# Is a Riemannian metric positive definite or positive semidefinite?

From Wikipedia

The Fisher information matrix is a N x N positive semidefinite symmetric matrix, defining a Riemannian metric on the N-dimensional parameter space,

But a Riemannian metric is defined to be a family of (positive definite) inner products.

So I was wondering whether a Riemannian metric is positive definite or positive semidefinite?

Thanks and regards!

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They need not refer to the same "metric." Take the Minkowski 3-space with form $x^2 + y^2 - z^2$ and the surface $z = \sqrt{1 + x^2 + y^2}.$ The form in three variables, restricted to (vectors tangent to) the indicated surface, actually is positive definite, the result being the hyperbolic plane. en.wikipedia.org/wiki/Hyperboloid_model You could be looking at an analogous situation. – Will Jagy May 4 '13 at 22:08

## 2 Answers

A Riemannian metric is positive definite and produces a Riemannian manifold. If you take the metric tensor to be positive semidefinite then you get what is called a semi-Riemannian manifold or pseudo-Riemannian manifold. With regards to the Fisher information metric on a statistical manifold it is usually assumed that the metric is positive definite. I would not trust Wikipedia. The standard reference on information geometry is Amari's Methods of Information Geometry.

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Thanks! "With regards to the Fisher information metric on a statistical manifold it is usually assumed that the metric is positive definite." Is Fisher information matrix positive definite or positive semidefinite then? – Tim May 4 '13 at 19:35
Tim, I posted a second answer in response to your comment so that I can fit in what I have to say. – Wintermute May 5 '13 at 15:03

It is usually assumed to be positive definite but you can have semi positive definite if you like. This second condition will change the manifold structure as I stated above. For example on the normal distribution the metric is given by $$\begin{pmatrix}\frac{1}{\sigma^2}&0\\ 0&\frac{2}{\sigma^2}\\ \end{pmatrix}$$ Since you have the restriction $\sigma > 0$ this matrix is always positive definite. In the case of the Weibull distribution the metric is given by $$\begin{pmatrix}\frac{k^2}{\lambda^2}&\frac{\gamma-1}{\lambda}\\ \frac{\gamma-1}{\lambda} & \frac{\gamma^2 -2\gamma +\frac{\pi^2}{6}+1}{\lambda^2}\\ \end{pmatrix}$$ Here is is possible to pick $\gamma$ so that the bottom right hand corner is 0. This will make the metric tensor semi positive definite. People will usually place a resriction so that you can't choose such a $\gamma$.

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Here is is possible to pick γ so that the bottom right hand corner is 0... Sure about that? – Did May 5 '13 at 15:18