Nonvanishing form in $\Omega^3(O(3))$ This is an old exam question which most of it I understand now due to the comments below. I still have two concerns. 

Define a $p$-form on $GL(n)$ as follows. 
  $$\Theta_p=tr(X^{-1}dX \wedge X^{-1}dX \wedge \cdots \wedge X^{-1}dX)$$
  (i) Restrict $\Theta_3$ to $O(3)$.  Writing the matrix $X$ with orthonormal column vectors $\mathbf{x_1, x_2,x_3}$, show that $X^{-1}dX$ is a skew symmetric matrix whose $i,j$th entry is $\mathbf{x}_i\cdot d\mathbf{x}_j$.
(ii) Show that at $X=I$, the forms $\mathbf{x_1} \cdot \mathbf{dx_2}, \mathbf{x_2 \cdot dx_3},$ and $\mathbf{x_3, \cdot dx_1}$ are linearly independent and $\Theta_3$ is nonzero. 
(iii) Deduce that since $(AX)^{-1}d(AX)=X^{-1}dX$, $\Theta_3$ is non vanishing at all points. 


(a) why can we just restrict $\Theta_3$ to $O(3)$? How do we know this form is still smooth? i.e. the induced map to $O(3) \rightarrow \wedge^*T^*O(3)$ is still smooth? 
(b) Why does the equality in (iii) show $X^{-1}dX$ is left invariant? 
I am also unsure what this means, since we are working with $1$-forms rather than vector fields. I suppose we are to show:
 $$(L_{A^{-1}}^*) X^{-1}dX_I = X^{-1}dX_A $$ 
 A: Yes, the wedge-product on $\Omega^\bullet(GL(n;\mathbb{R}))$ (or $\mathbb{C}$) and the matrix product on $\mathbb{R}^{n\times n}$ together defines a product on the tensor product of rings $\Omega^\bullet(GL(n;\mathbb{R}))\otimes_{\mathbb{R}}\mathbb{R}^{n\times n}$ which is denoted also by $\wedge$, and your formula is correct.  Here $X\colon GL(n;\mathbb{R})\to\mathbb{R}^{n\times n}$ is just the inclusion.
(1) $\Theta_p$ is a $p$-form on $GL(n)$ so it restricts (actually pulls back) to $O(n)\subset GL(n)$ under the usual inclusion.  Since $X^{-1}=X^T$ for $X\in O(3)$, the form $X^{-1}\,\mathrm{d}X$ is
$$
X^{-1}\,\mathrm{d}X=(\mathbf{x}_1\,\mathbf{x}_2\,\mathbf{x}_3)^T(\mathrm{d}\mathbf{x}_1\,\mathrm{d}\mathbf{x}_2\,\mathrm{d}\mathbf{x}_3)=(\mathbf{x}_i^T\mathrm{d}\mathbf{x}_j)=(\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j).
$$
To show this is skew-symmetric, remember $(X^T)\,\mathrm{d}X+\mathrm{d}(X^T)X=0$ on $O(3)$.  Taking the $(i,j)$th component yields:
$$
0=\sum_k(X^T)_{ik}(\mathrm{d}X)_{kj}+(\mathrm{d}X^T)_{ik}X_{kj}
=\sum_kX_{ki}\,\mathrm{d}X_{kj}+X_{kj}\,\mathrm{d}X_{ki}
=\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j+\mathbf{x}_j\cdot\mathrm{d}\mathbf{x}_i.
$$
(2) At identity, $\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j=\mathrm{d}X_{ij}$, so the three forms are linearly independent.  We have $\Theta_3$ is
$$
\sum_{i,j,k}(\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j)\wedge(\mathbf{x}_j\cdot\mathrm{d}\mathbf{x}_k)\wedge(\mathbf{x}_k\cdot\mathrm{d}\mathbf{x}_i)
$$
the summands are nonzero only if $i,j,k$ pairwise distinct, so this is sum over permutations of $1,2,3$.  Show that $(\mathbf{x}_i\cdot\mathrm{d}\mathbf{x}_j)\wedge(\mathbf{x}_j\cdot\mathrm{d}\mathbf{x}_k)\wedge(\mathbf{x}_k\cdot\mathrm{d}\mathbf{x}_i)$ is invariant under all permutations (which you should be able to do).  So we conclude, at identity,
$$
\Theta_3=6(\mathbf{x}_1\cdot\mathrm{d}\mathbf{x}_2)\wedge(\mathbf{x}_2\cdot\mathrm{d}\mathbf{x}_3)\wedge(\mathbf{x}_3\cdot\mathrm{d}\mathbf{x}_1)
\neq 0.
$$
(3) Since $(AX)^{-1}\,\mathrm{d}(AX)=X^{-1}A^{-1}A\,\mathrm{d}X=X^{-1}\,\mathrm{d}X$, the matrix of $1$-forms $X^{-1}\,\mathrm{d}X$ is left-invariant (on $GL(n)$).  Hence $\Theta_3$ is invariant on $GL(3)$ and hence on $O(3)$.  Together with (2), this shows $\Theta_3$ must be nonvanishing everywhere on $O(3)$.
A: HINT for (ii): The tangent space to O(3) at the identity matrix consists of all skew-symmetric matrices. If $A$ is in that tangent space, show that $(\mathbf x_i\cdot d\mathbf x_j)(A) = a_{ij}$.
Then recall that forms $\theta_1,\dots,\theta_k$ are linearly independent if and only if $\theta_1\wedge\dots\wedge\theta_k\ne 0$.
