Geometrical Equivalence and Action Type Geometrical Equivalence of Group Representations

TL;DR

This paper demonstrates that action-type geometrical equivalence and group geometrical equivalence do not necessarily imply overall geometrical equivalence of group representations, highlighting limitations in current equivalence criteria.

Contribution

It provides a counterexample showing the non-implication between different types of geometrical equivalences in group representations.

Findings

Counterexample disproves the implication from action-type to overall geometrical equivalence.

Highlights the independence of different geometrical equivalence notions in group representations.

Clarifies limitations in the criteria for classifying group representations based on geometrical equivalence.

Abstract

In this paper we present the example which proves that we can not conclude the geometrical equivalence of group representations from the corresponding action-type geometrical equivalence and group geometrical equivalence.

Equations

$$V\times G\ni \left(v,g\right)\to v\circ g\in V.$$

$$\alpha (v\circ g)=\alpha (v)\circ \beta (g)$$

$$(T_1,T_2{)}_{(V,G)}^{{}^{\prime }}=$$

$$\{(\alpha ,\beta )\in \mathrm{Hom}((XKF(Y),F(Y)),(V,G))|T_1\subseteq \ker (\alpha ),T_2\subseteq \ker (\beta )\}\text{.}$$

$$(T_1,T_2{)}_{(V,G)}^{{}^{\prime \prime }}=(\bigcap _{(\alpha ,\beta )\in (T_1,T_2{)}_{(V,G)}^{{}^{\prime }}}\ker (\alpha ),\bigcap _{(\alpha ,\beta )\in (T_1,T_2{)}_{(V,G)}^{{}^{\prime }}}\ker (\beta ))\subseteq$$

$$(XKF(Y),F(Y)).$$

$$(\underset{i=1}{\overset{n}{\bigwedge }}{w}_i)\Rightarrow w_0,$$

$$(V,G)\models ((\underset{i=1}{\overset{n}{\bigwedge }}{w}_i)\Rightarrow w_0).$$

$$T_{(V,G)}^{{}^{\prime }}=\{(\alpha ,\beta )\in \mathrm{Hom}((XKF(Y),F(Y)),(V,G))|T\subseteq \ker (\alpha )\}\text{.}$$

$$T_{(V,G)}^{{}^{\prime ▽}}=\bigcap _{(\alpha ,\beta )\in T_{(V,G)}^{{}^{\prime }}}\ker (\alpha )\subseteq XKF(Y).$$

$$stab(v)=\left\{g\in G|v\circ g=v\right\},$$

$$\ker (V,G)=\bigcap _{v\in V}stab(v).$$

$$v\circ \sigma (g)=v\circ g,$$

$$e_1\circ a=e_2,e_2\circ a=e_1,$$

$$e_1\circ b=e_1,e_2\circ b=e_2.$$

$$(x\circ y-x=0)\Rightarrow (y=1).$$

$$(V,G_1)\models ((x\circ y-x=0)\Rightarrow (y=1)),$$

$$(V,G_2)⊭((x\circ y-x=0)\Rightarrow (y=1))\text{,}$$

Full text

1 Introduction

All definitions of the basic notions of the universal algebraic geometry can be found, for example, in [5], [6], [7] and [8]. Also, there are fundamental papers [1], [4] and [2], [3].

Some problems of the universal algebraic geometry of many-sorted universal algebras were considered also in [10], [12], [13] and [14].

We consider the representations of groups over vector spaces over an arbitrary fixed field $K$. This field is fixed in all our considerations. In this article the representation of group is a pair $(V,G)$, where $V$ is a vector space over field $K$ and $G$ is a group. The signature of this algebraic object includes all the operations in the vector spaces $V$ (multiplication by every scalar $\lambda\in K$ we consider as unary operation), all the operations in the group $G$ and the operation of action of the $G$ on vector spaces $V$. We denote this operation by $\circ$:

[TABLE]

From here and below we will write briefly ”representation” instead ”representation of group”. The homomorphism $(\alpha,\beta)$: $(V,G)\rightarrow(W,H)$ from representation $(V,G)$ to representation $(W,H)$ is a pair, where $\alpha:V\rightarrow W$ is a linear mapping and $\beta:G\rightarrow H$ is homomorphism of groups, such that for every $v\in V$ and every $g\in G$ the equality

[TABLE]

holds. The reader can see that we consider representations as $2$-sorted universal algebras: the first sort is a sort of vectors of vector spaces and the second sort is a sort of elements of groups. This approach to the representations the reader can find also in [11], [15], [10], [13].

The variety of all the representations of groups over a fixed field $K$ will be denoted by $REP-K$.

As usually, we say that

Definition 1.1

The representation $W\left(X,Y\right)=\left(U\left(X,Y\right),H\left(X,Y\right)\right)$ is called the free representation generated by the sets $X$ and $Y$ if, for every $\left(V,G\right)\in REP-K$ and every mappings $f_{1}:X\rightarrow V$ and $f_{2}:Y\rightarrow G$, there is only one homomorphism of representations $\left(\alpha,\beta\right):\left(U\left(X,Y\right),H\left(X,Y\right)\right)\rightarrow\left(V,G\right)$, such that $\alpha_{\mid X}=f_{1}$, $\beta_{\mid Y}=f_{2}$.

