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This is the little brother of question 68071: elementary, simple-looking and probably much easier to answer. Of course, it is just a small part of question 68071, as anybody with $\lambda$-rings experience will see.

Let $n$ and $m$ be positive integers. Let $k$ be a field of characteristic $0$. Let $U$ be the canonical $n$-dimensional representation of $\mathrm{GL}_n\left(k\right)$. Let $V$ be the canonical $m$-dimensional representation of $\mathrm{GL}_m\left(k\right)$. A folk result I cannot prove tells me that

$\wedge^2\left(U\otimes V\right)\oplus \left(\wedge^2\left(U\right)\otimes\wedge^2\left(V\right)\right)^{\oplus 2} \cong U^{\otimes 2}\otimes\wedge^2 V \oplus \wedge^2 U\otimes V^{\otimes 2}$

(don't confuse direct powers with tensor powers in this equation; also, the usual precedence rules apply where $\oplus$ is seen as addition and $\otimes$ as multiplication) as representations of $\mathrm{GL}_n\left(k\right)\times \mathrm{GL}_m\left(k\right)$.

Since the representation theory of $\mathrm{GL}_n\left(k\right)\times \mathrm{GL}_m\left(k\right)$ is semisimple (or at least I hope so?), this isomorphism must somehow "factor" into isomorphisms of parts. Or not? I am far from feeling safe here.

Anyway, is there a nice explicit description of the above isomorphism, that shows us what it maps stuff to, like the Clebsch-Gordan formulae for $\mathrm{SL}_2\left(k\right)$ ?

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According to my calculations the two sides have different dimensions. –  Bruce Westbury Oct 13 '11 at 6:38
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The dimensions agree if you assume the unadorned exponent in $(-)^2$ indicates direct sum, e.g., $V^2$ means $V^{\oplus 2} = V \oplus V$ –  S. Carnahan Oct 13 '11 at 8:26
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This is EXACTLY what I meant by "don't confuse direct powers with tensor powers in this equation". I have been doing this myself too often... –  darij grinberg Oct 13 '11 at 13:00
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I plead: use the notation "\oplus 2" in the exponent! –  S123 Oct 13 '11 at 13:32
    
Thanks. I did not know this notation. –  darij grinberg Oct 13 '11 at 15:33
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3 Answers 3

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As long as 2 is invertible, we can use the isomorphism $W^{\otimes 2} \cong S^2 W \oplus \wedge^2 W$ to split $(U \otimes V)^{\otimes 2}$ in two different ways:

  1. $(S^2U \otimes S^2V) \oplus (\wedge^2 U \otimes \wedge^2 V)$ is the symmetric summand, hence isomorphic to $S^2(U \otimes V)$.

  2. $(S^2U \otimes \wedge^2 V) \oplus (\wedge^2 U \otimes S^2V)$ is the alternating summand, hence isomorphic to $\wedge^2(U \otimes V)$.

We use the decomposition of $\wedge^2(U \otimes V)$ to build up the right side of your equation:

  1. $(S^2U \otimes \wedge^2 V) \oplus (\wedge^2U \otimes \wedge^2 V) \cong U^{\otimes 2} \otimes \wedge^2 V$

  2. $(\wedge^2 U \otimes S^2V) \oplus (\wedge^2U \otimes \wedge^2 V) \cong \wedge^2 U \otimes V^{\otimes 2}$.

If you are only concerned with algebraic representations, then $GL_n \times GL_m$ is linearly reductive, so you get complete reducibility. I don't know what happens for representations of groups of field-valued points in general.

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Yes, I only meant algebraic representations. Many thanks! –  darij grinberg Oct 13 '11 at 13:00
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As I'm sure you know, this is an easy computation with symmetric polynomials/characters over the field $\mathbb{C}$. I will give a proof which shows how to leverage that fact and show that, for any commutative $\mathbb{Q}$-algebra $R$, and $U$ and $V$ any free $R$-modules, we have the required isomorphism. At the end, I'll try to make a general statement.


Lemma 1: Let $U$ a finite dimensional $\mathbb{C}$ vector space. For $\sigma$ a permutation in $S_d$, let $\sigma : U^{\otimes d} \to U^{\otimes d}$ act by permuting the factors according to $\sigma$. So we get a map $$\mathbb{C}[S_d] \to \mathrm{End}_{GL(U)}(U^{\otimes d}).$$ This map is surjective.

Proof: This is part of the statement of Schur-Weyl duality.

Corollary 2: Let $U$ and $V$ be finite dimensional $\mathbb{C}$ vector spaces. We have similar surjections: $$\mathbb{C}[S_d \times S_e] \to \mathrm{End}_{GL(U) \times GL(V)}(U^{\otimes d} \otimes V^{\otimes e})$$ and $$\mathrm{Mat}_{r \times s}( \mathbb{C}[S_e \times S_d]) \to \mathrm{Hom}_{GL(U) \times GL(V)} \left( \left(U^{\otimes d} \otimes V^{\otimes e}\right)^{\oplus r}, \left(U^{\otimes d} \otimes V^{\otimes e}\right)^{\oplus s} \right).$$

Now, let $U$ and $V$ be finite dimensional $\mathbb{Q}$-vector spaces. Observe that

$$\bigwedge\nolimits^{2}\left(U\otimes V\right)\oplus \left(\bigwedge\nolimits^2\left(U\right)\otimes\bigwedge\nolimits^2\left(V\right)\right)^{\oplus 2} \quad (\ast)$$ is naturally a quotient of $(U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 3}$. Even better, there is an idempotent $\pi_1$ in $\mathrm{Mat}_{3 \times 3}(\mathbb{Q}[S_2 \times S_2]$ such that $(\ast)$ is the image of $\pi_1$.

