Right translations of orthonormal bases.

[Kreizhn]

Homework Statement

Let G be a Lie group and L be its Lie algebra. Further assume that TG is equiped with a metric tensor g; that is, a (0,2)-covariant tensor that is symmetric and positive-definite on fibres of TG. Assume that \{ H_i\}_{i=1}^m are an orthonormal basis for for L. If R_X: G \to G is the right-translation operator acting as R_X(Y) = Y\cdot X with \cdot the group operation, show that for any X \in G the set \{ dR_X H_i \}_{i=1}^m form an orthonormal basis for T_X G.

The Attempt at a Solution

So via recent posts here, we know that dR_X H_i = H_i X. Consider the "inner-product" on T_X G given by \langle \cdot,\cdot \rangle. Then we want to show that
\langle H_i X, H_j X \rangle = \delta_{ij}
Now, I really don't think we can do anything more here without an explicit expression for the inner-product, or at least some consideration of what the group G is.

For my purposes, G is the unitary group G = \mathfrak U(N). Now normally when we do inner-products on Hilbert spaces, we can normally jump between arguments of the inner-product like
\langle H_i X, H_j X \rangle = \langle H_i X X^\dagger, H_j \rangle = \langle H_i, H_j \rangle

However, I'm really concerned about doing this, for two reasons. The first is that normally, the notion of unitarity requires that we work the inner product defined as an operator on a space and its dual. In this case, the inner-product is on two copies of T_X G. Since the metric is everywhere defined, I suppose we could say that G is a Riemannian manifold, in which case the musical isomorphisms give us a way of relating T_X^*G to T_XG and then it would be fine.

The second reason is "what is unitarity when we change the inner product?" In particular, we know what unitarity is for finite dimensions and L^2, but what does it mean on a Lie algebra? Does it make sense to have unitary operators on the space of skew-Hermitian matrices?

 

[Kreizhn]

Should I perhaps be using the Killing form for my inner-product?

 

[Dick]

Should I perhaps be using the Killing form for my inner-product?

Maybe. The Killing form is the (2,0) tensor that actually has a direct relation with the Lie algebra. But the Killing form isn't necessarily nondegenerate and it's certainly not positive definite. I'm a little puzzled what this is actually about.

 

[Kreizhn]

I might be making this more complicated than it has to be.

Essentially, I'm saying that we have an orthonormal basis for the Lie algebra. I want to show that the right-translate of those basis elements also forms an orthonormal basis in general.

I'm not sure if it can be done in general. I was hoping that maybe by specifying that the Lie group is \mathfrak U(N) and maybe using the (negative) Killing-form for a metric? Because finite dimensional unitaries form a compact group, the Killing-form would be negative-definite.

To make matters worse, while I've presented the problem in a Lie-group, Riemannian geometry framework, I actually want to consider the problem in a sub-Riemannian geometry However, since the Riemannian case is easier, I figure I should do that first.

 

[Dick]

Beats me. In the Riemannian case you've got an inner product and a notion of parallel transport, which let's you move vectors around in a way that's generally path dependent. Subriemannian, I don't know.

 

[Kreizhn]

But can we show that orthonormality is preserved in the Riemannian case?

 

[Dick]

But can we show that orthonormality is preserved in the Riemannian case?

Sure. Parallel transport is an isometry, isn't it?

 

[Kreizhn]

Um...I'm really not certain. Intuitively it seems like it should be, so I'll look into it. Even so, doesn't that just preserve the normality? Wouldn't we need it to also be "conformal," if that even makes sense in this framework. Would it be forced to be conformal? Let me look into these questions and I'll get back to you.

Edit: I realize now that because the isometry would have to preserve the induced metric, we don't need a notion of conformality.

Second Edit: I clearly should brush up on my Riemannian geometry, as for whatever reason I was working with a metric space definition of isometry rather than one which preserves the metric tensor.

 

[Kreizhn]

Well that was quick. It seems that so long as we use the Levi-Cevita connection then everything is okay. I don't see a problem with restricting ourselves to that connection.