Unital locally matrix algebras and Steinitz numbers

TL;DR

This paper investigates unital locally matrix algebras over a field, introducing Steinitz numbers to classify them and exploring the relationship between these numbers and the algebraic structure.

Contribution

It introduces Steinitz numbers as invariants for unital locally matrix algebras and analyzes their relationship with the algebra's structure.

Findings

Steinitz numbers classify unital locally matrix algebras

A relationship between Steinitz numbers and algebra structure is established

Provides a framework for understanding the structure of locally matrix algebras

Abstract

An $F$-algebra $A$ with unit $1$ is said to be a locally matrix algebra if an arbitrary finite collection of elements $a_1,$ $\ldots,$ $a_s $ from $ A$ lies in a subalgebra $B$ with $1$ of the algebra $A$, that is isomorphic to a matrix algebra $M_n(F),$ $n\geq 1.$ To an arbitrary unital locally matrix algebra $A$ we assign a Steinitz number $\mathbf{n}(A)$ and study a relationship between $\mathbf{n}(A)$ and $A$.

Equations

$$\prod _{p\in \mathbb{P}}{p}^{r_p},$$

$$\prod _{p\in \mathbb{P}}{p}^{r_p}\text{ and }\prod _{p\in \mathbb{P}}{p}^{k_p}$$

$$\prod _{p\in \mathbb{P}}{p}^{r_p+k_p},$$

$$I=\prod _{p\in P}{p}^{\infty }$$

$$\mathbf{n}(A)\cdot \mathbf{n}(B)=\mathbf{n}(A{\otimes }_{F}B).$$

$$\mathbf{n}(A)=\mathbf{n}(B).$$

$$\begin{gathered} \forall x_{ij},1\le i,j\le n,(({\wedge }_{i,j,t,s}{x}_{ij}{x}_{ts}={\delta }_{jt}{x}_{is}) \\ \wedge (x_{11}+x_{22}+\dots +x_{nn}=1)\to (1=0)), \end{gathered}$$

$$vw+wv=f(v,w)\cdot 1$$

$$Cl(\tilde{W},f)=M_{2^{\frac{n}{2}}}(F).$$

$$\begin{gathered} Clg(l,m)=⟨x_1,\dots ,x_m|x_i^l=1;x_i^{-1}{x}_j{x}_i=\xi x_j\text{ for }i<j; \\ {x}_i^{-1}{x}_j{x}_i={\xi }^{-1}{x}_j\text{ for }i>j;1\le i,j\le m⟩. \end{gathered}$$

$$\begin{gathered} Clg(l,I)=⟨x_i,i\in I|x_i^l=1;x_i^{-1}{x}_j{x}_i=\xi x_j\text{ for }i<j; \\ {x}_i^{-1}{x}_j{x}_i={\xi }^{-1}{x}_j\text{ for }i>j;i,j\in I⟩. \end{gathered}$$

$$Clg(l,m)\cong M_{l^{\frac{m}{2}}}(F);$$

$${\dim }_{F}Clg(l,m)=l^m.$$

$${\phi }_i(x_j)={\xi }^{{\delta }_{ij}}{x}_j,1\le j\le m,$$

$${\phi }_i^s(v)=\sum _{k=0}^{l-1}{\xi }^{sk}{x}_i^k{v}_k\in V,s=0,1,\dots ,l-1.$$

$$x_i^{-1}v{x}_i={\xi }^{-k_1-k_2-\cdots -k_{i-1}+k_{i+1}+\cdots +k_m}v.$$

$$\{\begin{array}{ll}{k}_2+\cdots +k_m& =0\mathrm{mod}l\\ -k_1+k_3+\cdots +k_m& =0\mathrm{mod}l\\ \dots \dots & \dots \dots \\ -k_1-k_2-\cdots -k_{m-1}& =0\mathrm{mod}l\end{array}$$

$$K=\left(\begin{array}[]{cccccc}0&1&1&\ldots&1&1\\ -1&0&1&\ldots&1&1\\ \vdots&&&&&\\ -1&-1&-1&\ldots&0&1\\ -1&-1&-1&\ldots&-1&0\\ \end{array}\right)$$

$$V=\{(a_1,a_2,\dots )|a_i\in \mathbb{C},\sum _{i=1}^{\infty }|a_i{|}^2<\infty \}.$$

