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I'm currently studying differential geometry on smooth manifolds using differential forms and I'm trying to apply it to what I have learned earlier about Lie groups, but something doesn't seem to quite work out.

More specifically, I'm looking at the smooth manifold of a Lie group and have a homomorphism to $\mathbb{R}^n$ from where I would like to pull back and push forward the differential geometric properties to the group. When I push-forward the coordinate vector fields $\partial_\mu$ I do get directional partial derivatives with respect to the generators, i.e. $\frac{\partial}{\partial G_\mu}$, so these seem to span the tangent space.

However from Lie group theory I know and understand that the generators $G_\mu$ themselves span the tangent space, and it's not quite clear to me how to match that with the partial derivatives. Is this a different definition of tangent vectors that is used here?

Next, I'm not sure how the coordinate differentials $dx_\mu$ are pulled back to the group to give sensible results. In the end what I would like to achieve is to transfer the lie algebra of the group to an algebra on the representation with vector fields and differentials. How is that done?

I'm probably missing something very fundamental here, but my attempts to understand only made my confusion greater. My textbooks are very unclear on these aspects.

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    $\begingroup$ I don't understand what you're asking. $\endgroup$ – Qiaochu Yuan Nov 16 '13 at 9:33
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    $\begingroup$ I'm asking how to make the standard definition of the tangent space on the Lie manifold using derivatives agree with the standard definition of the tangent space on the Lie group as generators from the Lie algebra. I don't see how they match up. $\endgroup$ – Dirk Dinther Nov 16 '13 at 10:48
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I'm a student in physics so my notation/terminology may be a bit off compared to what you're used to, but I'd like to give it a go. Note I'm only going to answer part of your question.

If you're talking about vector fields $\partial_{\mu}$ on the tangent space of your group manifold ($M(G)$) I don't believe they would have the Lie algebra of your group. In order to do that, I believe you would need to add a connection $\Gamma$ through use of the covariant derivative. This has the effect of adjusting your derivative to compensate for the geometry of the Group manifold. For example consider some covector field (which we can call a 1-form A). Then the Covariant derivative in component notation is stated as:

$$\nabla_{\mu}A_{\alpha}=\partial_{\mu}A_{\alpha}-\Gamma_{\mu\alpha}^{\beta}A_{\beta}$$

This might not seem very useful but something interesting happens when we consider the commutator of such derivatives.

$$[\nabla_{\mu},\nabla_{\nu}]A_{\alpha}-\nabla_{[\mu,\nu]}A_{\alpha}=A_{\beta}R_{\alpha\mu\nu}^{\beta}$$

Where $R_{\alpha\mu\nu}^{\beta}$ is non other than the Riemann curvature tensor. This is actually how Riemann originally defined the curvature tensor that bears his name. For the choice of coordinate vector fields for $\mu,\nu$ the second term on the left side vanishes and we have (All of these symmetries are on Wipedia's page ):

$$[\nabla_{\mu},\nabla_{\nu}]A_{\alpha}=A_{\beta}R_{\alpha\mu\nu}^{\beta}$$

In coordinate free notation, we can write:

$$R(\mu,\nu)\alpha=[\nabla_{\mu},\nabla_{\nu}]\alpha$$

The Curvature tensor also obeys the following properties:

$$(\nabla_{\mu}R)(\nu,w)+(\nabla_{\nu}R)(w,\mu)+(\nabla_{w}R)(\mu,\nu)=0$$

$$R(\mu,\nu)=-R(\nu,\mu)$$

Which means that:

$$[\nabla_{\mu}\nabla_{\nu}]=-[\nabla_{\nu},\nabla_{\mu}][\nabla_{\mu},\nabla_{\mu}]=0$$

$$[\nabla_{\mu},[\nabla_{w},\nabla_{\nu}]]+[\nabla_{\nu},[\nabla_{\mu},\nabla_{w}]]+[\nabla_{w},[\nabla_{\nu},\nabla_{\mu}]]=0$$

So the covariant derivatives form a Lie algebra. This shouldn't be too suprising because a Riemannian manifold is required to be locally isomorphic to euclidean space (say dimension=3 then locally SO(3)) however the curvature ends up acting as the structure factor (ie a topological obstruction to being a euclidean manifold) which in turn seems to determine the topological group of the manifold at large (in the case of a Lie manifold anyway). Or as Zee says (Group theory in a nutshell for physicists):

The structure constants determine the Lie algebra, which essentially determines the Lie group.

A final clue as to forming group elements from this comes from examining the exponentiated covariant derivative:

$$exp(u\nabla)exp(v\nabla)=exp(\{\mu\nabla+v\nabla\}+\frac{1}{2}[\mu\nabla,\nu\nabla]+.....)$$

Which is just an expansion of the Cambell Baker Hausdorf formula. This is how I see it the two concepts you mentioned connecting, but hey, I could be wrong.

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