Similarly, we can find $\pi_2$, an idempotent in $\mathrm{Mat}_{2 \times 2}(\mathbb{Q}[S_d \times S_e])$ such that the image of $\pi_2$ acting on $(U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 2}$ is $$U^{\otimes 2}\otimes\bigwedge\nolimits^2 V \oplus \bigwedge\nolimits^2 U\otimes V^{\otimes 2} \quad (\ast\ast)$$

We have a commutative diagram $$\begin{matrix} \mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_2 \times S_2]) & \rightarrow & \mathrm{Mat}_{2 \times 3}(\mathbb{C}[S_2 \times S_2]) \\ \downarrow & & \downarrow \\ \mathrm{Hom}_{GL(U) \times GL(V)} \left( (U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 3}, (U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 2} \right) & \rightarrow & \mathrm{Hom}_{GL(U_{\mathbb{C}}) \times GL(V_{\mathbb{C}})} \left( (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 3}, (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 2} \right) \end{matrix}$$ where for brevity I use a subscript $\mathbb{C}$ for tensor with $\mathbb{C}$. Call the vertical maps $\alpha$.

Let $A \subset \mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_d])$ be those maps $\phi$ such that $$\alpha(\phi) = \alpha(\phi \pi_1) = \alpha(\pi_2 \phi). \quad (\dagger)$$ Let $A_{\mathbb{C}}$ be the analogous subspace of $\mathrm{Mat}_{2 \times 3}(\mathbb{C}[S_d])$. Since $A$ and $A_{\mathbb{C}}$ are cut out by the same equations, we have $A_{\mathbb{C}} = A \otimes \mathbb{C}$.

The images of $\pi_1$ and $\pi_2$ are isomorphic over $\mathbb{C}$. Composing that isomorphism with the projection $(U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 3} \to (\ast)_{\mathbb{C}}$ and the injection $(\ast \ast)_{\mathbb{C}} \to (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 2}$, we get a map $\overline{\phi}$ in the bottom right hand corner of the diagram, obeying $\overline{\phi} = \overline{\phi} \alpha(\pi_1) = \alpha(\pi_1) \overline{\phi}$. Moreover, this map has rank $\dim (\ast) = \dim (\ast \ast)$.

Using Corollary 2, we can find $\phi$ with $\alpha(\phi) = \overline{\phi}$. So $\phi \in A_{\mathbb{C}}$ and $\alpha(\phi)$ has rank $\dim (\ast)$.

Crucial paragraph The subspace of $A_{\mathbb{C}}$ consisting of maps $\phi$ such that $\alpha(\phi)$ has rank $\dim (\ast)$ is Zariski open. And $A$ is Zariski dense in $A_{\mathbb{C}}$. (Any polynomial which vanishes at all rational inputs is zero.) So there is some $\phi \in A$ obeying $(\dagger)$. This $\phi$ must induce an isomorphism between $(\ast)$ and $(\ast \ast)$.

So we have shown that $(\ast) \cong (\ast \ast)$, and this isomorphism is induced by a matrix in $\mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_d])$. For any $\mathbb{Q}$ algebra $R$, this gives a corresponding isomorphism between the analogous representations of $GL(U \otimes R) \times GL(V \otimes R)$.


Wow, that took a lot longer than I expected it to. I dread the notation that would be needed to make this precise, but a theorem like the following should follow by similar arguments: Let $\mathcal{S}$ and $\mathcal{T}$ be two formulas made up from atoms $V_1$, $V_2$ ..., $V_r$ joined together direct sums, tensor products and Schur functors. For any commutative ring $R$ and sequence of positive integers $(n_1, \ldots, n_r)$, we can "plug in" $R^{n_i}$ for $V_i$ and get a $\prod GL_{n_i}(R)$ representation.

"Theorem" If $\mathcal{S}(\mathbb{C}, n_1, \ldots, n_r) \cong \mathcal{T}(\mathbb{C}, n_1, \ldots, n_r)$, then $\mathcal{S}(R, n_1, \ldots, n_r) \cong \mathcal{T}(R, n_1, \ldots, n_r)$ holds for any $\mathbb{Q}$-algebra $R$ and there is a "universal formula" for the isomorphism as a $\mathbb{Q}$-linear combination of reordering various tensor products.

When I set out to prove this, I expected to only need that all positive integers $\leq \max(n_1, \ldots, n_r)$ were invertible in $R$, not that $R$ was a $\mathbb{Q}$ algebra. But I couldn't get the Zariski density argument in the Crucial Paragraph to work without a ground field. I'm curious whether this was an artifact of the proof method, or a genuine obstacle.