$$f((a_1,a_2,\dots ))=\sum _{i=1}^{\infty }{a}_i^2\in \mathbb{C}.$$

$$g(w)=\sum _{i\in 1}^n{\alpha }_i^2.$$

$${\dim }_{F}A={\dim }_{F}B=\mathfrak{Card}I.$$

$$\mathbf{n}(A)=\mathbf{n}(B)=2^{\infty }.$$

$$Cl(V,f)=Cl(V,f{)}_{\bar{0}}+Cl(V,f{)}_{\bar{1}},$$

$$\bigcap _{i\ge 0}Cl(V_i)=\mathbb{C}\cdot 1.$$

$$a\in \bigcap _{i\ge 0}Cl(V_i),a\notin \mathbb{C}\cdot 1.$$

$$B_1=\{w_1-w_1^{\prime },\dots ,w_k-w_k^{\prime }\}\subset V_1.$$

$$v^{\perp }=\{v^{\prime }\in V|f(v,v^{\prime })=0\}.$$

$$V=\mathbb{C}v+v^{\perp }.$$

$$C(v)=vCl(v^{\perp },f{)}_{\bar{0}}+Cl(v^{\perp },f{)}_{\bar{0}}.$$

Full text

The second author was partially supported by the grant 346300 for IMPAN from the Simons Foundation and the matching 2015-2019 Polish MNiSW fund

Mathematics Subject Classification: 03C05, 03C60, 11E88

1. Introduction

Let $F$ be a ground field. Throughout the paper we consider associative $F$–algebras with $1$. We say that an algebra $A$ is a locally matrix algebra if an arbitrary finite collection of elements $a_{1},$ $\ldots,$ $a_{s}\in A$ lies in a subalgebra $B$, $1\in B\subset A$, that is isomorphic to a matrix algebra $M_{n}(F),$ $n\geq 1.$

J. G. Glimm [5] parametrised countable dimensional locally matrix algebras (under the name uniformly hyperfinite algebras) with Steinitz or supernatural numbers. O. Bezushchak, B. Oliynyk and V. Sushchanskii [3] extended their parametrisation to regular relation structures. The idea of diagonal embeddings was introduced by A. E. Zalesskii in [10]. In a series of papers A. A. Baranov and A. G. Zhilinskii used Steinitz numbers to classify diagonal locally simple Lie algebras of countable dimension [1], [2].

In this paper we introduce a Steinitz number $\mathbf{n}(A)$ for a unital locally matrix algebra of arbitrary dimension and study the relation between $\mathbf{n}(A)$ and $A$.

In the section 3 we show that for locally matrix algebras $A,$ $B$ of arbitrary dimensions universal theories of $A,$ $B$ coincide if and only if $\mathbf{n}(A)=\mathbf{n}(B)$.

In the section 4 we consider examples of unital locally matrix algebras and find their Steinitz numbers. Finally, in the section 4 we give example of nonisomorphic locally matrix algebras $A,$ $B$ of uncountable dimension such that $\mathbf{n}(A)=\mathbf{n}(B)=2^{\infty}.$

2. Preliminaries

Let $n,m\geq 1$ and let $1\in M_{n}(F)\subset M_{m}(F)$. Let $B\cong M_{k}(F)$ be the centralizer of $M_{n}(F)$ in $M_{m}(F)$. Then (see [6]), $M_{m}\cong M_{n}(F)\otimes_{F}M_{k}(F)$, which implies m=nk.

Let $\mathbb{P}$ be the set of all primes. A *Steinitz * or supernatural number is an infinite formal product of the form

[TABLE]

where $r_{p}\in\mathbb{N}\cup\{0,\infty\}$. The product of two Steinitz numbers

[TABLE]

is a Steinitz number

[TABLE]

where we assume, that $k_{p}\in\mathbb{N}\cup\{0,\infty\}$, $t+\infty=\infty+t=\infty+\infty=\infty$ for all positive integers $t$. Denote by $\mathbb{SN}$ the set of all Steinitz numbers. Obviously, the set of all positive integers $\mathbb{N}$ is a subset of the set of all Steinitz numbers $\mathbb{SN}$. The elements of the set $\mathbb{SN}\setminus\mathbb{N}$ are called *infinite Steinitz * numbers.

The Steinitz number $v$ divides $u$ if there exists $w\in\mathbb{SN}$, such that $u=v\cdot w$. The divisibility relation $|$ makes $\mathbb{SN}$ into a partially ordered set with the greatest element

[TABLE]

and the least element $1$. Moreover, the poset $(\mathbb{SN},|)$ is a complete lattice.

