I was reading Schutz's book on General Relativity. In it, he says that a(n) $M \choose N$ tensor is a linear function of $M$ one-forms and $N$ vectors into the real numbers.
So does that mean the determinant of an $n \times n$ matrix is a $0 \choose n$ tensor because it is a function that maps the $n$ column vectors of the matrix to a real number (the value of the determinant)?
But then, the determinant also maps the $n$ column vectors of the matrix to the same real number (the value of the determinant).
So would the tensor representation of the determinant be different if you choose the map for the column vectors than the map for the row vectors?
Yes, in fact the determinant $\det:(\mathbb R^n)^n\to \mathbb R$ is (up to constant multiple) the only alternating $n$-multilinear map (i.e. alternating $n$-covariant tensor, see John Lee, Introduction to Riemannian Manifolds, 2nd edition (2018), pages 400 and following). See for instance this question for a proof of this fact.