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Weyl's unitarian trick says that for a representation $\rho:G \rightarrow GL(V)$, there is a scalar product on V, such that $\rho$ is unitarian. This scalar product looks like this: $(v,w):= \frac{1}{|G|} \sum_{g\in G}(\rho(g)v,\rho(g)w)_{0}$. With this every representation is completely reducible.
My question is now how this would work explicitly. Let's suppose I have a group G with elements $(g_1,g_2,..., g_n)$ and $\rho$ explicitly given. How would I then go on to decompose the representation? I suppose I need the other representations as well. Does this trick then only have applications in proofs etc. where the decomposition is not needed explicitly?

Quasar
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    Once you have an inner produt, you can take orthogonal complements. Take any representation $\rho$, pick a vector $v$. The representation generated by $v$ is a subrepresentation of $\rho$. If any of these subreps is a proper subrep, take its orthogonal complement and you've split $\rho$ as a direct sum of two subreps. Now keep going until $\rho$ is a sum of irreducibles. – PL. Aug 26 '16 at 16:05
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    @PL. In practice, for "many" representations $V$, picking a random vector will almost surely give you a cyclic vector---that is, you won't find proper submodules using your suggestion. Think about what happens already for a finite cyclic group and a representation $V=V_1 \oplus V_2$ that is a sum of two non-isomorphic one-dimensional representations: any vector not in one of the lines $V_1$ or $V_2$ generates all of $V$. – Stephen Aug 26 '16 at 19:36

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The existence of a positive definite Hermitian form is very useful to decompose your representation once you have produced a proper sub-representation; I don't believe it can be used effectively for producing sub-representations. However, in practice, the averaging idea used to produce the scalar product can be used to explicitly produce subrepresentations.

First think about how you might find $G$-invariant vectors in $V$: you would start with an arbitrary non-zero vector $v$ and then form the average $$\pi(v)=\frac{1}{|G|} \sum_{g \in G} gv.$$ In fact, this defines you an endomorphism $\pi$ of $V$ whose image is precisely the set $V^G$ of $G$-fixed vectors.

The same trick can be used to produce the other isotypic components: for each irreducible representation $U$ of $\mathbf{C} G$, you want to find $G$-module homomorphisms $U \to V$. You look in the space $\mathrm{Hom}_{\mathbf{C}}(U,V)$ of all linear maps, and realize that the $G$-module homomorphisms are precisely the linear maps that are fixed by the $G$ action $$(gf)(u)=gf(g^{-1}u).$$ So thinking of the space of linear maps as a representation in its own right, the averaging trick just used above produces for you all the $G$-module homomorphisms from $U$ to $V$. Of course, this requires that you have a model for $U$, but that's a separate question.

(A parenthetical comment: the ideas in the preceding paragraph can be used directly in $V$ as well. Thinking carefully about what we have done gives the usual formulas for primitive idempotents in terms of the irreducible characters.)

Stephen
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