Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a dimension $n$, the space of curvature operators is the space $S^2_B(\Lambda^2\mathbb{R}^n)$ of symmetric endomorphisms $R$ of $\Lambda^2\mathbb{R}^n$ which satisfy the first Bianchi identity :

Given $x,y,z,t\in\mathbb{R}^n$, $\langle R(x\wedge y),z\wedge t\rangle+\langle R(y\wedge z),x\wedge t\rangle+\langle R(z\wedge x),y\wedge t\rangle=0$

This the space where the curvature tensor of a Riemannian manifold lives.

It is naturally equipped with the following action of $O(n,\mathbb{R})$ :

$\langle g.R(x\wedge y),z\wedge t\rangle =\langle R(gx\wedge gy),gz\wedge gt\rangle$

which splits into irreducible components :

$S^2_B(\Lambda^2\mathbb{R}^n)=\mathbb{R}Id_{\Lambda^2\mathbb{R^n}}\oplus S^2_0\mathbb{R}^n\wedge id_{\mathbb{R}^n} \oplus \mathcal{W}$

where $\wedge$ is the Kulkarni-Nomizu product and $\mathcal{W}$ is the space of Weyl tensors, that is tensors in $S^2_B(\Lambda^2\mathbb{R}^n)$ whose traces are all zero.

In my research on Ricci flow, I am investigating Hamilton's ODE, which is ODE on $S^2_B(\Lambda^2\mathbb{R}^n)$. I am currently doing some numerical exploration, and haven't come with a good way of getting a "random" initial condition. I would like this choice to be invariant under the action of $O(n,\mathbb{R})$. This implies that we can treat each component of the previous decomposition separately. My question then splits in two subquestions :

  • Q1: To treat the first two parts of the decomposition, I just need to generate a random symmetric matrix on $\mathbb{R}^n$ in an $O(n,\mathbb{R})$ invariant way. I believe this is classic "random matrix theory", but I am unfortunately totally ignorant about this field. What would be an efficient algorithm to solve this problem ?
  • Q2: How to treat the Weyl part ? Can we design an efficient algorithm for generating a Weyl tensor ?

PS: I know that the first question must be already treated somewhere, but while googling "random symmetric matrix" gave me interesting information, I wasn't able to recover an algorithm from that.

PS2: If that helps, the software I'm using is FreeMat, an open source clone of matlab.

share|improve this question
1  
Instead of generating each irreducible piece separately, isn't it easier to just generate a random 4th order tensor and project it orthogonally into the space of curvature tensors? –  Deane Yang Jan 21 '13 at 5:33
    
That might surely do the job. However, I'm worried about the computational cost and the $O(n,\mathbb{R})$ invariance. Maybe the best solution is to generate a random element $S^2(\Lambda^2\mathbb{R}^n)$ in an $O(\Lambda^2\mathbb{R}^n)$ invariant way, and then to project it to $S^2_B(\Lambda^2\mathbb{R}^n)$. But this needs an answer to Q1 ! –  Thomas Richard Jan 21 '13 at 5:48
    
Q1: Generate a random symmetric matrix any way you want and then hit it with a random element of O(n): en.wikipedia.org/wiki/Orthogonal_matrix#Randomization –  John Wiltshire-Gordon Jan 21 '13 at 6:08
    
To generate random vectors or tensors, you need to choose a probability measure on the vector- or tensor-space. For example, if you want to generate a random vector on $\mathbb{R}^n$, you can just use $n$ independent Gaussian random variables. This will be $O(n)$-invariant. Presumably, to generate a random 4th order tensor, you need to use $n^4$ independent random Gaussian variables. This would be costly if you want to do it in dimension 10 but presumably dimension 3 and 4 would be interesting enough? –  Deane Yang Jan 21 '13 at 6:16
1  
@Thomas Richard All I mean is the following lemma: Suppose a compact group G acts on a probability space X (without preserving measures). Then we obtain a new measure on X which is G-invariant: to integrate a function on X, pull it back along the action map GxX-->X and integrate using the product measure. –  John Wiltshire-Gordon Jan 21 '13 at 7:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.