Steinitz numbers were introduced by Ernst Steinitz [9] in 1910 to classify algebraic extensions of finite fields.

Definition 1**.**

Let $A$ be an infinite dimensional unital locally matrix algebra over a field $F$ and let $D(A)$ be the set of all positive integers $n$, such that there is a subalgebra $A^{\prime}$, $A^{\prime}\subseteq A$, $A^{\prime}\cong M_{n}(F)$ and $1_{A}\in A^{\prime}$, where $1_{A}$ is the identity of $A$. Then the least common multiple of the set $D(A)$ is called the Steinitz or supernatural number $\mathbf{n}(A)$ of the algebra $A$.

The following proposition is straightforward.

Proposition 1**.**

Let $A$ and $B$ be unital locally matrix algebras. Then the algebra $A\otimes_{F}B$ is a unital locally matrix and

[TABLE]

Theorem 1** (See [2], [3], [5]).**

If $A$ and $B$ are unital locally matrix algebras of countable dimension then $A$ and $B$ are isomorphic if and only if $\mathbf{n}(A)=\mathbf{n}(B)$.

For the rest of this paper, we will write $M_{n}(F)\subseteq A$ if there is a subalgebra $A^{\prime}$, $1\in A^{\prime}\subseteq A$, such that $A^{\prime}\cong M_{n}(F)$.

3. Universally equivalent algebras

Let $A$ be an algebraic system (see [7]). The universal elementary theory $UTh(A)$ consists of universal closed formulas (see [7]) that are valid on $A$. The systems $A$ and $B$ of the same signature are universally equivalent if $UTh(A)=UTh(B)$.

Theorem 2**.**

Let $A$ and $B$ be unital locally matrix algebras. Then $A$ and $B$ are universally equivalent if and only if their Steinitz numbers $\mathbf{n}(A)$ and $\mathbf{n}(B)$ are equal, i.e.

[TABLE]

For proof of this theorem we need a lemma.

Lemma 1**.**

Let $A$ be a unital locally matrix algebra. The property $n\notin D(A)$ can be written as a universal closed formula.

Proof.

Consider the formula

[TABLE]

where $\delta_{jp}$ is the Kronecker delta, i.e. \delta_{jp}=\begin{cases}1,&\text{if j=p}\\ 0,&\text{ifj\neq p}\end{cases} and $x_{ij}$, $1\leq i,j\leq n$.

The universal closed formula (1) is true if and only if $M_{n}(F)\nsubseteq A$. ∎

Proof of Theorem 2.

First we recall that if some universal closed formula is true on an algebraic structure, then it is true on any algebraic substructure.

If unital locally matrix algebras $A$ and $B$ are universally equivalent, then by Lemma 1 $M_{n}(F)\nsubseteq A$ if and only if $M_{n}(F)\nsubseteq B$. This implies that $D(A)$ is equal to $D(B)$ and therefore $\mathbf{n}(A)=\mathbf{n}(B)$.

Let‘s now suppose that $\mathbf{n}(A)=\mathbf{n}(B)$. From the Malcev Local Theorem (see [7]) it follows, that the algebra $A$ is unitally embeddable in an ultraproduct of matrix algebras $M_{n}(F)$, $n\in D(A)$. If $\mathbf{n}(A)=\mathbf{n}(B)$, then $D(A)=D(B)$. Since every $M_{n}(F)$, $n\in D(A)$, is unitally embeddable in $B$, it follows that the algebra $A$ is unitally embeddable in an ultrapower of the algebra $B$.

As the algebra $A$ is unitally embeddable in an ultrapower of the algebra $B$, if some universal closed formula is true on $B$, then it is true on $A$.

Similar, the unital locally matrix algebra $B$ is unitally embeddable in an ultrapower of $A$, so if some universal closed formula is true on $A$, then it is true on $B$ and our theorem is proved. ∎

Remark 1**.**

It is important that identity is added to the signature. Without $1$ in the signature any two infinite dimensional locally matrix algebras are elementarily equivalent.

4. Examples

Let $V$ be a vector space over a field $F$, $charF\neq 2$. Firstly, we recall a construction of Clifford algebras.

A map $f:\ V\to F$ is called a quadratic form if the following conditions hold:

1. (1)

$f(\alpha v)=\alpha^{2}f(v)$ for every $\alpha\in F$ and $v\in V$; 2. (2)

$f(u,v)=f(u+v)-f(v)-f(u)$ is a bilinear form.

