Skip to main content
link fixed and formatting
Source Link
David Roberts
  • 35.4k
  • 11
  • 124
  • 348

You can try to define the determinant of an $n \times n$ matrix with entries in a bipermutative (or symmetric bimonoidal) category $R$ by an analogue of the usual signed sum of $n$-fold products. However, it will usually not be a monoidal functor, and the inclusion $BGL_1(R) \to K(R)$ does generally not admit a retraction, which you might expect to get from a determinant. There is a counterexample in

and a discussion of what more might be needed (to circumvent this obstruction) in

You can try to define the determinant of an $n \times n$ matrix with entries in a bipermutative (or symmetric bimonoidal) category $R$ by an analogue of the usual signed sum of $n$-fold products. However, it will usually not be a monoidal functor, and the inclusion $BGL_1(R) \to K(R)$ does generally not admit a retraction, which you might expect to get from a determinant. There is a counterexample in

and a discussion of what more might be needed (to circumvent this obstruction) in

You can try to define the determinant of an $n \times n$ matrix with entries in a bipermutative (or symmetric bimonoidal) category $R$ by an analogue of the usual signed sum of $n$-fold products. However, it will usually not be a monoidal functor, and the inclusion $BGL_1(R) \to K(R)$ does generally not admit a retraction, which you might expect to get from a determinant. There is a counterexample in

and a discussion of what more might be needed (to circumvent this obstruction) in

Links
Source Link
LSpice
  • 12.9k
  • 4
  • 45
  • 69

You can try to define the determinant of an $n \times n$ matrix with entries in a bipermutative (or symmetric bimonoidal) category $R$ by an analogue of the usual signed sum of $n$-fold products. However, it will usually not be a monoidal functor, and the inclusion $BGL_1(R) \to K(R)$ does generally not admit a retraction, which you might expect to get from a determinant. There is a counterexample in

Ausoni, Christian (D-BONN); Dundas, Bjørn Ian (N-BERG); Rognes, John (N-OSLO)
Divisibility of the Dirac magnetic monopole as a two-vector bundle over the three-sphere.
Doc. Math. 13 (2008), 795–801.

and a discussion of what more might be needed (to circumvent this obstruction) in

Kragh, Thomas (S-UPPS)
Orientations on 2-vector bundles and determinant gerbes.
Math. Scand. 113 (2013), no. 1, 63–82. 

You can try to define the determinant of an $n \times n$ matrix with entries in a bipermutative (or symmetric bimonoidal) category $R$ by an analogue of the usual signed sum of $n$-fold products. However, it will usually not be a monoidal functor, and the inclusion $BGL_1(R) \to K(R)$ does generally not admit a retraction, which you might expect to get from a determinant. There is a counterexample in

Ausoni, Christian (D-BONN); Dundas, Bjørn Ian (N-BERG); Rognes, John (N-OSLO)
Divisibility of the Dirac magnetic monopole as a two-vector bundle over the three-sphere.
Doc. Math. 13 (2008), 795–801.

and a discussion of what more might be needed (to circumvent this obstruction) in

Kragh, Thomas (S-UPPS)
Orientations on 2-vector bundles and determinant gerbes.
Math. Scand. 113 (2013), no. 1, 63–82. 

You can try to define the determinant of an $n \times n$ matrix with entries in a bipermutative (or symmetric bimonoidal) category $R$ by an analogue of the usual signed sum of $n$-fold products. However, it will usually not be a monoidal functor, and the inclusion $BGL_1(R) \to K(R)$ does generally not admit a retraction, which you might expect to get from a determinant. There is a counterexample in

and a discussion of what more might be needed (to circumvent this obstruction) in

Source Link
John Rognes
  • 9.3k
  • 45
  • 52

You can try to define the determinant of an $n \times n$ matrix with entries in a bipermutative (or symmetric bimonoidal) category $R$ by an analogue of the usual signed sum of $n$-fold products. However, it will usually not be a monoidal functor, and the inclusion $BGL_1(R) \to K(R)$ does generally not admit a retraction, which you might expect to get from a determinant. There is a counterexample in

Ausoni, Christian (D-BONN); Dundas, Bjørn Ian (N-BERG); Rognes, John (N-OSLO)
Divisibility of the Dirac magnetic monopole as a two-vector bundle over the three-sphere.
Doc. Math. 13 (2008), 795–801.

and a discussion of what more might be needed (to circumvent this obstruction) in

Kragh, Thomas (S-UPPS)
Orientations on 2-vector bundles and determinant gerbes.
Math. Scand. 113 (2013), no. 1, 63–82.