Notation : Here $O(k)$ denotes the orthogonal group, $V_k(\mathbb{R}^n)$ the Stiefel manifold of $k$ orthonormal-frames and define $\sigma : V_k(\mathbb{R}^n) \times V_k(\mathbb{R}^n) \longmapsto \mathbb{R}$ defined by $\sigma((v_1,\cdots,v_k),(v_1',\cdots v_k')) = \lvert \lvert \text{det}(v_i| v_j')_{i,j=1}^k \rvert \rvert$.
Note that $(v_i| v_j')$ is just a notation for the $ij-th$ element of the matrix, where given two vectors, $(v|w)$ denotes the dot product (in this case the standard one) of vectors $v,w$.
Let's give a right-action of $O(k)$ on $V_k(\mathbb{R}^n)$ defined as following : given $A = (a_{ij})_{ij} \in O(k)$ we have $(v_1,\cdots,v_k)\cdot A = (w_1,\cdots w_k)$ where $w_j = \sum\limits_{i=1}^k a_{ij}v_i$
I'd like to prove that $\sigma((v_1,\cdots,v_k)\cdot A),(v_1',\cdots,v_k')\cdot B)) = \sigma((v_1,\cdots,v_k),(v_1',\cdots,v_k'))$, i.e that $\sigma$ factors through $G_k(\mathbb{R}^n) \times G_k(\mathbb{R}^n)$.
I think the trick here is to show that $\sigma((v_1,\cdots,v_k)\cdot A),(v_1',\cdots,v_k')\cdot B)) = \lvert \text{det}(A) \text{det}(B)\rvert \cdot \sigma((v_1,\cdots,v_k),(v_1',\cdots v_k'))$ and then conclude since $ \lvert \text{det}(A) \text{det}(B)\rvert = 1$ since $A,B \in O(k)$.
Unfortunately I'm unable to show the first equality above, any help would be appreciated.
Let $V$ denote the matrix whose columns are $v_1,\dots,v_k$ and $V'$ the matrix whose columns are $v_1',\dots,v_k'$. We can write $$ \sigma((v_1,\dots,v_k),(v_1',\dots,v_k')) = |\det(V^TV')|, $$ where $V^T$ denotes the transpose of $T$. Similarly, if $(v_1,\cdots,v_k)\cdot A = (w_1,\cdots w_k)$ and $W$ denotes the matrix whose columns are $w_1,\dots,w_k$, then $$ W = VA. $$ With that, we have $$ \begin{align} \sigma((v_1,\cdots,v_k)\cdot A),(v_1',\cdots,v_k')\cdot B)) &= |\det[(VA)^T(V'B)]| \\ & = |\det[A^T(V^TV')B]| \\ & = |\det A^T| \cdot |\det (V^TV')| \cdot |\det B| \\ & = |\det(V^TV')| = \sigma((v_1,\cdots,v_k)),(v_1',\cdots,v_k'))), \end{align} $$ which was what we wanted.