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 $$ (\Omega_{\alpha\beta}^\gamma v)_{ij} = e_i\cdot F(\omega_{\alpha\beta}^\gamma\wedge e_j) = e_i\cdot(\omega_{\alpha\beta}^\gamma\times e_j). $$
As mentioned on other answers, the symbol represents the exterior product of two vectors. I think it is a slight abuse of language, because that is a skew-symmetric matrix representing the cross-product of those vectors (and in most references, you'll find the notation $[\omega\times]$ or $[\omega]_{\times}$ instead, even if I find Groves to be one of the best books on this topic).
Basically, $\Omega\,v = \omega \times v$, and in the context of inertial navigation (the topic of that book) the main reason why you want to use matrices here instead of just representing the operation between both vectors, is that this matrix is a representation of the so(3) Lie algebra, and if you represent the attitude with a SO(3) (rotation) matrix $R$, then the exponential map between the algebra and the group allows you to write the differential equation $\dot R = \Omega R$, where $\omega$ would be the pre-integrated gyroscope signal.
Another reason why it may be helpful to write the cross-product operation as a matrix, is that in navigation and control you often need to write Jacobian matrices, and the partial derivatives of the cross-product are (using the book's notation) $\frac{\partial}{\partial v} \omega\times v = [\omega \wedge]$ and $\frac{\partial}{\partial \omega} \omega\times v = -[v \wedge]$.
