Skip to main content
My iPad autocorrects "monoid" to "monoidal"!
Source Link
Evan Jenkins
  • 7.2k
  • 1
  • 36
  • 56

I believe that the Grothendieck group of a commutative monoidalmonoid can be viewed as a special case of the Int construction. Namely, you can think of a commutative monoidalmonoid $M$ as a discrete symmetric monoidal category (i.e., a symmetric monoidal category with no non-identity morphisms). This category admits a unique trace, and taking $\operatorname{Int} M$ just gives the Grothendieck group of $M$, thought of as a discrete symmetric monoidal category.

I believe that the Grothendieck group of a commutative monoidal can be viewed as a special case of the Int construction. Namely, you can think of a commutative monoidal $M$ as a discrete symmetric monoidal category (i.e., a symmetric monoidal category with no non-identity morphisms). This category admits a unique trace, and taking $\operatorname{Int} M$ just gives the Grothendieck group of $M$, thought of as a discrete symmetric monoidal category.

I believe that the Grothendieck group of a commutative monoid can be viewed as a special case of the Int construction. Namely, you can think of a commutative monoid $M$ as a discrete symmetric monoidal category (i.e., a symmetric monoidal category with no non-identity morphisms). This category admits a unique trace, and taking $\operatorname{Int} M$ just gives the Grothendieck group of $M$, thought of as a discrete symmetric monoidal category.

Source Link
Evan Jenkins
  • 7.2k
  • 1
  • 36
  • 56

I believe that the Grothendieck group of a commutative monoidal can be viewed as a special case of the Int construction. Namely, you can think of a commutative monoidal $M$ as a discrete symmetric monoidal category (i.e., a symmetric monoidal category with no non-identity morphisms). This category admits a unique trace, and taking $\operatorname{Int} M$ just gives the Grothendieck group of $M$, thought of as a discrete symmetric monoidal category.