# Maurer-Cartan Form on orthogonal group

I'm having trouble understanding an assignment regarding the Maurer-Cartan form on orthogonal matrices:

Let $\text{O}(n)\subset\text{GL}(n,\mathbb{R})$ be the matrix Lie group of orthogonal matrices. Given $A\in\text{O}(n)$ let $A_i$ denote the $i$-th column vector. Writing $A=(A_1,\ldots A_n)$, we see that the elements $A\in\text{O}(n)$ are precisely the orthonormal bases $\{A_1,\ldots A_n\}$ of $\mathbb{R}^n$. Let $\pi_i:\text{SO}(n)\to\mathbb{R}^n$ be the map given by $\pi_i(A)=A_i$. Define 1-forms $\omega^i_j\in\Omega^1(\text{O}(n))$ as follows: given $v\in T_A\text{O}(n)$ $$d\pi_i(v)=:\omega^j_i(v)A_j$$

The assignment then splits to 4 parts which eventually lead to one concluding that $\omega=(\omega^i_j)$ is in fact the Maurer-Cartan form on $\text{O}(n)$. I can go into more detail with this if the added context helps to understand my issue.

This is a homework assigment, so I would prefer no solutions. I'm just having trouble understanding the definition of $\omega=(\omega^i_j)$. A couple questions:

• Shouldn't the domain of $\pi_i$ be $\text{O}(n)$ instead of $\text{SO}(n)$? Is this a typo?
• What is $\omega^j_i(v)$? it seems apparent to me that $\omega^j_i(v)$ is just a scalar, but I know that, regarding manifolds, the "$j$'' could imply that one sum's along the terms $\omega^j_i(v)A_j$, with the Einstein convention.
• Lastly, if $\omega^j_i(v)$ were just a scalar, I believe that $j$ must equivalent to $i$, since it's evident to me that $d\pi_i(v)$ simply projects vectors in $\text{O}(n)$ onto their $i$th column. But then $\omega^j_i$ would not be defined for $j\neq i$.

It would be fantastic if I could get some help clarifying, and I would be glad to give more detail on the question if needed. I feel like half the work in differential geometry is understanding the notation.

Also, are there any texts which cover the Maurer-Cartan form thoroughly? Our texts (Lee - Introduction to Smooth Manifolds/D.Carmo - Differential Forms and Applications/Spivak - A Comprehensive Introduction to Differential Geometry) don't seem to cover it and I would really like a reference text.

This is an example of a very general construction for matrix groups. Each group element $g$ is a matrix that depends on on some parameters (the co-ordinates on the group manifold $G$) . Let $dg$ be the matrix whose entries are the exterior derivative of the entries with respect to the parameters, then the left-invariant Muarer-Cartan form is the object $g^{-1}dg$. It is a matrix-valued one form on the group manifold.

Although constructed from elements of the group, the matrices $g^{-1}dg$ lie in the Lie algebra (for example, in the orthogonal group $g$ obeys $g^T=g^{-1}$ while $g^{-1}dg$ evaluated on any vector in $T(G)$ is a skew-symmetric matrix) so we can write $g^{-1}dg = \omega^a \lambda_a$ where $\lambda_a$ form a basis for the Lie algebra. The scalar-valued one forms $\omega_a$ are also refered to as Maurer-Cartan forms and are dual to the Lie algebra: If $L_a$ is the left-invariant vector field on the group manifold corresponding to the basis element $\lambda_a$, then $\omega^a(L_b)=\delta^a_b$ at all points in $G$.

Yes, the domain should be $O(n)$.

Each $\omega_i^j$ is a one-form, so it sends a single argument to a number. The argument is a vector in some space (in this case, $T_A(O(n))$).

The formula given does include an implied summation over $j$, I believe. Otherwise it'd be bound in the left side and un-bound in the right, and that makes no sense. Also, your conjecture about $j$ is part right: it's an index that over the same values as $i$.

Alas, I have no recommendation for a good book that discusses M-C forms. Frankly, I've never really understood them myself. Fortunately, I've never needed to. :)

Answer to followup question: You ask "isn't $\omega_i^j = 0$ for $j \ne i$?", and I don't know. But I figure I can work it out with an example. Let's look at $SO(2)$, which consists of all matrices $$U(t) = \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix}$$

Since $t \mapsto U(t)$ defines a curve in this 1-dimensional group, we can take its derivative at $t = 0$ to find a nice example tangent vector at $A = U(0) = I$. I get $$U'(0) = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$$

(A slight generalization shows that the tangent space to $SO(n)$ at the identity consists of all skew-symmetric matrices, by the way.)

So now $A = I$ and $v$ is the matrix $U'(0)$ above. What about $\pi_1$ and $\pi_2$? They send a matrix to its first and second columns, respectively. The derivative of $\pi_1$ will therefore send $v$ to its first column, $\begin{bmatrix} 0\\1\end{bmatrix}$. And we need to express this in terms of the first and second column of the identity. I get this: \begin{align} d\pi_1(v) &=\begin{bmatrix} 0\\1\end{bmatrix}\\ &=0 \cdot \begin{bmatrix} 1\\0\end{bmatrix} + 1 \cdot \begin{bmatrix} 0\\1\end{bmatrix} \end{align} from which I conclude that $\omega_1^2 = 1$. A similar computation shows that $\omega_2^1 = -1$, but that $\omega_i^i = 0$ for $i = 1, 2$.

So your conjecture -- that $\omega_i^j$ is zero for $j \ne i$ -- turns out to be exactly wrong. :) And in fact I'll bet it's not too tough to show that the $\omega_i^j$ form a skew-symmetric matrix in general.

• Thank you! That definitely clears up a lot! One more question: If $$d\pi_i(v)=\omega^1_i(v)A_1+\cdots +\omega^2_i(v)A_n,$$ then wouldn't all $\omega^j_i(v)=0$ for all $j\neq i$? But then we can't have that, because if $\omega^j_i(v)\neq 0$, then $\omega^j_i(v)$ wouldn't take values in the skew-symmetric matrices. Since you said you're not very good with M-C forms, I understand if you're not sure. I just thought I would ask. Apr 4, 2015 at 22:09
• See additional answer. Thanks for asking. I should have worked this out for myself 35 years ago, but never did until now. :( Apr 5, 2015 at 13:56
• Thank you! I was definitely having a mental block about the former question, but you helped break through it quite nicely. I sincerely appreciate you taking the time out of your day to help me. Apr 5, 2015 at 14:45