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10

It is a theorem of Brion and, independently, of Vinberg that varieties with an open $B$-orbit (a.k.a. spherical varieties) have in fact only finitely many orbits. A shorter argument is due to Matsuki (see his ICM talk) and independently (using the same idea) by me (On the set of orbits $\ldots$). Thus multiplicity free spaces are visible. Kac implies that ...


9

By Pieri's formula, a partition with $2n$ elements in $n$ rows, corresponding to a representation of $GL_n$, occurs in this representation with multiplicity equal to the number of ways of obtaining that partition by starting with the empty partition and $n$ times adding two elements, no two in the same column. For the determinant squared, which corresponds ...


9

I don't know how helpful this will be either. But let me modify your question a bit to ask what is the correct analogue of the Galois group (of say a finite field or number field) in Hodge theory? It is certainly true that when $X$ is a smooth complex projective variety, $Gal(\mathbb{C}/\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$ acts on $H=H^i(X,\mathbb{C})$, by ...


9

I'm not sure how helpful this is, but anyway: in the étale cohomology of a variety over a finite field there is an action of $\mathrm{Gal}(\overline{\mathbf F}_q/\mathbf F_q)$ on $$ H^\bullet_{\text{ét}}(X \times_{\mathbf F_q} \overline{\mathbf F}_q,\mathbf Q_\ell) $$ and in the Betti cohomology of a complex variety there is a Galois action of $\mathrm{Gal}(\...


9

Tony Scholl gave a talk at a conference in Warwick in 2013 on exactly this topic (his talk was called "Remarks on monodromy and weights"). He explained how to formulate a precise version of weight-monodromy for arbitrary varieties over p-adic fields (not necessarily proper or smooth). The idea is that for any field $K$, and any finite-type $K$-scheme $X$, ...


9

You can bound the filtration length (assuming the WM conj) using the weight spectral sequence of Rapoport--Zink. This is a sp seq converging to $H^*(X_{\overline{k}})$; and if the WM conj holds, then the monodromy filtration coincides with the filtration induced by this spectral seq. (There's a nice account of this in Scholl's paper https://www.dpmms.cam.ac....


6

Put $A(x) =\sum_{n\le x} a_n$, and $B(x) =\sum_{n\le x} a_n\log x/n$. Then $$ B(x) = \int_1^x A(t)\frac{dt}{t}. $$ So information about $A(x)$ readily translates to information about $B(x)$ and there is no loss since integration (which makes things smoother) is involved. But to pass from $B(x)$ to $A(x)$ we need to differentiate, and based on the ...


5

The answer is yes. Let $e_1,\ldots, e_n$ be the standard basis of $V$. Consider the morphism $$ f \colon \det(V) \to V^{\otimes n} $$ given by $$ f(e_1 \wedge \cdots \wedge e_n) = \sum_{\sigma \in S_n}(-1)^{\varepsilon(\sigma)}e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)}, $$ where $S_n$ is the symmetric group on $n$ letters and $\varepsilon(\sigma)$ ...


5

Yes, the $\ell$-adic weight filtration is compatible with the weight filtration in mixed Hodge theory under the comparison isomorphism. These facts go back to Deligne, and are described in his announcement Poids dans la cohomologie des variétiés algébriques ICM 1974. Finding a detailed proof is bit harder though... Added remarks You can take a look at Huber'...


4

For constant coefficients, the comparison statement, along with a sketch, appears in Deligne's ICM talk, Poids dans la cohomologie.... For things to work the way you seem to want in your second paragraph, you're going to need a lot more structure for $F$ than what you've given. At the very least, when $F$ is a local system (lisse), you need the weights ...


4

Any element of $\mathfrak{h}^*$ is called weight. I don't know why in this case the author used definite article. As a non-native speaker I would use indefinite one, i.e. "a weight is called regular if ..." That sounds wrong. One can speak about weights which are integral (or dominant) with respect to some (positive) root system. Perhaps the members of your ...


4

To obtain the highest weight of a semisimple Lie algebra $\mathfrak{g}$ you first have to choose Cartan subalgebra $\mathfrak{h} \leq \mathfrak{g}$ and then set of positive roots (or alternatively choose a Borel subalgebra). Then the highest weight vectors of your representation $V$ are given by linear system $$ \rho(X) v = 0,\quad \forall X \in \bigoplus_{\...


4

There is an explicit operator that maps $L^2$ isometrically onto $H^{s,\mu}_2$ :$$I_{s,\mu}u(x)=(1+|x|^2)^{-\mu/2}(I-\Delta)^{-s/2}u(x)$$The (inverse) Fourier transform $k_s(x)$ of $(1+|\omega|^2)^{-s/2}$ is also well documented (Bessel functions etc), and then $I_{s,\mu}u(x)=\int (1+|x|^2)^{-\mu/2} k_s(x-y)u(y)\ dy$ is Hilbert-Schmidt $L^2\to L^2$ iff its ...


3

Just an addendum to Ricky's answer: the multiplicity is indeed 1 which can be proved as follows. An occurrence of ${\rm det}(V)^{\otimes 2}$ inside $({\rm Sym}^2(V))^{\otimes n}$ is the same thing as a nonzero joint multilinear ${SL}_n$-invariant of $n$ quadratic forms $Q^{(1)},\ldots,Q^{(n)}$ in $n$ variables. By the first fundamental theorem of classical ...


