## Derivations of Weyl algebras are inner

Posted: September 23, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , , ,

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 $(1). \ \Box$

## Centralizers in the first Weyl algebra (4)

Posted: April 16, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , ,

So we have proved that if $k$ is a field of characteristic zero and $f \in A_1(k)$ with $\deg f \geq 1,$ then $C(f;A_1(k))$ is a commutative domain and it is a free module of finite rank over $k[f].$ What can be said about the field of fractions $Q$ of $C(f;A_1(k))$? The next theorem shows that $Q$ has a very simple form.

Theorem 2. (Amitsur, 1957) Let $k$ be a field of characteristic zero and let $f \in A_1(k)$ with $\deg f \geq 1.$ Let $Q$ and $k(f)$ be the field of fractions of $C(f;A_1(k))$ and $k[f]$ respectively. Then $Q$ is an algebraic extension of $k(f)$ and $Q=k(f)[g],$ for some $g \in C(f;A_1(k)).$

Proof. First note that, by Theorem 1, $C(f;A_1(k))$ is a commutative domain and hence its field of fractions exists. Now let $g, d$ and $B$ be as defined in the proof of Theorem 1. We proved in that theorem that for every $h \in C(f;A_1(k))$ there exists some $0 \neq \mu(f) \in k[f]$ such that

$\mu(f)h \in B=k[f]+gk[f] + \ldots + g^{d-1}k[f]. \ \ \ \ \ \ \ \ (*).$

If in $(*)$ we choose $h=g^d,$ then we will get $g^d \in k(f) + gk(f) + \ldots + g^{d-1}k(f).$ So $g$ is algebraic over $k(f)$ and thus $k(f)[g]$ is a subfield of $Q.$ Also $(*)$ shows that $h \in k(f)[g],$ for all $h \in C(f;A_1(k))$ and thus $C(f;A_1(k)) \subseteq k(f)[g].$ Therefore $C(f;A_1(k)) \subseteq k(f)[g] \subseteq Q$ and hence $Q=k(f)[g]. \ \Box$

## Centralizers in the first Weyl algebra (3)

Posted: April 16, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , ,

We are now going to prove that $C(f;A_1(k))$ is a commutative $k$-algebra for all $f \in A_1(k)$ with $\deg f \geq 1.$ Recall that by $\deg f$ we mean the largest power of $y$ in $f.$ First a simple lemma.

Lemma 2. Let $k$ be a field of characteristic zero and let $f \in A_1(k)$ with $\deg f \geq 1.$ If $m \geq 0$ is an inetger, then the set $V_m$ consisting of all elements of $C(f;A_1(k))$ of degree at most $m$ is a finite dimensional $k$-vector space.

Proof. It is clear that $V_m$ is a $k$-vector space. The proof of finite dimensionality of $V_m$ is by induction over $m.$ If $m=0,$ then $V_m=k$ and there is nothing to prove. So suppose that $m \geq 1$ and fix an element $g \in V_m$ with $\deg g=m.$ Of course, if there is no such $g,$ then $V_m=V_{m-1}$ and we are done by induction. Now, let $h \in V_m.$ If $\deg h < m,$ then $h \in V_{m-1}$ and if $\deg h = m,$ then there exists some $a \in k$ such that $h - ag \in V_{m-1},$ by Lemma 1. Thus $V_m = kg + V_{m-1}$ and hence $\dim_k V_m = \deg_k V_{m-1} + 1$ and we are done again by induction. $\Box$

Theorem 2. (Amitsur, 1957) Let $k$ be a field of characteristic zero and let $f \in A_1(k)$ with $\deg f = n \geq 1.$ Then $C(f;A_1(k))$ is commutative.

