Pseudo-Riemannian tensor and Morse index

[Jimmy Snyder]

Let $g_{ij}$ be a tensor, where $0 \leq i,j \leq n$. The Morse index $\mu$ is the number of negative eigenvalues of g. On page 469 of Eberhard Zeidler's QFT III: Gauge Theory, it says that g is Riemannian if $\mu = 0$ and pseudo-Riemannian if $0 < \mu < n$. Is this correct? If so, what kind of tensor is it when $\mu = n$?

 

[I like Serena]

A Riemannian manifold requires the metric tensor to be positive-definite.
Positive-definite is equivalent to all eigenvalues being positive.
If one or more is negative, the tensor is indefinite.
It can still be used for a pseudo-Riemannian manifold.

The reason I can think of for a distinction for $\mu = n$, is that the tensor is negative-definite.

 

[Matterwave]

I'd imagine that if you had all negative eigenvalues, you could always just pick the negative of the metric and you'd have a Riemannian manifold again. I'm not sure such a sign would be physical from a practical point of view.

From a mathematical point of view, I'm not sure what it would mean.

 

[quasar987]

Let $g_{ij}$ be a tensor, where $0 \leq i,j \leq n$. The Morse index $\mu$ is the number of negative eigenvalues of g. On page 469 of Eberhard Zeidler's QFT III: Gauge Theory, it says that g is Riemannian if $\mu = 0$ and pseudo-Riemannian if $0 < \mu < n$. Is this correct? If so, what kind of tensor is it when $\mu = n$?

Are you certain you have this straight? What you call the Morse index is usually just called the signature of g. The signature is a well-defined concept for any symmetric bilinear form on a vector space. The Morse index on the other hand, is a number associated with a critical point p of a Morse function f on a manifold M. It is defined as the signature of the Hessian of f at p. The Hessian of f at p is the bilinear form whose matrix wrt some coordinate chart is the matrix of second partial derivatives. It is a symmetric bilinear form obviously so we may speak of its signature. (check) The Morse condition on f just means that this bilinear form is nondegenerate so it can have any signature btw 0 and n.

It just seems very strange to me why anyone would call the signature of a bilinear form the Morse index as there is nothing at all "Morsy" about it afaik.

 

[blue_raver22]

I can confirm that the statement on page 469 of Eberhard Zeidler's QFT III: Gauge Theory is correct. The Morse index, denoted by \mu, is a mathematical concept used to determine the number of negative eigenvalues of a tensor. In the case of a tensor g_{ij} with 0 \leq i,j \leq n, if \mu = 0, then the tensor g is considered to be Riemannian. This means that all of its eigenvalues are non-negative, indicating a positive-definite metric.

On the other hand, if 0 < \mu < n, then the tensor g is considered to be pseudo-Riemannian. This means that it has both positive and negative eigenvalues, indicating an indefinite metric. This type of tensor is commonly used in the study of General Relativity, where the metric tensor describes the curvature of spacetime.

Finally, when \mu = n, the tensor g is considered to be degenerate. This means that all of its eigenvalues are zero, indicating a singular metric. In this case, the tensor g cannot be used to define a metric and is not considered to be either Riemannian or pseudo-Riemannian.

In summary, the Morse index provides a useful tool for classifying tensors based on their eigenvalues and determining the type of metric they represent. It is an important concept in the study of Riemannian and pseudo-Riemannian geometry and has many applications in theoretical physics, particularly in the field of General Relativity.