Posts Tagged ‘Weyl algebra’

We defined the n-th Weyl algebra A_n(R) over a ring R in here.  In this post we will find the GK dimension of A_n(R) in terms of the GK dimension of R. The result is similar to what we have already seen in commutative polynomial rings (see corollary 1 in here). We will assume that k is a field and R is a k-algebra.

Theorem. {\rm{GKdim}}(A_1(R))=2 + {\rm{GKdim}}(R).

Proof. Suppose first that R is finitely generated and let V be a frame of R. Let U=k+kx+ky. Since yx = xy +1, we have

\dim_k U^n = \frac{(n+1)(n+2)}{2}. \ \ \ \ \ \ \ \ \ (*)

Let W=U+V. Clearly W is a frame of A_1(R) and

W^n = \sum_{i+j=n} U^i V^j,

for all n, because every element of V commutes with every element of U. Therefore, since V^j \subseteq V^n and U^i \subseteq U^n for all i,j \leq n, we have W^n \subseteq U^nV^n and W^{2n} \supseteq U^nV^n. Thus W^n \subseteq U^nV^n \subseteq W^{2n} and hence

\log_n \dim_k W^n \leq \log_n \dim_k U^n + \log_n \dim_k V^n \leq \log_n \dim_k W^{2n}.

Therefore {\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)), by (*), and we are done.

For the general case, let R_0 be any finitely generated k– subalgebra of R. Then, by what we just proved,

2 + {\rm{GKdim}}(R_0)={\rm{GKdim}}(A_1(R_0)) \leq {\rm{GKdim}}(A_1(R))

and hence 2+{\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)). Now, let A_0 be a k-subalgebra of A_1(R) generated by a finite set \{f_1, \ldots , f_m\}. Let R_0 be the k-subalgebra of R generated by all the coefficients of f_1, \ldots , f_m. Then A_0 \subseteq A_1(R_0) and so

{\rm{GKdim}}(A_0) \leq {\rm{GKdim}}(A_1(R_0))=2 + {\rm{GKdim}}(R_0) \leq 2 + {\rm{GKdim}}(R).


{\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R)

and the proof is complete. \Box

Corollary. {\rm{GKdim}}(A_n(R))=2n + {\rm{GKdim}}(R) for all n. In particular, {\rm{GKdim}}(A_n(k))=2n.

Proof. It follows from the theorem and the fact that A_n(R)=A_1(A_{n-1}(R)). \Box

Notation. Throughout this post we will assume that k is a field, V is a k-vector space, E=\text{End}_k(V) and \mathfrak{I} = \{f \in E : \ \text{rank}(f) < \infty \}. Obviously \mathfrak{I} is a two-sided ideal of E.

If \dim_k V = n < \infty, then E \cong M_n(k), the ring of n \times n matrices with entries in k, and thus E is a simple ring, i.e. the only two-sided ideals of E are the trivial ones: (0) and E. But what if \dim_k V = \infty ?  What can we say about the two-sided ideals of E if \dim_k V = \infty ?

Theorem 1. If \dim_k V is countably infinite, then \mathfrak{I} is the only non-trivial two-sided ideal of E.

Proof. Let J be a two-sided ideal of E and consider two cases.

Case 1. J \not \subseteq \mathfrak{I}. So there exists f \in J such that \text{rank}(f)=\infty. Let \{v_1, v_2, \ldots \} be a basis for V and let W be a subspace of V such that V = \ker f \oplus W. Note that W is also countably infinite dimensional because f(V)=f(W). Let \{w_1,w_2, \ldots \} be a basis for W. Since \ker f \cap W = (0), the elements f(w_1), f(w_2), \ldots are k-linearly independent and so we can choose g \in E such that gf(w_i)=v_i, for all i. Now let h \in E be such that h(v_i)=w_i, for all i. Then 1_E=gfh \in J and so J=E.

