This question was asked in my quiz on Lie Groups and I couldn't solve it. I tried it again at home but in vain. Now, I am posting it here as I want to learn this concept.
Question: Prove that $SL(2,\mathbb{R})$ is a Lie Group. What is the dimension of the vector space consisting of all left invariant vector fields on $SL(2 ,\mathbb{R})$?
I have proved it to be Lie group.
But I am struck in founding the dimension of the vector space.
A vector field X is a function that associate smoothly to every point p of G an element or vector $X_p$ of the tangent space of the group G (which in this case is also a manifold). So for every point p you have the vector $X_p \in T_p(G) $.
Vector field is invariant i.e. $X_{F(p)}(f)=X_p(f∘F)$, where F is a function that maps every point of G to another point of G, $f\in C^{\infty} G$ . If F is the function given by the left multiplication of the group, then you have a Left Invariant Vector Field.What does it means by "If F is the function given by the left multiplication of the group"? How can a function be given by left multiplication of a group? So, I am stuck on what it means to be left invariant.
For this question:
Each $p\in SL(2 ,\mathbb{R})$ is associated to an $X_p \in SL(2 ,\mathbb{R})$. Now, let F be a function that maps a point of $SL(2,\mathbb{R})$ to another point of $SL(2, \mathbb{R})$.
Kindly let me know what it means to be a left invariant vector fields and also give some hints on how to use that to find the dimension here.
I shall be really thankful.