Proof. Let $S$ and $\overline{S}$ be as defined in the proof of Theorem 1. As we mentioned in there, $\overline{S}$ is a cyclic subgroup of $\mathbb{Z}/n\mathbb{Z}$ of order $d,$ for some divisor $d$ of $n.$ Let $\overline{m}, \ m > 0,$ be a generator of $\overline{S}$ and choose $g \in C(f;A_1(k))$ such that $\deg g = m.$ Now let

$B = k[f] + gk[f] + \ldots + g^{d-1}k[f].$

Clearly $B \subseteq C(f;A_1(k)).$ Let $T = \{mi+nj: \ 0 \leq i \leq d-1, \ j \in \mathbb{Z}, j \geq 0 \}.$ So basically $T$ is the set of all non-negative integers which appear as the degree of some element of $B.$ Let $p \in S.$ Then $p \equiv mi \mod n,$ for some inetger $0 \leq i \leq d-1$ because $\overline{m}$ is a generator of $\overline{S}.$ Hence $p = mi + nj,$ for some integer $j.$ If $j \geq 0,$ then $p \in T$ and if $j < 0,$ then $0 \leq p \leq mi \leq m(d-1).$ Thus if $h \in C(f;A_1(k))$ and $\deg h > m(d-1),$ then $\deg h \in T.$ Let $V$ be the set of all elements of $C(f;A_1(k))$ of degree at most $m(d-1).$ By Lemma 2, $V$ is $k$-vector space and $\dim_k V = v < \infty.$ The claim is that

$C(f;A_1(k))=B + V. \ \ \ \ \ \ \ \ (*)$

Clearly $B+V \subseteq C(f;A_1(k))$ because both $B$ and $V$ are in $C(f;A_1(k)).$ To prove $C(f;A_1(k)) \subseteq B+V,$ let $h \in C(f;A_1(k)).$ We use inducton over $\deg h.$ If $\deg h=0,$ then $\deg h = \deg 1$ and hence $h \in k,$ by Lemma 1. If $\deg h \leq m(d-1),$ then $h \in V$ and we are done. Otherwise, $\deg h \in T$ and hence there exists some $h_1 \in B$ such that $\deg h = \deg h_1.$ Thus, by Lemma 1, there exists some $a_1 \in k$ such that $\deg(h - a_1h_1) < \deg h.$ Therefore by induction $h-a_1h_1 \in B+V$ and hence $h \in B+V$ because $a_1h_1 \in B.$ This completes the proof of $(*).$

Now let $h \in C(f;A_1(k))$ and let $0 \leq i \leq v = \dim_k V.$ Clearly $f^i h \in C(f;A_1(k))$ and hence

$f^ih - h_i \in B, \ \ \ \ \ \ \ \ (**)$

for some $h_i \in V.$ Since $\dim_k V=v,$ the elements $h_0, \ldots , h_v$ are $k$-linearly dependent and so $\sum_{i=0}^v a_ih_i=0$ for some $a_i \in k$ which are not all zero. It now follows from $(**)$ that $(\mu(f)h \in B,$ where $0 \neq \mu(f)=\sum_{i=0}^v a_if^i \in k[f].$ So we have proved that for every $h \in C(f;A_1(k))$ there exists some $0 \neq \mu(f) \in k[f]$ such that $\mu(f)h \in B.$ Let $h_1, h_2 \in C(f;A_1(k))$ and let $0 \neq \mu_1(f), \mu_2(f) \in k[f]$ be such that $\mu_1(f)h_1 \in B$ and $\mu_2(f)h_2 \in B.$ Then, since $B$ is clearly commutative, we have $\mu_1(f)h_1 \mu_2(f)h_2 = \mu_2(f)h_2 \mu_1(f)h_1.$ Therefore, since $k[f]$ is commutative and $h_1$ and $h_2$ commute with $f,$ we have $\mu_1(f) \mu_2(f)h_1h_2=\mu_1(f) \mu_2(f)h_2h_1.$ Thus, since $A_1(k)$ is a domain and $\mu_1(f), \mu_2(f) \neq 0,$ we have $h_1h_2=h_2h_1.$ Hence $C(f;A_1(k))$ is commutative. $\Box$

In part (4), which will be the last part, we will find the field of fractions of $C(f;A_1(k)).$

## Centralizers in the first Weyl algebra (2)

Posted: April 16, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , ,

Theorem 1. (Amitsur, 1957) Let $k$ be a field of characteristic zero and let $f \in A_1(k)$ with $\deg f =n \geq 1.$ Then $C(f;A_1(k))$ is a free $k[f]$-module of rank $d,$ where $d$ is a divisor of $\deg f.$

