Thursday, April 26, 2012

Asymptotic geometry of symmetric spaces II: $\mathbb H^2$ from $sl(2;\mathbb R)$

This is a sequel to the first part in the "Asymptotic geometry" series, found below. The goal here is to obtain the metric of the space of unit area ellipses starting from the Lie algebra $sl(2; \mathbb R)$. Instead of thinking about a space on which a Lie groups acts, we can start from "the opposite end", from Lie algebras. I will work explicitly with the defining and adjoint representations, although much could be done without any reference to a representation. I'll start with some generalities.

Symmetric spaces


This subsection will be a bit technical, but will hopefully become clearer in the next subsection. Consider a (semi-simple) Lie algebra $\mathfrak g$ associated to a Lie group $G$. The Lie bracket $\left[X,Y\right]$ can be used to define the adjoint representation $\ad X : Y \to \left[X,Y\right]$, which are formed of square matrices of size $dim(\mathfrak g)\times dim(\mathfrak g)$. The symmetric bilinear form \begin{equation} \kappa \left( X, Y \right) \doteq \tr \left( \ad X \ad Y \right) \end{equation} is known as the Killing form, which can be used as an inner product on the Lie algebra by setting $\langle X, Y\rangle \doteq \kappa \left( X, Y \right)$. Now try to find involutive automorphisms $\sigma$ of the Lie algebra, i.e. $\sigma^2 =1$. Then the Lie algebra decomposes according to the eigenvalues $\pm 1$ as $\mathfrak g = \mathfrak p + \mathfrak k$ with $\sigma (\mathfrak p) = - \mathfrak p$ and $\sigma (\mathfrak k) = \mathfrak k$. It's easy to show that the subsets are orthogonal, i.e. $\langle \mathfrak p, \mathfrak k\rangle=0$, and that they satisfy the commutation rules \begin{equation}\left\{ \begin{aligned} \left[\mathfrak p,\mathfrak p \right] &\subseteq \mathfrak k \\ \left[\mathfrak p,\mathfrak k \right] &\subseteq \mathfrak p\\ \left[\mathfrak k,\mathfrak k \right] &\subseteq \mathfrak k. \end{aligned} \right. \end{equation} This procedure is known as the Cartan decomposition. Note that $\mathfrak k$ is a subalgebra, but $\mathfrak p$ is not. If $K$ is the Lie group corresponding to $\mathfrak k$, then the homogeneous space $G/K$ is a symmetric space. We can denote elements in the symmetric space as a left coset $p K$, where $p$ is the exponential map of an element in the subset $\mathfrak p$. Then left actions on the symmetric space are well defined, $ L_g :p K \to g p K = p' k K = p' K$ ($L_g$ is just the left multiplication) for some $p'$ and $k$. There's an elegant formula for the Maurer-Cartan form (the expression for the differential is originally due to Helgason, I think). Suppose that the elements in the Lie algebra are sorted as $X_i, X_j, \ldots \in \mathfrak g$, $Y_\mu, Y_\nu, \ldots \in \mathfrak p$ and $Z_a, Z_b, \ldots \in \mathfrak k$. Then an element in the symmetric space can be written as $p = \mbox{EXP}\left( Y (\omega)\right) \doteq \mbox{EXP}\left( \omega_\mu Y_\mu\right)$, and the Maurer-Cartan form can be written in the form of the nontrivial formula (in a slightly less obscure notation than Helgason's, I hope; note that everything can of course also be done in the right invariant formalism) \begin{equation} p^{-1} dp = \left( \frac{1-e^{-\ad Y (\omega)}}{\ad Y (\omega)} \right) \left( Y(d\omega) \right). \end{equation} In components, \begin{equation} p^{-1} \partial_\mu p = \left( \frac{1-e^{-\ad Y (\omega)}}{\ad Y (\omega)} \right) \left( Y_\mu \right) = X_k \left( \frac{1-e^{-\ad Y (\omega)}}{\ad Y (\omega)} \right)_{k \mu} . \end{equation} It's also common to write $p^{-1} dp = X_i \otimes \Omega_i$, and call the one forms \begin{equation} \Omega_i = \left( \frac{1-e^{-\ad Y (\omega)}}{\ad Y (\omega)} \right)_{k \mu} d\omega^\mu \doteq \mathcal A_{k \mu} d\omega^\mu \label{matrix.gauge} \end{equation} the (left invariant) Maurer-Cartan forms. The left invariant metric is then \begin{equation} ds^2 \doteq \langle p^{-1} dp, p^{-1} dp \rangle = \Omega_i \kappa_{ij} \Omega_j, = d\omega^\mu \left( \mathcal A^T \kappa \mathcal A \right)_{\mu\nu} d\omega^\nu \label{metric.explicit}\end{equation} where $\kappa_{i j} = \langle X_i, X_j\rangle$. The above formula may however not be very useful in practice, since it involves exponentiations of the adjoint representation. The point here is that the metric can be defined and calculated by purely algebraic operations.

