Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

• Forgot your password?
• Apply for membership

1. • CommentRowNumber1.
   • CommentAuthorJ-B Vienney
   • CommentTimeMay 27th 2023
   • PermaLink
   Author: J-B Vienney
   Format: MarkdownItexAdded definition of a right module over a monoid <a href="https://ncatlab.org/nlab/revision/diff/module+over+a+monoid/9">diff</a>, <a href="https://ncatlab.org/nlab/revision/module+over+a+monoid/9">v9</a>, <a href="https://ncatlab.org/nlab/show/module+over+a+monoid">current</a>

   Added definition of a right module over a monoid

   diff, v9, current

2. • CommentRowNumber2.
   • CommentAuthorsbackman
   • CommentTimeJul 11th 2023
   • PermaLink
   Author: sbackman
   Format: TextI am not a category theorist, so forgive my ignorance, but is there a way to view convex cones in \mathbb{R}^n in this setting as modules over the monoid of non negative real numbers with polyhedral cones being the finitely generated examples of such cones?

   I am not a category theorist, so forgive my ignorance, but is there a way to view convex cones in \mathbb{R}^n in this setting as modules over the monoid of non negative real numbers with polyhedral cones being the finitely generated examples of such cones?
3. • CommentRowNumber3.
   • CommentAuthorJ-B Vienney
   • CommentTimeJul 11th 2023
   • PermaLink
   Author: J-B Vienney
   Format: MarkdownItexThe inclusion $\mathbb{R}_{\ge 0} \rightarrow \mathbb{R}$ is a semi-ring morphism. So, every $\mathbb{R}$-vector space becomes a $\mathbb{R}_{\ge 0}$-module by [[restriction of scalars]], in particular $\mathbb{R}^n$. Now, the convex cones in $\mathbb{R}^n$ are exactly the sub-$\mathbb{R}_{\ge 0}$-modules of $\mathbb{R}^n$. And the polyhedral cones are exactly the convex cones which are finitely generated as a $\mathbb{R}_{\ge 0}$-module.

   The inclusion $\mathbb{R}_{\ge 0} \rightarrow \mathbb{R}$ is a semi-ring morphism. So, every $\mathbb{R}$-vector space becomes a $\mathbb{R}_{\ge 0}$-module by restriction of scalars, in particular $\mathbb{R}^n$. Now, the convex cones in $\mathbb{R}^n$ are exactly the sub-$\mathbb{R}_{\ge 0}$-modules of $\mathbb{R}^n$. And the polyhedral cones are exactly the convex cones which are finitely generated as a $\mathbb{R}_{\ge 0}$-module.

4. • CommentRowNumber4.
   • CommentAuthorsbackman
   • CommentTimeJul 11th 2023
   • PermaLink
   Author: sbackman
   Format: TextGreat, thanks!

   Great, thanks!
5. • CommentRowNumber5.
   • CommentAuthorJ-B Vienney
   • CommentTimeJul 11th 2023
   • (edited Jul 11th 2023)
   • PermaLink
   Author: J-B Vienney
   Format: MarkdownItexBut note that I suppose that a convex cone is a subset $X$ of $\mathbb{R}^n$ such that $\alpha.x+\beta.y \in X$ if $\alpha,\beta \in \mathbb{R}_{\ge 0}$ and $x,y \in X$, so they are convex cones $X$ with $0 \in X$.

   But note that I suppose that a convex cone is a subset $X$ of $\mathbb{R}^n$ such that $\alpha.x+\beta.y \in X$ if $\alpha,\beta \in \mathbb{R}_{\ge 0}$ and $x,y \in X$, so they are convex cones $X$ with $0 \in X$.

6. • CommentRowNumber6.
   • CommentAuthorJ-B Vienney
   • CommentTimeJul 11th 2023
   • (edited Jul 11th 2023)
   • PermaLink
   Author: J-B Vienney
   Format: MarkdownItexYou would also have the same if you replace $\mathbb{R}_{\ge 0}$ by $\mathbb{R}_{\gt 0}$, but in this case you should note require a zero in the definition of semi-ring and module over a semi-ring.

   You would also have the same if you replace $\mathbb{R}_{\ge 0}$ by $\mathbb{R}_{\gt 0}$, but in this case you should note require a zero in the definition of semi-ring and module over a semi-ring.