiative product over $R$$R$.

It is easy to check that the commutator operator is bilinear(here the linear is $\mathbb{Z}-$$\mathbb Z-$linear, since $\mathrm{Ring}\cong \mathbb{Z}-\mathrm{Alg}$$\mathrm{Ring}\cong\mathbb Z-\mathrm{Alg}$). We will see why the bilinear property is important after we use tensor product to define algebra over a ring $R$$R$​.

It is also obviously that $[X,Y]=-[Y,X]$$[X,Y]=-[Y,X]$​. The non trivial things is the Jacobi identity.

Proposition. Jacobi identity

$$\begin{array}{}\text{(2)}& [X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0\end{array}$$

To remeber this, you could consider you have $[-,[-,-]]$$[-,[-,-]]$ and let $C_3$$C_3$ act on it, the sum of the elements equal to zero. Question: What is the connections between Lie bracket and $1+\omega +{\omega }^2=0$$1+\omega+\omega^2=0$?

Jacobi identity should be viewed as a kind of Leibniz law. Let us consider $\mathfrak{ad}:X⟼{\mathfrak{ad}}_X:=[X,-]$$\mathfrak{ad}:X\longmapsto \mathfrak{ad}_X:=[X,-]$

Since $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0⟹{\mathrm{ad}}_X([Y,Z])=-[Y,[Z,X]]-[Z,[X,Y]]$$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0\implies\mathrm{ad}_X([Y,Z])=-[Y,[Z,X]]-[Z,[X,Y]]$

i.e.

$$\begin{array}{}\text{(3)}& {\mathrm{ad}}_X([Y,Z])=-[Y,[Z,X]]-[Z,[X,Y]]=[Y,[X,Z]]+[[X,Y],Z]=[Y,{\mathrm{ad}}_{X}Z]+[{\mathrm{ad}}_{X}Y,Z]\end{array}$$

As we mentioned, $[-,-]$$[-,-]$ is a non-assotiative product, what if we consider $A\ast B:=[A,B]$$A*B:=[A,B]$?

Then $(3)$$(3)$ will become ${\mathrm{ad}}_{\mathrm{X}}(Y\ast Z)=Y\ast {\mathrm{ad}}_{\mathrm{X}}Z+{\mathrm{ad}}_{\mathrm{X}}Y\ast Z$$\mathrm{ad_X}(Y*Z)=Y*\mathrm{ad_X}Z+\mathrm{ad_X}Y*Z$​! That is, Leibniz law!

Proof.

$$\begin{array}{}\text{(4)}& \begin{array}{c}[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]\\ =X[Y,Z]-[Y,Z]X+Y[Z,X]-[Z,X]Y+Z[X,Y]-[X,Y]Z\end{array}\end{array}$$

Using the fact that $[A,B]=-[B,A]$$[A,B]=-[B,A]$, we get

$$\begin{array}{}\text{(5)}& \begin{array}{c}X[Y,Z]+[Z,Y]X+Y[Z,X]+[X,Z]Y+Z[X,Y]+[Y,X]Z=\\ XYZ-XZY+ZYX-YZX+YZX-YXZ+XZY-ZXY+ZXY-ZYX+YXZ-XYZ=0\end{array}\end{array}$$

Proposition. $\begin{array}{c}[{\partial }_i,z_j]={\delta }_{i,j}:=\{\begin{array}{l}1,i=j\\ 0,i\ne j\end{array}\ne 0.\end{array}$$[\partial_i,z_j]=\delta_{i,j}:=\begin{cases}1,i=j\\0,i\ne j\end{cases}\ne 0.$

Proof. By definition, we get that $[{\partial }_i,z_j]={\partial }_i\cdot z_j-z_j\cdot {\partial }_i$$[\partial_i,z_j]=\partial_i\cdot z_j-z_j\cdot\partial_i$.

$$\begin{array}{rl}[{\partial }_i,z_j]p(z)& =({\partial }_i{z}_j-z_j{\partial }_i)p(z)\\ & ={\partial }_i(z_{j}p(z))-z_j({\partial }_{i}p(z))\\ & ={\delta }_{ij}p(z)+z_j{\partial }_{i}p(z)-z_j{\partial }_{i}p(z)\\ & ={\delta }_{ij}p(z)◻\end{array}$$

Corollary. $A_n(\mathbb{K})$$A_n(\mathbb K)$ is not a commutative ring, $A_n(\mathbb{K})\cong \mathbb{K}[x_1,...,x_n,y_1,...,y_n]/⟨y_i{x}_j-x_j{y}_i-{\delta }_{i,j}⟩$$A_n(\mathbb K)\cong\mathbb K[x_1,...,x_n,y_1,...,y_n]/\langle y_ix_j-x_jy_i-\delta_{i,j}\rangle$.

