While reading about coordinate transformations. I came across this
$\Omega^{\gamma}_{\beta,\alpha}=[\omega^{\gamma}_{\beta,\alpha}\wedge]$
What does the caret (or wedge) mean? In the book it looks more like a caret than a wedge.
Taken from Groves, Paul D Principles of GNSS, Inertial, And Multisensor Integrated Navigation Systems 2nd ed p. 45.
Thank you in advance.
The $\Omega$ matrix acts the same as if you take the outer product of the $\omega$ vector. Symbolically $\Omega A = \omega \wedge A$
or
$\Omega = [ \omega \wedge]$
The exterior algebra is a way of representing oriented subspaces of a vector space. Given two vectors $v, w$, the quantity $v\wedge w$ is called a bivector and represents an oriented plane. In three dimensions (and only three dimensions), the space of bivectors is also three dimensional. If $e_1, e_2, e_3$ is a basis of vectors, then $$ 1,\; e_1,\; e_2,\; e_3,\; e_2\wedge e_3,\; e_3\wedge e_1,\; e_2\wedge e_3,\; e_1\wedge e_2\wedge e_3 $$ is a basis for the exterior algebra. If we let $F$ be the identification $$ F(e_2\wedge e_3) = e_1,\quad F(e_3\wedge e_1) = e_2,\quad F(e_2\wedge e_3) = e_3, $$ and extend linearly to all bivectors, then $F(v\wedge w) = v\times w$ is exactly the cross product of vectors $v$ and $w$.
So assuming $\omega_{\alpha\beta}^\gamma$ are vectors, then they're saying that $$ \Omega_{\alpha\beta}^\gamma v = F(\omega_{\alpha\beta}^\gamma\wedge v) = \omega_{\alpha\beta}^\gamma\times v $$ for any vector $v$ (which is perfectly fine since $\wedge$ and $F$ are linear), and so $\Omega_{\alpha\beta}^\gamma$ has components $$ (\Omega_{\alpha\beta}^\gamma v)_{ij} = e_i\cdot F(\omega_{\alpha\beta}^\gamma\wedge e_j) = e_i\cdot(\omega_{\alpha\beta}^\gamma\times e_j). $$
