How do the components of a cross product transform? Let $x^{j}$ and $y^{k}$ be the components of two vectors $x,y\in \mathbb{R}^{3}$. According to the way the compontents of $x$ and $y$ transform when we change the basis, we know they are $(1,0)$-tensors.
But how do the components of the cross product $x\times y$ transform when we change the basis? I know that they are given by $c_{i}:=\epsilon_{ijk}x^{j}y^{k}$ (Edit: I forgot to say that I used the Einstein notation here).
This object has an index below, so I ask myself if it is a $(0,1)$-tensor. But I'm pretty sure thats not true. 
So can someone tell me what object this is and how its components transform when we change the basis? 
 A: Well, you're correct, in some sense, the cross-product of two vectors is not really a vector of the same type. The language used to distinguish the two types is:


*

*polar vectors: flip sign of component under coordinate inversion

*axial vectors: do not flip sign of component under inversion (also known as pseudovector see this wikipedia article)


All of this is in three dimensions. That is key, because what really underlies this is the correspondence between one and two forms in three dimension. Hodge duality allows us to exchange:
$$ adx+bdy+cdz \leftrightarrows a dy \wedge dz + a dz \wedge dx+ a dx \wedge dy$$
Of course, these are not tensors of the same type. We have type $(1,0)$ vs type $(2,0)$. Again, Hodge duality allows us to identify these. So, incidentally, there are also identifications of $p$ and $n-p$ forms in higher dimensions which generalize this willful obfuscation of polar and axial vectors.
Of course, exchanging $a\partial_x+b\partial_y+c\partial_z$ for $adx+bdy+cdz$ is also exploiting another isomorphism. In Euclidean space it doesn't seem like much, but, beware in physics when the Minkowski metric lurks about, there are pesky minus signs which appear from "raising" or "lowering" the indices. All of this is nicely captured by the so-called musical isomorphisms which use $\sharp$ or $\flat$ to denote the raising or lowering operation in a coordinate free fashion. See this wikipedia article or this useful MSE post
A: You'd best think of the cross product by first defining the following:
Let $\varepsilon:\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3\rightarrow\mathbb{R}$ be a (0,3)-type tensor that is totally antisymmetric. Such tensors form a one-dimensional subspace within the space of (0,3)-type tensors, therefore any such tensor (except the zero tensor) differ only by a scalar factor.
As such, the $\varepsilon$ tensor is uniquely determined by its value on a full set of basis vectors.
Let $E=\{e_1,e_2,e_3\}$ be the standard basis of $\mathbb{R}^3$, and let us define this basis to be positively oriented. Then let us define $\varepsilon$ to be $$ \varepsilon(e_1,e_2,e_3)=1. $$ We can think of the list $(e_1,e_2,e_3)$ to represent a positively oriented unit cube in $\mathbb{R}^3$, thus $\varepsilon$ maps oriented volumes to lists of vectors, since we want the volume of a positively oriented unit cube to be 1, and we know that oriented volumes of paralelepipeds are antisymmetric and multilinear in their edges.
If you let $\varepsilon$ act on the basis vectors in any permutation, rather than the positive permutation only, you get $$ \varepsilon(e_i,e_j,e_k)=\epsilon_{ijk} $$ obviously.
Now, letting $x,y$ and $z$ be arbitrary vectors, $\varepsilon(x,y,z)$ is linear in $z$, thus the map $z\mapsto\varepsilon(x,y,z)$ is a linear functional that depends on $x$ and $y$, and by the Riesz representation theorem there exists a unique $x\times y\in\mathbb{R}^3$ vector for which $$ \langle x\times y,z\rangle=\varepsilon(x,y,z).$$
Now, in components, the linear functional mentioned above is $\epsilon_{ijk}x^jy^k$, and the vector, which is the cross product is $g^{ij}\epsilon_{jkl}x^ky^l$.
However in $\mathbb{R}^3$, using the standard cartesian coordinates, $g^{ij}=\delta^{ij}$, thus the two are numerically equivalent.
