Skip to main content
MathOverflow

Return to Question

edited title
Source Link
willie
willie
  • 499
  • 2
  • 5

Does the oriental inject intosimplex map to the cube?

For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<\dots<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$, in which $1$ appears $i$ many times. In $n$-category theory the totally ordered set with $n+1$ elements and the powerset of the set with $n$ elements admit the following categorifications called the $n$-th oriental $O(n)$ and the lax $n$-cube $\boxtimes^n$. These are $n$-categories, whose 1-truncations are the ordered set with $n+1$ elements and the powerset of the set with $n$ elements, respectively. Is there a $n$-functor $O(n) \to \boxtimes^n$ that sends $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$,higher analogue of this fact in which $1$ appears $i$ many timesn-category theory?

Does the oriental inject into the cube?

For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<\dots<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$, in which $1$ appears $i$ many times. In $n$-category theory the totally ordered set with $n+1$ elements and the powerset of the set with $n$ elements admit the following categorifications called the $n$-th oriental $O(n)$ and the lax $n$-cube $\boxtimes^n$. These are $n$-categories, whose 1-truncations are the ordered set with $n+1$ elements and the powerset of the set with $n$ elements, respectively. Is there a $n$-functor $O(n) \to \boxtimes^n$ that sends $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$, in which $1$ appears $i$ many times?

Does the simplex map to the cube?

For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<\dots<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$, in which $1$ appears $i$ many times. Is there a higher analogue of this fact in n-category theory?

Better MathJax usage. If you use `...` instead of `\ldots` or `\cdots` or `\dots` in full-fledged LaTeX, you get inferior results. And other details.
Source Link
edit approved Dec 2, 2023 at 22:08
Community Bot
edit approved Dec 2, 2023 at 22:08
Community Bot
  • 1
  • 2
  • 3

For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<...<n\}$$\{0<1<\dots<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < ...<0<1<...1)$$(0 < \dots<0<1<\dots<1)$, in which 1$1$ appears $i$ many times. In n$n$-category theory the totally ordered set with $n+1$ elements and the powerset of the set with n$n$ elements admit the following categorifications called the $n$-th oriental $O(n)$ and the lax n$n$-cube $\boxtimes^n$. These are n$n$-categories, whose 1-truncations are the ordered set with $n+1$ elements and the powerset of the set with n$n$ elements, respectively. Is there a n$n$-functor $O(n) \to \boxtimes^n$ that sends $i$ to the increasing sequence $(0 < ...<0<1<...1)$$(0 < \dots<0<1<\dots<1)$, in which 1$1$ appears $i$ many times?

For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<...<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < ...<0<1<...1)$, in which 1 appears $i$ many times. In n-category theory the totally ordered set with $n+1$ elements and the powerset of the set with n elements admit the following categorifications called the $n$-th oriental $O(n)$ and the lax n-cube $\boxtimes^n$. These are n-categories, whose 1-truncations are the ordered set with $n+1$ elements and the powerset of the set with n elements, respectively. Is there a n-functor $O(n) \to \boxtimes^n$ that sends $i$ to the increasing sequence $(0 < ...<0<1<...1)$, in which 1 appears $i$ many times?

For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<\dots<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$, in which $1$ appears $i$ many times. In $n$-category theory the totally ordered set with $n+1$ elements and the powerset of the set with $n$ elements admit the following categorifications called the $n$-th oriental $O(n)$ and the lax $n$-cube $\boxtimes^n$. These are $n$-categories, whose 1-truncations are the ordered set with $n+1$ elements and the powerset of the set with $n$ elements, respectively. Is there a $n$-functor $O(n) \to \boxtimes^n$ that sends $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$, in which $1$ appears $i$ many times?

Source Link
willie
willie
  • 499
  • 2
  • 5

Does the oriental inject into the cube?

For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<...<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < ...<0<1<...1)$, in which 1 appears $i$ many times. In n-category theory the totally ordered set with $n+1$ elements and the powerset of the set with n elements admit the following categorifications called the $n$-th oriental $O(n)$ and the lax n-cube $\boxtimes^n$. These are n-categories, whose 1-truncations are the ordered set with $n+1$ elements and the powerset of the set with n elements, respectively. Is there a n-functor $O(n) \to \boxtimes^n$ that sends $i$ to the increasing sequence $(0 < ...<0<1<...1)$, in which 1 appears $i$ many times?