I am trying to understand the space of all orthogonal tensors, I aksed a more general version of this question here but with no solution yet found. The solutions for order-$2$ tensors are clear so thus the simplest case is a $2\times2\times2$ tensor. This means if the $2\times2\times2$ tensor is denoted by $a_{i,j,k}$, then the following three equations must hold:

$$ \sum_{i=1}^2 \sum_{j=1}^2 a_{i,j,1} \overline{a}_{i,j,2} = 0 $$

$$\sum_{i=1}^2 \sum_{k=1}^2 a_{i,1,k} \overline{a}_{i,2,k} = 0 $$

$$\sum_{j=1}^2 \sum_{k=1}^2 a_{1,j,k} \overline{a}_{2,j,k} = 0 $$

Except this time instead of considering the most general case of the $2\times2\times2$ tensor being complex, I want to consider a simpler version. That is where $a_{1,1,1}$, $a_{2,1,1}$, $a_{1,2,1}$, $a_{1,1,2}$ are positive reals and the rest of the entries are complex values. How can one characterize the space of solutions to these three equations? I do believe that the solution space should be 6 dimensional.

I am trying to understand the space of all orthogonal tensors, I asked a more general version of this question here but with no solution yet found. I want to look at the simplest case first, namely a $2\times2\times2$ tensor. This means if the $2\times2\times2$ tensor is denoted by $a_{i,j,k}$, then the following three equations must hold:

$$ \sum_{i=1}^2 \sum_{j=1}^2 a_{i,j,1} \overline{a}_{i,j,2} = 0 $$

$$\sum_{i=1}^2 \sum_{k=1}^2 a_{i,1,k} \overline{a}_{i,2,k} = 0 $$

$$\sum_{j=1}^2 \sum_{k=1}^2 a_{1,j,k} \overline{a}_{2,j,k} = 0 $$

Except this time instead of considering the most general case of the $2\times2\times2$ tensor being all complex, I want to consider a simpler version. That is where $a_{1,1,1}$, $a_{2,1,1}$, $a_{1,2,1}$, $a_{1,1,2}$ are positive reals and the rest of the entries are complex values. How can one characterize the space of solutions to these three equations? I do believe that the solution space should be 6 dimensional.

I am trying to understand the space of all orthogonal tensors, I aksed a more general version of this question here but with no solution yet found. The solutions for order-$2$ tensors are clear so thus the simplest case is a $2\times2\times2$ tensor. This means if the $2\times2\times2$ tensor is denoted by $a_{i,j,k}$, then the following three equations must hold:

$$ \sum_{i=1}^2 \sum_{j=1}^2 a_{i,j,1} \overline{a_{i,j,2}} = 0 $$

$$\sum_{i=1}^2 \sum_{k=1}^2 a_{i,1,k} \overline{a_{i,2,k}} = 0 $$

$$\sum_{j=1}^2 \sum_{k=1}^2 a_{1,j,k} \overline{a_{2,j,k}} = 0 $$

Except this time instead of considering the most general case of the $2\times2\times2$ tensor being complex, I want to consider a simpler version. That is where $a_{1,1,1}$, $a_{2,1,1}$, $a_{1,2,1}$, $a_{1,1,2}$ are positive reals and the rest of the entries are complex values. How can one characterize the space of solutions to these three equations? I do believe that the solution space should be 6 dimensional.

I am trying to understand the space of all orthogonal tensors, I aksed a more general version of this question here but with no solution yet found. The solutions for order-$2$ tensors are clear so thus the simplest case is a $2\times2\times2$ tensor. This means if the $2\times2\times2$ tensor is denoted by $a_{i,j,k}$, then the following three equations must hold:

$$ \sum_{i=1}^2 \sum_{j=1}^2 a_{i,j,1} \overline{a}_{i,j,2} = 0 $$

$$\sum_{i=1}^2 \sum_{k=1}^2 a_{i,1,k} \overline{a}_{i,2,k} = 0 $$

$$\sum_{j=1}^2 \sum_{k=1}^2 a_{1,j,k} \overline{a}_{2,j,k} = 0 $$

Except this time instead of considering the most general case of the $2\times2\times2$ tensor being complex, I want to consider a simpler version. That is where $a_{1,1,1}$, $a_{2,1,1}$, $a_{1,2,1}$, $a_{1,1,2}$ are positive reals and the rest of the entries are complex values. How can one characterize the space of solutions to these three equations? I do believe that the solution space should be 6 dimensional.