Is $A_1^i \overline{A_1^j} A_1^k \overline{A_1^{\ell}} = \overline{A_i^1} A_j^1 A_1^k \overline{A_1^{\ell}} = \delta_i^k \delta_j^{\ell}$ true?

Question

Let $A = (A_i^j)$ be a unitary matrix. That is, $A^{-1} = \overline{A}^t$.

Is it true that $$A_1^i \overline{A_1^j} A_1^k \overline{A_1^{\ell}} = \overline{A_i^1} A_j^1 A_1^k \overline{A_1^{\ell}} = \delta_i^k \delta_j^{\ell}?$$

Answer 1

In the more general case of orthogonal matrices the equation already doesn't hold. Suppose $A$ to be the rotation matrix by an angle $\theta\in(0,\frac{\pi}{2})$ which is $$\begin{pmatrix} \cos{\theta}&-\sin{\theta}\\ \sin{\theta}& \quad\cos{\theta} \end{pmatrix}$$ This matrix is orthogonal and for $i,k=1$ and $j,\ell=2$ we get $$A_1^i\overline{A_1^j}A_1^k\overline{A_1^{\ell}}=\sin^2(\theta)\cos^2(\theta)\neq 0$$ and $$\overline{A_i^1}A_j^1A_1^k\overline{A_1^{\ell}}=-\sin^2(\theta)\cos^2(\theta)\neq 0.$$ Therefore both terms are not equal. In the case of $\theta=\frac{\pi}{3}$ neither of terms equals 1 meaning they both can't be of the form $\delta_j^k\delta_j^{\ell}$, assuming you mean the dirac delta.