When does $T^{ij} = T_{ij}$? Suppose we have some tensor with components $T^{ij}$. Then suppose that we also have $T_{ij}$. 
When would $T^{ij}T_{ij} = (T^{ij})^2 = (T_{ij})^2$?
 A: Note; I'll use the summation convention throughout here.
In the context of differential geometry, the indices on tensorial objects are raised and lowered with the metric on the space (manifold) being studied.  So for example
$$
  T^i_{\phantom i j} = g^{ik}T_{kj} 
$$
and
$$
  T^{ij} = g^{ik}g^{jl}T_{kl}
$$
Notice that if the metric is simply that of Euclidean space, namely if $g_{ij} = \delta_{ij}$, then raising and lowering does not change the numerical values of tensor components.  In particular, one would have
$$
  T^{ij} = T_{ij}
$$
Notice that in expressions like
$$
  T^{ij}T_{ij}
$$
both indices are being summed over, where as in the expressions
$$
  (T^{ij})^2, \qquad (T_{ij})^2
$$
one usually (this is actually a matter of notational preference) doesn't intend for their to be any implied summation, so typically its notationally safe to assume that
$$
  T^{ij}T_{ij}\neq (T^{ij})^2, \qquad T^{ij}T_{ij}\neq (T_{ij})^2
$$
but if the metric satisfies $g_{ij} = \delta_{ij}$, then it is true that
$$
  (T^{ij})^2=(T_{ij})^2
$$
A: i,j run only over space indexes and we are working in realms of special relativity then all of them will have same signature take metric to be g=diag(1(time),-1(x),-1(y),-1(z)). Hence does not make a diff if you write it in top or bottom. And in fact (i,j) are latin symbols which when used suggest space indexes. And ($\mu,\nu$) i.e. greek indices when written suggest space and time indexes
