3
$\begingroup$

In physics papers, particularly those by Aspinwall, or textbooks, I encounter things like

$$\mathcal{M} \simeq O(\Gamma_{4,20})\setminus O(4,20)/((O(4)\times O(20))$$

For instance, this is from https://arxiv.org/abs/hep-th/9707014. Am I correct in understanding that the numerator is $O(\Gamma_{4,20})$ folllowed by a set theoretic subtraction of $O(4,20)$?

If I wanted to compute the dimension of $\mathcal{M}$, I know I'd have to subtract the dimension of the numerator from the dimension of the denominator (which is just $(4\times 3/2) + (20 \times 19/2)$). What is the dimension of the numerator?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ That rather looks like a double coset to me, i.e. you're quotienting a left action of $\mathrm{O}(\Gamma)$ and a right action of $\mathrm{O}(4)\times\mathrm{O}(20)$ out of $\mathrm{O}(4,20)$ $\endgroup$
    – ACuriousMind ♦
    Jun 6, 2016 at 19:20
  • $\begingroup$ Ah, thank you! So does that mean that if one writes $G_1\setminus H / G_2$, then the dimension will be $dim(H) - dim(G_1) - dim(G_2)$? $\endgroup$
    – leastaction
    Jun 6, 2016 at 19:22
  • 1
    $\begingroup$ The group acting from the left is a discrete group (dimension 0) consisting of transformations of a lattice, similar to a matrix group with integer entries. As for the dimension, it will be at least that, but may be higher if the actions are not faithful/effective. $\endgroup$
    – doetoe
    Jun 6, 2016 at 20:30
  • $\begingroup$ It is useful to remember that the dimension of $\mathcal{M} \simeq O(\Gamma_{p,q})\setminus O(p,q)/((O(p)\times O(q))$ is just $pq$. $\endgroup$
    – Antoine
    Jun 7, 2016 at 12:09

2 Answers 2

Reset to default
3
$\begingroup$

This is not really an answer (the answer is ACuriousMind's comment: this is a double coset space), but it may help to consider the construction of the moduli space of elliptic curves, as this can be done in the same way but is very easy.

Every complex elliptic curve is obtained as $\Bbb C$ modulo a lattice. Scaling the lattice by a complex number gives an isomorphic curve, so you can scale in such a way that the lattice is generated by 1 and by $\tau\in\Bbb H$, the complex upper half plane (this is similar to gauge fixing). Not all $\tau$ give different lattices: two sets of generators give the same lattice if they are related by an element of $GL_2(\Bbb Z)$ (there is some residual gauge freedom). Since we work with oriented bases, we can restrict to $SL_2(\Bbb Z)$. It is not hard to see that the action of $SL_2(\Bbb Z)$ on a basis $1,\tau$ corresponds to an action on $\Bbb H$ by Möbius transformations

$$\begin{pmatrix}a & b\\ c & d\end{pmatrix}\tau = \frac{a\tau + b}{c\tau + d}$$

This gives us the moduli space as a quotient

$$\mathcal M \cong SL_2(\Bbb Z)\backslash \Bbb H$$

In general, if you have a space (possibly with some extra structure like a Riemannian metric) on which some group of automorphisms acts transitively (i.e. every point can be mapped to every point) by some mapping of the whole space onto itself, then this space can be written as the quotient of this group by the stabilizer (i.e. the subgroup fixing a given point) of any point. This is the orbit-stabilizer theorem.

In our example, the complex upper half plane $\Bbb H$ has an obvious complex structure as a subset of $\Bbb C$, and its group of holomorphic automorphisms is $SL_2(\Bbb R)$ acting by Möbius transformations, except that $+I$ and $-I$ do the same thing, and the automorphisms are really $SL_2(\Bbb R)/\langle-I\rangle = PSL_2(\Bbb R)$. The stabilizer of the point $i$ is $SO_2(\Bbb R)\subset PSL_2(\Bbb R)$, so that

$$\mathcal M \cong SL_2(\Bbb Z)\backslash PSL_2(\Bbb R)/SO_2(\Bbb R)$$

Improve this answer
$\endgroup$
4
  • $\begingroup$ Do you mean $\mathcal{M} \simeq SL_2(\mathbb{Z})/\mathbb{H}$ instead? $\endgroup$
    – leastaction
    Jun 10, 2016 at 4:20
  • 2
    $\begingroup$ No. If you prefer to write the forward slash you could write $\mathbb H/SL_2(\mathbb Z)$, but many people find it more natural to write the quotient how I did in the case of a left action. In general $X/G$ and $G\X$ are quotients of $X$, whose elements are subsets of $X$. $\endgroup$
    – doetoe
    Jun 10, 2016 at 5:42
  • $\begingroup$ Oh I see. I guess then in your first equation, what you write as just $SL_2(\mathbb{Z})\setminus\mathbb{H}$ could also be written as $\mathbb{H}/SL_2(\mathbb{Z})$. $\endgroup$
    – leastaction
    Jun 10, 2016 at 5:52
  • 1
    $\begingroup$ Yes, that is quite common. The expression at the end, where we take the quotient under a compatible left and right action simultaneously would look less readable though as $(X/H)/G$, more so because here the actions are really by left and right matrix multiplication. $\endgroup$
    – doetoe
    Jun 10, 2016 at 6:04
2
$\begingroup$

There are two things going on. One is modulo that is the forwards slash / and the other is set-minus $\setminus$ the backwards slash. The $$ G_4(20) = \frac{O(4,20)}{O(4)\times O(20)} $$ is the Grassmanian space defined by $4$-planes. the group $O(\Gamma_{4,20})$ is an orthogonal group over the unimodular transformations, a bit like saying $O(n,\mathbb Z)$, where Aspinwall introduces this on the two-torus earlier in the paper. The moduli space is this orthogonal group "minus" these Grassmanians.

Improve this answer
$\endgroup$
0

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.