Let V and W be any two subspaces of $(\mathbb{C}^d)^{\otimes n}$ such that there exists two irreducible and non-isomorphic representations $\rho\_V: G \to GL(V)$ and $\rho\_W: G \to GL(W)$. Does this imply: V and W are orthogonal subspaces?


I am new to representation theory, so apologies if this is basic knowledge. I could not find a good proof/counterexample for this fact, but something along the lines is used in Proposition 1 in [Stembridge - Orthogonal sets of Young symmetrizers](https://doi.org/10.1016/j.aam.2009.08.004).
In this link, it is shown that Young symmetrizers corresponding to two different shapes are orthogonal by using the fact that they project onto irreducible and non-isomorphic representations.



Orthogonality of irreducible and non-isomorphic representations
I wasn't sure what the question meant; to make sense of it, $V$ and $W$ would have had to be subspaces of a common space with some sort of form with respect to which we could measure orthogonality, and two abstract representations need not be presented in this way.


Your reference to [Stembridge - Orthogonal sets of Young symmetrizers](https://doi.org/10.1016/j.aam.2009.08.004) suggested that you were thinking of, say, complex representations of a finite group $G$. In this setting, the group algebra $\mathbb C[G]$ has a $V$-isotypic subspace (consisting of the matrix coefficients $g \mapsto \langle v^\*, g\cdot v\rangle$ for $v \in V$ and $v^\* \in V^\*$) and a $W$-isotypic subspace (analogously defined). These are orthogonal for the usual ($G$-equivariant) inner product on $\mathbb C[G]$, by the [Schur orthogonality relations](https://en.wikipedia.org/wiki/Schur_orthogonality_relations).


EDIT: You have updated the question to indicate that you just want $V$ and $W$ to be two subspaces of $(\mathbb C^d)^{\otimes n}$, presumably with the usual inner product inherited from its natural identification with $\mathbb C^{d n}$. (I'm not sure what the purpose of the separate $d$ and the $n$ parameters is.) In this generality, the answer can be ‘no’ in all but the most trivial cases: consider, for example, the cyclic group of order $2$ acting trivially on a line $V$ in $(\mathbb C^2)^{\otimes1}$, and non-trivially on a non-orthogonal line $W$ in $(\mathbb C^2)^{\otimes1}$. (This can even be extended, as you probably meant, to an action of the cyclic group on $(\mathbb C^2)^{\otimes1}$ that makes its actions on $V$ and $W$ subrepresentations.)


If you added, as you probably meant to do, the requirement that $G$ actually act on $(\mathbb C^d)^{\otimes n}$, preserving some inner product $\langle\cdot, \cdot\rangle$ there (not necessarily the standard one!), and that the representations of $G$ on $V$ and $W$ make them subrepresentations of this big $d n$-dimensional representation (as is the case for finite groups $G$, again, by the [Schur orthogonality relations](https://en.wikipedia.org/wiki/Schur_orthogonality_relations), when we embed abstract representations $G$-equivariantly in $\mathbb C[G] \cong (\mathbb C^{\lvert G\rvert})^{\otimes 1}$, equipped with its standard ‘$L^2$’ inner product), *then* the answer is once again ‘yes’, that the spaces of $V$ and $W$ are orthogonal *with respect to this $G$-fixed inner product*. (I emphasise the need for consistency in the choice of inner product because *every* action of a finite group on a complex vector space preserves *some* inner product there. For example, if we chose the lines in the example of the previous paragraph to be $V = \operatorname{Span}(1, 1)$ and $W = \operatorname{Span}(1, 0)$, then the inner product with squared norm $(x, y) \mapsto 2\lvert x\rvert^2 - 4\operatorname{Re}(x\overline y) + 5\lvert y\rvert^2$ is preserved by the cyclic group of order $2$, and, indeed, $V$ and $W$ are orthogonal with respect to this inner product even though they are not with respect to the ‘standard’ pairing.)


I will now prove the modified statement. For every $v \in V$, the [Riesz representation theorem](https://en.wikipedia.org/wiki/Riesz_representation_theorem) produces a unique element $T(v) \in W$ such that $\langle T(v), w\rangle$ equals $\langle v, w\rangle$ for all $w \in W$. We have that $T : V \to W$ is $G$-equivariant (because $G$ preserves the inner product), so, since $V$ and $W$ are irreducible and non-isomorphic, $T$ must be identically $0$. Then $\langle v, w\rangle = \langle T(v), w\rangle$ equals $0$ for all $v \in V$ and $w \in W$.


