Is the cross product a tensor or pseudotensor? I am confused as to the tensor nature of the cross product. Maybe I'm misunderstanding what I'm reading. But I seem to be getting conflicting messages.
 A: The definition of 'tensor' is often different to physicists and mathematicians. To a mathematician a tensor is a multilinear object - an element of a tensor product space. A cross product is a vector, therefore it's a tensor.
To a physicist it's particularly an object which transforms tensorially under changes of coordinates, ie, with one copy of the coordinate transformation matrix per index. A cross product doesn't transform like a vector in the standard $(x,y,z)$ basis, but instead like a thing called a pseudovector.
Pseudovectors and pseudotensors can also be represented as antisymmetric elements of a tensor product space. Cross products act like antisymmetric elements of $ \mathbb{R}^3 \otimes \mathbb{R}^3 $, which is a tensor product space, and when written in this space (with basis $(x \wedge y, y \wedge z, z \wedge x)$) they do transform tensorially. It's just when you write them as elements of $\mathbb{R}^3$ that they don't.
The coordinates of $a \times b$ are given by: $a_i b_j - a_j b_i$, which has two indices and therefore must transform with two copies of a coordinate transformation. However, when you map write pseudovectors as vectors, you map 'ij' to a single index ($x \times y \rightarrow z$)  using the Hodge star operation, even though it should still transform somehow with two copies of the coordinate transformation. Hence, it does not transform 'tensorially'.
Exterior algebra clears this all up. Cross products should be written in the basis $(x \wedge y, y \wedge z, z \wedge x)$, and transform indices with $\wedge^2 R$ if coordinates transform as $R$.
See also this question.
A: The cross product is a bilinear vector valued mapping hence is a tensor.