The quadratic form $f:\ V\to F$ is nondegenerate if and only if the bilinear form $f(u,v)$ is nondegenerate.

The Clifford algebra $Cl(V,f)$ is a unital algebra generated by the vector space $V$ and $1$ and defined by relations $v^{2}=f(v)\cdot 1$ for all $v\in V$. Hence,

[TABLE]

holds for any $v,w$ from the vector space $V$ in the Clifford algebra $Cl(V,f)$.

Let $\{v_{i}\}_{i\in I}$ be a basis of the vector space $V$. Assume, that the set of indexes $I$ is ordered. Then all possible ordered products $v_{i_{1}}v_{i_{2}}\ldots v_{i_{k}}$, $i_{1}<i_{2}<\ldots<i_{k}$, and $1$ (that can be defined as the empty product) is a basis of the Clifford algebra $Cl(V,f)$.

Obviously, if $\dim_{F}V=n$, then $\dim_{F}Cl(V,f)=2^{n}$. If $V$ is an infinite dimensional vector space, then $\dim_{F}V=\dim_{F}Cl(V,f)$.

Onward, we assume that the field $F$ is algebraically closed and the quadratic form $f$ is nondegenerate.

Theorem 3**.**

[6*]**

Let $Cl(V,f)$ be the Clifford algebra defined by a nondegenerate quadratic form $f$ on $V$ and $\dim_{F}V=n<\infty$. If the number $n$ is even, then the Clifford algebra $Cl(V,f)$ is isomorphic to the matrix algebra $M_{2^{\frac{n}{2}}}(F)$. If the number $n$ is odd, then the algebra $Cl(V,f)$ is isomorphic to the direct sum of matrix algebras $M_{2^{\frac{n-1}{2}}}(F)\oplus M_{2^{\frac{n-1}{2}}}(F)$.*

Theorem 4**.**

Let $V$ be an infinite dimensional vector space. Then the Clifford algebra $Cl(V,f)$ is locally matrix and $\mathbf{n}(Cl(V,f))=2^{\infty}$.

Proof.

Assume, that $S$ is a finite subset of the Clifford algebra $Cl(V,f)$. As $S$ is finite, there is a finite dimensional subspace $W$ of the vector space $V$, such that $S\subseteq Cl(W,f)$.

For any finite dimensional subspace $W$ of a vector space $V$ there is a subspace $\tilde{W}$ of $V$, such that the following conditions hold:

1. (1)

$\dim_{F}\tilde{W}$ is even; 2. (2)

$W\subseteq\tilde{W}$; 3. (3)

the restriction of the form $f$ to $\tilde{W}$ is nondegenerate.

Assume now, that $\dim_{F}\tilde{W}=n$. Then $S\subseteq Cl(W,f)\subseteq Cl(\tilde{W},f)$ and

[TABLE]

Therefore, by the definition of a locally matrix algebra and the Steinitz number corresponding to it, the Clifford algebra $Cl(V,f)$ is locally matrix and $\mathbf{n}(Cl(V,f))=2^{\infty}$. ∎

To obtain more examples we consider a generalization of Clifford algebras.

Choose a positive integer $l>1$ that is coprime with characteristic of the ground field $F.$ Let $\xi\in F$ be an $l$-th primitive root of $1.$ Consider the generalized Clifford algebra

[TABLE]

Such algebras in a more general form were considered in [8].

Similarly, for an arbitrary ordered set $I$ we consider

[TABLE]

Theorem 5**.**

1) For an even $m$ we have

[TABLE]

2) for an odd $m$ we have $Clg(l,m)\cong\underbrace{M_{l^{\frac{m-1}{2}}}(F)\oplus\cdots\oplus M_{l^{\frac{m-1}{2}}}(F)}_{l}.$

Proof.

Using the Groebner–Shirshov technique [4] we see that ordered monomials $x_{1}^{k_{1}}\cdots\cdots x_{m}^{k_{m}},$ $0\leq k_{i}\leq l-1,$ $1\leq k_{i}\leq m,$ form a basis of the algebra $Clg(l,m).$ In particular,

[TABLE]

For an arbitrary $1\leq i\leq m$ the mapping

[TABLE]

extends to an automorphism of the algebra $Clg(l,m).$

Lemma 2**.**

Let $V$ be a subspace of $Clg(l,m)$ that is invariant under $\text{Aut }Clg(l,m).$ Then $V$ is spanned by all ordered monomials lying in $V.$

Proof.

