Suppose that $G$ is a Lie group and $g \in G$ is a generic element. What does the notation "$dg$" refer to? Is it the differential of the function $G \to G$ given by left (or right) multiplication by $g$? Should I take an expression for $g$ in local coordinates and then look at the differential one forms $dx_i$ in some way?

The notes I'm using give this as an example: If $G=GL(n, \mathbb{R})$ and $A = (a_{ij}) \in GL(n, \mathbb{R})$, then $dA = (da_{ij})$.

  • $\begingroup$ Think of $g$ as the continuous map $g: G\to G$, given by $h\mapsto g\cdot h$, left multiplication by $g$. Then $dg$ is nothing but the differential of this map (i.e. 1-form). $\endgroup$ – Hamed May 1 '16 at 21:45
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    $\begingroup$ @Hamed: I think you are not respecting the conceptual difficulty of the question. A priori, the differential of multiplication by $g$ at a point $x \in G$ is a linear map $dg_x : T_x(G) \to T_{gx}(G)$. You need to say more about how you're extracting a 1-form from this. One option is to pick a trivialization of the tangent bundle of $G$ (e.g. by left or right multiplication), so that you can interpret $dg$ as a $\mathfrak{g}$-valued $1$-form. But you should make the choice of trivialization explicit, e.g. it really does matter whether you trivialize using left or right multiplication. $\endgroup$ – Qiaochu Yuan May 1 '16 at 21:49

I think the main case of use for this notation is for matrix groups. As in your example, if $G\subset GL(n,\mathbb R)$ is a closed subgroup, then you can view the inclusion of $G$ as a smooth function $g:G\to M_n(\mathbb R)$. Taking the exterior derivative, you get $dg\in\Omega^1(G,M_n(\mathbb R))$ and your example is exactly this form for $G=GL(n,\mathbb R)$. This is mainly used to express the Maurer-Cartan forms (which encode the trivializations of $TG$ by left or right translations). Since these translations are the restrictions of linear maps, it follows that the left Maurer Cartan form can be written as $g^{-1}dg$ (i.e. to compute its action on a tangent vector, apply $dg$ and then multiply from the left by the inverse matrix of $g$). For the right Maurer Cartan form, one similarly obtians $(dg)g^{-1}$. I think these notations for the Maurer Cartan forms are sometimes also used for general Lie groups, but in my opinion this is rather confusing.


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