I have an elementary question regarding the exterior derivative which confuses me. I am reading Theorem 2.1.13 of the book Aspects of Multivariate Statistical Theory by Robb J Muirhead. The part that confuses me is as follows.
$h_1, h_2,\dots,h_m,h_{m+1},\dots,h_n$ are column matrices of the same dimension of variables. Equation (22), as shown pasted below, then writes $$\bigwedge_{j=m+1}^n\bigwedge_{i=1}^m h_j'dh_i,$$ where $'$ indicating transpose of a matrix. Each $dh_i$ appears repeatedly in the exterior product over $j$. Should that imply this expression vanishes? Yet, the author does not seem to claim so.
Below is the aforementioned theorem and the part of the proof I am confused over.
In the proof, from the sentence "First consider the matrix ..." to Eq. (22), $H$ is taken as a matrix without imposing the condition $H'H=I$.
As an example, let $n=3,\, m=1$ and $H=[h_1,h_2,h_3]$ where $h_i$ is of dimension $3$. The double exterior derivative product in Eq. (22) becomes $$\big(h_{1,2}\,dh_{1,1}+h_{2,2}\,dh_{2,1}+h_{3,2}\,dh_{3,1}\big)\wedge \big(h_{1,3}\,dh_{1,1}+h_{2,3}\,dh_{2,1}+h_{3,3}\,dh_{3,1}\big).$$ The associative rule does apply, right? Is the following identity wrong? Why? $$h_{1,2}\,dh_{1,1}\wedge h_{1,3}\,dh_{1,1}=h_{1,2}h_{1,3}(dh_{1,1}\wedge dh_{1,1}).$$
I would appreciate an elucidation.
No, definitely not. Remember that $h_i$ is a vector in $\Bbb R^n$, and so $dh_i$ will have linearly independent entries in each slot. In particular, $H_1$ (in the source's notation) is an $n\times m$ matrix with orthonormal column vectors. Note that this giant wedge product will be an $m(n-m)$-form built out of all the $dx_{ki}$, where $x_{ki}$ is the $k$th entry of the column $h_i$.
Also, note that the order (and range) of the author's wedge product is different from yours. He has $1\le i\le m$ and then, for each $i$, $i+1\le j\le n$. Regarding your example, here, then $n=3$ and $m=1$. Now $h_1$ is a unit vector in $3$-space, so it varies over the $2$-sphere; we expect to get a nonzero $2$-form here. And, in fact, we do: Remember that what you've written makes no sense. You need $h'_2dh_1 = dh_1\cdot h_2$, which is the dot product of $h_2$ with $dh_1$; since $h_2$ and $h_3$ are orthogonal, they give a basis for the tangent plane of the sphere at the point $h_1$, and this wedge product is precisely the area $2$-form for the sphere. (You are, I think, forgetting about the vector nature of everything here.)