It is easy to prove [11], that $\left(U\left(X,Y\right),H\left(X,Y\right)\right)=(XKF(Y),F(Y))$, where $F(Y)$ is the free group with the free set of generators $Y$, $KF(Y)$ is the group ring over this group and $XKF(Y)=\bigoplus\limits_{x\in X}xKF(Y)$ is the free $KF(Y)$-module with the free basis $X$.

2 Basic notions of the algebraic geometry of representations.

From here and below we suppose that the sets $X$ and $Y$ of generators of the free representations $(XKF(Y),F(Y))$ are finite. Every equation in the algebraic geometry of representation is equivalent to the equation $v=0$, where $v\in XKF(Y)$ or to the equation $f=1$, where $f\in F(Y)$ for arbitrary $X$ and $Y$. So, in algebraic geometry of representations we can consider the system of equations $T=(T_{1},T_{2})$, where $T_{1}\subseteq XKF(Y)$, $T_{2}\subseteq F(Y)$, or, briefly, $(T_{1},T_{2})\subseteq(XKF(Y),F(Y))$. If we will resolve this system of equations in the representation $\left(V,G\right)\in REP-K$, then the set $\mathrm{Hom}((XKF(Y),F(Y)),(V,G))$ has for us the role of the affine space. The solution of the system $(T_{1},T_{2})$ in $\left(V,G\right)$ will be the set

[TABLE]

[TABLE]

The algebraic $(V,G)$-closure of the system $(T_{1},T_{2})$ will be the set

[TABLE]

[TABLE]

This is the maximal system of equations, which has the same solutions as the system $(T_{1},T_{2})$.

Definition 2.1

Let $(V_{1},G_{1}),(V_{2},G_{2})\in REP-K$. We say that $(V_{1},G_{1})$ and $(V_{2},G_{2})$ are geometrically equivalent if $(T_{1},T_{2})_{(V_{1},G_{1})}^{{}^{\prime\prime}}=(T_{1},T_{2})_{(V_{2,},G_{2})}^{{}^{\prime\prime}}$, for every $(T_{1},T_{2})\subseteq(XKF(Y),F(Y))$ and every $X$ and $Y$. We use the notation $(V_{1},G_{1})\sim(V_{2},G_{2})$.

The notion of geometric equivalence can be defined in arbitrary variety of universal algebras, as one-sorted, as many-sorted. The reader can consult in [5].

In our theory

Definition 2.2

The logic formulas which have the form:

[TABLE]

where $w_{i}$ can be either $(v_{i}=0)$ or $(f_{i}=1)$, for $v_{i}\in XKF(Y)$ or $f_{i}\in F(Y)$,$\ 0\leq{i\leq{n}}$, $n\in\mathbb{N}$, are called quasi-identities.

Definition 2.3

Let $(V,G)\in REP-K.$ We say that $(V,G)$ fulfills the quasi-identity (2.1) if, for every $(\alpha,\beta)\in\mathrm{Hom}((XKF(Y),F(Y)),(V,G))$ such that $\alpha(v_{i})=0$, when $w_{i}$ is $(v_{i}=0)$, or $\beta(f_{i})=1$, when $w_{i}$ is $(f_{i}=1)$, $1\leq{i\leq{n}}$, we have that $\alpha(v_{0})=0$, when $w_{0}$ is $(v_{0}=0)$, or $\beta(f_{0})=1$, when $w_{0}$ is $(f_{0}=1)$. We denote:

[TABLE]

By [9, Theorem 2] we have the

Proposition 2.1

Let $(V_{1},G_{1}),(V_{2},G_{2})\in REP-K$ and $(V_{1},G_{1})\sim(V_{2},G_{2})$ then $(V_{1},G_{1})$ and $(V_{2},G_{2})$ fulfill same quasi-identities.

By [5, Proposition 13] we also have the

Proposition 2.2

Let $\Theta$ some variety of universal algebras, $H_{1},H_{2}\in\Theta$ are finitely generated. Then $H_{1}\sim H_{2}$ if and only if exist injections $H_{1}\hookrightarrow H_{2}^{I_{2}}$ and $H_{2}\hookrightarrow H_{1}^{I_{1}}$, where $I_{1},I_{2}$ are some sets of indexes and $H_{1}^{I_{1}},H_{2}^{I_{2}}$ corresponding Cartesian powers of algebras $H_{1}$ and $H_{2}$.

In [10] the action type algebraic geometry of representations was also considered. This geometry was elaborated on in order to avoid the influence of the algebraic geometry of the acting group on the algebraic geometry of representation.

In this geometry we consider only system of action type equations, i.e., system of equations which have form $T\subseteq XKF(Y)$. The set of solutions of this system in the representation $(V,G)$ is the set

[TABLE]

The action type $(V,G)$-closure of the system of equations $T$ is a set

[TABLE]

This is the maximal system of action type equations, which has the same solutions as the system $T$.