Proof. Suppose that $S$ is the set of all integers $m \geq 0$ for which there esists some $g \in C(f;A_1(k))$ such that $\deg g = m.$ Clearly $S$ is a submonoid of $\mathbb{Z}.$  For any $m \in S$ let $\overline{m}$ be the image of $m$ in $\mathbb{Z}/n\mathbb{Z}$ and put $\overline{S}=\{\overline{m}: \ m \in S \}.$ Since $\overline{S}$ is a submonoid of a finite cyclic group, it is a cyclic subgroup and hence $d=|\overline{S}|$ divides $|\mathbb{Z}/n\mathbb{Z}|=n.$ Let $S=\{\overline{m_i}: \ 1 \leq i \leq d \},$ where $m_1=0$ and, in general, each $m_i$ is chosen to be non-negative and the smallest member of its class $\overline{m_i}.$ That means if $m \equiv m_i \mod n$ and $m \geq 0,$ then $m \geq m_i.$ For any $1 \leq i \leq d,$ let $g_i \in C(f;A_1(k))$ with $\deg g_i=m_i.$ So we can choose $g_1$ to be any element of degree zero in $C(f;A_1(k)).$ We choose $g_1=1.$ To complete the proof of the theorem, we are going to show that, as a $k[f]$-module, $g_1, \ldots , g_d$ generate $C(f;A_1(k))$ and $g_1, \ldots , g_d$ are linearly independent over $k[f].$ We first show that $C(f;A_1(k))=\sum_{i=1}^d g_ik[f].$ Clearly $\sum_{i=1}^d g_ik[f] \subseteq C(f;A_1(k)$ because $f, g_i \in C(f;A_1(k)),$ for all $1 \leq i \leq d.$ Now let $g \in C(f;A_1(k)$ and suppose that $\deg g = \ell.$ If $\ell = 0,$ then $\deg g = \deg 1$ and hence, by Lemma 1, $g \in k \subseteq g_1k[f] \subseteq \sum_{i=1}^d g_ik[f].$ If $\ell \geq 1,$ then $\overline{\ell}=\overline{m_j},$ for some $j.$ We also have $\ell \geq m_j$ by minimality of $m_j.$ Thus $\ell=m_j+nu$ for some integer $u \geq 0.$ Therefore $\deg g = \ell = m_j+nu=\deg g_jf^u.$ Now both $g$ and $g_jf^u$ are obviously in $C(f;A_1(k)).$ So if $s$ and $t$ are the leading coefficeints of $g$ and $g_jf^u,$ respectively, then by Lemma 1, $s=at$ for some $a \in k.$ Therefore $\deg(g - ag_if^u) \leq \ell - 1$ and, since $g - ag_if^u \in C(f;A_1(k)),$ we can apply induction on $\deg g$ to get $g - ag_jf^u \in \sum_{i=1}^d g_i k[f].$ Thus $g \in \sum_{i=1}^d g_i k[f].$ It remains to show that $g_1, \ldots , g_d$ are linearly independent over $k[f].$ Suppose, to the contrary, that

$g_1 \mu_1(f) + \ldots + g_d \mu_d(f)=0, \ \ \ \ \ \ \ \ (*)$

for some $\mu_i(f) \in k[f]$ and not all $\mu_i(f)$ are zero. Note that if $i \neq j$ and $\mu_i(f) \mu_j(f) \neq 0,$ then $\deg (g_i \mu_i(f)) \equiv m_i \mod n$ and $\deg (g_j \mu_j(f)) \equiv m_j \mod n.$ Since $i \neq j,$ we have $m_i \not\equiv m_j \mod n$ and hence  $\deg (g_i \mu_i(f)) \neq \deg (g_j \mu_j(f)).$ Thus the left hand side of $(*)$ is a polynomial of degree $\max \{\deg(g_i \mu_i(f)): \ g_i \mu_i(f) \neq 0 \}$ and so it cannot be equal to zero. $\Box$

In part (3) we will prove that $C(f;A_1(k))$ is commutative.