$\mathbb H^2 \simeq P_2$ from $sl(2;\mathbb R)$


The Lie algebra $\mathfrak g \doteq sl(2;\mathbb R)$ can be defined as the set of traceless two by two matrices with elements \begin{equation} X(\omega) \doteq \left( \begin{array}{cc} -\omega_2 & \omega_1 + \omega_3 \\ \omega_1-\omega_3 & \omega_2 \end{array} \right) \doteq \sum\limits_{i=1}^{3} \omega_i X_i. \label{definingrep} \end{equation} In this particular basis the matrices $X_i$ in fact satisfy the Lie algebra $so(1,2;\mathbb R)$, \begin{equation} \left\{ \begin{aligned} \left[X_1,X_2 \right]&= 2 X_3 \\ \left[X_2,X_3 \right]&= -2 X_1\\ \left[X_3,X_1 \right]&= -2 X_2, \end{aligned}\label{comms.so21} \right. \end{equation} which is just a realization of the well known isomorphism $sl(2;\mathbb R) \simeq so(1,2;\mathbb R)$ (note how flipping the sign in e.g. the first commutator brings the algebra to the familiar $so(3;\mathbb R)$ Lie algebra). Now try to find all involutive automorphisms $\sigma$ of the Lie algebra. It can be shown that in this case there is only one such mapping, \begin{equation} \sigma : X \to -X^T. \end{equation} It is then straightforward to see that $\mathfrak p = \left\{ X_1, X_2 \right\}$ and $\mathfrak k = \left\{ X_3 \right\}$. Note that $K=\exp (\mathfrak k) = SO(2;\mathbb R)$. Using radial coordinates $\left(\omega_1, \omega_2 \right) = \left( r \sin \phi, r \cos \phi \right)$, we have \begin{equation} \small{p \doteq \exp \left(\omega_1 X_1 + \omega_2 X_2 \right) = \left( \begin{array}{cc} \cosh r -\cos \phi \sinh r & \sin \phi \sinh r \\ \sin \phi \sinh r & \cosh r +\cos \phi \sinh r \end{array} \right)}, \end{equation} which is identical to the matrix in eq. (10) in part I. The matrix $\mathcal A$ in eq. \eqref{matrix.gauge} can be computed explicitly as follows. The $\ad$ basis can be written as in the defining representation in eq. \eqref{definingrep}, \begin{equation} \ad X(\omega) \doteq 2\left( \begin{array}{ccc} 0 & \omega _3 & -\omega _2 \\ -\omega _3 & 0 & \omega _1 \\ -\omega _2 & \omega _1 & 0 \end{array} \right) = \sum\limits_{i=1}^{3} \omega_i \ad X_i. \end{equation} It is straightforward to verify that e.g. $\ad X_1 (X_2) = 2 X_3$, where now $X_2 = \left(0,1,0\right)^T$, $X_3 = \left(0,0,1\right)^T$ and the $\ad$ representation acts in the usual way. The matrix $\mathcal A$ can be expanded in power series (in fact the power series is the definition), \begin{equation} \mathcal A = \left( \frac{1-e^{-\ad Y (\omega)}}{\ad Y (\omega)} \right) = \sum\limits_{n=0}^{\infty} \frac{1}{(n+1)!}\left(-\ad Y(\omega) \right)^n ,\end{equation} remembering that $Y(\omega) \in \mathfrak p$. By using the Cayley-Hamilton theorem, the right hand sum can be written in finitely many powers of the matrix as \begin{equation} c_0 (\omega) +c_1 (\omega) \ad Y(\omega) + c_2 (\omega) \ad Y(\omega)^2. \end{equation} The task of finding the coefficients $c_n (\omega)$ is straightforward, although slightly tedious (NOTE TO SELF: I once wrote a Mathematica function, similar to the MatrixExp function, that does the trick automatically... must remember to post it somewhere and insert a link here). Anyway, the result is \begin{equation} \scriptsize{ \mathcal A = \left( \begin{array}{ccc} \frac{\sinh (2 r) \cos ^2(\phi )}{2 r}+\sin ^2(\phi ) & \frac{\left(r-\frac{1}{2} \sinh (2 r)\right) \sin (\phi ) \cos (\phi )}{r} & \frac{\sinh ^2(r) \cos (\phi )}{r} \\ \frac{\left(r-\frac{1}{2} \sinh (2 r)\right) \sin (\phi ) \cos (\phi )}{r} & \frac{\sinh (2 r) \sin ^2(\phi )}{2 r}+\cos ^2(\phi ) & -\frac{\sinh ^2(r) \sin (\phi )}{r} \\ \frac{\sinh ^2(r) \cos (\phi )}{r} & -\frac{\sinh ^2(r) \sin (\phi )}{r} & \frac{\sinh (2 r)}{2 r} \end{array} \right)}. \end{equation} Using the definition in eq. \eqref{metric.explicit} will then yield the same metric as in part I (up to a factor of 2): \begin{equation}ds^2 = dr^2 + \sinh^2 (r) d\phi^2.\label{metricP2} \end{equation}

