Identifying Lorentz Covariant Equations Statement: $\phi , A^{\mu}, T^{\mu \nu}$ are a Lorentz scalar, vector, and tensor. Which of the following equations are Lorentz covariant.
a. $\phi = A_{0}$
b. $\phi = A^{\mu}A_{\mu}$ 
c. $\phi = A_{0}A^{0}$
d. $\phi = T_{\mu \nu}T^{\mu \nu}$
e. $T_{\mu \nu} = T^{\nu \mu}$
f. $T_{\mu \nu} = T_{\nu \mu}$
g. $T^{\mu \nu} = A^{\mu}+A^{\nu}$
h. $T_{\mu \nu} = -T_{\mu \nu}$
i. $T_{\mu}^{\nu}=-T_{\nu}^{\mu}$
j. $T_{\mu \nu} = A^{\mu} A^{\nu}$
k. $\phi = det( T^{\mu \nu})$
l. $\phi = det( T_{\mu}^{\nu})$
From my (limited) understanding of Lorentz covariance I would identify b. and d. as Lorentz covariant, but I'm having trouble understanding how I would go about determining in general whether an equation is Lorentz covariant. I would appreciate any recommendations for non-group theory reading materials on this, or just help in general. Thanks in advance.
 A: a. $\phi = A_{0}$
This asserts "The scalar $\phi$ is equal to the zeroth component of the vector $\bf A$ in the given reference frame." Not LC.
b. $\phi = A^{\mu}A_{\mu}$
says "The scalar $\phi$ is equal to the scalar product of the vector $\bf A$ with itself." This is LC.
c. $\phi = A_{0}A^{0}$
says "The scalar $\phi$ is equal to the product of the zeroth covariant component of the vector $\bf A$ with the zeroth contravariant component of the vector $\bf A$." Not LC.
d. $\phi = T_{\mu \nu}T^{\mu \nu}$
says "the scalar $\phi$ is equal to the complete contraction of the second rank tensor $\bf T$ with itself". This is LC.
e. $T_{\mu \nu} = T^{\nu \mu}$
says "the covariant components of $\bf T$ are equal to the swapped contravariant components of $\bf T$, in the given reference frame".
Not LC
f. $T_{\mu \nu} = T_{\nu \mu}$
says "$\bf T$ is a symmetric second rank tensor." This is LC.
g. $T^{\mu \nu} = A^{\mu}+A^{\nu}$
says "it so happens that, in the given reference frame, the components of the second rank tensor $\bf T$ are equal to a sum of pairs of components of the vector $\bf A$" Not LC
h. $T_{\mu \nu} = -T_{\mu \nu}$
says $2 T_{\mu \nu} = 0$ and therefore the tensor $\bf T$ is zero. This is LC
But perhaps you meant $T_{\mu \nu} = -T_{\nu \mu}$.
This says "$\bf T$ is antisymmetric". This is LC
i. $T_{\mu}^{\nu}=-T_{\nu}^{\mu}$
This one is ambiguous because the order of the indices is not clear, but in any case it is not LC. It seems to be saying that the tensor changes sign if one index is raised and the other lowered, which might be true in some given reference frame but it is not a well-formed tensor expression.
j. $T_{\mu \nu} = A^{\mu} A^{\nu}$
This would be LC if the up/down placement of the indices agreed, but as they do not, it is not LC.
k. $\phi = det( T^{\mu \nu})$
A determinant will usually change from one frame to another, so this is not LC.
l. $\phi = det( T_{\mu}^{\nu})$
Like k.
If in doubt, you can check the above by applying a coordinate transformation to each part of any given equation, and see if you get back the same equation. But the quick way is simply to learn the rules of well-formed tensor expressions.
To get an intuition about this, consider the case of a rotation. This often makes things obvious. For example, if some invariant scalar $\phi$ was equal to a given component of a vector in some given frame, then after rotating the coordinate axes the component will change but the scalar will not. So that takes care of (a), and (c) is similar. In case (e) note that contravariant components transform differently to covariant components. And so on.
A: b,d,f,h,i are Lorentz covariant (LC) equations.
Equations that equate a scalar to a scalar, or V to V, or T to T are LC.
Complete contraction of a vector or a tensor forms a scalar.
e and j hold any Lorentz system because the Lorentz transformation is linear, where indices up and down don't matter.
But in any theory where covariant and contravariant tensors differed, e and j would not be covariant.  Because of this, I think it is best not to call them 'Lorentz covariant' since it seems to change the meaning of covariant in that case.
k and l are a bit tricky. A definition of the determinant is
$\epsilon'^{\mu'\nu'\rho'\sigma'}{\rm Det[L]}=L^{\mu'}_{\mu} L^{\nu'}_{\nu} L^{\rho'}_{\rho} L^{\sigma'}_{\sigma}\epsilon^{\mu\nu\rho\sigma}.$
That makes it look like a scalar, so k and l should be considered covariant.
A: As someone said in the comments, you should explain what your understanding of Lorentz covariance is. I'm going to assume the Wikipedia point of view.
Any equation you (rightfully) write with a scalar on one side is Lorentz covariant. In fact, a scalar is independent of the coordinates used. So, assuming that the equations are right, a, b, c, d, k, l are all covariant.  
To extend this to tensors of nonzero rank, an equation is Lorentz covariant if it transforms consistently: if on the left you have a type (h,k) tensor, so you must have on the right side: both sides have to transform the same way. With this definition, f, g, h would all be covariant. For this same reason e, j would be wrong: to transform one side you'd need two direct matrices, on the other side two inverse matrices.
For i, since on both sides are present type (1,1) tensors, it's better to manually transform them to check:
${_{\mu}}^{\nu}=−{_{}}^{\mu}$
Now I'm going to use the transformation matrix $\Lambda$ and its inverse $\Lambda^{-1}$, transforming the left side and checking the effects of the transformation on the right side.
${_{}}^{\nu}\Lambda^{\lambda}_{\nu}{\Lambda^{-1}}^{\mu}_{\sigma}=−{_{}}^{\mu}\Lambda^{\lambda}_{\nu}{\Lambda^{-1}}^{\mu}_{\sigma}$
${}_{\sigma}^{\lambda}=−{_{}}^{\mu}\Lambda^{\lambda}_{\nu}{\Lambda^{-1}}^{\mu}_{\sigma}$
${}_{\sigma}^{\lambda}g^{\nu\nu}g_{\lambda\lambda}g^{\sigma\sigma}g_{\mu\mu}=−{_{}}^{\mu}\Lambda^{\lambda}_{\nu}{\Lambda^{-1}}^{\mu}_{\sigma}g^{\nu\nu}g_{\lambda\lambda}g^{\sigma\sigma}g_{\mu\mu}$
${}_{\lambda}^{\sigma}g^{\nu\nu}g_{\mu\mu}=−{_{}}^{\mu}\Lambda^{\nu}_{\lambda}{\Lambda^{-1}}^{\sigma}_{\mu}$
${}_{\lambda}^{\sigma}g^{\nu\nu}g_{\mu\mu}=−{}_{\lambda}^{\sigma}$
Hence, since the transformed equation is wrong, the equation isn't covariant.