Definition 2.4

Let $(V_{1},G_{1}),(V_{2},G_{2})\in REP-K$. We say that $(V_{1},G_{1})$ and $(V_{2},G_{2})$ are action type geometrically equivalent if $T_{(V_{1},G_{1})}^{{}^{\prime\triangledown}}=T_{(V_{2},G_{2})}^{{}^{\prime\triangledown}}$, for every $T\subseteq XKF(Y)$ and every $X$ and $Y$. We use the notation $(V_{1},G_{1})\sim_{at}(V_{2},G_{2})$.

By [10, Corollary 2 from Proposition 4.2], if two representations $(V_{1},G_{1})$ and $(V_{2},G_{2})$ are geometrically equivalent then they are action type geometrically equivalent.

If $(V,G)\in REP-K$, for every $v\in V$ we can consider the stabilizer of $v$, defined by

[TABLE]

and the $\ker(V,G)$, defined by

[TABLE]

It is easy to check that $\ker(V,G)$ is a normal subgroup of $G$. We denote by $\tilde{G}$ the quotient group $G/\ker(V,G)$ and by $\sigma$ the natural epimorphism $\sigma:G\rightarrow G/\ker(V,G)$. It also is easy check that we obtain the representation $(V,\tilde{G})$ over the vector spaces $V$ if we define the action of the group $\tilde{G}$ over the vector spaces $V$ thusly

[TABLE]

where $v\in V$,$\ g\in G$.

Definition 2.5

The representation $(V,\tilde{G})$ is called the faithful image of the representation $(V,G)$.

By [10, Corollary 4 from Theorem 5.1], we have the

Proposition 2.3

Every representation $(V,G)\in REP-K$ is action type geometrically equivalent to its faithful image $(V,\tilde{G})$.

3 The relation between geometrical equivalence and action type

geometrical equivalence ofgroup representations.

In this Section we will discuss the following question:

Problem 1

Let $(V_{1},G_{1}),(V_{2},G_{2})\in{REP-K}$. Can we conclude the geometrical equivalence $(V_{1},G_{1})\sim(V_{2},G_{2})$ from the action type geometrical equivalence $(V_{1},G_{1})\sim_{at}(V_{2},G_{2})$ and the geometrical equivalence of groups $G_{1}\sim{G_{2}}$?

The negative answer to this question was remarked by [10, Remark 5.1], but no counterexample was presented. This question has a sense, because in the action type algebraic geometry of representations we only consider the specific systems of equations which have form $T\subseteq XKF(Y)$ and the specific form of the algebraic closure (2.2). By this restriction we avoid the influence of the algebraic geometry of action groups. But one question which comes naturally is whether we are losing some important information about representations by this restriction. The negative answer to the Problem 1 shows that we indeed lose some information.

Theorem 3.1

Let $(V_{1},G_{1}),(V_{2},G_{2})\in{REP-K}$. From conditions

$(V_{1},G_{1})\sim_{at}(V_{2},G_{2})$* and* 2. 2.

$G_{1}\sim{G_{2}}$**

we can not conclude $(V_{1},G_{1})\sim(V_{2},G_{2})$.

Proof. We consider a vector space $V$ over arbitrary field $K$, such that $\dim(V)=2$, with the basis $\{e_{1},e_{2}\}$. We also consider the groups $G_{1}=\left\langle a\right\rangle\cong\mathbb{Z}_{2}$ and $G_{2}=\left\langle a\right\rangle\times\left\langle b\right\rangle\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. We define the action of $G_{1}$ and $G_{2}$ over $V$ thusly:

[TABLE]

[TABLE]

Hence, we obtain the representations $(V_{1},G_{1})$ and $(V_{2},G_{2})$. $\ker(V,G_{2})=\left\langle b\right\rangle\cong\mathbb{Z}_{2}$. $G_{2}/\ker(V,G_{2})\cong G_{1}=\left\langle a\right\rangle\cong\mathbb{Z}_{2}$. Therefore, $(V,\widetilde{G_{2}})$ the faithful image of the representation $(V,G_{2})$ is isomorphic to the representation $(V_{1},G_{1})$. So, by Proposition 2.3 $(V,G_{1})\sim_{at}(V,G_{2})$.

The injections $G_{1}\hookrightarrow G_{2}$ and $G_{2}\hookrightarrow G_{1}\times G_{1}$ exist. Therefore, by Proposition 2.2, $G_{1}\sim{G_{2}}$.

Now, we consider the quasi-identity

[TABLE]

We have that

[TABLE]

because $\ker(V,G_{1})=\left\{1\right\}$, and

[TABLE]

because $\ker(V,G_{2})\neq\left\{1\right\}$. It means by Proposition 2.1 that $(V_{1},G_{1})\nsim(V_{2},G_{2})$.

4 Acknowledgements

We acknowledge the support of Coordenação de Aperfeiçoamento de Pessoal de Nível Superior - CAPES (Coordination for the Improvement of Higher Education Personnel, Brazil).

We are thankful to Prof. E. Aladova for her important remarks, which helped a lot in writing this article.