Monday, April 16, 2012

Asymptotic geometry of symmetric spaces I: The space of unit area ellipses is the hyperbolic space $\mathbb H^2$

This blog article was supposed to be a transcript of sorts of a seminar talk I gave about my work at Aalto university math department's differential geometry seminar, but I noticed that I was actually rewriting the whole thing from start to finish. This version is primarily aimed at physicists or people who think mathematics in the traditional proof/theorem sense is not a very good method for an introductory exposition (such as myself). I decided to post the whole thing in four separate parts, since it's not very easy to manage big HTML+LaTeX posts... Hopefully I will get some feedback which I may use in the later parts!

The purpose of this first part (out of four) is to give a very easy going description of the geometry of the space of unit area ellipses (with center of mass at the origin), also known as $P_2 = SL(2;\mathbb R)/SO(2; \mathbb R)$, which is isomorphic to the hyperbolic space $\mathbb H^2$. Much of the material may seem to have been pulled out of a magicians hat, but will hopefully become clearer in part II, where I will present the problem a bit more traditionally and carefully by starting from the Lie algebra $sl(2; \mathbb R)$. In part III I will talk about the asymptotic geometry of $P_2$, and the conformal structure on its boundary รก la Fefferman and Graham (see e.g. Graham & Zworski). I will also try to point how and why these "asymptotic conformal symmetries" play an important role in the so-called gauge/gravity dualities. In part IV I will discuss my own work on the asymptotic geometry, conformal structures and the infinite dimensional Lie algebra of asymptotic symmetries on the boundary of $P_3$ (see here for the arXiv version; I must note that I was a bit hasty with the publication and I will try to elaborate the results a bit further in the blog version), which can be understood as a space of unit volume ellipsoids in $\mathbb R^3$. This is a five dimensional noncompact symmetric space of constant negative Ricci scalar curvature, which however is not isomorphic to $\mathbb H^5$.


Figure 1: Some ellipses. Black is the circle, Red is $\mathcal E (1/2,5)$ and green is $\mathcal E (1,3/2)$ (see below for the definition of $\mathcal E$).


