4
$\begingroup$

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 $dA = (da_{ij})$.

Improve this question
$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ @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
1
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.