I have seen similar post asking for interpretation of the Maurer-Cartan form, but I am still struggling to understand it, so let me try to work a specific example and pose a specific question.

Let $G$ be the group containing all matrices of the form

\begin{bmatrix} x &y \\ 0 &1/x \end{bmatrix}

where $x \neq 0$.

I read that the Maurer-Cartan form is defined as $\Omega = g^{-1} dg$ so I computed

$\Omega= \begin{bmatrix} 1/x &-y \\ 0 &x \end{bmatrix} \begin{bmatrix} dx &dy \\ 0 &-dx/x^2 \end{bmatrix}= \begin{bmatrix} dx/x &ydx/x^2+dy/x \\ 0 &-dx/x \end{bmatrix}$

So, for example, at the point $g= \begin{bmatrix} 5 &3 \\ 0 &1/5 \end{bmatrix} $ we have $\Omega_g = \begin{bmatrix} dx/5 &3dx/25+dy/5 \\ 0 &-dx/5 \end{bmatrix} $

But I read that at a specific point $g\in G$, the entity $\Omega_g$ is supposed to be a linear map from $T_gG$ to $T_eG$. Looking at what I computed above, I don't see how $\Omega_g$ is a linear map on $T_gG$. For example, what is $\Omega_g (\frac{\partial}{\partial x})$?

I'm also interested in the claim that $\Omega$ is left-invariant. I am able to show that the 1-forms appearing as entries of $\Omega$ are each left-invariant via explicit computation.

--Proof:

enter image description here

However, I have seen the following "direct" proof, applicable in general, which appears similar in spirit:

--Proof:

enter image description here

However, because $dh$ is not a 1-form in the usual sense, and we are not really working in some coordinates, I am not sure how I am supposed to interpret this "proof" besides recognizing the analogy. Why is the above proof true?

As for background, you may assume I understand Lee's Smooth Manifold text (2nd edition) chapters 1-14. Lee's treatment of 1-form is always real-valued, so this different kind of 1-form is throwing me off.

  • If you want to have a different perspective, the Maurer Cartan form is the equivalent of the derivative for the multiplicative calculus (look it up!). – geodude Jan 13 '15 at 10:26
  • It might help you to figure out what $T_gG$ actually is. Since $G$ is a subgroup of the $2\times 2$ matrices (which is an $\Bbb R^4$), we identify $T_gG$ with a $2$-dimensional subspace of the $2\times 2$ matrices. Note, moreover, that $T_gG = L_{g*}T_eG = L_{g*}\mathfrak g$. – Ted Shifrin Jan 13 '15 at 17:45
  • @TedShifrin I get it now. I was confused because I kept thinking that $G$ is two-dimensional, so I kept thinking elements of $T_g G$ as $2\times 1$ column vectors – suncup224 Jan 14 '15 at 8:00
up vote 1 down vote accepted

The matrix $dg$ is an element of $T_gG$ and is mapped to $T_eG$ by simple multiplication from the left by $g^{-1}$ which is of course a linear operation.

The Lie algebra of your group are the matrices of the form $$\begin{bmatrix}a & b\\ 0 & -a \end{bmatrix}$$ and in your example the resulting matrix $$\begin{bmatrix}dx/5 & 3dx/25+dy/5\\ 0 & -dx/5 \end{bmatrix}$$ is indeed in this form and belongs to $T_eG$.

To compute $\Omega_g(\partial_x)$, simply put $dx=1$ and $dy=0$ since these are the coordinates for the basis vector $\partial_x$.

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