Case 2. (0) \neq J \subseteq \mathfrak{I}. Choose 0 \neq f \in J and suppose that \text{rank}(f)=n \geq 1. Let \{v_1, \ldots , v_n \} be a basis for f(V) and extend it to a basis \{v_1, \ldots , v_n, \ldots \} for V. Since f \neq 0, there exists s \geq 1 such that f(v_s) \neq 0. Let f(v_s) = b_1v_1 + \ldots + b_nv_n and fix an 1 \leq r \leq n such that b_r \neq 0. Now let g \in \mathfrak{I} and suppose that \text{rank}(g)=m. Let \{w_1, \ldots , w_m \} be a basis for g(V) and for every i \geq 1 put g(v_i)=\sum_{j=1}^m a_{ij}w_j. For every 1 \leq j \leq m define \mu_j, \eta_j \in E as follows: \mu_j(v_r)=w_j and \mu_j(v_i)=0 for all i \neq r, and \eta_j(v_i)=b_r^{-1}a_{ij}v_s for all i. See that g=\sum_{j=1}^m \mu_j f \eta_j \in J and so J=\mathfrak{I}. \ \Box

Exercise. It should be easy now to guess what the ideals of E are if \dim_k V is uncountable. Prove your guess!

Definition. Let n \geq 1 be an integer. A ring with unity R is called n-simple if for every 0 \neq a \in R, there exist b_i, c_i \in R such that \sum_{i=1}^n b_iac_i=1.

Remark 1. Every n-simple ring is simple. To see this, let J \neq (0) be a two-sided ideal of R and let 0 \neq a \in J. Then, by definition, there exist b_i,c_i \in R such that \sum_{i=1}^n b_iac_i=1. But, since J is a two-sided ideal of R, we have b_iac_i \in J, for all i, and so 1 \in J.

It is not true however that every simple ring is n-simple for some n \geq 1. For example, it can be shown that the first Weyl algebra A_1(k) is not n-simple for any n \geq 1.

Theorem 2. If \dim_k V = n < \infty, then E is n-simple. If \dim_k V is countably infinite, then E/\mathfrak{I} is 1-simple.

Proof. If \dim_k V = n, then E \cong M_n(k) and so we only need to show that M_n(k) is n-simple. So let 0 \neq a =[a_{ij}] \in M_n(k) and suppose that \{e_{ij}: \ 1 \leq i.j \leq n \} is the standard basis for M_n(k). Since a \neq 0, there exists 1 \leq r,s \leq n such that a_{rs} \neq 0. Using a = \sum_{i,j}a_{ij}e_{ij} it is easy to see that \sum_{i=1}^n a_{rs}^{-1}e_{ir}ae_{si}=1, where 1 on the right-hand side is the identity matrix.  This proves that E is n-simple. If \dim_k V is countably infinite, then, as we proved in Theorem 1, for every f \notin \mathfrak{I} there exist g,h \in E such that gfh=1_E. That means E/\mathfrak{I} is 1-simple. \Box

Remark 2. An n-simple ring is not necessarily artinian. For example, if \dim_k V is countably infinite, then the ring E/\mathfrak{I} is 1-simple but not artinian.

Let k be a field. We proved here that every derivation of a finite dimensional central simple k-algebra is inner. In this post I will give an example of an infinite dimensional central simple k-algebra all of whose derivations are inner. As usual, we will denote by A_n(k) the n-th Weyl algebra over k. Recall that A_n(k) is the k-algebra generated by x_1, \ldots , x_n, y_1, \ldots , y_n with the relations x_ix_j-x_jx_i=y_iy_j-y_jy_i=0, \ y_ix_j-x_jy_i= \delta_{ij}, for all i,j. When n = 1, we just write x,y instead of x_1,y_1. If \text{char}(k)=0, then A_n(k) is an infinite dimensional central simple k-algebra and we can formally differentiate and integrate an element of A_n(k) with respect to x_i or y_i exactly the way we do in calculus.  Let me clarify “integration” in A_1(k). For every u \in A_1(k) we denote by u_x and u_y the derivations of u with respect to x and y respectively. Let f, g, h \in A_1(k) be such that g_x=h_x=f. Then [y,g-h]=0 and so g-h lies in the centralizer of y which is k[y]. So g-h \in k[y]. For example, if f = y + (2x+1)y^2, then g_x=f if and only if g= xy + (x^2+x)y^2 + h(y) for some h(y) \in k[y]. We will write \int f \ dx = xy+(x^2+x)y^2.

Theorem. If \text{char}(k)=0, then every derivation of A_n(k) is inner.