Let $v=\Sigma_{k=0}^{l-1}x_{i}^{k}v_{k}\in V,$ where $v_{k}$ lies in the subalgebra generated by $x_{1},$ $\ldots,$ $x_{i-1},$ $x_{i+1},$ $\ldots,$ $x_{m}.$ Then

[TABLE]

Viewing these $l$ equalities as a system of equations in $v_{0},$ $x_{i}v_{1},$ $\ldots,$ $x_{i}^{l-1}v_{l-1}$ with the Vandermonde determinant $\det|\xi^{ij}|\neq 0$ we represent each $x_{i}^{k}v_{k}$ as a linear combination of $v,$ $\varphi_{i}(v),$ $\ldots,$ $\varphi_{i}^{l-1}(v).$ Hence each $x_{i}^{k}v_{k}$ lies in $V.$

Repeating this argument several times we get the assertion of lemma. ∎

Corollary 1**.**

The algebra $Clg(l,m)$ does not contain proper ideals that are invariant under $\text{Aut }Clg(l,m).$

Corollary 2**.**

The algebra $Clg(l,m)$ is semisimple.

Now our aim is to find monomials $x_{1}^{k_{1}}\cdots x_{m}^{k_{m}}$ that lie in the center $C$ of $Clg(l,m).$ By Lemma 2 these monomials span $C.$ Let $v=x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m}^{k_{m}}\in C.$ We have

[TABLE]

Hence

[TABLE]

If $m$ is even then the $m\times m$ matrix

[TABLE]

over $\mathbb{Z}/l\mathbb{Z}$ is invertible. Therefore in this case $v=1$ and $C=F\cdot 1.$ This proves the part 1) of the theorem.

If $m$ is odd then the rank of $K$ is $m-1$ and the system (2) has $l$ solutions $(i,-i,i,-i,\ldots,i),$ $i=0,1,\ldots,l-1.$ Hence $\dim_{F}C=l.$ This implies that the algebra $Clg(l,m)$ is a direct sum of $l$ simple summands, $Clg(l,m)=A_{1}\oplus\cdots\oplus A_{l}.$ Automorphisms of $Clg(l,m)$ permute the ideals $A_{1},$ $\ldots,$ $A_{l}.$ By Corollary 1 of the Lemma 2 they are conjugate under $\text{Aut }Clg(l,m),$ hence isomorphic. Since $\dim_{F}Clg(l,m)=l\cdot\dim_{F}A_{1}=l^{m},$ it follows that $\dim_{F}A_{1}=l^{m-1},$ $A_{1}\cong M_{l^{\frac{m-1}{2}}}(F).$ This proves the part 2) of the theorem. ∎

Theorem 6**.**

The Steinitz number of a unital locally matrix algebra $Clg(l,I)$, where the set $I$ is infinite, is $l^{\infty}$.

Proof.

For an arbitrary finite subset $S$ of $Clg(l,I)$ there exists a finite subset $J\subset I$, such that $S\subset Clg(l,J).$ Without loss of generality, we can assume that $\mathfrak{Card}J=m$ is an even number.Then by Theorem 5 $Clg(l,J)\cong M_{l^{\frac{m}{2}}}(F).$ This completes the proof of the theorem. ∎

Theorem 7**.**

If $\tau=p_{1}^{k_{1}}p_{2}^{k_{2}}p_{3}^{k_{3}}\ldots$ is a Steinitz number, such that $k_{j}=\infty$ for some positive integer $j$, then for any infinite dimension $\alpha$ there is a unital locally matrix algebra $A$, such that $\dim_{F}A=\alpha$ and $\mathbf{n}(A)=\tau$.

Proof.

Let $k_{j}=\infty$. Let $\tau^{\prime}=p_{1}^{k_{1}}p_{2}^{k_{2}}\ldots p_{j-1}^{j-1}p_{j+1}^{j+1}\ldots$. In [3] it is proved that there exists a countable dimensional locally matrix algebra $A^{\prime}$, such that $\mathbf{n}(A^{\prime})=\tau^{\prime}$.

