In order to prove the invariance of the trace of a tensor under the transformation $\tilde{T}^{i,j}=\Sigma_{k,l} O^i_kO^j_lT^{k,l}$ where $O\in SO(3)$
I have to prove that $\Sigma_{k,l} O^i_kO^i_l=\delta_{kl}$ but I cannot see how to prove this... any good reference/help?
NOTE: I know that by definition of orthogonal matrix $\Sigma_{k,l} O^i_kO^l_i=\delta_{kl}$
but in what I have written above both the orthogonal matrices have the same line fixed by $i$ (pay attention to the position of indices)
This is confusing. Everything would look more natural if you worked with tensors of type $(1,1)$, that is, $T_i^{\,j}$. As you formulate the question, one should first ask what is the trace of a $(2,0)$-tensor? In order to define it, you need a metric $g_{ij}$. Then you may want to prove $$ \sum_{ij} T^{ij} g_{ij}=\sum_{ij} \tilde{T}^{ij} g_{ij}. $$ Rewriting this, you get $$ \sum_{ij} T^{ij} g_{ij}=\sum_{ijkl} T^{kl} O_k^{\,\,i} O_l^{\,\,j} g_{ij} $$ which reduces to $$ \sum_{ij} O_k^{\,\,i} \, g_{ij} \, O_l^{\,\, j}=g_{kl}. $$ This is a reasonable definition of an orthogonal transformation. Of course, if $g_{ij}=g^{ij}=\delta_{ij}$ then everything reduces to the Euclidean case and the "matrix" $O$ being orthogonal. But then there is no reason to distinguish upper and lower indices.