A group seen as a category

Obsolete entry, go to this one.

Every group \(G\) is a category. It has only one object, \(G\) itself, and the morphisms corresponds to group elements.

Now, consider any set \(X\), and their bijections. We have other category.

A funtor from the former to the latter is nothing else that a group action of \(G\) over \(X\)! In this sense, we get a linear representation when \(X\) is a vector space and we shrink the image of the functor to their linear automorphisms (instead of all the bijections).

And a natural transformation between such two functors is a \(G\)-equivariant map.

Comments

Post a Comment

Popular posts from this blog

About the definition of angle, from Geometric Algebra

Image
In GA, bivectors represent  2-dimensional directions. Any simple bivector can be written as \[ \theta e_1e_2 \] with \(\theta \in \mathbb{R}\) and \(\{e_i\}\) unitary and orthogonal vectors. The unitary vectors specify the 2- direction itself, and \(\theta\) is the size (analogous to the length of a vector, that is a 1-dimensional direction). Given two vectors \(a\) and \(b\) (that we take to be unitary, wlog) we say that the simple bivector \(\theta e_1e_2\) is the angle formed by them if \[ ab=e^{\theta e_1 e_2} \] where we take as a definition \[ e^A=1+A+A^2 /2+\cdots=\lim_N (1+\frac{A}{N})^N \] Real analysis let us assure that this expression is well defined when \(A\) is a 2-blade. In fact, since \(e_1e_2e_1e_2=-1\), we can write \[ e^{\theta e_1 e_2}=cos(\theta)+e_1 e_2sin(\theta) \] where \(cos\) and \(sin\) are defined by their power series. Oserve that \(b=a e^{\theta e_1 e_2}\) and then \[ b=a(cos(\theta)+e_1 e_2sin(\theta))=a\cdot cos(\theta)+ ae_1e_2 \cdot sin(\theta) \]
Read more

Spherical, hyperbolic and parabolic geometry

Image
Let's consider the space \(\mathbb{M}^4\), that is, the manifold \(\mathbb{R}^4\) with coordinates \((x,y,z,t)\) but with the pseudo-Riemannian metric given in each tangent space by \[ ds^2=dx^2+dy^2+dz^2-dt^2. \] This metric is left invariant by the group \(O(3,1)\) when acting on \(\mathbb{M}^4\). It also leaves invariant the cone \[ t^2-x^2-y^2-z^2=0 \] so is logical that we pay attention to it. Now, following TRTR from Penrose page 423, we can consider the family of hyperplanes inside \(\mathbb{M}^4\) given by \[ z+t+\lambda(t-z)=2 \] intersecting our cone in different 2-dimensional submanifolds \(P_{\lambda}\) of \(\mathbb{M}^4\). Here we have a picture from that book, where we consider only \((x,z,t)\) for purposes of drawability. If we analyse the case \(\lambda=0\), we observe that we obtain the submanifold \(P_0\), \(E\) in the picture with a shape of a parabola, given by the embedding \begin{align*}   \phi \colon \mathbb{R}^2 &\longrightarrow \mathbb{M}^4\\   (u_1,u_2
Read more

General covariance and contravariance (new version)

Image
Suppose that a group \(G\) acts transitively on a space \(X\). A transformation \(T\in G\) can be thought of indistinctly as either "moving" the objects in \(X\) or as a change of "point of view" in the following way: Suppose we have a comfortable, mathematically speaking, space with a distinguished point, \((S,s_0)\). We are going to think of \((\mathbb Z,0)\), for easy. Suppose also a bijection \(b_1:S\to X\) (in the line of reasoning of the notes homogeneous space#Intuitive approach and basis and change of basis ). It is better to think in terms of \(b_1^{-1}\). That is, for every \(x\in X\) we have a kind of "coordinates" \(s\) for \(x\), given by \(s=b_1^{-1}(x)\in S\). Think, for example, of a vector space \(V\) and the isomorphism \(b_1:\mathbb R^2 \to V\) (fixing a basis in \(V\)). We can think of this as our initial point of view , and we can imagine as if we were located at \(x_0=b_1(0)\in X\) with some devices to take measurements that let u
Read more