Although this question is highly related to my past question, but I'd like to understand general theory of symmetric and antisymmetric reps. of a finite group so let me post this question.
As far as I know, the Kronecker (or direct product) representation is defined as follows;
Let ${\mathscr R}_{\alpha}$ and ${\mathscr R}_{\beta}$ are irreducible representations of a finite group ${\mathscr G}$. ${\mathscr R}_{\alpha, \beta}$ acts on vector spaces $V_{\alpha}, V_{\beta}$ by their representation matrices $M^{[\alpha, \beta]}(g)$, respectively. I denote the basis of $V_{\alpha}, V_{\beta}$ by $\{\left|i\right\rangle\}$, $\{\left|s\right\rangle\}$ $(i = 1, 2, \dots, d_{\alpha}\equiv {\rm dim} {\mathscr R}_{\alpha}, j=1, 2, \dots, d_{\beta} )$.
Then the Kronecker representation ${\scr R}_{\alpha \times \beta}$ is the representation of ${\mathscr G}$ on the product space $V \otimes V$, which satisfies
\begin{align} \left|A\right\rangle &\rightarrow M^{[\alpha \times \beta ]}_{AB} \left|B\right\rangle \\ &\equiv M^{[\alpha]}_{ij} M^{[\beta]}_{st} \left|j\right\rangle \otimes \left|t\right\rangle, \end{align} where I suppressed the summation symbols.
Since the Kronecker representation is also a representation of $\scr{G}$, it can be decomposed into sum of irreducible representations. This can be done explicitly by calculation of Clebsch-Gordan coefficients.
Next, let us assume the case two irreducible representations are exactly same, ${\scr R}_{\alpha} = {\scr R}_{\beta}$. Then the product space itself has two invariant subspaces under the group action, namely symmetric and antisymmetric product space. The basis of them are given by \begin{align} \frac{1}{2} \left( \left|i\right\rangle \otimes \left|s\right\rangle + \left|s\right\rangle \otimes \left|j\right\rangle\right) ~~(i \leq s), \\ \frac{1}{2} \left( \left|i\right\rangle \otimes \left|s\right\rangle - \left|s\right\rangle \otimes \left|j\right\rangle\right) ~~(i < s). \end{align} The dimension of spaces $S(V^2), \Lambda(V^2)$ spanned by these basis are $ \frac{d(d+1)}{2} $ and $ \frac{d(d-1)}{2} $, respectively($d = d_{\alpha} = d_{\beta}$).
My question is: How can we determine the character of the Kronecker rep. which acts on the symmetric subspace (i.e. the character of the symmetric product representation)?
Some textbooks states it is given by $\chi^{[\alpha \times \alpha]}_{[\rm sym]}(g) = \frac{1}{2} (\chi^2(g) + \chi(g^2))$, but I don't understand it now.
In fact, the derivation of this formula which I considered is; On the symmetric subspace $S^2(V)$, $\left|A \right\rangle$ takes the form of $\frac{1}{2} \left( \left|j\right\rangle \otimes \left|s\right\rangle + \left|s\right\rangle \otimes \left|t\right\rangle\right)$. Then $M^{[\alpha \times \alpha]}_{AB, {\rm [sym]}}$ should be defined as \begin{align} \left|A\right\rangle &\rightarrow M^{[\alpha \times \alpha ]}_{AB} \left|B\right\rangle \\ &\equiv M^{[\alpha]}_{ij} M^{[\alpha]}_{st} \frac{1}{2} \left( \left|j\right\rangle \otimes \left|t\right\rangle + \left|t\right\rangle \otimes \left|j\right\rangle \right) \\ &= \frac{1}{2} \left( M^{[\alpha]}_{ij} M^{[\alpha]}_{st} + M^{[\alpha]}_{it} M^{[\alpha]}_{sj} \right) \frac{1}{2} \left( \left|j\right\rangle \otimes \left|t\right\rangle + \left|t \right\rangle \otimes \left|j\right\rangle \right). \end{align} The last equality follows from the symmetry under $ j \leftrightarrow t$.
The character of $M^{[\alpha \times \alpha]}_{AB, {\rm [sym]}}$ is defined as \begin{align} \chi^{[\alpha \times \alpha]}_{\rm [sym]} (g) \equiv \sum_{A} M^{[\alpha \times \alpha]}_{AA, {\rm [sym]}}. \end{align}
Then, in terms of the original irreducible reps., this quantity is expressed as \begin{align} \chi^{[\alpha \times \alpha]}_{\rm [sym]} (g) = \frac{1}{2} \sum_{i \leq s} \left( M^{[\alpha]}_{ii} M^{[\alpha]}_{ss} + M^{[\alpha]}_{is} M^{[\alpha]}_{si} \right). \end{align}
The point here is the fact that the summation region is $i \leq s$, not arbitrary. If we don't care about this constraint, then $S^2(V)$ has $d^2$ basis, but this is wrong I think.
However, the problem is that if we forget $i \leq s$ and calculate the summation for arbitrary $i, s$, we obtain the formula which I quoted.
Where was the mistake(s) in what I wrote?
Actually, I derived the term $M^{[\alpha]}_{is} M^{[\alpha]}_{si}$ by hand and this might mean $\chi^2(g) = \chi(g^2)$, which cannot be true in general so definitely I made a mistake somewhere.
Sorry for the long sentences.
Suppose the character on the original space is $\chi$.
Suppose the eigenvalues of $g$ on the original space are $\lambda_1,\dots,\lambda_d$.
Then the eigenvalues of $g^2$ on the original space are $\lambda_1^2,\dots,\lambda_d^2$.
Also the eigenvalues of $g$ on the symmetric square are $\lambda_1^2,\dots,\lambda_d^2$ and $\lambda_i \lambda_j$ for $1\leqslant i<j \leqslant d$.
So the character of $g$ on the symmetric square is $$ \sum_{i=1}^{d}\lambda_i^2+ \sum_{1\leqslant i<j \leqslant d}\lambda_i\lambda_j $$ and this sum is equal to $$ \frac{1}{2}\left((\sum_{i=1}^{d}\lambda_i)^2 +\sum_{i=1}^{d}\lambda_i^2\right) $$ which is just $$ \frac{1}{2}\left(\chi(g)^2+\chi(g^2)\right). $$