Ask Your Question

Is SageManifolds adequate to work with homogeneous Riemannian manifolds?

asked 2019-04-08 11:47:03 +0200

emiliocba gravatar image

updated 2019-04-08 23:34:36 +0200

slelievre gravatar image

I have just discovered the SageManifolds Project, which computes several objects from differential geometry. I admit I haven't studied in details the tutorials, because before spending several hours doing that I want to be sure that it is going to help me in my purpose.

I want to compute the Riemann curvature tensor of compact homogeneous Riemannian manifolds. Roughly speaking, each of those spaces has the following ingredients:

  1. The differential manifold is given by $M=G/K$, where $G$ is a compact Lie group and $K$ is a closed subgroup of $G$.

  2. At the Lie algebra level, $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ with $\mathfrak{p}$ an $\textrm{Ad}(K)$-invariant subspace of $\mathfrak{g}$. The tangent space $T_{eK}G/K$ is naturally identified with $\mathfrak{p}$.

  3. An $\textrm{Ad}(K)$-invariant inner product $\langle \cdot,\cdot\rangle$ on $\mathfrak{p}$.

Thus, the Riemannian metric on $G/K$ is obtained by translating the inner product $\langle \cdot,\cdot\rangle$ on $T_{eK}G/K \simeq \mathfrak{p}$ to any $T_{gK}G/K$ by the map $xK\mapsto gxK$ (which becomes an isometry).

The main point with these spaces is that the whole geometry is determined by $\mathfrak g$ and $\langle\cdot,\cdot\rangle$, and it is not necessary to deal with charts.

In particular, one determines any curvature object (Riemann curvature tensor, Ricci tensor, Scalar curvature, etc) only at the point $eK$.

In most of the examples that I quickly see in the tutorials of SageManifolds begins by defining charts.

How can I work on a homogeneous Riemannian manifold without defining charts?

It would be very useful to count with a simple example, say $G=SU(2)$, $K={1}$, and the inner product on $\mathfrak p=\mathfrak g$ has orthonormal basis ${aX_1,bX_2,cX_3}$ where $a,b,c$ are positive numbers and $$ X_1 = \begin{bmatrix} i & 0 \newline 0 & -i \end{bmatrix}, \qquad X_2 = \begin{bmatrix} 0 & 1 \newline -1 & 0 \end{bmatrix}, \qquad X_3 = \begin{bmatrix} 0 & i \newline i & 0 \end{bmatrix}. $$

edit retag flag offensive close merge delete

1 Answer

Sort by ยป oldest newest most voted

answered 2019-04-08 13:24:42 +0200

eric_g gravatar image

updated 2019-04-08 13:25:17 +0200

In the current setting, all calculations on manifolds end as calculations on charts. Nothing specific to homogeneous manifolds has been implemented yet. Would you be interested in developing such a part? Note that the implementation of Lie groups has started, see here.

edit flag offensive delete link more


Well, I could try to help with the mathematical background, but my program skills are pretty poor, so somebody should optimize every algorithm

emiliocba gravatar imageemiliocba ( 2019-04-08 18:21:31 +0200 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower


Asked: 2019-04-08 11:47:03 +0200

Seen: 332 times

Last updated: Apr 08 '19