Let $I$ be an ordered set of cardinality $\alpha$, let $A^{\prime\prime}=Clg(p_{j},I)$. Then $\mathbf{n}(A^{\prime\prime})=p_{j}^{\infty}$ and $\mathbf{n}(A^{\prime}\otimes A^{\prime\prime})=\tau.$ It is easy to see that $\dim_{F}A^{\prime}\otimes_{F}A^{\prime\prime}=\alpha$. ∎

A Steinitz number $\tau=\prod_{p\in\mathbb{P}}p^{r_{p}}$ is called locally finite if $r_{p}<\infty$ for any $p\in\mathbb{P}$.

Conjecture 1**.**

If $\tau$ is a locally finite Steinitz number, $A$ is a unital locally matrix algebra and $\mathbf{n}(A)=\tau$, then $A$ is countable dimensional.

5. Non isomorphic algebras with equal Steinitz numbers

We will construct two nonisomorphic unital locally matrix algebras of uncountable dimension, such their Steinitz numbers are equal.

Thus we show that Theorem 2 in the case of uncountable dimension is not true. Consider the vector space:

[TABLE]

Let $f$ be the quadratic form:

[TABLE]

The vector space $V$ has uncountable dimension.

Assume that $I$ is a set of indexes, whose cardinality $\mathfrak{Card}I$ is equal to the dimension of the vector space $V$. Let now $W$ be a complex vector space with basis $w_{i},i\in I$. Assume that $g$ is the quadratic form on $W$ determined for arbitrary $w=\alpha_{1}w_{i_{1}}+\alpha_{2}w_{i_{2}}+\ldots+\alpha_{n}w_{i_{n}}$, $\alpha_{i}\in\mathbb{C}$ by the rule:

[TABLE]

Define Clifford algebras $A=Cl(V,f)$ and $B=Cl(W,g)$. As follows from their constructions, algebras $A$ and $B$ are unital locally matrix and

[TABLE]

In addition, from Theorem 4 it follows, that

[TABLE]

Theorem 8**.**

Clifford algebras $A$ and $B$ are not isomorphic.

To prove this theorem we need some lemmas.

Let $V$ be a vector space with a quadratic form $f$. Then there is the natural $\mathbb{Z}/2\mathbb{Z}$-gradation of the Clifford algebra $Cl(V,f)$:

[TABLE]

where subspaces $Cl(V,f)_{\bar{0}}$ and $Cl(V,f)_{\bar{1}}$ are spanned by all even $\underbrace{V\cdot V\cdot\ldots\cdot V}_{2i}$ and all odd $\underbrace{V\cdot V\cdot\ldots\cdot V}_{2i+1}$ products respectively.

Lemma 3**.**

Let $V=V_{0}\supset V_{1}\supset V_{2}\supset\ldots$ be a descending chain of subspaces $dim_{\mathbb{C}}(V/V_{i})<\infty$, $i\geq 0$, and $\bigcap_{i\geq 0}V_{i}=(0)$. Then

[TABLE]

Proof.

Let us show that without loss of generality we can assume the space $V$ to be countable dimensional. Indeed, let

[TABLE]

For an arbitrary ${i\geq 0}$ there exists a finite dimensional subspace $W_{i}\subset V_{i}$ such that $a\in Cl(W_{i})$. Let $V^{\prime}=\sum_{i\geq 0}W_{i}$, $dim_{\mathbb{C}}V^{\prime}=\varkappa_{0}$, and let $V^{\prime}_{i}=V_{i}\cap V^{\prime}$. Then $V^{\prime}=V^{\prime}_{0}\supset V^{\prime}_{1}\supset\ldots$, $\bigcap_{i\geq 0}V^{\prime}_{i}=(0)$ and $\bigcap_{i\geq 0}Cl(V^{\prime}_{i})\neq\mathbb{C}\cdot 1$.

Assume therefore that the space $V$ is countable dimensional. Let $U=\{u_{0},u_{1},\ldots\}$ be a basis of $V$.

Without loss of generality we assume that $u_{0}\notin V_{1}$.

For any finite dimensional subspace $V^{\prime}\subset V$ there clearly exists $n\geq 1$ such that $V^{\prime}\cap V_{n}=(0).$ We will construct an ascending chain of finite subsets $U_{0}\subset U_{1}\subset\ldots$ of the basis $U$ and an increasing sequence of integers $n_{1}<n_{2}<\ldots$ such that

1. (1)

$u_{0},u_{1},\ldots,u_{k}\in U_{k}$, 2. (2)

$U_{k}$ is a basis of the space $V$ modulo $V_{n_{k+1}}$, for any $k\geq 0$.