In general, we have:

$$\begin{array}{}\text{(6)}& \begin{array}{c}[{\partial }^I,z_J]=\{\begin{array}{l}0,I\ne J\\ I!,I=J\end{array}\end{array}\end{array}$$

Here $I,J$$I,J$ is multi-index. If $I=(i_1,i_2,...,i_n)$$I=(i_1,i_2,...,i_n)$, then ${\partial }^I={\partial }_1^{i_1},...,{\partial }_n^{i_n}$$\partial^I=\partial_1^{i_1},...,\partial_n^{i_n}$. The $I!:=i_1!...i_n!$$I!:=i_1!...i_n!$. The length of the multi-index is defined as $\begin{array}{c}\alpha =\sum _{k=1}^n{i}_k\end{array}$$\alpha=\sum_{k=1}^n i_k\\$.

Well, using the fact that $[{\partial }_i,z_j]={\delta }_{i,j}$$[\partial_i,z_j]=\delta_{i,j}$ we can write any element as the canonical form:

$$\begin{array}{}\text{(7)}& \mathcal{D}=\sum _{i\in (I,J)}{z}^I{\partial }^J\end{array}$$

That is, an polynomial coefficient differential operator.

For example, consider ${\partial }_1^2\cdot z_1^2$$\partial_1^2\cdot z_1^2$. Observe that $[{\partial }_1^2,z_1^2]=2={\partial }_1^2\cdot z_1^2-z_1^2\cdot {\partial }_1^2$$[\partial_1^2,z_1^2]= 2=\partial_1^2\cdot z_1^2-z_1^2\cdot\partial_1^2$, hence ${\partial }_1^2\cdot z_1^2=2+z_1^2\cdot {\partial }_1^2$$\partial_1^2\cdot z_1^2=2+z_1^2\cdot\partial_1^2$.

Connections with differential equations.

Let us consider the ring of homomorphic functions on $\mathrm{\Omega }\subseteq {\mathbb{C}}^n$$\Omega\subseteq\mathbb C^n$, denote as $H(\mathrm{\Omega })$$H(\Omega)$, which is a sheaf of ring.

Then there exists a natural way to define a $A_n(\mathbb{C})-$$A_{n}(\mathbb C)-$module structure on the sheaf $H(\mathrm{\Omega })$$H(\Omega)$.

Each

$$\begin{array}{}\text{(8)}& \mathcal{D}f=g\end{array}$$

Gives you a differential equation.

Now let us define another kind of algebra, motivate by my paper (PDF) ODE: An Algebraic Approach (researchgate.net) and Weyl Algebra.

Consider $\mathbb{C}[e^{\lambda z_i},{\partial }_1,...,{\partial }_n]$$\mathbb C[e^{\lambda z_i},\partial_1,...,\partial_n]$. Here $\lambda$$\lambda$ run over all the complex number.

Then

$$\begin{array}{}\text{(9)}& \begin{array}{c}[{\partial }_i,e^{\lambda z_j}]=\{\begin{array}{l}\lambda e^{z_i},i=j\\ 0,i\ne j\end{array}\end{array}\end{array}$$

Hence we have

$$\begin{array}{}\text{(10)}& {\partial }_i\cdot e^{\lambda z_i}-e^{\lambda z_i}\cdot {\partial }_i=\lambda \cdot e^{\lambda z_i}⟹{\partial }_i\cdot e^{\lambda z_i}-\lambda \cdot e^{\lambda z_i}=e^{\lambda z_i}\cdot {\partial }_i⟹({\partial }_i-\lambda )\cdot e^{\lambda z_i}=e^{\lambda z_i}\cdot {\partial }_i\end{array}$$

i.e.

$$\begin{array}{}\text{(11)}& ({\partial }_i-\lambda )=e^{\lambda z_i}\cdot {\partial }_i\cdot e^{-\lambda z_i}\end{array}$$