4 added 828 characters in body; added 334 characters in body; deleted 14 characters in body

I don't think that

Note: As explained below, there is a clash of nomenclature between what you write Morita calls a Maurer--Cartan form and what Cartan introduced (which is quite rightdescribed in the wikipedia page, say).

This is in response to one of Anirbit's comments. In Morita's book Geometry of Differential Forms, he calls any left-invariant form on $G$ a Maurer--Cartan form. I don't think that this is standard. For me, as my answer above, the Maurer--Cartan form is Lie algebra valued. The two notions of Maurer--Cartan forms can of course be reconciled. Choose a basis $(e_i)$ for $\mathfrak{g}$ and a canonical dual basis $e^i$ for $\mathfrak{g}^*$. Let $\omega^i$ be the left-invariant one-form which agrees with $e^i$ at the identity. Then $\omega = \sum_i \omega^i e_i$ is what I have been calling the (left-invariant) Maurer--Cartan form.

While I'm at it, let me explain the nature of my factors of $2$, since that seems also to be in dispute. For me the wedge product is defined as follows $\alpha \wedge \beta := \alpha \otimes \beta - \beta \otimes \alpha$, without a factor of $\frac12$.

3 Why the missing brackets?!

I don't think that what you write is quite right.

First of all there are two Maurer-Cartan forms: left-invariant and right-invariant. They are one-forms with values in the Lie algebra. If we identify the Lie algebra (=left-invariant vector fields) with the tangent space at the identity, then the left-invariant MC form $\omega$ is such that acting on a vector field $\xi$ on $G$ gives for all $g \in G$, $$\omega(\xi)_g = (L_g)_*^{-1} \xi_g,$$ where $L_g$ means left multiplication by $g\in G$. There is a also a right-invariant one-form defined similarly but using right multiplication.

Now suppose that $\xi$ is a left-invariant vector field on $G$. This means that $$\xi_g = (L_g)_* \xi_e,$$ where $\xi_e$ is the value of $\xi$ at the identity $e\in G$. In that case, $$\omega(\xi)_g = (L_g)_*^{-1} (L_g)_* \xi_e = \xi_e,$$ which is constant, since it does not depend on $g$.

Now, as you point out, if $X$ and $Y$ are left-invariant vector fields, then it is immediate that $\omega$ satisfies the structure equation: $$d\omega(X,Y) = -\omega([X,Y]).$$

Now choose a basis $(e_i)$ for the Lie algebra, so that we can write $\omega = \sum_i \omega^i e_i$, where the $\omega^i$ are one-forms on $G$. Notice that $\omega(e_i)=e_i$, whence $\omega^j(e_i) = \delta^j_i$.

Applying the structure equation to $X=e_i$ and $Y=e_j$ you see that, on the one hand, $$d\omega(e_i,e_j)=-\omega([e_i,e_j]) = - [e_i,e_j] = - f_{ij}{}^k e_k,$$ whence $$d\omega^k(e_i,e_j) = f_{ij}{}^k.$$ But this is precisely the result of applying $$-\tfrac12 \sum_{i,j} f_{ij}{}^k \omega^i \wedge\omega^j$$ on $e_i$ and $e_j$, hence the identity $$d\omega^k = -\tfrac12 \sum_{i,j} f_{ij}{}^k \omega^i \wedge\omega^j.$$

To write down explicitly the Maurer-Cartan forms, it is not hard. You have to compute the derivative of $L_g$ in your chosen coordinates. It is particularly easy if the group $G$ is a matrix group, in which case you have $\omega_g = g^{-1}dg$ and again you have to compute this in your favourite coordinates for $G$.

I just realised that I forgot to answer the bit about the second form of the structure equation. That equation is usually confusing at first because the notation hides the fact that $[\omega,\omega]$ also involves the wedge product of one-forms. By definition, $[\omega,\omega]$ is the Lie-algebra valued 2-form on $G$ whose value on vector fields $X,Y$ is given by $$[\omega,\omega](X,Y) = [\omega(X),\omega(Y)] - [\omega(Y),\omega(X)] = 2 [\omega(X),\omega(Y)].$$ If you now take $X=e_i$ and $Y=e_j$, left-invariant vector fields, you see that $$-\tfrac12 \omega,\omega [\omega,\omega](e_i,e_j) = -[e_i,e_j] = -\sum_k f_{ij}{}^k e_k,$$ agreeing again with $d\omega(e_i,e_j)$.

2 added 698 characters in body

I don't think that what you write is quite right.

First of all there are two Maurer-Cartan forms: left-invariant and right-invariant. They are one-forms with values in the Lie algebra. If we identify the Lie algebra (=left-invariant vector fields) with the tangent space at the identity, then the left-invariant MC form $\omega$ is such that acting on a vector field $\xi$ on $G$ gives for all $g \in G$, $$\omega(\xi)_g = (L_g)_*^{-1} \xi_g,$$ where $L_g$ means left multiplication by $g\in G$. There is a also a right-invariant one-form defined similarly but using right multiplication.

Now suppose that $\xi$ is a left-invariant vector field on $G$. This means that $$\xi_g = (L_g)_* \xi_e,$$ where $\xi_e$ is the value of $\xi$ at the identity $e\in G$. In that case, $$\omega(\xi)_g = (L_g)_*^{-1} (L_g)_* \xi_e = \xi_e,$$ which is constant, since it does not depend on $g$.

Now, as you point out, if $X$ and $Y$ are left-invariant vector fields, then it is immediate that $\omega$ satisfies the structure equation: $$d\omega(X,Y) = -\omega([X,Y]).$$

Now choose a basis $(e_i)$ for the Lie algebra, so that we can write $\omega = \sum_i \omega^i e_i$, where the $\omega^i$ are one-forms on $G$. Notice that $\omega(e_i)=e_i$, whence $\omega^j(e_i) = \delta^j_i$.

Applying the structure equation to $X=e_i$ and $Y=e_j$ you see that, on the one hand, $$d\omega(e_i,e_j)=-\omega([e_i,e_j]) = - [e_i,e_j] = - f_{ij}{}^k e_k,$$ whence $$d\omega^k(e_i,e_j) = f_{ij}{}^k.$$ But this is precisely the result of applying $$-\tfrac12 \sum_{i,j} f_{ij}{}^k \omega^i \wedge\omega^j$$ on $e_i$ and $e_j$, hence the identity $$d\omega^k = -\tfrac12 \sum_{i,j} f_{ij}{}^k \omega^i \wedge\omega^j.$$

To write down explicitly the Maurer-Cartan forms, it is not hard. You have to compute the derivative of $L_g$ in your chosen coordinates. It is particularly easy if the group $G$ is a matrix group, in which case you have $\omega_g = g^{-1}dg$ and again you have to compute this in your favourite coordinates for $G$.

I just realised that I forgot to answer the bit about the second form of the structure equation. That equation is usually confusing at first because the notation hides the fact that $[\omega,\omega]$ also involves the wedge product of one-forms. By definition, $[\omega,\omega]$ is the Lie-algebra valued 2-form on $G$ whose value on vector fields $X,Y$ is given by $$[\omega,\omega](X,Y) = [\omega(X),\omega(Y)] - [\omega(Y),\omega(X)] = 2 [\omega(X),\omega(Y)].$$ If you now take $X=e_i$ and $Y=e_j$, left-invariant vector fields, you see that $$-\tfrac12 \omega,\omega = -[e_i,e_j] = -\sum_k f_{ij}{}^k e_k,$$ agreeing again with $d\omega(e_i,e_j)$.