Proof. I will prove the theorem for n=1, the idea of the proof for the general case is similar. Suppose that \delta is a derivation of A_1(k). Since \delta is k-linear and the k-vector space A_1(k) is generated by the set \{x^iy^j: \ i,j \geq 0 \}, an easy induction over i+j shows that \delta is inner if and only if there exists some g \in A_1(k) such that \delta(x)=gx-xg and \delta(y)=gy-yg. But gx-xg=g_y and gy-yg=-g_x. Thus \delta is inner if and only if there exists some g \in A_1(k) which satisfies the following conditions

g_y=\delta(x), \ \ g_x = -\delta(y). \ \ \ \ \ \ \ (1)

Also, taking \delta of both sides of the relation yx=xy+1 will give us

\delta(x)_x = - \delta(y)_y. \ \ \ \ \ \ \ \ (2)

From (1) we have \delta(x) = - \int \delta(y)_y \ dx + h(y) for some h(y) \in k[y]. It is now easy to see that

g = - \int \delta(y) \ dx + \int h(y) \ dy

will satisfy both conditions in (1). \ \Box

A non-linear automorphism of A_n(k). Let k be a field. For any u, v \in A_n(k) we let [u,v]=uv-vu. So the realtions that define A_n(k) become [x_i,x_j]=[y_i,y_j]=0, \ [y_i,x_j]=\delta_{ij}, for all i,j.

Lemma 1. Let f,g \in k[x_1, \cdots , x_n] and 1 \leq r,s \leq n. Then

1) [fy_r,g] = f \frac{\partial{g}}{\partial{x_r}}.

2) [fy_r,gy_s] = f \frac{\partial{g}}{\partial{x_r}}y_s - g \frac{\partial{f}}{\partial{x_s}}y_r.

Proof. An easy induction shows that y_r x_r^{\ell} = x_r^{\ell}y_r + \ell x_r^{\ell -1} for all \ell. Applying this, we will get that if h = x_1^{\alpha_1} \cdots x_n^{\alpha_n}, then y_r h = \frac{\partial{h}}{\partial{x_r}} + hy_r. So, since every element of k[x_1, \cdots , x_n] is a finite linear combination of monomials in the form h, we will get

y_r g = \frac{\partial{g}}{\partial{x_r}} + gy_r, \ \ \ \ \ \ \ \ \ \ \ \ (*)

for all g \in k[x_1, \cdots , x_n]. Both parts of the lemma are straightforwad results of (*). \Box

Notation. Let n \geq 2 and fix an integer 1 \leq m < n. For every 1 \leq i \leq m choose f_i \in k[x_{m+1}, \cdots , x_n] and put f_{m+1} = \cdots = f_n = 0.

Lemma 2. For any 1 \leq r,s,t \leq n we have \frac{\partial{f_r}}{\partial{x_s}} \cdot \frac{\partial{f_t}}{\partial{x_r}} = 0.

Proof. If r > m, then f_r = 0 and we are done. If r \leq m, then x_r will not occur in f_t and so \frac{\partial{f_t}}{\partial{x_r}} = 0. \ \Box

Now define the maps \varphi : A_n(k) \longrightarrow A_n(k) and \psi : A_n(k) \longrightarrow A_n(k) on the generators by

\varphi (x_i) = x_i + f_i, \ \varphi(y_i)= y_i - \sum_{r=1}^n \frac{\partial{f_r}}{\partial{x_i}}y_r


\psi (x_i)=x_i-f_i, \ \psi(y_i)=y_i + \sum_{r=1}^n \frac{\partial{f_r}}{\partial{x_i}}y_r,

for all 1 \leq i \leq n and extend the definition homomorphically to the entire A_n(k) to get k-algebra homomorphisms of A_n(k). Of course, we need to show that these maps are well-defined i.e. the images of x_i,y_i under \varphi and \psi satisfy the same relations that x_i, y_i do. Before that, we prove an easy lemma.

Lemma 3. \varphi(f) = \psi(f)=f for all f \in k[x_{m+1}, \cdots , x_n].

Proof. Let f = \sum c_{\alpha} x_{m+1}^{\alpha_{m+1}} \cdots x_n ^{\alpha_n}, where c_{\alpha} \in k and \alpha_i \geq 0. Then

\varphi(f) = \sum c_{\alpha} (x_{m+1} + f_{m+1})^{\alpha_{m+1}} \cdots (x_n + f_n)^{\alpha_n}.

But by our choice f_{m+1} = \cdots = f_n = 0 and thus \varphi(f)=f. A similar argument shows that \psi(f)=f. \ \Box

Lemma 4. The maps \varphi and \psi are well-defined.

