Doubt in Lorentz Transformation I've tried to do the following exercise:

Show that $\sum_{\mu} D^{\mu\mu}$ and $\sum_{\mu}D_{\mu\mu}$ are not invariant under Lorentz transformations but $\sum_{\mu} D^{\mu}_{\mu}$ are.

I've had the following doubts:
I had to write to matrix for Lorentz Transformations because the object under consideration had two index, for example: $\sum_{\mu}\Lambda^{\mu}_{\nu}\Lambda^{\theta}_{\gamma}D^{\mu\mu}$
right?
The following sequence is wrong: $\sum_{\mu}\Lambda^{\mu}_{\mu}\Lambda^{\mu}_{\mu}D^{\mu\mu}$ ?
I don't have much experience with index, so please if anyone could help me I appreciate.
 A: Omitting the sum signs and using some general properties of Lorentz transformations and traces in the following. Lorentz transformations  fulfill $$\Lambda_{\,\alpha}^{ \mu } g_{\mu\nu} \Lambda_{\,\beta}^{ \nu}=g_{\alpha \beta}.$$ 
Recalling the definition of the trace $\mathrm{ Tr}(A_{\alpha}^{\,\beta})=A_{\alpha}^{\,\alpha} $ one can see from above that $\Lambda^{-1}\Lambda =1$ or in components $\Lambda_{\,\alpha}^{ \mu }\Lambda_{\,\mu}^{ \beta} =(\Lambda_{\alpha}^{\, \mu })^{-1}\Lambda_{\,\mu}^{ \beta} = \Lambda_{\,\mu}^{ \beta}(\Lambda_{\alpha}^{\, \mu })^{-1} = \delta^{\beta}_{\alpha}  $.
Now in your case $D^{\mu}_{\,\mu}$ can be written as $\mathrm{Tr}(D^{\mu}_{\,\nu})$. Now Lorentz transformations can be applied on each index separately giving $$\mathrm{Tr}(D^{\mu}_{\,\nu}) \rightarrow \mathrm{Tr}(\Lambda_{\,\mu}^{ \alpha}D^{\mu}_{\,\nu}\Lambda_{\,\beta}^{ \nu})=  \Lambda_{\,\mu}^{ \alpha}D^{\mu}_{\,\nu}\Lambda_{\,\alpha}^{ \nu} =\Lambda_{\,\alpha}^{ \nu}\Lambda_{\,\mu}^{ \alpha}D^{\mu}_{\,\nu} =(\Lambda_{\alpha}^{\, \nu})^{-1}\Lambda_{\,\mu}^{ \alpha}D^{\mu}_{\,\nu}= \delta^{\nu}_{\mu}D^{\mu}_{\,\nu}= D^{\mu}_{\,\mu} $$
So $D^{\mu}_{\,\mu}$ is indeed invariant under the transformations. Similarly one could probably show that the other $D$-terms do not stay invariant.
A: The trace of a matrix is in general invariant for change of basis (whether or not that change is a Lorentz transformation). In fact one has as a general property, for the matrix $A$
$$
\textrm{tr}(A) = \textrm{tr}(UU^{-1}A)=\textrm{tr}(U^{-1}AU) = \textrm{tr}(A')
$$
the trace being defined as $\textrm{tr}(A)= \sum_{j\in J} \alpha^j A(e_j) = \sum_{j\in J} A^{j}_{\phantom{j}j}$, with $\left\{\alpha^j\right\},\left\{e_j\right\}$ being the dual and standard basis, respectively.
On the other hand the contributions $\sum_{j\in J}A^{jj}$ (or $A_{jj}$, respectively) need not satisfy any particular property, and they in fact do not.
The general trick hint is that whenever the sum index appears as covariant and contravariant index and is summed over, the result is generally invariant by change of basis. This is because covariant indeces are components of the dual bases that transform with the change of base matrix and the contravariant indeces are components of the standard vector bases, that transform with the inverse of the change of base matrix. Summing over all the contributions they cancel each other since $U^{-1}U=1$.
A: Not sure to what extent one is supposed to solve homework problems on this site, but one can at least help.
The question reveals a number of different aspects that can be addressed. The first is probably the meaning of upper and lower indices, respectively called contravariant and covariant indices. They transform differently. (There is some comprehensive material on the Internet that one can read to understand what they mean.)
The next is the notion of the Einstein summation convention. It is a convention stating that whenever one has a pair of lower and upper indices that are summed over, one can ignore the summation sign
$$\sum_{\mu} A_{\mu} A^{\mu} \equiv A_{\mu} A^{\mu} $$
So when you see such repeated indices they are summed over. You'll notice that this convention is often tacitly used, as in some of the answers provided here.
Another aspect is that one can use the metric tensor to lower or raise indices
$$ A_{\mu} = A^{\nu} g_{\nu\mu} . $$
Without doing the actual problem, I'll try to point you in the right direction. The key is to use the relationship for Lorentz transformations given in one of the answers
$$ g_{\mu\nu} \Lambda^{\mu}_{\ \alpha} \Lambda^{\nu}_{\ \beta} = g_{\alpha\beta} . $$
together with the raising and lower of indices with the aid of the metric tensor to show what you need to show. Hope this helps.
