Geometric difference between T_{a}^{b} and T^{a}_{b} tensors

[quickAndLucky]

What is the difference between ${T{_{a}}^{b}}$ and ${T{^{a}}_{b}}$ ? Both are (1,1) tensors that eat a vector and a dual to produce a scalar. I understand I could act on one with the metric to raise and lower indecies to arrive at the other but is there a geometric difference between the objects to which these are components?

 

[andrewkirk]

I don't think there's a geometric difference. The difference is that, where $V$ and $V^*$ are the underlying vector space and its dual, $T_a{}^b$ is a function from $V\times V^*$ to $\mathbb R$ and $T^a{}_b$ is a function from $V^*\times V$ to $\mathbb R$. Or if we think of the tensor as being a function that takes two arguments, $T_a{}^b$ takes a vector as its first argument and $T^a{}_b$ takes a dual vector as its first argument.

 

[Orodruin]

Generally, they are different (1,1) tensors (you could imagine sidestepping #2 by defining the obvious equivalence between functions from $V\times V^*$ and functions from $V^* \times V$). It is certainly not true that $T^a_{\phantom ab} \omega_a V^b = T^{\phantom ba}_b \omega_a V^b$ except in the case where $T^a_{\phantom ab} g_{ac}$ is symmetric.

 

[sweet springs]

$T_{\ \ \ \nu}^\mu=g_{\nu\xi}T^{\mu\xi}$ and

$T^{\ \nu}_\mu=g_{\mu\xi}T^{\xi\nu}$

are different in general. Obviously when T is symmetric

$T^\mu_{\ \ \ \nu}=T^{\ \nu}_\mu$ and they can be denoted as $T^{\nu}_\mu$.