Matrix representation on d dimensional vector space

[tfhub]

Hi, i have been going through some elementray reading on group theory.
if g(θ) is a group element parameterized by the continuous variable θ. g(θ i )| θ i =identity,

if D(θ i ) is a matrix representation on a d dimensional vector space V . What is D(θ i )| θ i =0 ?

 

[jvicens]

Hello,

Thank you for sharing your question about group theory. To start, let's define some terms for those who may not be familiar with group theory.

A group is a mathematical structure that consists of a set of elements and a binary operation that combines any two elements to form a third element. The elements in a group can be numbers, matrices, or other mathematical objects, and the binary operation can be addition, multiplication, or other operations.

Now, let's break down your question. You have a group element, g(θ), that is parameterized by the continuous variable θ. This means that θ can take on any value within a certain range, and g(θ) will still be a valid element in the group. Your first statement says that g(θ i )| θ i =identity, which means that when θ is equal to θ i , the group element g(θ i ) is equal to the identity element of the group.

Moving on to the second part of your question, you mention a matrix representation, D(θ i ), on a d-dimensional vector space V. This means that for each value of θ i , there is a corresponding matrix D(θ i ) that represents the group element g(θ i ) in the d-dimensional vector space V. The matrix D(θ i ) is called a representation of the group element g(θ i ).

Finally, your question asks about the value of D(θ i ) when θ i =0. This means that you want to know the representation of the group element g(0) in the d-dimensional vector space V. Since g(0) is the identity element of the group, the representation of g(0) in any d-dimensional vector space V will also be the identity matrix, which is a matrix with 1s on the diagonal and 0s everywhere else.

I hope this helps to clarify your question. If you have any further questions, please don't hesitate to ask.