Tag Info

Hot answers tagged

4

Yes. This is a corollary of Schur-Weyl duality. You need at least the additional assumption that your symmetric monoidal category is enriched over $k$-vector spaces. In general I don't see any reason to expect that the action of $k[S_n]$ is faithful; consider, for example, the special case where we only look at $1$-dimensional vector spaces. Sometimes. The ...


4

The problem is with #2. The rank of a Lie algebra is the dimension of any of its Cartan subalgebras (all of which are isomorphic). None of them need to contain diagonal matrices. You can see one such subalgebra computed for $SO(2n + 1, \mathbb{C})$ in this paper: Structure Theory of Semisimple Lie Groups - Knapp


4

Since $\tau$ is of order 3, the minimal polynomial of the action $T \colon W \to W$ of $\tau$ is a factor of $x^3 - 1 = (x - 1)(x - \omega)(x - \omega^2)$. Hence the minimal polynomial of $T$ has no multiple root and therefore $W$ is a semisimple $\mathbb{C}[T]$-module, which means that $W$ is spanned by its eigenvectors (Linear algebra!).


4

I'll assume all of your $V_i$ are in fact the same vector space $V$, or else I don't understand what your notation means. Then $S_n$ acts on $G \times \dots \times G$ by permuting the factors (in your example $G = GL(V)$ but this argument applies more generally), and your semidirect product is a wreath product $$G \wr S_n = (G \times \dots \times G) \rtimes ...


3

The following argument works provided that we make the extra assumption $p\neq2$. Assume contrariwise that there is a $G$-equivariant homomorphism $s:V\to W$ such that $s(w)=w$ for all $w\in W$. Let us fix two elements $x,x'\in X, x\neq x'$. Because the action of $G$ is doubly transitive there exists an element $g\in G$ such that $g\cdot x= x'$ and $g\cdot ...


2

The problem is that the property "diagonal matrices", or "antisymmetric matrices" depends on the basis of the underlying vector space. The standard way to represent the Lie algebra $\mathfrak{so}(n)$ faithfully by matrices, is by antisymmetric matrices. However, we may also represent $\mathfrak{so}(n)$ faithfully by matrices which are not skew-symmetric. ...


1

To elaborate on mt_'s comment, one can more generally construct representations of $G$ on function spaces, by which I mean a vector space $V$ consisting of functions (lying in some appropriate class -- finitely supported, continuous, square-integrable, smooth, tempered, etc) $f:G\rightarrow W$, where $W$ is a vector space over some field $k$. Then the space ...


1

The operator you are defining is $\pi_\phi(a)$ for a fixed $a$. The inequality you posted shows that $\pi_\phi(a)$ is bounded on the subspace of elements of the form $b+N_\phi$, for all $b$. A bounded operator is continuous.


1

Hint: a matrix of the form $A = \begin{bmatrix}a&b&0\\c&d&0\\0&0&1\end{bmatrix}$ has a natural action on $\Bbb R^2 \oplus \Bbb R$ as: $\left(\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\right)^T,0\right) + ((0,0),z)$ and has the same determinant as the $2\times 2$ matrix in the upper-left.


1

The circle group has a natural action by rotations in a real vector space of dimension $2$. Here your have dimension$~3$, and you must essentially choose to waste that extra dimension. More precisely, there has to be a line through the origin whose vectors are fixed by all linear operators of your representation (this can be formally proved, though I won't ...


1

I think I know what's wrong. It turns out I was confused about the statement on the geometric side. In the Ginzburg/Chriss text, the irreducible representations are not exactly the Borel-Moore homology of the fibers but the image of Borel-Moore homology of some tubular neighborhood. If one takes such a neighborhood the two-dimensional representation I ...


1

The discrepancy comes from the fact that there are two different affine Hecke algebras, one for the weight lattice and one for the root lattice. The root lattice version is a subalgebra of the weight lattice version. Concretely, the root lattice version is the one appearing in your method 1. It may be realised as the quotient of the group algebra of the free ...



Only top voted, non community-wiki answers of a minimum length are eligible