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Question about notation used in writing the moduli space in string theory

+ 2 like - 0 dislike
105 views

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?

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user leastaction
asked Jun 6, 2016 in Theoretical Physics by leastaction (425 points) [ no revision ]
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 $\mathrm{O}(4,20)$

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user ACuriousMind
Ah, thank you! So does that mean that if one writes $G_1\setminus H / G_2$, then the dimension will be $dim(H) - dim(G_1) - dim(G_2)$?

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user leastaction
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.

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user doetoe
Thank you @doetoe!

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user leastaction
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 $pq$.

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user user40085

2 Answers

+ 2 like - 0 dislike

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

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

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user doetoe
answered Jun 6, 2016 by doetoe (110 points) [ no revision ]
Thank you @doetoe for a comprehensive answer!

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user leastaction
+ 2 like - 3 dislike

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.

This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user Lawrence B. Crowell
answered Jun 6, 2016 by Lawrence B. Crowell (590 points) [ no revision ]

This is wrong. There is not set-minus but one left quotient and one right quotient, as explained in the comments to the question and in doetoe answer.

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