Given math model to model relations using a tensor, how do we put restriciton on tensor so that it can capture symmetric relations The goal is to model real life relations between stuff and people.
Say we have sets $E$, $R$ and functions $h:E \to V$, $t :E \to W$ and $r:R \to U$, where $V,W,U$ are finite dimentional vector spaces with potentially distinct dimentions but all with the same underlying field. 
The idea is that $E$ is the set of stuff and people, $R$ is the possible kind of relations between them.  For example, thing is $E$ could be John Smith, table, star_wars etc. Thins in $R$ could be "is_fighting_with", "lives_in", "in_love_with" etc. The functions $h,t,r$ are just mapping that let us represent the stuff ans relations.  Here relations is a tuple in $(a,r,b)\in E\times R\times E$. So because stuff $a$ could be in the head of the tuple or the tail of the tuple, so we have two maps $h,t$ for that. 
We can then tell how strong the relation r is between a and b by using a tensor product with a predefined tensor $\chi$:
$\chi\times_1 h(a)\times_2r(r) \times_3 t(b)$.
Here $\times_i$ is the mode_i product.
You see the idea? And the value of the tensor product will give the indication of stronngess of the relation between a and b. 
Now some relations are symmetric. For example, "mutualy_in_love","slept_with","as_smart_as". Let's denote the relaiton by "equals_to", well in this case, we want to put restriction on $\chi$ such that $\chi\times_1 h(i)\times_2r(equals-to) \times_3 t(j)=\chi\times_1 h(j)\times_2r(equals-to) \times_3 t(i)$ for all $i,j \in \mathcal{E}$.
SO my question is: what is the restriction we put on $\chi$
 A: The short answer is that 

The restriction you desire is on $\chi$ is that $\chi: \text{Sym}^2(V)^* \times U \longrightarrow Z$; i.e. that $\chi$ be a function on on the symmetric bilinear functions $\text{Sym}^2(V)^* \times U$. 

But technically the restriction you desire is on $r$ and not on $\chi$. The idea is the following:

$r$ (or $R$) is itself a tensor, in particular, it is a bilinear
  tensor and $\chi$ is a trilinear tensor. The relations being symmetric forces the tensor $r$ to be symmetric, $\chi$ only "implicitly inherits its symmetry from $r$."  

Intuitive Explanation
If $R$ is the set of relations then $R \subset \mathcal{P}(E \times E)$, where $\mathcal{P}$ is the powerset operator (if $R$ were just one relation we would have $R \subset E \times E$); therefore it makes sense to think of $R$ as having "one more extra dimension than a regular relation." A relation is nothing more than a boolean (i.e. $0-1$) matrix; our set of relations is nothing more than a boolean (i.e. $0-1$) "3-D array." In other words if $r' \in R$ and $e,e' \in E $ and $er'e'$ is true then the most reasonable value to give to (the coordinate) $r(r')_{e,e'}$ is one, likewise if $er'e'$ is false then $r(r')_{e,e'}=0$. If all of the $r' \in R$ are symmetric then $r(r')_{e,e'} = r(r')_{e',e}$ and therefore $\chi $ can defined as any (3-linear) function on "symmetric 3-D arrays" with the dimensions of $r$. Therefore symmetry is a restriction on the 3-D array $r$ and not on $\chi$; as a matter of fact we will eventually dispense with the superfluous function/tensor $\chi$ altogether.
Rigorous Explanation
Let's make this definition more rigorous: 


*

*enumerate each $e \in E$ by a number in $\{1,...,|E|\}$, 

*likewise enumerate each $r' \in R$ by a number in $\{1,...,|R|\}$, 

*let $v_i$ be a basis for $V$ and $u_i$ be a basis for $U$, where $\dim (V) = |E|$ and $\dim (U) = |R|$

