Timeline for An isomorphism of 2-Schur modules
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| S Mar 13 at 13:27 | history | suggested | The Amplitwist | CC BY-SA 4.0 |
fixed broken link to Wikipedia
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| Mar 13 at 11:15 | review | Suggested edits | |||
| S Mar 13 at 13:27 | |||||
| Oct 16, 2011 at 20:15 | comment | added | darij grinberg | Ah, I see now. My question about $K_0$ was targeting the characteristic-$p$ (or general-ring) case; I see now that it is rather off-topic here. | |
| Oct 16, 2011 at 18:21 | comment | added | David E Speyer | No, this doesn't work for infinite length modules. But finite dimensional reps of $GL_n(K)$ are definitely finite length! I can't decipher the $\Lambda$-ring terminology, so I haven't been looking at the more general questions. Regarding your second point, $\phi$ is fixed. This is a nonhomogenous linear equation in $\psi$. | |
| Oct 16, 2011 at 18:12 | comment | added | darij grinberg | Your argument about equivalent filtrations - does it work for infinite-length modules as well? I am not sure whether the general case I want to consider in question 68071 has the finite-length property, and even if it does, it does not feel right to me to make use of it. -- Neither am I sure that $\alpha\left(\phi\psi\right)=\alpha\left(\pi_1\right)$ is a linear equation. Isn't it rather multilinear? | |
| Oct 15, 2011 at 16:24 | comment | added | David E Speyer | I don't have any good ideas about getting into finite characteristic. I am still curious about the question of whether the result can be extended from $\mathbb{Q}$-algebras to $\mathbb{Z}[1/N]$ algebras, where $N$ is some explicit constant. | |
| Oct 15, 2011 at 16:23 | comment | added | David E Speyer | 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. | |
| Oct 15, 2011 at 16:17 | comment | added | David E Speyer | 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. | |
| Oct 15, 2011 at 5:07 | comment | added | darij grinberg | ... 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? | |
| Oct 15, 2011 at 5:03 | comment | added | darij grinberg | ... 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 ... | |
| Oct 15, 2011 at 5:00 | comment | added | darij grinberg | 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 ... | |
| Oct 13, 2011 at 20:26 | history | answered | David E Speyer | CC BY-SA 3.0 |