7
$\begingroup$

Given a map $f:M_n(k)\to k$ (with $k$ some field) such that $f(AB)=f(A)f(B)$ for all matrices $A$ and $B$, is it necessarily the case that $f$ factors through the determinant, i.e. does there exist a multiplicative map $g:k\to k$ such that $f=g\circ\det\,$? Are constraints on $k$ necessary?

A simple corollary would be that nonzero multiplicative maps on subgroups of the general linear group $GL_n(k)$ factor through multiplicative maps on the units $k^\times\to k^\times$.

Two definitions of $\det$ I'm aware of (written with our setting in mind):

  • The unique alternating mulilinear map (of column vectors in a matrix) sending $I$ to $1_k$.
  • The trace of the map induced by $A$ on the $n$th exterior power $\mathrm{Alt}^nk^n$.

By Gaussian elemination, any multiplicative map $f$ from matrices to the base field is determined by its values on the matrices representing elementary row operations and upper triangular matrices.

One stumbling block is that it seems hard, in general, to fully characterize the multiplicative maps on the base field $k$. With a finite field it would just be integer powers and the zero map, but with the reals you get all (positive) real powers too, and some funky stuff may occur with other fields.

  • 2
    For a relatively wide range of fields $k$ and integers $n$ we have that $SL_n(k)$ is the commutator subgroup of $GL_n(k)$. Your mapping $f$ sends all the commutators to $1_k$, so given that the determinant distinguishes the cosets in $GL_n(k)/SL_n(k)$, the answer to your question seems to be *YES*, whenever $GL_n(k)'=SL_n(k)$.2012-06-15
  • 0
    @JyrkiLahtonen It's always the case that $GL'=SL$, right? A nonzero multiplicative map $GL\to k$ descends to a homomorphism $GL/SL\to k^\times$, and $\det$ descends to an isomorphism $GL/SL\cong k^\times$, hence the $f$ will factor through $\det$, so that takes care of that. How do we extend this to all of $M_n$ - Zariski topology maybe (which I know next to nothing about)?2012-06-15
  • 0
    No, it's not always true that $SL$ is the commutator subgroup of $GL$. For example, if $k=\mathbb{F}_2$, we have that $SL_2(k)=GL_2(k)\simeq S_3$, but $S_3$ is not its own commutator subgroup.2012-06-15
  • 0
    Ah I see. $SL=\mathrm{ker}\,\det$ and in general $GL'\subseteq SL$.2012-06-15
  • 0
    Extending the argument to singular matrices is tricky, but might work. Aren't all the rank $n-1$ matrices in the same orbit of $GL_n(k)$? So if $f$ takes a non-zero value on one of them, it will do so on all of them. But we get lower rank matrices as products of diagonal rank $n-1$ matrices and can keep going, so $f(A)\neq0$ for all $A$. But if $f(\mathrm{diag}(1,1,\ldots,1,0))\neq0$, then we run into a well-definedness problem, because we can multiply that matrix with $\mathrm{diag}(1,1,\ldots,1,x)$ without changing the product. It seems to me that then $f$ has to be constant $=1$???2012-06-15
  • 0
    @Jyrki: I've had a flash of geometric insight as to why your rank claims are true, and your argument looks correct to me. If you edit it into your answer I will be happy to accept it.2012-06-26
  • 0
    I'm afraid I'm no longer certain about the rank $n-1$ matrices forming a single orbit. With $GL_n$ acting from the left each orbit contains a unique matrix in reduced row echelon form. From the right I guess we get the transpose. May be 2-sided action? Anyway, I haven't given up on this.2012-07-01

1 Answers 1