Proof. I will only prove the lemma for \varphi because the proof for \psi is identical. Since f_i \in k[x_1, \cdots , x_n], we have \varphi(x_i) \in k[x_1, \cdots , x_n], for all i, and thus \varphi(x_i) and \varphi(x_j) commute. The relations [\varphi(y_i), \varphi(x_j)] = \delta_{ij} follow from the first part of Lemma 1 and Lemma 2. The relations [\varphi(y_i), \varphi(y_j)]=0 follow from the second part of Lemma 1 and Lemma 2. \Box.

Theorem. The k-algebra homomorphisms \varphi and \psi are automorphisms.

Proof. We only need to show that \varphi and \psi are the inverse of each other. Lemma 3 gives us \varphi \psi(x_i) = \psi \varphi(x_i)=x_i and Lemma 2 with Lemma 3 will give us \varphi \psi(y_i)=\psi \varphi (y_i)=y_i, for all i. \ \Box

Let R be a ring and let n \geq 0 be an integer. The n-th Weyl algebra over R is defined as follows. First we define A_0(R)=R. For n \geq 1, we define A_n(R) to be the ring of polynomials in 2n variables x_i, y_i, \ 1 \leq i \leq n, with coefficients in R and subject to the relations

x_ix_j=x_jx_i, \ y_iy_j=y_jy_i, \ y_ix_j = x_jy_i + \delta_{ij},

for all i,j, where \delta_{ij} is the Kronecker delta. We will assume that every element of R commutes with all 2n variables x_i and y_i. So, for example, A_1(R) is the ring generated by x_1,y_1 with coefficients in R and subject to the relation y_1x_1=x_1y_1+1. An element of A_1(R) is in the form \sum r_{ij}x_1^iy_1^j, \ r_{ij} \in R.. It is not hard to prove that the set of monomials in the form

x_1^{\alpha_1} \ldots x_n^{\alpha_n}y_1^{\beta_1} \ldots y_n^{\beta_n}

is an R-basis for A_n(R). Also note that A_n(R)=A_1(A_{n-1}(R)). If R is a domain, then A_n(R) is a domain too. It is straightforward to show that if k is a field of characteristic zero, then A_n(k) is a simple noetherian domian.

Linear automorphisms of A_n(k). Now suppose that k is field. Define the map \varphi : A_1(k) \longrightarrow A_1(k) on the generators by \varphi(x_1)=ax_1+by_1, \ \varphi(y_1)=cx_1+dy_1, \ a,b,c,d \in k. We would like to see under what condition(s) \varphi becomes a k-algebra homomorphism. Well, if \varphi is a homomorphism, then since y_1x_1=x_1y_1+1, we must have


Simplifying the above will give us (ad-bc)y_1x_1=(ad-bc)x_1y_1 + 1 and since y_1x_1=x_1y_1+1, we get ad-bc=1.  We can now reverse the process to show that if ad-bc=1, then \varphi is a homomorphism. So \varphi is a homomorphism if and only if ad-bc=1. But then the map \psi : A_1(k) \longrightarrow A_1(k) defined by

\psi(x_1)=dx_1 - by_1, \ \psi(y_1)=-cx_1+ay_1

will also be a homomorphism and \psi = \varphi^{-1}. Thus \varphi is an automorphism of A_1(k) if and only if ad-bc=1. In terms of matrices, the matrix S=\begin{pmatrix} a & b \\ c & d \end{pmatrix} defines a linear automorphism of A_1(k) if and only if \det S=1.

We can extend the above result to A_n(k), \ n\geq 1. Let S \in M_{2n}(k), a 2n \times 2n matrix with entries in k. Let {\bf{x}}=[x_1, \ldots , x_n, y_1, \ldots , y_n]^T and define the map \varphi: A_n(k) \longrightarrow A_n(k) by {\bf{x}} \mapsto S {\bf{x}}. Clearly \varphi is a k-algebra homomorphism if and only if \varphi(x_i), \varphi(y_i) satisfy the same relations that x_i,y_i do, i.e.

\varphi(x_i)\varphi(x_j)=\varphi(x_j) \varphi(x_i), \ \varphi(y_i) \varphi(y_j)=\varphi(y_j) \varphi(y_i),  \ \varphi(y_i) \varphi(x_j)=\varphi(x_j) \varphi(y_i) + \delta_{ij}, \ \ \ \ \ \ \ \ \ (1)

for all 1 \leq i,j \leq n. Let I_n \in M_n(k) be the identity matrix and let {\bf{0}} \in M_n(k) be the zero matrix. Let J=\begin{pmatrix} {\bf{0}} & I_n \\ -I_n & {\bf{0}} \end{pmatrix}. Then, in terms of matrices, (1) becomes

SJS^T=J. \ \ \ \ \ \ \ \ \ \ (2)

 Clearly if S satisfies (2), then S is invertible and thus \varphi will be an automorphism. So (2) is in fact the necessary and sufficient condition for \varphi to be an automorphism of A_n(k).

