Converting contravariant (or covariant) tensors to mixed tensors

Sirs,

As a physicist studying General Relativity and Quantum Field Theory, I feel like I have my head wrapped around upstairs/downstairs notation fairly well. However, my questions is as follows:

Is it possible to convert a completely contravariant tensor such as $T^{\mu\nu}$ into a mixed variance tensor, such as $T^{\mu}_{\nu}$ using the Minkowski metric (or any other metric, for that matter)? If so, how would the signs of the entries change when converting from $T^{\mu\nu}$ to $T^{\mu}_{\nu}$?

To clarify a bit, I already understand that if you have a contravariant vector (which I understand is just a 1-dimensional tensor) such as $A^{\rho}$ and you hit it on the left with the Minkowski metric $g_{\rho\mu}$ you get:

$g_{\rho\mu}A^{\rho} = A_{\mu}$

where:

$A^{\rho} = {A^{0}, A^{1}, A^{2}, A^{3}}$

and:

$A_{\mu} = <A_{0}, A_{1}, A_{2}, A_{3}> = <A^{0}, -A^{1}, -A^{2}, -A^{3}>$

I guess, to put it another way, how does all of these sign conventions spell out when you list out the element of $T^{\mu}_{\nu}$ explicitly, if all of the entries of the original $T^{\mu\nu}$ were positive?

Most texts on upstairs/downstairs notation seem to either not address the signs of the entries of mixed tensors at all, or just address the sign convention for vectors and take mixed tensors as granted and proceed.

$T^a{}_b = T^{ac} g_{cb}$ is just a matrix product if you write it out in components: $$\begin{pmatrix} T^0{}_0 & T^0{}_1 & T^0{}_2 & T^0{}_3 \\ T^0{}_0 & T^0{}_1 & T^0{}_2 & T^0{}_3 \\ T^0{}_0 & T^0{}_1 & T^0{}_2 & T^0{}_3 \\ T^0{}_0 & T^0{}_1 & T^0{}_2 & T^0{}_3 \end{pmatrix} = \begin{pmatrix} T^{00} & T^{01} & T^{02} & T^{03} \\ T^{10} & T^{11} & T^{12} & T^{13} \\ T^{20} & T^{21} & T^{22} & T^{23} \\ T^{30} & T^{31} & T^{32} & T^{33} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} .$$ So the entries in the first column keep their signs, the rest are negated.
Similarly if you lower the left index ($T_a{}^b = g_{ac} T^{cb}$), but then it's the first row that is unchanged; you can write that out for yourself.
• One should note that in this example, Hans seem to have identified the metric $g_{cb}$ with the Minkowski metric, and in the case of General Relativity, this would not (in general) have the form of the diagonal matrix Hans used. – Muphrid Aug 18 '16 at 0:27
• I did not read it that way. One part of the question reads, "Is it possible to convert...using the Minkowski metric (or any other metric, for that matter)?" It is true that the OP uses $g_{\mu \nu}$ to mean the Minkowski metric at one point as well, however, but I do not feel that the question is confined only to the case where the spacetime metric is the Minkowski metric. – Muphrid Aug 19 '16 at 1:36