# How to write down explictly the isomorphism of two finite dimensional representation of compact groups?

When dealing with finite dimensional representations over $\mathbb C$ of a compact group $G$, Character Theory provides us with a convenient way to determine whether two representations are isomorphic. Suppose that $W$, $V$ are two finite dimensional representations of $G$, and suppose we can compute their characters. Is there a convenient way to write down an explict isomorphism between $W$ and $V$?

My above question might be too vague. Let's look at an example, which motivates my question.

Let $V$ be a finite-dimensional vector space, show that there are natural isomorphisms $\oplus_{j~even,j=0}^\infty S^{j}(V)\otimes E^{n-j}(V)=\oplus_{j~odd,j=0}^\infty S^j(V)\otimes E^{n-j}(V), n>0.$ where $S^{j}(V)$ is the degree $j$ part of the symmetric algebra of $V$, $E^{j}(V)$ is the degree $j$ part of the exterior algebra of $V$. Naturality can be interpreted as "isomorphic as representation of $GL(V)$".

I've tried to directly write down an isomorphism but didn't manage it. I observed that "isomorphic as repn of $GL(V)$" is equivalent to isomorphic as repn of $SU(V)$ (when chosen a metric) since $GL(V)$ and $SU(V)$ generate the same subalgebra in $End(V)$, by density theorem. Now Character Theory apply and we can easily compute the character of both sides. For examples, if $\lambda_1,...,\lambda_n$ are the eigenvalues of the linear operator $T$, then the eigenvalues of the induces operator on $E^j(V)$ are the $\lambda_{i_1}\cdots\lambda_{i_j},i_1< \ldots < i_j$. In this way, we can prove the existence of an expected isomorphism.

Now my question is, can we unwrap the Character Theory to yield an explict formula for the promised isomorphism? Of course, it will also be appreciated if you can write down a formula without the aid of Character Theory.

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""isomorphic as repn of GL(V)" is equivalent to isomorphic as repn of SU(V) (when chosen a metric) since GL(V) and SU(V) generate the same subalgebra in End(V), by density theorem." This is not true: you can twist any rep of $GL(V)$ by a power of the determinant; without changing the restriction of the representation to $SL(V)$. – Johannes Ebert Jan 29 '13 at 9:33
Perhaps you mean $U(V)$. – S. Carnahan Jan 29 '13 at 11:20
You could ask your general question for trivial representations, which is about showing two vector spaces with the same dimension are isomorphic. Of course we usually do that by picking bases of each. The $G$-analogue is to break into irreps, first. Now we want to isomorph two irreps $V,W$ of the same type. An irrep is linearly spanned by the $G$-orbit of the high weight vector, so once we find high weight vectors in $V,W$ (unique up to scale) and correspond them, $G$-equivariance tells us how to correspond their orbits, and linearity does the rest. – Allen Knutson Jan 29 '13 at 11:25

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