Let $U_{0}$ be a maximal subset of $U$ such that $U_{0}$ is linearly independent modulo $V_{1}$ and $u_{0}\in U_{0}$. Let $n_{1}=1$.

Suppose that subsets $U_{0}\subset U_{1}\subset\ldots U_{k}\subset U$ and integers $1=n_{1}<\ldots<n_{k+1}$ have been selected; the subset $U_{k}$ is a basis of $V$ modulo $V_{n_{k+1}}$; $u_{0},\ldots,u_{k}\in U_{k}$.

Let $l\geq k+1$ be a smallest integer such that $u_{l}\notin U_{k}$; $u_{0},u_{1},\ldots,u_{l-1}\in U_{k}$. There exists an integer $n_{k+2}>n_{k+1}$ such that $U_{k}\cup\{u_{l}\}$ is linearly independent modulo $V_{n_{k+2}}$. Let $U_{k+1}$ be a maximal subset of $U$ that is linearly independent modulo $V_{n_{k+2}}$ and $U_{k}\cup\{u_{l}\}\subseteq U_{k+1}$. Clearly, $U_{k+1}$ is a basis of $V$ modulo $V_{n_{k+2}}$ and $u_{0},\ldots,u_{k+1}\in U_{k+1}$. Considering the subspaces $V_{n_{k}}$ instead of $V_{k}$, $k\geq 1$, we can assume that $U_{k}$ is a basis of $V$ modulo $V_{k+1}$, $k\geq 0$.

Now we will construct a new ordered basis of $V$. Let $B_{0}=U_{0}$. Let $U_{1}=U_{0}\cup\{w_{1},\ldots,w_{k}\}$. For each $w_{i}$, $1\leq i\leq k$, there exists an element $w^{\prime}_{i}\in SpanU_{0}$, such that $(w_{i}-w^{\prime}_{i})\in V_{1}$. Let

[TABLE]

Clearly, $Span(B_{0}\cup B_{1})=SpanU_{1}.$ Continuing in this way we construct disjoint finite set $B_{1},B_{2},\ldots$ such that $B_{n}\subset V_{n}$ and $Span(B_{0}\cup B_{1}\cup\ldots B_{n})=SpanU_{n}$, $n\geq 0$. Hence $B=\bigcup_{i\geq 0}B_{i}$ is a basis of $V$.

Let all elements in $B_{j}$ be greater than all elements in $B_{i}$, $i<j$. For each $i\geq 0$ elements in $B_{i}$ are ordered in an arbitrary way.

Strictly ordered products of elements from $B$ form a basis of the Clifford algebra $Cl(V,f)$. Ordered products $b_{1}b_{2}\ldots b_{t}$, where $b_{1},\ldots,b_{t}\in B\cap V_{i}$, form a basis of $Cl(V_{i},f)$. This implies the assertion of the Lemma. ∎

Let now $v$ be an element of $V$, such that $f(v)\neq 0$. Define a set

[TABLE]

It is clear, that

[TABLE]

By $C(v)$ we denote the centralizer of the element $v$ in the Clifford algebra $Cl(V,f)$, i.e. $C(v)=\{a\in Cl(V,f)\ |\ v\cdot a=a\cdot v\}$.

Lemma 4**.**

1. (A)

If $v\in V$, $f(v)=1,$ then

[TABLE] 2. (B)

Let $n$ be an odd number. Then for the elements $v_{i}=(a_{i1},a_{i2},\ldots,a_{ii},\ldots)$, $1\leq i\leq n$, such that a_{ij}=\begin{cases}1,\text{if j=i}\\ 0,\text{if j\neq i}\end{cases}, we have

[TABLE]

Proof.

The equality (3) implies, that any element from $Cl(V,f)$ is equal to a sum of products of the element $v$ and some elements from $v^{\perp}$.

Define the operator $U_{v}:Cl(V,f)\to Cl(V,f)$, via $x\mapsto vxv$. As $f(v)=1$ we have

[TABLE]

Hence the operator $U_{v}^{2}$ has two eigenvalues $-1$ and $1$. For any vector $u$ from $v^{\perp}$ we have

[TABLE]

So $uv=-vu$. As $f(v)=1,$ we obtain the equality

[TABLE]

Hence any vector from $v^{\perp}$ is an eigenvector of the operator $U_{v}$ with the eigenvalue $-1$.