Symmetries of the space of unit area ellipses (UAE)


What exactly is a symmetry? We could describe a ball being symmetric under rotations about it's center. Here's an equation for a (unit) ball in \(\mathbb R^2\), i.e. a circle: \begin{equation} x^2 + y^2 = 1.\label{circleeq} \end{equation} The equation \eqref{circleeq} is invariant under a rotation by an angle \(\phi \), \begin{equation} \left( \begin{array}{c} x' \\ y' \end{array} \right) = \left( \begin{array}{c} x \cos\phi + y \sin\phi \\ -x \sin\phi +y \cos\phi\end{array} \right) \label{rotation}. \end{equation} I will define the matrix that does the rotation as (the symbol \( \doteq\) is here used to define a notation) \begin{equation} k(\phi) \doteq \left( \begin{array}{cc} \cos \phi & \sin\phi \\ -\sin\phi & \cos\phi \end{array} \right) \in SO(2;\mathbb R). \end{equation} (I apologize for not making a distinction between a representation and a Lie group element... hopefully that will not cause confusion) I'll just denote the circle as \begin{equation} S^1 =\left\{ (x,y) \in \mathbb{R}^2 \left| x^2+y^2=1 \right. \right\}. \label{circle} \end{equation} The invariance under rotations \eqref{rotation} can then be expressed as \begin{equation} k(\phi)\cdot S^1 = \left\{ (x,y)\in \mathbb{R}^2 \left| {x'}^2+{y'}^2=1 \right. \right\} = S^1 . \end{equation} We could say that the rotation maps the circle into another circle, which is another solution to the equation, albeit identical. Symmetries may in fact be defined as transformations that map the solution space to itself. Here the solution space is zero dimensional, since there is only one solution, which admittedly makes the example rather boring... We could instead consider the space of all UAEs (never mind about what the equation would be, we just need the space). An UAE can be obtained by acting on the unit area circle by a linear transformation which preserves the area, like this: \begin{equation} \left( \begin{array}{c} x' \\ y' \end{array} \right) \doteq \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) \ \ , \ \ a d-b c =1 .\label{speciallinear} \end{equation} We can call the matrix \(g \), after which the right hand side area preserving condition can be expressed simply as \( \det g = 1 \). Invariance of the area is a simple consequence of invariance of the volume form, \( dx' \wedge dy' = dx \wedge dy \) (the volume form is just the integration measure in e.g. \( \int\! f(x,y)dx dy \)). Two by two matrices with unit determinant form the Lie group of type \( SL(2; \mathbb R)\). Let's consider a particular matrix in \( SL(2; \mathbb R)\) of the following type: \begin{equation} h(r) \doteq \left( \begin{array}{cc} e^{-r} & 0 \\ 0 & e^{r} \end{array} \right) . \end{equation} The exponential notation is simply a matter of convenience and will be useful below. Its effect is to shrink the \(x\) coordinate by a factor \( e^{-r}\) and to stretch the y coordinate by \( e^{r}\) (for positive \(r\)), therefore it deforms the circle into a unit area ellipse, explicitly \begin{equation} h(r)\cdot S^1 = \left\{ (x,y) \in \mathbb{R}^2 \left| e^{-2r} x^2+ e^{2r} y^2=1 \right. \right\} . \end{equation} On the other hand, we also have the rotation matrix \( k(\phi)\). Clearly we can get any ellipse by acting on $S^1$ with $k(\phi) h(r)$. It is in fact possible and convenient to express any matrix \(g\) with \( \det g = 1\) in a combination of these kinds of two matrices by using a polar decomposition, \begin{equation} g = k(\phi/2)h(r)k(-\phi/2) k(\phi') \doteq p(r,\phi) k(\phi'), \end{equation} with \begin{equation} p(r,\phi) = \left( \begin{array}{cc} \cosh r -\cos \phi \sinh r & \sin \phi \sinh r \\ \sin \phi \sinh r & \cosh r +\cos \phi \sinh r \end{array} \right).\label{symmspace} \end{equation} Now we can define any ellipse as \begin{equation} \mathcal E (r,\phi) \doteq p(r,\phi) \cdot S^1, \end{equation} since $ k(\phi') \cdot S^1 = S^1$. In fact, we wouldn't need the circle at all: for $r>0$, there's a one to one correspondence between all ellipses and the elements $p(r,\phi)$. The origin in the space of ellipses is the circle $S^1$, which corresponds to $p(0,\phi)$. It is indeed perhaps more intuitive to consider the origin as a point rather than a circle... Now $g$ acts on the space of ellipses by multiplication from the left: $\mathcal E \to g \mathcal E = g p \cdot S^1$. But note that we can always rearrange $g p(r,\phi) = p(r',\phi') k(\phi'')$, and since $k \cdot S^1 = S^1$, the transformation preserves the form $p\cdot S^1$ of the space of ellipses. This is of course just a realization of a homogeneous space: $SO(2)$ is the isotropy group of the circle, and the space of ellipses can be identified with $SL(2)/SO(2)$ (I will always assume the field to be $\mathbb R$ and will therefore often drop it).

