# Tag Info

4

This is making things too hard. This is the same as a $\mathfrak g$ action on $\mathbb C[G]$, which is given by differentiation by right invariant vector fields. Extending to $U(\mathfrak g)$ is given by taking the associated differential operator. The pairing you want is applying this operator and then evaluating at $e$.

3

First of all, let me point out that for fixed $\zeta$, this incuded representation will have $3$ distinct irreducible components, not $3(q-1)$; you get the $q-1$ from allowing $\zeta$ to vary. (You probably know that, but just to make sure). You've got more-or-less the right idea of how to do it. There's a slight issue though, which I'll explain shortly. ...

3

Proof in the case $n$ is odd By Sylow's 1st theorem there is a subgroup $H \vartriangleleft G$ of order (at least) 2 × 2 × 2 = 8. This could be one of five possibilities: $\mathbb{Z}_8, \mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ and additionally $D_8 \simeq \mathbb{Z}_4 \rtimes \mathbb{Z}_2, Q = ... 3 Since this is over$\mathbb{C}$, a representation is uniquely determined by its character. Thus, it suffices to show that the character of$V\otimes \mathbb{C}G$is a multiple of the character of$\mathbb{C}G$. But this is trivial, since the character of$\mathbb{C}G$just has value$|G|$at the identity and$0$elsewhere, so the character of$V\otimes ...

2

When $U$ is finite dimensional the answer is yes. Think of $U^\ast$ and $H \otimes U^\ast$ as $\hom(U, k)$ and $\hom(U, H)$ respectively. Then the map $U^\ast \to H \otimes U^\ast$ is given by $\phi \mapsto (\mathrm{id}_H \overline\otimes\phi)\circ\delta$, where here $f \overline\otimes g$ is the map $a \otimes b \mapsto f(a)g(b)$.

2

Let $A=\mathbb{C}[x]$ and consider the regular representation of $A$ acting on itself. Every nonzero element generates a subrepresentation isomorphic to $A$, so this definitely does not have any irreducible subrepresentations.

2

Okay well partial answer: the answer should is the minimum dimension of a faithful real representation, certainly it can't be better than this. Morally, you just choose a generic point in the space and take a convex hull of that point along with the other $|G|-1$ elements of its $G$ orbit. Unfortunately this may have extra symmetries, take for example the ...

2

I'm afraid that I don't really understand what you're asking, but it seems to me that you're confused. Let me try to make a few comments that might shed some light on the matter. A representation $(\rho,V)$ is a vector space $V$, equipped with a $G$-action $\rho$. A morphism of representations $(\rho,V)\rightarrow(\rho',V')$ is then a morphism of vector ...

2

Normally this kind of thing is verified by the "what else?" argument, that is, what are the chances you really have two different canonical maps in that square? That said, it could be comforting to know it's possible to really check such a thing. So: Let me change your $\oplus_i$ to $\oplus_j$, to avoid collision of notation. We have to show that for every ...

2

You can think of the first argument given by Jyrki Lahtonen as an application of the $p$-group fixed point theorem, which just says that if a (finite) $p$-group $G$ acts on a (finite) set $X$ then $$|X^G| \equiv |X| \bmod p$$ and in particular if $|X|$ isn't divisible by $p$ then the action has a fixed point. It's surprisingly useful to have a name for ...

2

The number of elements in $V$ is a power of $p$. Because $H$ is a $p$-group, its orbits on $V$ have sizes that are also powers of $p$. Therefore the number of singleton orbits is a multiple of $p$. Because the zero vector forms a singleton orbit, there must be others. Part two? There are finitely many entries in the matrices representing elements of $H$. ...

2

In fact, it turns out that there are only finitely many finite groups with a given number of conjugacy classes. This is not hard to prove and it's not too hard to find all of them with up to four conjugacy classes, say: http://groupprops.subwiki.org/wiki/There_are_finitely_many_finite_groups_with_bounded_number_of_conjugacy_classes

1

Yes, the inclusion of a factor into a direct product $M \to M \oplus N$ defined by $m \mapsto (m, 0)$ is always an injective homomorphism no matter what modules you choose for $M$ and $N$. In particular, you can inject $M \to M \oplus R[G]^n$ not just for some $n$, but in fact for any $n$. If $M$ and $N$ are $G$-modules then it would be a good exercise for ...

1

No. Look at $\mathfrak{sl}_3$. The simple roots are $\alpha_1,\alpha_2$ with $$(\alpha_1,\alpha_1)=2\;\;\;\mbox{and}\;\;\;(\alpha_1,\alpha_2)=-1.$$ Write $\Lambda_1=a\alpha_1+b\alpha_2$ and solve the system $$\begin{cases}(\Lambda_1,\alpha_1)=1\\(\Lambda_1,\alpha_2)=0\end{cases}$$ to get $\Lambda_1=\frac{2}{3}\alpha_1+\frac{1}{3}\alpha_2$ (similarly, ...

1

If we work over $\mathbb C$ then the number of irreducible representations of a group is equal to the number of conjugacy classes of elements in that group. For an abelian group this is simply the number of elements in the group. Also in an abelian group every subgroup is normal, so the number of normal subgroups is equal to the number of subgroups. I ...

1

There is a difference, which is not subtle. It's right there in the definition: the physics definition has the additional condition that the complex conjugate $\overline{V}$ of the representation $V$ isn't isomorphic to it. This happens whenever any value of the character $$\chi_V(g) = \text{tr}(\pi(g))$$ is not real, since the character of $\overline{V}$ ...

1

It is really only a statement about $\def\P#1{{{\mathfrak p}^{#1}}}\hat 1_\P k$, where $$\hat I_{\P {-k} }(x) = \int_{F^{n-1}}I_{ \P {-k} } (y) \psi ( y^tx) \, dy = \int_{(\P {-k})^{n-1}}\psi ( y^tx) \, dy ,$$ and follows from a general fact about integration over (compact) topological groups (with Haar measure): Fact: Suppose $\psi: G \mapsto \mathbb ... 1 There are groups for which no such polyhedron exists, and hence there is no minimum dimension. For example, consider a group$G$of order$|G|=2^{\mathfrak{c}}$. If$G$is the group of symmetries of a polyhedron in$n$-dimensional Euclidean space for some$n\in\mathbb{N}$, then$G$is a subgroup of$\operatorname{Sym}(\Bbb{R}^n)\$, the group of symmetries of ...

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