Left invariant vector fields of a lie group

[fliptomato]

Comment: My question is more of a conceptual 'why do we do this' rather than a technical 'how do we do this.'

Homework Statement

Given a lie group G parameterized by x_1, ... x_n, give a basis of left-invariant vector fields.

Homework Equations

We have a basis for the vector fields at the identity, namely the Lie algebra: v_1, ..., v_n. For a general group element g, we can write g^{-1}<br /> \frac{\partial g}{\partial x_i}.

The Attempt at a Solution

Apparently the procedure is to write v_i = A^{ij}g^{-1}<br /> \frac{\partial g}{\partial x_j}, where the A^{ij} are just coefficients. Then we can somewhat magically read off the left invariant vector fields w_i via:

w_i = A^{ij}\frac{\partial}{\partial x_j}

Where I have assumed a sum over repeated indices (though haven't been careful with upper or lower indices).

Why is this the correct procedure? I'm especially concerned about this thing \frac{\partial g}{\partial x_j}... is it an element of the tangent space or is it dual to the tangent space?

Thanks for any thoughts,
Flip

 

[fliptomato]

Or perhaps a more basic question. For a matrix lie group, the elements of the lie algebra are also matrices. Like any manifold, the tangent space is spanned by the basis vectors \frac{\partial}{\partial x^{ij}}|_{e}. What matrices do each of these basis vectors correspond to?

For a general element of the lie group g, how does \frac{\partial}{\partial x^{ij}}|_{e} differ from \frac{\partial g}{\partial x^{ij}}|_{e}?

 

[Hurkyl]

Presumably,

\frac{\partial g}{\partial x^i}

is supposed to be the partial derivative (in the i-th direction) of the multiplication-by-g automorphism of G. This notation is poor, though, since it doesn't indicate whether that's right-multiplication-by-g or left-multiplication-by-g.


Whatever's being done, it's just working through the definition of left-invariant. If L_g is left-multiplication-by-g, then it has a derivative L_{g*}, and a vector field v is invariant under all such diffeomorphisms iff

<br /> v(gh) = L_{g*}(h)(v(h))<br />

(recall that L_{g*} is a linear map from the tangent space at h to the tangent space at gh) If we set h = e we get

v(g) = L_{g*}(e)(v(e)),

which gives us the value of v at every point g of G.


One thing that makes the lie group interesting is that we can use right translation to give us canonical isomorphisms between all of the tangent spaces, and can thus identify each of them with the corresponding lie algebra. (We can get a different correspondence by using left translation) That seems to be rolled into this derivation too.


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For the matrix Lie group embedded in R^mxm, the tangent space at the identity is the set of all matrices I+v, where v is an element of the corresponding matrix Lie algebra. (I is the identity matrix) Of course, to turn this affine space into a vector space, we take I to be the origin.

The tangent space at any particular point g is the set of all matrices of the form g(I+v). (Which is the same as the set of all matrices of the form (I+v)g)