Let $\mathcal A$ be a Type $II_1$ von Neumann algebra, equipped with a finite trace $\tau: \mathcal A \to \mathbb C$. Further, denote by $\mathcal A^\times \subset \mathcal A$ the group of invertible elements. It is well-known that $\tau$ induces a positive-valued determinant function $\det: \mathcal A^\times \to \mathbb R^+$, that is, among many other properties, multiplicative, i.e \begin{equation} \det(AB) = \det(A)\det(B). \end{equation} Question: Does there exist a non-trivial, multiplicative determinant function $\det: \mathcal A^\times \to \mathbb C$, such that $\det(\mathcal A^\times) \cap (\mathbb C \setminus \mathbb R_{\geq 0}) \neq \emptyset$. If not, what exactly goes wrong ?
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5$\begingroup$ For $z$ on the unit circle, there exist unitaries $U,V$ such that $UVU^{-1}V^{-1} = zI$, so any determinant sends the unit circle to $1$, and thus the nonzero scalars to the positive reals. [This follows from the irrational and rational rotation algebras being embeddable in the hyperfinite W* factor, hence in every type II finite factor.] Also, the unitary group of a type II$_f$ factor is simple (hence perfect), so any determinant will send the whole thing to $1$. Then finish with polar decomposition.... There is likely a uniqueness argument too. $\endgroup$– David HandelmanCommented Jun 13, 2018 at 14:22
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$\begingroup$ And of course I meant the unitary group modulo its centre (the scalars on the unit circle) is simple, not the unitary group itself, and the first statement then yields that the unitary group is perfect. $\endgroup$– David HandelmanCommented Jun 14, 2018 at 22:20
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