automorphism group

In mathematics, the automorphism group of an object ''X'' is the group consisting of automorphisms of ''X'' under composition of morphisms. For example, if ''X'' is a finite-dimensional vector space, then the automorphism group of ''X'' is the group of invertible linear transformations from ''X'' to itself (the general linear group of ''X''). If instead ''X'' is a group, then its automorphism group $\operatorname(X)$ is the group consisting of all group automorphisms of ''X''.

Especially in geometric contexts, an automorphism group is also called a symmetry group. A subgroup of an automorphism group is sometimes called a transformation group.

Automorphism groups are studied in a general way in the field of category theory.

Examples

If ''X'' is a set with no additional structure, then any bijection from ''X'' to itself is an automorphism, and hence the automorphism group of ''X'' in this case is precisely the symmetric group of ''X''. If the set ''X'' has additional structure, then it may be the case that not all bijections on the set preserve this structure, in which case the automorphism group will be a subgroup of the symmetric group on ''X''. Some examples of this include the following:
*The automorphism group of a field extension $L/K$ is the group consisting of field automorphisms of ''L'' that fix ''K''. If the field extension is Galois, the automorphism group is called the Galois group of the field extension.
*The automorphism group of the projective ''n''-space over a field ''k'' is the projective linear group $\operatorname_n(k).$
*The automorphism group $G$ of a finite cyclic group of order ''n'' is isomorphic to $(\mathbb/n\mathbb)^\times$, the multiplicative group of integers modulo ''n'', with the isomorphism given by $\overline \mapsto \sigma_a \in G, \, \sigma_a(x) = x^a$. In particular, $G$ is an abelian group.
*The automorphism group of a finite-dimensional real Lie algebra $\mathfrak$ has the structure of a (real) Lie group (in fact, it is even a linear algebraic group: see below). If ''G'' is a Lie group with Lie algebra $\mathfrak$, then the automorphism group of ''G'' has a structure of a Lie group induced from that on the automorphism group of $\mathfrak$.

If ''G'' is a group acting on a set ''X'', the action amounts to a group homomorphism from ''G'' to the automorphism group of ''X'' and conversely. Indeed, each left ''G''-action on a set ''X'' determines $G \to \operatorname(X), \, g \mapsto \sigma_g, \, \sigma_g(x) = g \cdot x$, and, conversely, each homomorphism $\varphi: G \to \operatorname(X)$ defines an action by $g \cdot x = \varphi(g)x$. This extends to the case when the set ''X'' has more structure than just a set. For example, if ''X'' is a vector space, then a group action of ''G'' on ''X'' is a ''group representation'' of the group ''G'', representing ''G'' as a group of linear transformations (automorphisms) of ''X''; these representations are the main object of study in the field of representation theory.

Here are some other facts about automorphism groups:
*Let $A, B$ be two finite sets of the same cardinality and $\operatorname(A, B)$ the set of all bijections $A \mathrel B$. Then $\operatorname(B)$, which is a symmetric group (see above), acts on $\operatorname(A, B)$ from the left freely and transitively; that is to say, $\operatorname(A, B)$ is a torsor for $\operatorname(B)$ (cf. #In category theory).
*Let ''P'' be a finitely generated projective module over a ring ''R''. Then there is an embedding $\operatorname(P) \hookrightarrow \operatorname_n(R)$, unique up to inner automorphisms.

In category theory

Automorphism groups appear very naturally in category theory.

If ''X'' is an object in a category, then the automorphism group of ''X'' is the group consisting of all the invertible morphisms from ''X'' to itself. It is the unit group of the endomorphism monoid of ''X''. (For some examples, see PROP.)

If $A, B$ are objects in some category, then the set $\operatorname(A, B)$ of all $A \mathrel B$ is a left $\operatorname(B)$- torsor. In practical terms, this says that a different choice of a base point of $\operatorname(A, B)$ differs unambiguously by an element of $\operatorname(B)$, or that each choice of a base point is precisely a choice of a trivialization of the torsor.

If $X_1$ and $X_2$ are objects in categories $C_1$ and $C_2$, and if $F: C_1 \to C_2$ is a functor mapping $X_1$ to $X_2$, then $F$ induces a group homomorphism $\operatorname(X_1) \to \operatorname(X_2)$, as it maps invertible morphisms to invertible morphisms.

In particular, if ''G'' is a group viewed as a category with a single object * or, more generally, if ''G'' is a groupoid, then each functor $F: G \to C$, ''C'' a category, is called an action or a representation of ''G'' on the object $F(*)$, or the objects $F(\operatorname(G))$. Those objects are then said to be $G$-objects (as they are acted by $G$); cf. $\mathbb$-object. If $C$ is a module category like the category of finite-dimensional vector spaces, then $G$-objects are also called $G$-modules.

Automorphism group functor

Let $M$ be a finite-dimensional vector space over a field ''k'' that is equipped with some algebraic structure (that is, ''M'' is a finite-dimensional algebra over ''k''). It can be, for example, an associative algebra or a Lie algebra.

Now, consider ''k''-linear maps $M \to M$ that preserve the algebraic structure: they form a vector subspace $\operatorname_(M)$ of $\operatorname(M)$. The unit group of $\operatorname_(M)$ is the automorphism group $\operatorname(M)$. When a basis on ''M'' is chosen, $\operatorname(M)$ is the space of square matrices and $\operatorname_(M)$ is the zero set of some polynomial equations, and the invertibility is again described by polynomials. Hence, $\operatorname(M)$ is a linear algebraic group over ''k''.

Now base extensions applied to the above discussion determines a functor: namely, for each commutative ring ''R'' over ''k'', consider the ''R''-linear maps $M \otimes R \to M \otimes R$ preserving the algebraic structure: denote it by $\operatorname_(M \otimes R)$. Then the unit group of the matrix ring $\operatorname_(M \otimes R)$ over ''R'' is the automorphism group $\operatorname(M \otimes R)$ and $R \mapsto \operatorname(M \otimes R)$ is a group functor: a functor from the category of commutative rings over ''k'' to the category of groups. Even better, it is represented by a scheme (since the automorphism groups are defined by polynomials): this scheme is called the automorphism group scheme and is denoted by $\operatorname(M)$.

In general, however, an automorphism group functor may not be represented by a scheme.

See also

* Outer automorphism group
* Level structure, a technique to remove an automorphism group
* Holonomy group

Notes

Citations

References

*
*
*
*
*
{{refend

External links

*https://mathoverflow.net/questions/55042/automorphism-group-of-a-scheme

Group automorphisms