3

Let $X$ be an arbitrary object in $\mathcal{C}$. I write $\{ W, F \}$ for the limit of $F$ weighted by $W$. By definition, $$\mathcal{C} (X, \{ W, F \}) \cong [\mathcal{I}, \textbf{Set}] (W, \mathcal{C} (X, F))$$ naturally in $X$. If we have $U \Rightarrow W$ (note the direction!) and $F \Rightarrow G$ then functoriality of $W$ and $F$ on the RHS gives a map ...


2

For $\mu$ a weight, let $||\mu||_1$ denote the one-norm of $\mu$ (the sum of the absolute values of its entries) and let $Z(\mu)$ be the number of zero coordinates of $\mu$. Let $k\geq0$ and $1\leq p\leq n$. Write $r(\mu)=(k+p-||\mu||_1)/2$. If $r(\mu)$ is a non-negative integer, then \begin{align*} m_{\pi_{\Lambda_{k,p}}}(\mu) &= \sum_{j=1}^{p} (-1)^{j-...


2

Just evaluate $\langle \lambda, \alpha \rangle$ for all roots $\alpha$. It is sufficient to test the equality for positive roots only. Also, the weight $\lambda$ is regular if and only if it has trivial stabilizer in the Weyl group. For example, if you write $\lambda$ in Bourbaki's "$\epsilon$-basis" then regularity in $A_n$-case means that no two ...


2

Consider $\mathcal A = \{\{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}\}$. Then $sc(\{1,3\}) - sc(\{1,4\}) = sc(\{2,3\}) - sc(\{2,4\})$. In particular, it's impossible to have $sc(\{1,3\}) > sc(\{1,4\})$ but $sc(\{2,3\}) < sc(\{2,4\})$.


2

The following should give an example where you can't achieve every ranking. Let $n=4$ and let $\mathcal{A} = \{ (1,2),(1,3),(1,4),(2,3),(2,4),(3,4)\} = [4]^{(2)}$. I claim that we can't achieve any ranking which starts $(1,2) > (3,4) > \ldots$. Indeed, since $sc_w(1,2) > sc_w(3,4)$ if follows that $\max \{ w(1),w(2) \} > \min \{ w(3),w(4)\}$ ...


2

I think you're overcomplicating things. You have showed that: (a) As a vector space we have the following decomposition: $$ M = \bigoplus_{[\nu]\in\mathfrak{h}^*/\Lambda_r} M^{[\nu]}. $$ (b) Furthermore, each $M^{[\nu]}$ is a submodule of $M$. From this it follows directly that the direct sum above is a decomposition of modules. Edit to clarify: Note that ...


2

I believe not: let $f$ be any colouring and take a maximal equivalence relation $\sim$ on $\omega$ with the property that $m\sim n$ implies $f(\{m,n\})=0$. Note that $\sim$ can be extreme: the identity relation if $f$ is constant with value $1$, and $\sim$ is $\omega^2$ of $f$ is constant with value $0$. For the corresponding partition $P$ we have $W^{(1)}_P=...


2

Submodules/quotients have the same (edit: generalized) infinitesimal character. Now use Harish-Chandra theorem. EDIT: Definitely an overkill. Easier solution is in the comments of the question.


2

I only find one paper (a book chapter, not a book itself) with the indicated title, Arthur - An introduction to the trace formula, and I can't find in it the sentences you quote, so it's hard to speak exactly to your questions. Could you give the exact reference? There are two natural choices of how to parameterise standard parabolics: by the simple roots ...


2

See proof of Theorem 6.11 of Representations of semisimple Lie algebras in the BGG category $\mathcal{O}$ by James E. Humphreys. This theorem proves what you want in the case $\mathfrak{p}$ is a Borel subalgebra ($\mathfrak{p} = \mathfrak{b}$). You just need existence of projective covers in parabolic setting, statement about $\mathrm{Ext}^1$ (I think both ...


2

In Humphreys, weights are defined in section 0.7 "Representations". He notes that in the finite case all weights are integral, but that this is not true for infinite dimensional representations. Note that by definition, if $V_\lambda \neq 0$, then $\lambda$ is a weight of $V$. Since for any $\lambda\in\mathfrak{h}^*$ there are modules V with $V_\lambda\neq ...


1

Your idea is pretty much spot on; this is the area of spectral graph theory. Often the graph Laplacian is used rather than its adjacency matrix -- the Laplacian is defined as $L = D - A$ where $A$ is the adjacency matrix and $D$ is a diagonal matrix whose $(i,i)$ entry is the sum of the weights of edges out of vertex $i$. The smallest eigenvalue of the ...


1

Since $W$ is a real vector space, $L_X$ is a real operator, so if we write $w_j=u_j+\sqrt{-1}v_j$, then $L_X \bar{w}_j=L_X u_j - \sqrt{-1}L_X v_j=\overline{L_X w_j}=\overline{\Theta_j(X)w_j}=-\Theta_j(X)\bar{w}_j$.


1

I don't think there is other reason. I think in general group representation theory you consider regular elements as those which have trivial stabilizer. Then you should perhaps speak about "regular with respect to affine action". People sometimes abuse notation and talk about character $\lambda$ when in fact they mean $\chi_\lambda$. Sometimes you can ...


1

There is infinitely many linkage classes each containing some $\Phi^+_I$-dominant elements. But since any module from $\mathcal{O}$ is finitely generated it will decompose only into finitely many modules from $\mathcal{O}_\chi$. I hope I understood your question correctly. There are some problems, e.g. $\mathfrak{sl}_2$ has just one simple root and $t\alpha ...


1

This is true for general modules, you don't need anything specific about nilpotent cohomology. Every homomorphism from a $F(\lambda)$ is uniquely determined by it's highest weight vector.


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