## Centralizers in the first Weyl algebra (1)

Posted: April 16, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , ,

In this post I am going to look at the centralizer of non-central elements in the first Weyl algebra over some field $k$ of characteristic zero. Recall that the first Weyl algebra $A_1(k)$ is defined to be the $k$-algebra generated by $x$ and $y$ with the relation $yx = xy+1.$ It then follows easily that $yr = ry + r',$ for every $r \in k[x],$ where $r' = \frac{dr}{dx}.$ It is easily seen that the center of $A_1(k)$ is $k.$ Also, every non-zero element of $A_1(k)$ can be uniquely written in the form $\sum_{i=0}^n r_i y^i,$ where $r_i \in k[x]$ and $r_n \neq 0.$ We call $n$ the degree of $f.$ It is easy to see that $A_1(k)$ is a domain. For every $f \in A_1(k),$ we will denote by $C(f;A_1(k))$ the centralizer of $f$ in $A_1(k).$ The goal is to show that if $f \notin k,$ then $C(f;A_1(k))$ is a commutative algebra and also a free $k[f]$-module of finite rank. This result is due to Amitsur.

Remark 1. If $r \in k[x],$ then $y^nr = \sum_{i=0}^n \binom{n}{i}r^{(i)}y^{n-i},$ where $r^{(i)}$ means the $i$-th derivative of $r$ with respect to $x.$ This follows easily by induction and the fact that $yr=ry+r'.$

Remark 2. If $f \in k[x],$ then $C(f;A_1(k))=k[x].$ This is easy to see: clearly $k[x] \subseteq C(f;A_1(k)).$ Conversely, if $g = r_ny^n + r_{n-1}y^{n-1} + \ldots + r_0, \ r_i \in k[x], \ r_n \neq 0$ commutes with $f$ and $n \geq 1,$ then comparing the coefficients $y^{n-1}$ in both sides of $fg=gf$ will give us $nr_nf'=0,$ which is a contradiction. Thus $n=0$ and so $g \in k[x].$

So, by the above remark, we only need to find the centralizer of an element of $A_1(k)$ in the form $f = \sum_{i=0}^n r_iy^i, \ r_i \in k[x], \ n = \deg f \geq 1.$

Lemma 1. Let $k$ be a field of characteristic zero and let $f = r_n y^n + \ldots + r_0 \in A_1(k), \ n \geq 1, \ r_n \neq 0.$ Let $g=s_my^m + \ldots + s_0, \ s_m \neq 0,$ and $h=t_my^m + \ldots + t_0, \ t_m \neq 0,$ be two elements of $C(f;A_1(k)).$ Then $s_m=a t_m,$ for some $a \in k.$

Proof.  Since $yr=ry+r',$ for any $r \in R=k[x],$ induction on $\ell$ shows that $y^{\ell}r = ry^{\ell} + \ell r'y^{\ell - 1} + \ldots,$ for any integer $\ell \geq 1.$ Therefore the coefficient of $y^{n+m-1}$ in $fg$ and $gf$ are $nr_ns_m' + r_ns_{m-1}+r_{n-1}s_m$ and $ms_mr_n' + s_mr_{n-1} + s_{m-1}r_n,$ respectively. Thus, since $fg=gf,$ we must have

$nr_ns_m' + r_ns_{m-1}+r_{n-1}s_m = ms_mr_n' + s_mr_{n-1} + s_{m-1}r_n.$

Hence, since $R$ is commutative, we will have

$nr_ns_m'=mr_n's_m. \ \ \ \ \ \ \ \ (1)$

A similar arguemnt shows that $fh=hf$ implies that

$nr_nt_m'=mr_n't_m. \ \ \ \ \ \ \ \ (2)$

Now, multiplying both sides of (1) by $t_m$ and both sides of (2) by $s_m$ and then subtracting the resulting identities will give us $nr_n(t_ms_m'-s_mt_m')=0.$ Thus

$t_ms_m'-s_mt_m'=0, \ \ \ \ \ \ \ \ (3)$

because $R$ is a domain, $r_n \neq 0$ and the characteristic of $k$ is zero. Now look at $R$ as a subalgebra of the field of rational functions $k(x).$ Then, since $s_m \neq 0,$ by (3) we have $(s_m/t_m)'=0$ and hence $s_m/t_m \in k,$ i.e. $s_m=at_m,$ for some $a \in k. \ \Box$

In part (2) we will prove that $C(f;A_1(k))$ is a free $k[f]$-module of finite rank.

## Weyl algebras; definition & automorphisms (2)

Posted: January 25, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: ,

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

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

and

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

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 $i. \ \Box$

## Weyl algebras; definition & automorphisms (1)

Posted: January 24, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , ,

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

$\varphi(y_1)\varphi(x_1)=\varphi(x_1)\varphi(y_1)+1.$

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 $\det S=1.$