13
votes
Accepted
Issue with a "proof" for Maschke's Theorem
You seem to be assuming that there is a unique projection onto a subspace of a vector space. That is not true. For instance, if $W$ is two-dimensional space with basis $\{e_1,e_2\}$, then for any ...
- 316k
9
votes
Accepted
Isaacs Character Theory - exercise 4.11
For Question 1, to show that $t = |G|/q$, it is sufficient to show that all involutions in $G$ are conjugate. There is a hint on how to do that in the book.
If not, choose two non-conjugate ...
- 84.8k
8
votes
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If F < E are fields, how is it possible for a representation X, to be irreducible as an F-representation, but reducible as an E-representation?
I think this is based in a misunderstanding of how we can convert between $E$ and $F$ vector spaces. For concreteness, let's work with $\mathbb{C}$ and $\mathbb{R}$, but you'll see that the same idea ...
- 32.6k
8
votes
There are no irreducible representations of dimension less than $7$ for the Lie algebra $\mathfrak g_2$.
One possibility is to use the fact that the formal character of an irreducible representation is invariant under the action of the Weyl group. In the case of $\mathfrak{g}_2$ the Weyl group has order ...
- 128k
8
votes
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Regular Representation of infinite groups
For an abstract infinite group $G$ the definition is the same as in the finite case; over a field $k$ you can always consider the action of the group algebra $k[G]$ on itself by left multiplication (...
- 399k
7
votes
Accepted
Representations of abelian groups
It is not true in the infinite-dimensional case. An irreducible representation of an abelian group $A$ over a field $k$ is the same thing as a simple module over the commutative $k$-algebra $k[A]$. ...
- 399k
7
votes
Accepted
Perfect groups whose character degrees square divide its order
If you extend to perfect groups, this is pretty easy. Let $G$ be your favourite simple group, and let $A$ be a very large abelian group. A semidirect product $X=A\rtimes G$ has character degrees ...
- 10.7k
7
votes
Motivation for representation theory
Have you tried googling to find motivatations for the subject? Some are here, here, and here. Historical background for the subject is here. The question that first led Frobenius to develop ...
- 39.6k
6
votes
If F < E are fields, how is it possible for a representation X, to be irreducible as an F-representation, but reducible as an E-representation?
Let's apply your logic to the example, since clearly something must go wrong.
Here $F = \mathbb{R}$, $E = \mathbb{C}$, $X: C_3 \rightarrow GL_2(\mathbb{R})$ given by $g \mapsto X(g) = \begin{pmatrix} ...
- 727
6
votes
Accepted
Dimension of the space of equivariant homomorphisms.
There are no factorials involved in the dimension of $\mathrm{Hom}_{\Bbb C[G]}(V,V)$. Let's just use Schur's lemma a bunch of times and the fact that $\mathrm{Hom}(V,U\oplus W)=\mathrm{Hom}(V,U)\oplus\...
- 17k
6
votes
Accepted
Two-dimensional representation of $Q_8$ is not "real"
By the sounds of it you want an explanation rather than a proof, because the proof is just the fact that such a representation has to be irreducible, and there is only one by counting the squares of ...
- 10.7k
6
votes
Possible degrees of group characters
Yes, for instance the symmetric group $S_{n+1}$ has an irreducible representation of degree $n$. Precisely, take the canonical permutation representation $V$ of $S_{n+1}$; it has degree $n+1$, and we ...
- 24.2k
6
votes
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Subadditivity of the multiplicity in a composition series?
The following certainly works for modules, and should therefore also work in $\mathsf{C}$ by the Freyd–Mitchell embedding theorem:
\begin{align*}
\newcommand{\im}{\mathrm{im}}
[\im(ϕ + ψ) : S]
&...
- 15.7k
6
votes
Why is a linear representation of a quiver a functor?
Functors must be compatible with identities and with composition of morphisms.
So if $Q$ is any quiver and $F$ is a functor from $\mathrm{FrCat}(Q)$ into another category $\mathcal{C}$, then
$F(1_x) =...
- 15.7k
6
votes
Accepted
In what sense are Pauli matrices "intertwiners"?
I'm pretty sure that your objection is exactly right. That is, the Pauli matrices are literally a/the standard basis for the (complexified?!) Lie algebra $\mathfrak sl(2)$. Yes, $\mathbb R^3$ with ...
