Is the determinant the only multiplicative matrix function? Is there a matrix invariant or property that is multiplicative, i.e.,
$$f(AB) = f(A) f(B)$$
other than the determinant? In addition, some matrix norms are submultiplicative, but is there a supermultiplicative property? 
 A: With the exception of $GL_2(\mathbb{F}_2)$, the commutator subgroup of $GL_n(k)$ is $SL_n(k)$ (for $k$ a field). So any multiplicative map from $GL_n(k)$ to an abelian group factors through determinant. This is with no hypotheses on continuity.
EDIT Johannes asks about noninvertible matrices. Let $M$ be an abelian monoid and $\mathrm{Mat}_{n \times n}(k) \to M$ a multiplicative map. Then $GL_n$ must map to the group of units of $M$, so $SL_n(k)$ must map to $1$ by the above (except for $n=2$, $k = \mathbb{F}_2$). Now, if $X$ and $Y \in \mathrm{Mat}_{n \times n}(k)$ are noninvertible matrices of the same rank then there are matrices $U$ and $V$ in $SL_n(k)$ with $UXV=Y$. So our function must be constant on functions of the same rank. Let $e_r$ be its value on matrices of rank $r$. Now, rank $n-1$ idempotents exist. So $e_{n-1}^2=e_{n-1}$. But any noninvertible matrix is a product of rank $n-1$ matrices, so $e_{n-k}$ is a power of $e_{n-1}$ and we deduce $e_{n-1} = e_{n-2} = \cdots = e_1 = e_0$. In short, a multiplicative map to an abelian monoid must take the same value on all noninvertible matrices.
A: It depends on what is the target space. Linear representations of ${\bf Gl}_n(k)$ do satisfy $\rho(AB)=\rho(A)\rho(B)$ by definition, and they often extend in a natural way to ${\bf M}_n(k)$.
On another hand, invariant scalar functions $f$ are necessarily of the form $f(A)=(\det A)^s$, $(\det A)_+^s$, $(\det A)_-^s$ or $|\det A|^s$.