*and let $h: E \rightarrow  V$ send the element $e_i$ to $v_i$
then we define the bilinear function $r : V \times V \rightarrow U$ (i.e. "3-D" matrix/array) as 
\begin{equation}
r(v_{i} , v_j  ) = \sum_{k}r_{i,j}^ku_k
\end{equation}
where the coefficients are defined by 
\begin{equation}
r_{i,j}^k = \begin{cases} 1 & \text{if } e_i r_k' e_ j \text{ is true}\\
0 & \text{if } e_i r_k' e_ j \text{ is false}  
\end{cases};
\end{equation}
this is well defined by linearity since a (multi-)linear mapping is uniquely defined by its behavior on a basis, in this case the basis is $b_i \otimes b_j \in V\bigotimes V$, see the universal mapping property for a "category theory" explanation. Technically you also need to know the universal property for direct sums of vector spaces. Intuitively $V$ is the direct sum of the lines $L_i = \{c v_i \ | \ c \in \mathbb{F} \}$ i.e. $V = \bigoplus_i L_i$; and $V \otimes V$ is the space defined by the direct sum of the planes $L_i \bigotimes L_j  = \{ c (e_i \otimes e_j) \ | \ c \in \mathbb{F} \}$; i.e. $V \otimes V = \bigoplus_{i,j} L_i \bigotimes L_j$, and $r$ being a symmetric tensor is the same as saying that $r$ does not depend on the orientation of the planes $L_i \otimes L_j$; i.e. $r$ has no "right-hand rule." 

The restriction that the relations be symmetric is a restriction on the tensor $r$. It is nothing more than the rule that $r(b_{i} , b_j  )= r(b_{j} , b_i  )$ for all $i,j$; i.e. $r$ is a symmetric bilinear tensor. We express by saying that $r \in \text{Sym}^2(V)^* \times U$. 

You are now free to define $\chi: \text{Sym}^2(V)^* \times U \longrightarrow Z$ (where $X^{*}$ is the dual space of $X$ and $Z$ is some other vector space) however you please; in particular, the succinct notation $\chi: \text{Sym}^2(V)^* \times U \longrightarrow Z$ expresses the fact that $\chi$ is a function on the "set of symmetric relations" $\text{Sym}^2(V)^* \times U$.
A Generalization Beyond 0-1/true-false
It seems the reason you want to define some extra parameter $\chi$ is that you would like to like to generalize the relations $r \in R$ to have values other than $\{0,1\}$. This can be done directly with $r$ as well, there is no reason to introduce a superfluous $\chi$. Let me elaborate; in the definition of $r^k_{i,j}$ we could have instead written 
\begin{equation}
r_{i,j}^k =  
\alpha \ \ \ \ \text{      if the relation } e_i r_k 'e_ j \text{  is $\alpha$ strong}
\end{equation}
where $\alpha$ is a scaler, i.e. $\alpha \in \mathbb{F}$, that measures the strength of the relation $e_i r_k' e_ j$. Nothing else in the definition needs to be modified since $r$ was already defined to be a bilinear symetric tensor. The $\chi$ you are looking for is the $\alpha $ above.
Computer Science Explanation
Just in case you may be asking "But Pedro I really like the Function $\chi$ and I don't care for all of this abstraction, can you do it with the $\chi$?" the answer is no. The rationale is that introducing the parameter $\chi$ into "I don't know, say a machine-learning algorithm" would increase the complexity of the algorithm by a large linear factor. The rationale is the following: if you have a 4-D array $\chi$ that is a function of $\left(V^* \right)^{2} \times U$ (or $E^2 \times R$) then you will have four nested for-loops, as opposed to the three nested for-loops for the simpler $r^k_{i,j}$. Dropping the $\chi$ altogether has practical computational implications. As a matter of fact, we can take this further; it is a well-known fact that although the general vector space, $\left(V^{*}\right)^2$, of bilinear forms $T : V^{2} \rightarrow \mathbb{F} $ has dimension $[\dim (V)]^2$ we have that the vector space, $\text{Sym}^2\left(V\right)^*$, of symmetric bilinear forms $T : V^{2} \rightarrow \mathbb{F} $ has dimension $\binom{\dim (V)}{2}$, which further cuts the number of operations by more than half.