- 49.8k
5
votes
Validating a Character Table for a Given Finite Group
Adding to the excellent answer of @BrauerSuzuki I wanted to comment on the point:
For $H≤G$, one gets a permutation character $π$ from the (transitive) action of $G$ on the set of (left) cosets by (...
- 51
5
votes
Accepted
How to construct the zero matrix in GAP
First off, approach 3 happens because the rows are identical so change in one happens in others. See the GAP manual here for an explanation.
Approach 2 tries to use a specialised function from the ...
- 6,862
5
votes
Accepted
What is the relationship between (highest) weight vectors on $\mathfrak{g}/\mathfrak{p}$ and elements of the fundamental rep of $\mathfrak{g}$?
Firstly, yes (conjugacy classes of) parabolic subalgebras are in 1-to-1 correspondence with subsets of the simple roots as you say and we denote these with crossed nodes on our Dynkin (or Satake) ...
- 3,287
5
votes
Accepted
Proving an identity regarding character of irreducible representation
Theorem Let $\chi \in Irr(G)$, and $x,y \in G$, then
$$\chi(x)\chi(y)=\frac{\chi(1)}{|G|}\sum_{z \in G}\chi(xy^z)$$
Before proving this theorem we need to set notation and an observation. Write $\...
- 45.7k
5
votes
There are no irreducible representations of dimension less than $7$ for the Lie algebra $\mathfrak g_2$.
If you are willing to use that $\mathfrak g_2$ is simple, then you can use the fact that any non-zero representation is injective and that $\mathfrak g_2$ has dimension $14$. So for a representation ...
- 19.2k
5
votes
Accepted
When can the basis of a Lie algebra always be made Hermitian?
To write things in mathematical language and notation here, we are considering an $n$-dimensional matrix Lie algebra $\mathfrak{g} \subseteq \mathfrak{gl}_d(\mathbb{C})$ (to a mathematician a "...
- 399k
5
votes
Contradictions about simple modules of matrix rings
In a commutative ring, nilpotent elements annihilate every simple module, because any nilpotent element belongs to the Jacobson radical.
This is generally false for noncommutative rings. For instance, ...
- 235k
5
votes
Accepted
Determine the group structure from its character table: a group of order $24$ as an example
As you know that $P$ is normal, all that is left to prove that $Q_1 P = G$ and $Q_1 \cap P =1$, using the characterization of internal semidirect products. Both of these follow from order ...
- 17k
5
votes
Accepted
Endomorphisms of a Lie algebra representation
Things are not quite as easy. If you have non-zero eigenvectors for different weights, then the sum of these spaces is not an eigenspace. Moreover, any homomorphism of $\mathfrak g$-modules preserves ...
- 19.2k
5
votes
representation about Lie algebra and Lie group
Well for a start we cannot write every element of $G$ as a matrix nor as an exponential. Lie algebras can always be written as matrices by Ado's theorem but that is not true for Lie groups. Of course ...
- 3,287
5
votes
Accepted
The character of the alternating sum of the exterior powers of the standard representation
If $V$ is a representation of a group $G$ of degree $d$, $g\in G$ and $p\geq0$, then the character of $\Lambda^pV$ evaluated at $g$ is $$\chi_{\Lambda^pV}(g)=\sigma_p(\lambda_1,\dots,\lambda_m),$$ ...
5
votes
Accepted
Tensor product of Representations of Lie algebras
Your second definition just doesn't define a Lie algebra representation at all; a Lie algebra representation must in particular be a linear map, and what you've written down isn't linear.
Conceptually ...
- 399k
5
votes
Representation of $\mathrm{Diff}(M)$ onto $C^\infty(M)$
The corresponding Lie algebra representation "ought to be" the action of vector fields by the Lie derivative. The sign is supposed to be there. The issue, as far as I can tell, is that there ...
- 399k
5
votes
Accepted
Is the alternating group $ A_8 $ a subgroup of the exceptional lie group $G_2$?
$G_2$ is famously the group of symmetries of an antisymmetric trilinear form in its 7 dimensional representation.
The 7 dimensional representation $V$ of $A_8$ does not preserve any antisymmetric ...
- 10.7k
5
votes
Accepted
Perfect Groups with faithful complex irreps
No. Take PerfectGroup(245760,4) (this requires GAP 4.12 :-) ), generated by
...
- 17k
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