Disclaimer: I am not a mathematician, just a young neuroscientist trying to understand a paper. So forgive me if I have horribly misunderstood something.

So in order to understand this paper (that involves using a Riemannian kernel for a support vector machine: https://hal.archives-ouvertes.fr/hal-00820475/document) I am trying to understand some of the maths behind part of the methods. I am still not entirely sure what a differentiable manifold is, despite having watched a number of video lectures on the topic (It's a topological space which is locally homomorphic to Euclidean space - with continous transfer functions?). But more importantly: the paper states that the space of symmetric positive definite matrices forms a differentiable manifold. This seems important as the we need to calculate the distances between covariance matrices (which by definition live in the SPD space) in order for the SVM to perform well.

So, to me, it seems that the author's of the paper achieved impressive SVM classification results because their similarity (or distance) measure was more suited to the space in which they were working. ie a Riemannian kernel in a differentiable manifold. Is it just a happy coincidence that the space of SPD matrices forms this.. mathematical object within which we can perform better similarity measures?

I suppose my entire confusion can be summed up as: Why does a collection of a specific type of matrix (SPD) form this other topological space with specific properties that make it more efficient for measuring distances between points (matrices) in the space?

  • 1
    $\begingroup$ I am sorry, SPD matrices really sounds funny. $\endgroup$ May 8 '18 at 19:16
  • 1
    $\begingroup$ Hahaha whoops.. $\endgroup$
    – bidby
    May 8 '18 at 19:23

The set of symmetric matrices is a finite dimensional vector space. Any finite dimensional vector space is a manifold (just choose a basis to obtain a global chart). Let us denote the set of symmetric $n\times n$ matrices by $S$.

Now to be a positive definite matrix you will need that all the eigenvalues are positive. But this is an open condition: If a matrix $M$ has strictly positive (real part) eigenvalues, so do the eigenvalues of all nearby matrices. This means that SPD matrices form an open subset of the space of symmetric matrices. Open subsets of manifolds are manifolds.

  • $\begingroup$ Does your last statement hold for differentiable manifolds too? $\endgroup$
    – Călin
    May 13 '19 at 12:13
  • 1
    $\begingroup$ Yes, almost by definition. $\endgroup$
    – Thomas Rot
    May 13 '19 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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