Orthogonality makes sense without character theory. There's an inner product on the space of representations given by $\dim \operatorname {Hom}(V, W)$. By Schur's lemma the irreps are an orthonormal basis. This is "character orthogonality" but without the characters.

How to recover the usual version from this conceptual version? Notice $\dim \operatorname{Hom}(V,W) = \dim \operatorname{Hom}(V \otimes W^*, 1)$ where $1$ is the trivial representation. So in order to make the theory more concrete you want to know how to pick off the trivial part of a representation. This is just given by the image of the projection $\frac1{|G|} \sum_{g\in G} g$.
The dimension of a space is the same as the trace of the projection onto that space, so $$ \def\H{\rule{0pt}{1.5ex}H} \dim \operatorname{Hom}(V \otimes W^*, 1) = \operatorname{tr}\left(\frac1{|G|} \sum_{g\in G} {\large \rho}_{\small V \otimes W^*}(g)\right) = \frac1{|G|} \sum_{g\in G} {\large\chi}_{V}(g)\ {\large\chi}_{W}\left(g^{-1}\right) \\ $$ using the properties of trace under tensor product and duals.

Orthogonality makes sense without character theory. There's an inner product on the space of representations given by $\dim \operatorname {Hom}(V, W)$. By Schur's lemma the irreps are an orthonormal basis. This is "character orthogonality" but without the characters.

How to recover the usual version from this conceptual version? Notice $$\dim \operatorname{Hom}(V,W) = \dim \operatorname{Hom}(V \otimes W^*, 1)$$ where $1$ is the trivial representation. So in order to make the theory more concrete you want to know how to pick off the trivial part of a representation. This is just given by the image of the projection $\frac1{|G|} \sum_{g\in G} g$.
The dimension of a space is the same as the trace of the projection onto that space, so $$ \def\H{\rule{0pt}{1.5ex}H} \dim \operatorname{Hom}(V \otimes W^*, 1) = \operatorname{tr}\left(\frac1{|G|} \sum_{g\in G} {\large \rho}_{\small V \otimes W^*}(g)\right) = \frac1{|G|} \sum_{g\in G} {\large\chi}_{V}(g)\ {\large\chi}_{W}\left(g^{-1}\right) \\ $$ using the properties of trace under tensor product and duals.

Orthogonality makes sense without character theory. There's an inner product on the space of representations given by dim Hom(V, W). By Schur's lemma the irreps are an orthonormal basis. This is "character orthogonality" but without the characters.

How to recover the usual version from this conceptual version? Notice dim Hom(V,W) = dim Hom(V \otimes W*, 1) where 1 is the trivial representation. So in order to make the theory more concrete you want to know how to pick off the trivial part of a representation. This is just given by the image of the projection 1/#G \sum_g g. The dimension of a space is the same as the trace of the projection onto that space, so dim Hom(V \otimes W*, 1) = tr(1/#G \sum_g \rho_{V \otimes W*}(g)) = 1/#G \sum_g \chi_V(g) \chi_W(g^-1) using the properties of trace under tensor product and duals.

Orthogonality makes sense without character theory. There's an inner product on the space of representations given by $\dim \operatorname {Hom}(V, W)$. By Schur's lemma the irreps are an orthonormal basis. This is "character orthogonality" but without the characters.

How to recover the usual version from this conceptual version? Notice $\dim \operatorname{Hom}(V,W) = \dim \operatorname{Hom}(V \otimes W^*, 1)$ where $1$ is the trivial representation. So in order to make the theory more concrete you want to know how to pick off the trivial part of a representation. This is just given by the image of the projection $\frac1{|G|} \sum_{g\in G} g$.
The dimension of a space is the same as the trace of the projection onto that space, so $$ \def\H{\rule{0pt}{1.5ex}H} \dim \operatorname{Hom}(V \otimes W^*, 1) = \operatorname{tr}\left(\frac1{|G|} \sum_{g\in G} {\large \rho}_{\small V \otimes W^*}(g)\right) = \frac1{|G|} \sum_{g\in G} {\large\chi}_{V}(g)\ {\large\chi}_{W}\left(g^{-1}\right) \\ $$ using the properties of trace under tensor product and duals.