A 2n \times 2n matrix which satisfies (2) is called symplectic. See that if S is a 2 \times 2 matrix, then S is symplectic if and only if \det S=1.

Throughout k is a field, \text{char}(k)=0 and A is a k-algebra.

Example 2. Suppose that \delta is a derivation of A which is not inner. If A is \delta-simple, then B=A[x;\delta] is simple. In partcular, if A is simple, then A[x; \delta] is simple too.

Proof. Suppose, to the contrary, that B is not simple. So B has some non-zero ideal I \neq B. Let n be the minimum degree of non-zero elements of I. Let J be the set of leading coefficients of elements of I. Clearly J is a left ideal of A because I is an ideal of B. To see that J is also a right ideal of A, let a \in J and b \in A. Then there exists

f=ax^n + \text{lower degree terms} \in I.

But, by Remark 5

fb = abx^n + \text{lower degree terms} \in I

and so ab \in J, i.e. J is also a right ideal. Next, we’ll show that J is a \delta-ideal of A:

if a_n \in J, then there exists some f(x)=\sum_{i=0}^n a_ix^i \in I. Clearly xf - fx \in I, because I is an ideal of B. Now

xf - fx=\sum_{i=0}^n xa_i x^i - \sum_{i=0}^n a_ix^{i+1}=\sum_{i=0}^n (a_ix +\delta(a_i))x^i - \sum_{i=0}^n a_i x^{i+1}

= \sum_{i=0}^n \delta(a_i)x^i.

So \delta(a_n) \in J, i.e. J is a non-zero \delta-ideal of A. Therefore J=A, because A is \delta-simple. So 1 \in J, i.e. there exists g(x)=x^n + b_{n-1}x^{n-1} + \cdots + b_0 \in I. Finally let a \in A. Now, ga - ag, which is an element of I, is a polynomial of degree at most n-1 and the coefficient of x^{n-1} is b_{n-1}a - ab_{n-1} + n \delta(a), which has to be zero because of the minimality of n. Thus, since \text{char}(k)=0, we may let c=\frac{b_{n-1}}{n} to get \delta(a)=ca-ac. That means \delta is inner. Contradiction! \Box

Lemma. If A is simple and \delta = \frac{d}{dx}, then A[x] is \delta-simple.

Proof. Let I \neq \{0\} be a \delta-ideal of A[x]. Let f=\sum_{i=0}^n a_ix^i, \ a_n \neq 0, be an element of I of the least degree. Suppose n > 0. Then, since I is a \delta-ideal, we must have \frac{df}{dx}=\sum_{i=0}^{n-1}ia_ix^{i-1} \in I. Hence na_{n-1}=0, by the minimality of n, and thus a_n=0 because \text{char}(k)=0.  This contradiction shows that n=0 and so f=a_0 \in A \cap I. Hence Aa_0A \subseteq I because I is an ideal of A[x] and A \subset A[x]. But A is simple and so Aa_0A = A, i.e. 1 \in Aa_0A \subseteq I and thus I=A[x]. \Box

Definition 5. The algebra A[x][y, \frac{d}{dx}] is called the first Weyl algebra over A and is denoted by \mathcal{A}_1(A). Inductively, for n \geq 2, we define \mathcal{A}_n(A)=\mathcal{A}_1(\mathcal{A}_{n-1}(A)) and we call \mathcal{A}_n(A) the nth Weyl algebra over A.

Example 3. If A is simple, then \mathcal{A}_n(A) is simple for all n. In particular, \mathcal{A}_n(k) is simple.

Proof. By Remark 3 in part (1), \delta = \frac{d}{dx} is a non-inner derivation of A[x]. So, \mathcal{A}_1(A) is simple by the above lemma and Example 2 in part (1). The proof is now completed by induction over n. \Box

Example 4. In this example we do not need to assume that \text{char}(k)=0. Let A be a simple ring and k its center. Let B be a simple k-algebra but the center of B may or may not be k. The first part of the corollary in this post shows that A \otimes_k B is simple. This is another way of constructing new simple rings from old ones.