Undergrad General Linear Group GL(n) on Vector Spaces and canonical pairing invariance

  • Thread starter Thread starter jv07cs
  • Start date Start date
  • Tags Tags
    Vector spaces
Click For Summary
The discussion focuses on the action of the general linear group GL(n) on vector spaces and their duals. It clarifies that GL(n) acts on the n-dimensional space V only after selecting an ordered basis, which also defines an ordered dual basis for V*. The canonical pairing between vectors and covectors is shown to be GL(n) invariant through the associative property of matrix multiplication. Specifically, the transformation of covectors and vectors under an invertible matrix demonstrates this invariance in the context of the pairing. Ultimately, the group GL(V) acts on V by linear automorphisms, reinforcing the definition of invariance in the canonical pairing.
jv07cs
Messages
44
Reaction score
2
Does anyone have a reference that explains how the general linear group GL(n) acts on vector spaces and dual spaces? Furthermore, I would like to understand why the canonical pairing ##\langle\cdot, \cdot\rangle: V \times V^* \to \mathbb{F}##, ##(v,\alpha) \mapsto \langle\alpha,v \rangle := \alpha(v)##, is GL(n) invariant.
 
Physics news on Phys.org
In my opinion,

GL(n) acts, not on V, but on k^n, so it acts on the n dimensional space V only after choosing an ordered basis for V, (and thus also an ordered dual basis for V*).

Then if a is a covector in V*, represented by a row vector, and v is a vector in V, represented by a column vector, and if M is an invertible nxn matrix, then M takes a to aM and takes v to Mv, hence the fact that the matrix product aMv is associative, i.e. (aM)v = a(Mv), expresses the GL(n) - invariance of the pairing taking <v,a> in VxV*, to the dot product (a.v).

The group that acts on V, is called GL(V), and is by definition the group of linear automorphisms of V, hence it acts on V by definition, exactly as above, i.e. M in GL(V) takes v to M(v), and takes a to the composition aoM. Hence as above, (aoM)(v) = a(M(v)), shows the invariance, which here is actually a definition.
 
Last edited:
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K