# Showing that the moduli stack of triangles is equivalent to the quotient stack $[\tilde{T}/S_3]$

$\require{AMScd}$ A good reference for this question is found in the first chapter in the unfinished book Algebraic Stacks by Behrend, Conrad, Edidin, Fulton, Fantechi, Göttsche, and Kresch. (http://www.math.uzh.ch/index.php?file&key1=5158), but I'll try and give all the definitions myself.

Definitions:

Define a family of triangles to be a fibre bundle $V\xrightarrow{p}S$ with a continuously varying metric (i.e., a continuous function $V\times_SV\to\mathbb R_{\ge0}$ that restricts to a metric for each fibre) such that each fibre is isometric to a plane triangle. The moduli stack of triangles is the category whose objects are families of triangles and whose morphisms are Cartesian squares \begin{CD} V' @>>> V\\@VVp'V @VVpV\\ S' @>f>> S \end{CD}

Now let $\tilde{T}$ be the set $$\{(a,b,c)\in\mathbb R^3\colon a,b,c>0, a+b>c,a+c>b,b+c>a\}$$ And let $\tilde{Y}$ be a family of triangles over $\tilde{T}$ such that the fibre over each $(a,b,c)\in > T$ is a plane triangle with side lengths $a,b,c$. So $\tilde{T}$ is a moduli space for all labelled triangles and $\tilde{Y}$ is a universal family over it.

We have an action of $S_3$ on $\tilde{T}$ which permutes the three axes. This induces an action by isometries on the set of all fibres of $\tilde{T}$.

Form the category (quotient stack) $[\tilde{T}/S_3]$. The objects of this category are diagrams $$\tilde{T}\xleftarrow{f} E\xrightarrow{p}B$$ where $E\xrightarrow{p}B$ is a principal $G$-bundle (a fibre bundle with fibre $S_3$ (with the discrete topology) and a continuous right action of $S_3$ on $E$), and $f$ is a $G$-equivariant map. The morphisms in this category are diagrams that look a bit like houses, with the bottom square Cartesian.

I want to show that the two categories are equivalent, via an equivalence that commutes with the natural functors to $\mathbf{Top}$ (so an isomorphism of stacks). This corresponds to our intuition that we can form the moduli stack of triangles by taking the moduli stack of ordered triangles and forming the quotient by the action of $S_3$.

I'm having problems defining a functor between these categories. Given a family $V\xrightarrow{p}S$, I want to form a principal $S_3$-bundle over $S$ that has a $G$-invariant map to $\tilde{T}$. This means finding a $6$-sheeted cover of $S$.

My attempt:

Given a family of triangles $V\xrightarrow{p}S$, let $$E=\{(u,s,g)\;\vert\; u\in\tilde{T}, g\colon \tilde{Y}_u\to V_s\textrm{ is an isometry}\}$$

This has natural maps to $S$ and to $\tilde{T}$ and every fibre has six elements, corresponding to the six possible orderings of the sides.

However, we haven't yet defined topology on this space, since I haven't specified a space for the isometries $g$ to live in. And I'm not sure how to. I've tried considering certain function spaces with the compact-open topology, but I don't want to include any local compactness assumptions on the space $S$.

How can we define an appropriate topology on this space to get the desired functor? How can we show that the resulting space $E$ is a fibre bundle over $B$? Or is there a better way of doing this?

• Could you add bibliographical information for the link you included? Th author, at the very least! Commented Apr 4, 2015 at 18:15
• I believe it's the first chapter of the book by Behrend, Conrad, Edidin, Fulton, Fantechi, Göttsche, and Kresch (whew!). Hopefully this gets finished someday.
– Hoot
Commented Apr 4, 2015 at 18:37
• It is indeed. I'll add that now. Commented Apr 4, 2015 at 20:32

It turns out that I was wrong to try and think about using the compact-open topology (especially given that our functions $g$ don't live in any natural ambient space) - the product topology, identifying the map $g$ with its graph, is much more natural.

So we could view $E$ as

$$E=\left\{\left(u,s,\left((a,g(a))\colon a\in \tilde{Y}_u\right)\right)\;\middle\vert\; u\in\tilde{T}, g\colon \tilde{Y}_u\to V_s\textrm{ is an isometry}\right\}$$

But this is a bit complicated, and is essentially equivalent to the much simpler

$$E_V=\left\{\left(x, (\alpha_1,\alpha_2,\alpha_3)\right)\in S\times V^3\;\middle\vert\; \alpha_1,\alpha_2,\alpha_3\textrm{ are the vertices of the triangle }V_x \right\}$$

This now has obvious maps to $\tilde{T}$ and to $S$. Moreover, it is a fibre bundle with fibre homeomorphic to $S_3$ and there is a natural action of $S_3$ on $E$:

$$\left(s,\left(\alpha_1,\alpha_2,\alpha_3\right)\right).\sigma= \left(s,\left(\alpha_{\sigma(1)},\alpha_{\sigma(2)},\alpha_{\sigma(3)}\right)\right)$$