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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.

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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 $GL_n(k)'=SL_n(k)$. – Jyrki Lahtonen Jun 15 '12 at 7:41
@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)? – anon Jun 15 '12 at 7:50
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. – Jyrki Lahtonen Jun 15 '12 at 7:53
Ah I see. $SL=\mathrm{ker}\,\det$ and in general $GL'\subseteq SL$. – anon Jun 15 '12 at 7:56
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 $=1$??? – Jyrki Lahtonen Jun 15 '12 at 8:30

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 $GL_n(k)'=SL_n(k)$, and we restrict the domain of $f$ to invertible matrices.

Lemma in N. Jacobson, Basic Algebra I, p. 377, tells us that $SL_n(k)$ is its own commutator subgroup (and obviously then also equal to $GL_n(k)'$) unless $n=2$ and $|k|=2$ or $3$. In the case $n=2$, $k=\mathbb{F}_2$ we have that $GL_2(k)=SL_2(k)\simeq S_3$ meaning that $GL_2(k)'$ is an index two subgroup of $SL_2(k)$. Using the ideas of the proof of this result it is easy to see that in the case $k=\mathbb{F}_3$ we do get that $GL_2(k)'=SL_2(k)$ (but $SL_2(k)$ is not its own commutator subgroup).

So we know that the restriction of your mapping $f$ to $GL_n(k)$ must factor via the determinant except possibly in the case $n=2,k=\mathbb{F}_2$.

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I had a vague recollection that the list of exceptions to $GL_n(k)'=SL_n(k)$ is finite, but had to look it up. Sorry about that. – Jyrki Lahtonen Jun 15 '12 at 8:11
Even given the failure of the commutator argument in the case $n=2$, $k=\mathbb{F}_2$ it is difficult to see, how we could get functions other than the determinant. This is because an index two subgroup must be mapped to 1, but the field $k$ has no elements of order 2 :-) – Jyrki Lahtonen Jun 15 '12 at 8:33

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