Geometry


What is the distance between two ellipses $p(r,\phi)$ and $p(r',\phi')$, or rather, the metric tensor in the space of ellipses? Clearly the metric would need to be invariant under the Lie group $SL(2)$, so we might try to apply the transformations on an arbitrary two dimensional metric and demand invariance. This would lead to a system of PDEs to be solved, which is never fun. It's in fact possible to find the (unique) metric in a more elegant way. For this we need the left invariant Maurer-Cartan form $p^{-1} dp$, i.e. take the differential of $p$ and multiply by the inverse of $p$ from the left. Explicitly, \begin{equation} p^{-1} dp = \left( \begin{array}{cc} -\cos \phi & \sin \phi \\ \sin \phi & \cos \phi \end{array} \right) dr \\ - \sinh r \left( \begin{array}{cc} \sin \phi \cosh r& \cos \phi \cosh r - \sinh r \\ \cos \phi \cosh r + \sinh r & -\sin \phi \cosh r \end{array} \right) d\phi \end{equation} Then, as if by a miracle, we can define the metric as \begin{equation}ds^2 \doteq \frac{1}{2} \tr \left( p^{-1} dp \cdot p^{-1} dp \right) = dr^2 + \sinh^2 (r) d\phi^2.\label{metricP2} \end{equation} Compare this to the Euclidean metric on $\mathbb R^2$ in radial coordinates: \begin{equation} ds^2 = dr^2 + r^2 d\phi^2. \end{equation} The metrics are asymptotically equivalent as $r$ approaches zero, but the metric of eq. \eqref{metricP2} grows exponentially for large $r$. Then the distance between ellipses with same angle is $\mbox{dist}\! \left( p(r_1,\phi), p(r_2,\phi)\right) = |r_1-r_2|$ and the distance between ellipses with same "radius" is $\mbox{dist} \left( p(r,\phi_1), p(r,\phi_2)\right) = |\phi_1-\phi_2| \sinh r$. We can express the metric in possibly slightly more familiar coordinates by \begin{equation}\left\{ \begin{aligned} \xi &= \frac{\sin \phi \sinh r}{\cosh r + \cos \phi \sinh r}\\ \eta &=\frac{1}{\cosh r + \cos \phi \sinh r} \end{aligned}\label{coords.poincare} \right. \end{equation} which results in the metric \begin{equation} ds^2 = \frac{1}{\eta^2} \left(d\xi^2 + d \eta^2 \right), \end{equation} which is the familiar Poincare upper half place metric of the hyperbolic space $\mathbb H^2$.

That's all for now. In the next part I'll discuss symmetric spaces via a Cartan decomposition of the Lie algebra $sl(2; \mathbb R)$.

Friday, April 6, 2012

Hi!

OK so the idea here is to write about all kinds of math and physics related stuff and provide supplementary information and introductory posts about my own work. I'll also convert seminar talks etc. into blog posts here and whatever else springs to my mind. My desire to start a blog started after I noticed that I screwed up slightly in one of my publications, which I wanted to correct (it's not a big enough thing to warrant an erratum).

Anyway, time for some MathJax testing:

 $$g^{-1}dg = X_i \mathcal {A^{i}}_j d\omega^j$$
Inline math test \( \int dx \).

Yay! I'm ready to go!

Like the background image? I made it myself by copy/pasteing some "fanciest" formulae from my notes (on a tablet PC), rotated and projected it and did some other manipulations. Nice, eh? ;)