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Thanks a lot. Thoughts like these were hovering in my mind for some days now, but I see clearer after having read your reply. And maybe I have some light to shed as well: First of all, $\mathbb C$ is a red hering, since Schur duality works just as well over $\mathbb Q$. Second, we can replace your Crucial Paragraph by something that seems more in line with Nature: Just as you introduced your isomorphism $\overline{\phi}$ and its lift $\phi$, you could call $\overline{\psi}$ the inverse of $\overline{\phi}$ and lift it to some $\psi$. There is a mildly annoying hurdle in the way, namely ... –  darij grinberg Oct 15 '11 at 5:00
    
... the fact that $\overline{\phi}\overline{\psi}=\mathrm{id}$ does not yield $\phi\psi=1$, but fortunately this is an artefact of small-dimensional vector spaces, since for $U$ and $V$ sufficiently high-dimensional, $\alpha$ is actually an isomorphism. And once the right $\phi$ and $\psi$ are found for high-dimensional $U$ and $V$, they will work both for smaller-dimensional $U$ and $V$ (by projection) and for even-higher-dimensional $U$ and $V$ (by projection again, and by the injectivity of $\alpha$). This is a trick known from the construction of the universal polynomials required to ... –  darij grinberg Oct 15 '11 at 5:03
    
... define the notion of special $\lambda$-rings. I assume that this is enough to make your "Theorem" a Theorem, although I fear I will fight shy of writing it up until a better language is found for these ideas. What this does NOT help with is the case of characteristic $p$, which indeed offers additional complications: There are no isomorphisms, but only equality in $K_0$. And here is a question I have never dared ask: What does equality in $K_0$ actually mean? Are two modules equal in $K_0$ if they possess equivalent filtrations? Or just the direct sums of these modules with something else? –  darij grinberg Oct 15 '11 at 5:07
    
Hi Darij! If two modules are equivalent in $K_0$ the they posses filtrations with the same simple quotients (not necessarily in the same order). Let $A$ be the free abelian group on isomorphism classes of simple modules. For any finite length module $M$, let $\delta(M) \in A$ be the sum of the elements in a Jordan-Holder filtration. By the Jordan-Holder theorem, $\delta(M)$ is well defined. It is easy to see that $\delta$ extends to a linear map $K_0 \to A$. In particular, if $[M] = [N]$ in $K_0$, then $\delta(M) = \delta(N)$, as promised. –  David Speyer Oct 15 '11 at 16:17
    
Regarding $\phi$ and $\psi$, fix $\phi$ as in my answer. Consider $\psi$ such that $\alpha(\phi \psi) = \alpha(\pi_1)$ and $\alpha(\psi \phi) = \alpha(\pi_2)$. Since the modules are isomorphic over $\mathbb{C}$, these equations are solvable in matrices with entries in $\mathbb{C}[S_d \times S_e]$. Since these are inhomogenous linear equations with coefficients in $\mathbb{Q}$, there must be some solution in matrices with entries in $\mathbb{Q}[S_d \times S_e]$ as well. The above equations are inherited in any $\mathbb{Q}$ algebra, so the inverse is given by a universal formula as desired. –  David Speyer Oct 15 '11 at 16:23
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This is not an answer, but a comment to S. Carnahan's answer:

Let me write down the isomorphism $\left(S^2 U \otimes \wedge^2 V\right) \oplus \left(\wedge^2 U \otimes S^2 V\right) \to \wedge^2\left(U\otimes V\right)$ explicitly: It sends $\left(uu^{\prime}\otimes v\wedge v^{\prime}, p\wedge p^{\prime} \otimes qq^{\prime}\right)$ to $\dfrac12\left(\left(u\otimes v\right)\wedge\left(u^{\prime}\otimes v^{\prime}\right) - \left(u\otimes v^{\prime}\right)\wedge\left(u^{\prime}\otimes v\right) + \left(p\otimes q\right)\wedge\left(p^{\prime}\otimes q^{\prime}\right) + \left(p\otimes q^{\prime}\right)\wedge\left(p^{\prime}\otimes q\right)\right)$ (where both $\wedge$ and multiplication operators bind more strongly than $\otimes$). The inverse isomorphism sends $\left(u\otimes v\right)\wedge\left(u^{\prime}\otimes v^{\prime}\right)$ to $\left(uu^{\prime}\otimes v\wedge v^{\prime}, u\wedge u^{\prime}\otimes vv^{\prime}\right)$.

Note that the inverse isomorphism is fraction-free. Hence, the isomorphism $\wedge^2\left(U\otimes V\right)\oplus \left(\wedge^2\left(U\right)\otimes\wedge^2\left(V\right)\right)^{\oplus 2} \to U^{\otimes 2}\otimes\wedge^2 V \oplus \wedge^2 U\otimes V^{\otimes 2}$ is fraction-free as well (but its inverse probably not).

As it comes to generalizing to $\wedge^n$, I assume the inverse isomorphism is the better bet.

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