Skip to main content
MathOverflow

Return to Question

added 35 characters in body
Source Link
H1ghfiv3
H1ghfiv3
  • 1.3k
  • 7
  • 14

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 $\det: \mathcal A^\times \to \mathbb C$ attaining values in $\mathbb C \setminus \mathbb R_{\geq 0}$, such that ?$\det(\mathcal A^\times) \cap (\mathbb C \setminus \mathbb R_{\geq 0}) \neq \emptyset$. If not, what exactly goes wrong ?

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 $\det: \mathcal A^\times \to \mathbb C$ attaining values in $\mathbb C \setminus \mathbb R_{\geq 0}$ ? If not, what exactly goes wrong ?

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 ?

deleted 3 characters in body
Source Link
Denis Serre
Denis Serre
  • 52.4k
  • 10
  • 146
  • 300

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 subgroupgroup 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 $\det: \mathcal A^\times \to \mathbb C$ attaining values in $\mathbb C \setminus \mathbb R_{\geq 0}$ ? If not, what exactly goes wrong ?

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 subgroup 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 $\det: \mathcal A^\times \to \mathbb C$ attaining values in $\mathbb C \setminus \mathbb R_{\geq 0}$ ? If not, what exactly goes wrong ?

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 $\det: \mathcal A^\times \to \mathbb C$ attaining values in $\mathbb C \setminus \mathbb R_{\geq 0}$ ? If not, what exactly goes wrong ?

Source Link
H1ghfiv3
H1ghfiv3
  • 1.3k
  • 7
  • 14

Obstruction to the existence of a complex-valued determinant function

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 subgroup 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 $\det: \mathcal A^\times \to \mathbb C$ attaining values in $\mathbb C \setminus \mathbb R_{\geq 0}$ ? If not, what exactly goes wrong ?