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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
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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
    
As a general rule, if $V=\oplus_i V_i$, $W=\oplus_j W_j$ are decompositions into $G$-irreducible components, then $\mathrm{Hom}(V,W)=\oplus_{i,j}\mathrm{Hom}(V_i,W_j) =\oplus_{i,j}(V_i)^*\otimes W_j$ and for the $G$-equivariant homomorphisms we have ${\mathrm{Hom}}_G(V,W)=\oplus_{i,j}((V_i)^*\otimes W_j)^G$ where Schur s lemma tells us that $((V_i)^*\otimes W_j)^G$ is zero$ if $V_i$, $W_j$ are not equivalent and one-dimensional otherwise. So the problem is equivalent to finding fixed vectors of a representation. –  Claudio Gorodski Jan 29 '13 at 12:38
    
Look up "Koszul complex" –  Bruce Westbury Jan 29 '13 at 12:41

2 Answers 2

I know this fact from a homological algebra background, so forgive me if I am using a language that is foreign to you.

Easy Lemma: Let If $(X,d)$ be an exact chain complex of vector spaces over a field of characteristic $0$ $K$ and $e:X \to X$ a contraction, i.e. $de +ed =c $, with $0 \neq c \in K$. Then $d+e :\bigoplus X_{2j} \to \bigoplus X_{2j-1}$ is an isomorphism.

If $e$ and $d$ are explicit, this is reasonably explicit. The appropriate chain complex for your problem is constructed as follows. Fix a vector space $V$ of finite dimension, write $E^p := \Lambda^{p} V^{\ast}$ and $S^q := Sym^q V^{\ast}$. Let $R^{\ast} := \bigoplus_{p,q} E^p \otimes S^q$. This is a graded commutative algebra if you give $E^p$ the degree $p$ and $S^q$ the degree $2q$.

There are canonical, mutually inverse, isomorphism $d:E^1 \to S^1$ and $e:S^1 \to E^1$. Extend $d$ to all of $R$ by the property $d(xy)= (dx)y + (-1)^{deg(x) } x (dy)$ and the requirement that $d(S^q)=0$. Do the same with $e$ (but here $e(E^p)=0$). The formulas $d^2=e^2=0$ hold.

Explicit formulas are

$$ d(v_1 \wedge \ldots v_p)\otimes (w_1 \ldots w_q) = \sum_{i=1}^{p} (-1)^{i-1} (v_1 \wedge v_{i-1} \wedge v_{i+1} \ldots v_p \otimes (dv_i w_1 \ldots w_q)) $$

and

$$ e(v_1 \wedge \ldots v_p)\otimes (w_1 \ldots w_q) = (-1)^p \sum_{i=1}^{q} (ew_i \wedge v_1 \wedge \ldots v_p \otimes ( w_1 \ldots w_{i-1} w_{i+1} \ldots w_q)). $$

Now on the piece $E^p \otimes S^q$, the formula $de+ed=p+q$ holds, as you prove using the product formulae and induction. The sequence (maps given by $d$)

$$ 0\to E^n \to E^{n-1} \otimes S ^1 \to \ldots E^1 \otimes S^{n-1} \to S^n \to 0. $$

is a chain complex and $e$ is a contraction, as in the above easy lemma. The sum $d+e$ is your desired isomorphism.

Remark: the construction is completely natural and therefore $GL(V)$-equivariant.

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NB. $p+q$ is not a constant non-zero element of $K$, so the easy lemma needs to be slightly modified. Moreover, the hypothesis that the characteristic of $K$ is zero is unnecessary in the lemma, but crucial for the case you study. –  ACL Jan 29 '13 at 10:52
    
But the differentials $d$ and $e$ preserve the number $p+q$ (make a picture of the bigraded algebra to see this). –  Johannes Ebert Jan 29 '13 at 17:06

I find easier to construct "the" isomorphism between the direct sums, for variable $n$, of the pair of spaces you have.

We consider the algebraic De Rham complex $C$ of the affine space~$V$, that is: for every $p$, $C^p$ is the space of differential forms of degree $p$ with polynomial coefficients and $C$ is the direct sum of the spaces $C^p$. I let $C^+$ be the direct sum of the $C^p$, for $p>0$. The exterior derivation $d$ maps $C^p$ to $C^{p+1}$; let $Z^p$ be its kernel (space of closed forms of degree $p$). It is known (we assume that the characteristic of $K$ zero) that the cohomology of the complex $(C,d)$ is zero in strictly positive degrees. In other words, $d$ induces isomorphisms $C^p/Z^p\simeq Z^{p+1}$. Moreover, $C^0$ is the ring of polynomials and $Z^0=K$ (constant polynomials). In differential topology, the obvious contraction from the affine space to the origin furnishes canonical antiderivatives to closed forms. One can check that it preserves forms with polynomial coefficients. We thus get linear maps $I\colon Z^{p+1}\to C^p$ such that $d\circ I = \mathrm{id}$ and $C^p=Z^p\oplus I(Z^{p+1})$ for every $p\geq 0$.

Let $C_e$ and $C_o$ be the subspace of $C$ consisting of differential forms whose coefficients are even, respectively odd, polynomials. One checks that $d(C_e)\subset C_o$, $d(C_o)\subset C_e$, $I(C_e^+)\subset C_o$ and $I(C_o^+)\subset C_e$. Consequently, $$ C_e = Z_e \oplus I(Z_o^+) = K \oplus Z_e^+ \oplus I(Z_o^+) $$ and $$ C_o = Z_o\oplus I(Z_e^+)=Z_o^+\oplus I(Z_e^+). $$

Let $u\colon C_e\to C_o$ be the map given by $0$ on $K$, $I$ on $Z_e^+$ and $d$ on $I(Z_o^+)$, let $v\colon C_o\to C_e$ be the map given by $I$ on $Z_o^+$ and $d$ on $I(Z_e^+)$. Then $u\circ v$ is the identity and $v\circ u$ is the projection from $C_e$ to $Z_e^+\oplus I(Z_o^+)$ with kernel $K$. They induce the isomorphisms you are looking for.

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