geometric representation theory
representation, 2-representation, ∞-representation
Grothendieck group, lambda-ring, symmetric function, formal group
principal bundle, torsor, vector bundle, Atiyah Lie algebroid
Eilenberg-Moore category, algebra over an operad, actegory, crossed module
Be?linson-Bernstein localization?
An action
of a group $G$ on a set $X$ is called regular if it is both transitive and free, that is, if for any pair of elements $x,y \in X$, there is exactly one group element $g \in G$ such that $g * x = y$.
Suppose $G$ acts transitively on $X$ by $* : G \times X \to X$, and suppose moreover that this action is faithful. Then $G$ acts freely (and hence regularly) on $X$ if and only if the group $Aut_G(X)$ of $G$-equivariant automorphisms (i.e., bijections $\phi : X \to X$ commuting with the action of $G$) acts transitively (and hence regularly) on $X$.
First we show that $Aut_G(X)$ acts freely on $X$. Suppose $\phi \in Aut_G(X)$ is such that $\phi(x) = x$ for some $x\in X$, and let $y\in X$ be arbitrary. By the assumption that $G$ acts transitively, there is a $g \in G$ such that $y = g*x$. But then $G$-equivariance implies that
Since this holds for all $y\in Y$, $\phi$ must be equal to the identity $\phi = id_X$, and therefore $Aut_G(X)$ acts freely on $X$.
Next, suppose that $G$ also acts freely on $X$, and let $x,y \in X$ be arbitrary. Then we can define a $G$-equivariant automorphism $\phi$ such that $\phi(x) = y$ by
where for each $z$, $g_z$ is the unique group element such that $z = g_z*x$. Conversely, suppose that $Aut_G(X)$ acts transitively on $X$, and let $x\in X$, $g\in G$ such that $g*x = x$. By the assumption, for any $y \in X$, there exists $\phi \in Aut_G(X)$ such that $\phi(x) = y$, from which it follows that
Since $g*y = y$ for all $y \in X$, therefore $g = 1$ by the assumption that $G$ acts faithfully on $X$.
The action of $G$ on itself by multiplication $\cdot : G \times G \to G$ (on the left or on the right) is a regular action, called the (left or right) regular representation of $G$.
If one views a combinatorial map $M$ as the transitive action of a certain group of permutations, then $M$ represents a regular map (Siran 2006) just in case this action is regular. For example, the five Platonic solids may be represented as regular combinatorial maps.
We discuss regular actions via homotopy type theory.
Since doing group representation theory in homotopy type theory corresponds to working in the context of a delooped group in homotopy type theory, the regularity of an action is naturally expressed there. Transitivity ensures that all the points in the homotopy quotient are connected by equivalences, while freeness means that the space of equivalences between two points is itself contractible. Hence if $\ast \colon \mathbf{B} G \vdash X(\ast): Type$ corresponds to a regular action, then the quotient $\sum_{\ast: \mathbf{B} G} X(\ast)$ is contractible.
Restriction to 1-groups is unnecessary here, and we say
An ∞-action of an ∞-group is a regular $\infty$-action if its homotopy quotient is contractible.
For any $G$-action (∞-action) $X \colon BG \to U$, its automorphism group is (see at automorphism ∞-group in HoTT)
and
by $\tilde{X}(P,-) \coloneqq P(\ast)$.
If $X$ is regular, then $\tilde{X}$ is regular.
First, we need to argue that $X(\ast)$ is merely inhabited. Since $X$ is regular, we have $\sum_{(b:BG)} X(b)$ contractible. This gives a center of contraction $(b,x)$. Now, since $B G$ is connected, it follows that $\|b=\ast\|$. Since we are proving the mere proposition $\|X(\ast)\|$, we get to use $b=\ast$. Now we obtain $\|X(\ast)\|$.
Next, to show that $\tilde{X}$ is regular we need to show that $\tilde{X}$ has a contractible total space. The dependent sum type $\sum_{(b : BAut_G(X))} \tilde{X}(b)$ is equivalent to $\sum_{(P:BG \to U)} \|P=X\| \times P(\ast)$. Contractibility is a mere proposition, and we have $\|X(\ast)\|$, so we get to use a point $x:X(\ast)$. This gives us a center of contraction $(X,refl,x)$ of the total space of $\tilde{X}$.
Now let $P : BG \to U$, let $\| P = X \|$, let $p_0 : P(\ast)$. To show regularity, it suffices to find a term of type
This type is equivalent to showing that there are
Now we use that $X(b)$ is equivalent to $b=\ast$ (we get this fact from regularity, together with a point $x:X(\ast)$). Since we need this particular fiberwise equivalence, it suffices to show that
$\sum_{b:BG} P(b)$
is contractible. Now this is a mere proposition, so we can eliminate $t : \| P = X \|$ to obtain the proof.
If $X$ is a principal homogeneous space on $G$, in the sense that the type $\sum_{(g:G)} g_\ast(x)=y$ is contractible for all $x,y:X(\ast)$, and $\tilde{X}$ is regular, then $X$ is regular.
Again, we first show that $X(\ast)$ is merely inhabited. The total space of $\tilde{X}$ has center of contraction $(P,p_0)$. Since $\|P=X\|$ and since we are proving a mere proposition, we get to use $P=X$. Now $\|X(\ast)\|$ follows from $p_0:P(\ast)$. The regularity of $X$ is a mere proposition, so we get to use $x_0:X(\ast)$. This gives us the center of contraction $(\ast,x_0)$. It remains to show that
$\prod_{(b:BG)} \prod_{(x:X(b))} \sum_{(\alpha : b=\ast)} \mathrm{trans}(\alpha,x) = x_0$.
Of course, it would suffice to prove the stronger statement
$\prod_{(b:BG)} \prod_{(x:X(b))} \mathrm{isContr} (\sum_{(\alpha : b=\ast)} \mathrm{trans}(\alpha,x) = x_0)$.
However, now we get to use that $BG$ is connected. Therefore it suffices to show that
$\prod_{x:X(*)} isContr (\sum_{\alpha : G} \mathrm{trans}(\alpha,x) = x_0)$
This holds by assumption.
Jozef Siran, “Regular Maps on a Given Surface: A Survey”, Topics in Discrete Mathematics, 2006. (pdf)
Group Properties Wiki: Regular group action.
Last revised on April 24, 2018 at 09:16:55. See the history of this page for a list of all contributions to it.