A question about determinants an multilinear algebra

Question

Let $E$ a vector space of dimension $n$ over a field $\Gamma$ and consider the space $L(E;E)$ of linear transformations.

Assume that $F$ is a function from $L(E;E)$ to $\Gamma$ satisfying $F(\phi \circ \psi )= F(\phi)F(\psi)$.

Prove that $F$ can be written in the form $F(\phi) = f(\det \phi)$ where $f: \Gamma \to \Gamma $ is a mapping such that $f(\lambda \mu) = f(\lambda)f(\mu)$.

The suggestion is: let $\{e_{i}\}$ with $i=1,\dots ,n$ be a basis for $E$ and define the transformations $\psi_{ij}$ and $\varphi_{i}$ by:

$\psi_{ij}(e_{v}) = e_{v}$ if $v \neq i$ or $\psi_{ij}(e_{v}) = e_{i} + \lambda e_{j}$ if $v = i$ where $i,j = 1, \dots , n$.

And $\varphi_{i}(e_{v}) = e_{v}$ if $v \neq i$ or $\varphi_{i}(e_{v}) = \lambda e_{i}$ if $v=i$, where $i = 1, \dots,n$.

The idea is first prove that $F(\psi_{ij})= 1$ and $F(\varphi_{i})$ is independent of $i$ but i don't know how. I tried to do some things with the base, but it came to nothing, I also tried to prove that it $\psi_{ij}$ is its own inverse and then apply the property of $F$, i tried to use representation matrix but this doesn't help.

Any suggestion please? Thanks for your help! Greetings from Colombia

Answer 1

First notice that $F(0)$ and $F({\rm id}_E)$ are (like the image by $F$ of any projection) equal to $0$ or $1.$

If $F({\rm id}_E)=0$ then $F$ is the zero map. If $F(0)=1$ then $F$ is constantly equal to $1.$ Excluding from now on these two cases, we have $F(0)=0$ and $$F({\rm id}_E)=1.$$ From this we readily derive that for every $\phi\in GL(E),$ $F(\phi^{-1})=F(\phi)^{-1}$ and therefore, $F(\phi^{-1}\psi\phi)=F(\psi),$ in particular:

$F(\varphi_{i,\lambda})$ is independent of $i.$

Therefore, defining $f:\Gamma\to\Gamma$ by $$f(\lambda):=F(\varphi_{n,\lambda}),$$ we have $F(\varphi_{i_1,\lambda_1})\dots F(\varphi_{i_k,\lambda_k})=f(\lambda_1)\dots f(\lambda_k)=f(\lambda_1\dots\lambda_k),$ in particular $$f(0)=0$$ (since $F(\varphi_{1,0})\dots F(\varphi_{n,0})=F(\varphi_{1,0}\dots \varphi_{n,0})=F(0)=0$).

Besides, by the Steinberg relations, every $\psi_{i,j,\lambda}$ is a commutator in $GL(E)$ hence

$F(\psi_{i,j,\lambda})=1.$

Since the $\varphi_{i,\lambda}$'s and $\psi_{i,j,\lambda}$'s generate $GL(E),$ and the two multiplicative maps $F$ and $f\circ\det$ coincide on them, we conclude: $$\forall\phi\in GL(E)\quad F(\phi)=f(\det\phi).$$

Finally, for every $\phi\in L(E)\setminus GL(E),$ there exists a projection $P$ of rank $n-1$ (thereby conjugate to $\varphi_{n,0}$) such that $P\circ\phi=\phi,$ hence $F(\phi)=F(P)F(\phi)=f(0)F(\phi)=0,$ i.e. again: $F(\phi)=f(\det\phi).$

Answer 2

This is just some of my thought.

Since the dimension is finite, we could represent $\phi$ and $\varphi$ as matrices $A$ and $B$ and consider $F$ as a matrix function mapping from $\Gamma^{n^2}$ to $\Gamma$. Let $I$ be the identity matrix and $O$ be the zero matrix.

1. Since $F(A) = F(AI) = F(A)F(I)$, we have $F(I)=1$.

2. Since $F(OA) = F(O)F(A) = F(O)$ for any $A$, the nontrivial case would be $F(O)=0$. (Otherwise $F$ would be a constant function).

3. Consider an invertible matrix $M$, and assuming $F(M)$ $\neq$ 0. We then have $F(MM^{-1}) = F(M)F(M^{-1}) = 1$, and hence $1/F(M) =F(M^{-1})$.

4. Now, consider the Jordan decomposition of a matrix $M = PJP^{-1}$ and assume $F(P) \neq 0$. We have $F(M) = F(P)F(J)F(P^{-1}) = F(J)$, so the function $F$ only depends on the Jordan form of the matrix.

5. Let $D_J$ be the diagonal matrix consisted by only the diagonal elements of $J$. Since Jordan blocks are nilpotent, we have which and by $F(J^k) = F(D_J^k) = F(J)^k = F(D_J)^k$ for $k>n$, and hence $F$ only depends on the diagonal elements of the Jordan form.

6. Consider diagonal matrices represented by $D=diag(d_1,...,d_n)$. Let $D_1 = diag(d_1,1,...,1)$, $D_2 = diag(1,d_2,1,...,1)$ and so on. Note that $D = D_1 D_2...D_n$, and hence $F(D) = F(D_1)F(D_2)...F(D_n) \equiv f(det(D))$.

7. For any matrix $M=PJP^{-1}$, we have $F(M)=F(J)=F(D_J)=f(det(D_J))$. Also we know that $det(M) = det(J) = det(D_J)$. Hence, $F(M)= f(det(M))$.

8. $f(pq) = f(det(pqI))= F(pqI) = F(pI)F(qI) = f(det(pI))f(det(qI)) = f(p)f(q)$.

There are some issue when we assume $F(P)\neq 0$. If for an invertible $P$ we have $F(P) = 0$, then $F(P^{-1})$ is not defined (alternatively, if 0 has an inverse in this field $\Gamma$, we have $0=00^{-1}=1$ and the field structure is trivial), so we may assume $F(P)\neq 0$.