Let now $u_{1},u_{2},\ldots,u_{n}$ be elements from the set $v^{\perp}$. For the reasons given above it follows that if $n$ is even, then $u_{1}u_{2}\ldots u_{n}$ is an eigenvector of the operator $U_{v}$ with the eigenvalue $1$. Therefore the element $u_{1}u_{2}\ldots u_{n}$ of $Cl(V,f)$ is an element of the centralizer $C(v)$. Note, that if $n$ is odd, then $u_{1}u_{2}\ldots u_{n}$ is an eigenvector of the operator $U_{v}$ with the eigenvalue $-1$, and so $u_{1}u_{2}\ldots u_{n}$ is not an element of $C(v)$.

Similarly, if $n$ is even, then $vu_{1}u_{2}\ldots\cdot u_{n}$ is an eigenvector of the operator $U_{v}$ with the eigenvalue $1$ and $vu_{1}u_{2}\ldots u_{n}$ lies in $Cl(V,f)$ in the centralizer $C(v)$. If $n$ is odd, then $vu_{1}u_{2}\ldots u_{n}$ is not an element of $C(v)$. Therefore

[TABLE]

and the first statement is proved.

As for any $i$ we have $f(v_{i})=1$, the proof of the second statement follows from the first statement by induction. ∎

Let $\mathcal{V}$ be the set $\{v_{i}=(\underbrace{0,\ldots,0,1,}_{i}0,\ldots),i\geq 1\}$.

Proposition 2**.**

The centralizer of the set $\mathcal{V}$ in the algebra $A=Cl(V,f)$ is $\mathbb{C}\cdot 1.$

Proof.

Consider a centralizer of the set $\mathcal{V}:$

[TABLE]

The subspace $C(\mathcal{V})$ of the algebra $Cl(V,f)$ is $\mathbb{Z}/2\mathbb{Z}$–graded, i.e.

[TABLE]

Assume, that $a\in C_{\bar{1}}$. Then

[TABLE]

Let $n$ be an odd number such, that $2m+1<n.$ By Lemma 4, (B) $a\in v_{1}v_{2}\ldots v_{n}Cl(v_{1}^{\perp}\cap v_{2}^{\perp}\cap\ldots\cap v_{n}^{\perp},f)_{\bar{0}}$. Let $\{w_{j}\}_{j\in J}$, be an ordered basis of the subspace $v_{1}^{\perp}\cap v_{2}^{\perp}\cap\ldots\cap v_{n}^{\perp}$. Then $v_{1},\ldots,v_{n},w_{j}$, $j\in J$ is a basis of $V$. The set of ordered products of elements of this basis is a basis of $Cl(V,f)$. Elements $v_{1}\ldots v_{n}w_{j_{1}},\ldots w_{j_{k}}$, $j_{1}<\ldots<j_{k}$, lie in this basis. Hence,

[TABLE]

Therefore, $a=0$ and $C_{\overline{1}}=(0).$ So, $C(\mathcal{V})=C_{\overline{0}}$. According to Lemma 4, (B) $C_{\overline{0}}=\cap_{i\geq 0}Cl(v_{i},f)_{\overline{0}}$. Thus Lemma 3 implies $C(\mathcal{V})=\mathbb{C}\cdot 1.$ ∎

Proposition 3**.**

For an arbitrary countable subset $X$ of the algebra $B=Cl(W,g)$ the centralizer of $X$ in $B$ is different from $\mathbb{C}\cdot 1.$

Proof.

Let $X$ be a countable subset of the algebra $B=Cl(W,g).$ There exists a countable subset $I_{0}\subset I$, such that $X\subset Cl(W_{0},g),$ where $W_{0}=\text{Span}(w_{i},i\in I_{0}).$ So, for any indexes $i,$ $j\in I\setminus I_{0},$ $i\neq j,$ the element $w_{i}w_{j}$ belongs the centralizer of $X.$ ∎

Proof of Theorem 8.

By Proposition 2 the centralizer of the set $\mathcal{V}$ in the Clifford algebra $A$ is $\mathbb{C}\cdot 1.$ But by Proposition 3 the centralizer of $X$ in the Clifford algebra $B$ is different from $\mathbb{C}\cdot 1$ for an arbitrary countable subset $X$ of $B$. Therefore, $A$ and $B$ are not isomorphic. ∎

6. Acknowledgement

We are grateful to E. Zelmanov for helpful discussions and valuable comments.