Are all pseudovectors in 3D cross products of vectors? I have been taught that all cross-products of two vectors in 3D are pseudovectors because they don't change direction under a parity transformation. But, are must all pseudovectors be cross products of vectors or can do pseudovectors that can't be written as cross products exist?
 A: I hope this naïve handwaving could give an insight. Suppose we have a non-zero pseudovector $\mathbf p$ (else, $\mathbf0 = \mathbf0\times\mathbf0$, and $\mathbf0$ is both a vector and a pseudovector). Find a non-zero vector $\mathbf q$ orthogonal to it, and a non-zero vector $\mathbf r$ othogonal to both of them—it should be possible as the space is three-dimensional. Then $\mathbf q\times\mathbf r$ is a nonzero pseudovector orthogonal to both $\mathbf q,\mathbf r$, and it should lie on a line $\mathbf p$ spans. So there should exist a scalar $s$ such that $\mathbf p = s(\mathbf q\times\mathbf r)$, so then $s\mathbf q$ and $\mathbf r$ (or $\mathbf q$ and $s\mathbf r$ etc.) are desired vectors.
Note that possible sign changes don’t affect orthogonality and value of $s$.
A: "can [do] pseudovectors that can't be written as cross products exist" needs clear definition.  If I have a pseudovector $\vec a$ I can complete an orthogonal triad with $\vec b$ and $\vec c$, then choose the lengths of $\vec b$ and $\vec c$ so $b \times \vec c=\vec a$.  I don't think that answers your question.  Another way to get a pseudovector is to multiply a vector by a pseudoscalar, but that just pushes the issue down the line.  Where do we get a pseudoscalar?  One way is by the triple product of three vectors, but that involves a cross product.  I can certainly define an object $\vec A$ to be a pseudovector and have certain components in a particular reference system.  Can that be written as a cross product by the process I described?
