Do Metric Tensors Always Have Inverses?

[jv07cs]

I am reading about musical isomorphisms and for the demonstration of the index raising operation from the sharp isomorphism, we have to multiply the equation by the inverse matrix of the metric. Can we assume that this inverse always exists? If so, how could I prove it?

 

[anuttarasammyak]

In the cases of relativity, transformation of spacetime coordinates from one Frame of Reference to another should have inverse transformation which corresponds to another to one. But I have no idea of inverse of metric tensor. How do we use it ?

You say about music and I am sorry to have scarce knowledge of metric tensor in music. If is is just an ordinary matrix, its determinant would not be zero to have an inverse matrix.

 

[jv07cs]

In the cases of relativity, transformation of spacetime coordinates from one Frame of Reference to another should have inverse transformation which corresponds to another to one. But I have no idea of inverse of metric tensor. How do we use it ?

You say about music and I am sorry to have scarce knowledge of metric tensor in music.

The musical isomorphisms are just isomorphisms between the vector space and its dual that lead us to the operations of lowering and raising index. The term musical is because we use the symbols flat $\flat$ and sharp $\sharp$ to denote them (kind of an analogy with "flat" meaning lower in pitch and "sharp" meaning higher in pitch, and the idea of lowering and raising indices).

In this specific case we take a vector space endowed with a metric g and define the following flat isomorphism:
$$^\flat: V \rightarrow V^*$$ $$v \mapsto v^\flat$$

where $v^\flat \in V^*$ is defined as:
$$v^\flat(w) = g(v,w)$$
We can show that this is indeed an isomorphism and we get to the relation $v_i = g_{ij}v^j$, the lowering index operation, where $v_i$ are the components of $v^\flat$.

The $\sharp$ isomorphism is then defined as the inverse of the flat isomorphism: $\sharp = \flat^{-1}$ and we can arrive at the raising index operation by basically multiplying both sides of $v_i = g_{ij}v^j$ by the inverse matrix $g^{ik}$ and we arrive at $v^k = g^{ik}v_i$.

My question is regarding this process. Because for us to multiply by the inverse matrix we first have to assume that there is one and my question is if we can state that a metric always has an inverse? Or if, for this demonstration, one should include "assuming that g has an inverse"?

 

[anuttarasammyak]

Thanks for explaining in detail.
I see it and assume that
g_{ik}g^{kj}=\delta_i^j
g_{ik}=g_{ki},g^{kj}=g^{jk}
If so as I noted in #2 determinant g of {g_ik} should not be zero to have {g^kj}, e.g. g=-1 for SR.

But I am afraid my primitive knowledge does no good to your high mathematics.

 

[jv07cs]

Thanks for explaining in detail.
I see it and assume that
g_{ik}g^{kj}=\delta_i^j
g_{ik}=g_{ki},g^{kj}=g^{jk}
If so as I noted in #2 determinant g of {g_ik} should not be zero to have {g^kj}, e.g. g=-1 for SR.

But I am afraid my primitive knowledge does no good to your high mathematics.

I just found out that there is a theorem that states that, for finite dimensional vector spaces, a "bilinear form is degenerate if and only if the determinant of the associated matrix is zero", which would mean that the matrix associated with a non-degenerate bilinear form is non-singular. Since the metric tensor is a symmetric non-degenerate bilinear form, it will always have an inverse. This answers my question.

Thank you very much for the replies.

 

[jbergman]

I am reading about musical isomorphisms and for the demonstration of the index raising operation from the sharp isomorphism, we have to multiply the equation by the inverse matrix of the metric. Can we assume that this inverse always exists? If so, how could I prove it?

In pseudo-riemannian geometry the metric is non-degenerate, so yes, an inverse always exists.

 

[haushofer]

Isn't this inverse defined by identifying the vector space with its double dual?

 

[jbergman]

Isn't this inverse defined by identifying the vector space with its double dual?

Not sure your question, but if you have a basis and a dual basis, then you can identify the double dual basis and have a way to map vectors and covectors to each other. However, this correspondence depends on a choice of basis. In other words, the inverse will vary with a different basis, I believe.

With a metric tensor, you can map covectors and vectors to each other in a basis independent way via the musical isomorphisms.

 

[WWGD]

I think the non-degeneracy of the tensor in its matrix representation implies its invertibility.