Raising and Lowering Indices and metric tensors

[Physicist97]

The metric tensor has the property that it can raise and lower indices, but this is on the assumption that it (the metric) is symmetric. If we were to construct a metric tensor that was non-symmetric, would it still raise and lower indices?

 

[WWGD]

I think raising and lowering follow from the fact that $T^k_{l+1}V$ ( k covariant, (l+1) contravariant) is canonically isomorphic to $(V^{*} \times V^{*} \times...\times V^{*}) \times V\times \times...\times V$ (l copies of $V^{*}$, k copies of $V$). So I think you only need the metric to be positive definite for the canonical isomorphism to hold.

 

[Fredrik]

If $v\in V$, and $g:V\times V\to\mathbb R$ is a metric, we can denote the map $u\mapsto g(v,u)$ from $V$ into $\mathbb R$ by $g(v,\cdot)$. This is an element of $V^*$. The map $v\mapsto g(v,\cdot)$ is an isomorphism from $V$ to $V^*$. Since a metric is symmetric, the map $v\mapsto g(\cdot,v)$ is the same isomorphism from $V$ to $V^*$. The problem with a non-symmetric "metric" is that these would be two different isomorphisms from $V$ to $V^*$.

 

[Matterwave]

You can make whatever definition of raising/lowering indices you want. But if the definition is not useful, people might not use it. Using a non-symmetric "metric" may not be very useful, but if you can find a use for it, then go ahead and develop the machinery! :D

 

[WWGD]

Well, yes, but my point that the symmetric positive-definite form allows you to produce a natural isomorphism ( as k-linear maps)
between the initial tensor and the contracted one. This is what allows you to contract indices: you are substituting a tensor by an
isomorphic copy of it.