# Raising and Lowering Indices and metric tensors

1. Jun 17, 2015

### 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?

2. Jun 17, 2015

### 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.

3. Jun 18, 2015

### Fredrik

Staff Emeritus
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 $V^*$.

4. Jun 20, 2015

### 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

5. Jun 21, 2015

### 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.