2
$\begingroup$

A semi-simplicial complex (see Moore's 1958 paper: Semi-simplicial complexes and Postnikov systems, available at http://www-math.mit.edu/~hrm/kansem/moore-semi-simplicial-complexes.pdf, cf MathSciNet) consists of the following

  1. A set $X=\cup_{q\ge0}X_q$, where the $X_q$ are disjoint sets (an element of $X_q$ is called a $q$-simplex of $X$),

  2. functions $\partial_i:X_{q+1}\to X_q$, $i=0,\dots,q+1$, called face operators, and

  3. functions $s_i:X_q\to X_{q+1}$, $i=0,\dots,q$, called degeneracy operators.

The face and degeneracy operators are assumed to satisfy the relations $$\partial_i\partial_j=\partial_{j-1}\partial_i, i<j,$$ $$s_is_j=s_{j+1}s_i, i\le j,$$ $$\partial_is_j=\partial_{j+1}s_j=\text{identity},$$ $$\partial_is_j=s_{j-1}\partial_i, i<j,$$ and $$\partial_is_j=s_j\partial_{i-1}, i>j+1.$$

Let $\underline{C}$ be a small category. Its nerve (see Quillen's 1973 paper: Higher algebraic K-theory I), denoted $N\underline{C}$, is the (semi-) simplicial set whose $p$-simplices are the diagrams in $\underline{C}$ of the form $$X_0\to X_1\to\cdots\to X_p.$$ The $i$-th face (resp. degeneracy) of this simplex is obtained by deleting the object $X_i$ (resp. replacing $X_i$ by $id: X_i\to X_i$) in the evident way.

Question: what is the nerve of the Ising category?

Here we take the Ising category as the fusion category part of Ising modular tensor category (MTC), see 5.3.4 in p.37 of On classification of modular tensor categories. We only demand the same fusion rule as the Ising MTC.

enter image description here

$\endgroup$
7
  • 15
    $\begingroup$ The question might improve if you add a similarly detailed definition of the Ising category. $\endgroup$ Nov 6, 2021 at 20:31
  • 2
    $\begingroup$ What is the Ising category? There is more than one and they are not equivalent. $\endgroup$ Nov 6, 2021 at 21:44
  • 4
    $\begingroup$ A small remark: if I understand what you wrote, what you call semisimplicial complexes are more commonly called simplicial sets $\endgroup$ Nov 7, 2021 at 10:53
  • 1
    $\begingroup$ Thanks, to answer the definition of Ising category --- Here we take it as the fusion category of Ising modular tensor category (MTC), see 5.3.4 in p.37 of arxiv.org/pdf/0712.1377.pdf. We only demand the same fusion rule as the Ising MTC. If you think the fusion category of Ising MTC is not enough, we can take more constraints from the MTC. Thank you! $\endgroup$
    – wonderich
    Nov 11, 2021 at 13:41
  • 2
    $\begingroup$ Might I suggest deleting everything above the question "what is the nerve of the Ising category?" I would posit that anyone who could answer this question will know what a simplicial set and the nerve construction are, and it only serves to obscure the relevant information (for instance, I started out assuming that a semi-simplicial complex was a ncatlab.org/nlab/show/semi-simplicial+set) $\endgroup$
    – David Roberts
    Nov 11, 2021 at 14:48

1 Answer 1

6
$\begingroup$

The nerve construction only cares about the underlying category of $\mathit {Ising}$ (it doesn't care about the structure of fusion category, even less about the structure of modular tensor category). There are three anyon types, so what you get is just $\mathit{Vec}^3=\mathit{Vec}\oplus\mathit{Vec}\oplus\mathit{Vec}$, where $\mathit{Vec}$ denotes the category of finite dimensional complex vector spaces.

Since taking nerves commutes with cartesian products of categories, we are reduced to the question: what is the nerve of $\mathit{Vec}$? Now, since $\mathit{Vec}$ has an initial object (namely the zero vector space), the nerve of $\mathit{Vec}$ is a contractible simplicial set.

So the upshot is that the nerve of $\mathit{Vec}$ is not very interesting.

Somewhat more interesting is the nerve of the groupoid of finite dimensional vector spaces. In that case, you get something that has the homotopy type of $\coprod_{n\in\mathbb N} BU(n)$.

$\endgroup$
2
  • $\begingroup$ Excuse me: could you clarify: What is the nerve of the groupoid of Ising category? Thank you! $\endgroup$
    – wonderich
    Nov 26, 2021 at 20:43
  • $\begingroup$ It's $\Big(\coprod_{n\in\mathbb N} BU(n)\Big)^3=\coprod_{(n_1,n_2,n_3)\in\mathbb N^3} BU(n_1)\times BU(n_2)\times BU(n_3)$. The numbers $n_1,n_2,n_3$ count how many copies of $1$, of $\sigma$, and of $\psi$ you have. $\endgroup$ Nov 26, 2021 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.