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May 28, 2009, 11:18 pm • # 1

Does there exist a function $f : \mathcal{M}_n(F) \to F$ such that $f(AB) = f(A) f(B)$ that does not factor through the determinant, in the sense that there exists a multiplicative function $g$ such that $f(A) = g(\det A)$?

What if $F = \mathbb{C}$ and $f$ is required to be continuous?

What if $f$ is required to be polynomial in the entries of $A$?

What if $f$ is required to be invariant under conjugation by a permutation matrix?

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 loup blanc Posts: 1752Location: Tahiti May 29, 2009, 4:49 am • # 2  1) If $f: GL_n(F)\rightarrow{F}^*$ is a morphism then see: http://www.artofproblemsolving.com/Foru ... 0&t=218429 I have not reread my first post but I think that it shows that $f(A)=\psi(det(A))$ where $\psi$ is a morphism of $F^*$. 2) Now let $g: \mathcal{M}_n(F)\rightarrow{F}$ be a morphism. Case 1: $g(I)=0$ then $g=0$ and we are done. Case 2: $g(I)=1$. If $A$ is invertible then necessarily $g(A)\not=0$. Let $B$ be a non invertible matrix. Case 2.1: $g(0_n)=0$. If $N$ is nilpotent then $g(N)=0$. $B$ is product of nilpotent matrices (Sullivan,Product of nilpotent matrices,Linear and multilinear algebra,vol 56,3/2008/ p. 311-317). This result is valid for $n\geq{3}$ and I think, for any field. Therefore $g(B)=0$. We are done. Case 2.2: $g(0_n)=1$. There exists $B'$ s.t. $BB'=0$;then $g(B)\not=0$. Thus if $U$ is idempotent then $g(U)=1$. $B$ is a product of idempotent (for any field): see http://www.artofproblemsolving.com/Foru ... 0&t=271857 thus $g(B)=1$. Let $P$ be an invertible matrix. $g(PB)=g(P)=1$. Therefore $g=1$ and we are done. Conclusion: $g$ factors through the determinant without supplementary hypothesis. PS: it remains to solve the case 2.1 with $n=2$. I go to bed. _________________Le vieux loup perd ses poils mais garde son vice.Hard work beats talent when talent fails to work hard.
 loup blanc Posts: 1752Location: Tahiti May 29, 2009, 9:39 pm • # 3  The case 2.1 and $n=2$. The result of Sullivan is still valid because $\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&1\end{...$. In fact in its theorem he speaks about 4 nilpotent matrices,that explains the condition $n\geq{3}$. _________________Le vieux loup perd ses poils mais garde son vice.Hard work beats talent when talent fails to work hard.

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