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5

It turns out that the conjugacy class of an even permutation $g \in S_n$, $n > 1$, decomposes into smaller conjugacy classes in $A_n$ (always into two classes) iff the cycle decomposition of $g$ is into odd cycles of distinct length. So, if $n > 2$ is odd, then the $n$-cycles split into two conjugacy classes. If $n > 2$ is even, then the $(n - ...


3

It sounds like you're describing the quotient space $\mathbb{R}^n/S_n$. Some people call this the $n^{th}$ "symmetric power" (of $\mathbb{R}$), although be a little careful with that terminology because it can be used to refer to two other related but different constructions. This quotient is not a manifold, but can be thought of as an orbifold. For ...


3

As coffeemath has remarked in a comment, the function must be a bijection and thus a permutation in the symmetric group $S_{10}$. We can decompose it into cycles, and its order $30$ is the least common multiple of its cycle lengths. Thus its cycle lengths are among the divisors of $30$ not greater than $10$: $1$, $2$, $3$, $5$, $6$ and $10$. If there's a ...


2

There are at least two possible proofs, one of them by enumeration and another one using the exponential formula. By enumeration, first choose the elements to go on each cycle (multinomial coefficient): $$\frac{N!}{\prod_j (j!)^{P_j}}$$ Each of these elements generates $j!/j$ cycles (in placing labels on a directed cycle all orbits under the ...


2

By straightforward calculation, we see that \begin{align*} 1&\rightarrow 3\rightarrow 2\rightarrow 4\rightarrow 8\rightarrow 10\rightarrow 9\rightarrow1&&(7\mbox{-cycle)}\\ 5&\rightarrow 5&&(1\mbox{-cycle)}\\ 6&\rightarrow 7\rightarrow 6&&(2\mbox{-cycle)} \end{align*} and thus ...


2

Suppose $\sigma: H\to\mathrm{GL}(W)$ is a complex irreducible representation of $H$. Call the trace $\chi$. Then character theory says that $\sum_{h\in H}\chi(h)$ equals $|H|$ if $\sigma$ is trivial and $0$ otherwise. Additionally, the sum $\sum_{h\in H}\sigma(h)$ is an interwtining operator, and Schur's lemma says the intertwining operators of a complex ...


1

$\sum_{1\le i,j\le d_\rho} |\hat{f}(\rho)_{i,j}|^2 = ||\hat{f}(\rho)||^2$ is just the definition of the Frobenius norm of the matrix $\hat{f}(\rho)$.


1

Yes there is an easy formula. With respect to the usual inner product on what you call the symmetric algebra, for which the Schur functions form an orthonormal basis, the monomials $p_\mu$ in the power sums for an orthogonal (but not orthonormal) basis. This means that the coefficient of $p_\mu$ when expressing some element $a$ of the algebra in this basis ...


1

First, let me see if I understood correctly: Suppose $A=\{a_1,\ldots,a_n\}$ and $B=\{b_1,\ldots,b_n\}$ are two sets of the same cardinality $n$. Denote by $S_A$ the symmetric group in $A$, that is, the group of all possible bijections from $A$ to $A$. Similarly, define $S_B$ for $B$. Consider now the group $X=S_A\times S_B$. Given $(p,q)\in X$, we define ...


1

It's important that on the first page the paper states that, without loss of generality, all of their irreducible representations are unitary, so that $\rho(g)^\dagger=\rho(g)^{-1}$ for all $g\in G$. By the way, the beginning of section $3$ is in error. The sum $\sum_{h\in H}\rho(h)$ is not a projection, because it is not normalized correctly. Instead, ...


1

You have the action of $S_n$ on itself by conjugation. The equivalence classes, or orbits, under an action by conjugation are called conjugacy classes. What you need to show is that two elements in $S_n$ are conjugate if and only if they share the same cycle type. You can find a proof of that on page 126 of Dummit and Foote under Proposition 11, for one. ...


1

You've misread the notation. (12) is not an element of H - (12)(34) is. It sends 1234 to 2143. Then (14)(23) sends that to 3412, which is also (13)(24).


1

Here is the pyritohedron; let's call it $P$. It should be clear that its symmetry group is a subgroup of the symmetry group of the cube, which I assume you know is $\operatorname{Sym}(4) \times C_2$, the product of the symmetric group of degree four and a cyclic group of order $2$. As a reminder, this is because rotations of the cube permute the $4$ ...


1

What you call "exponentiation" means conjugacy: $\phi$ is a homomorphism $G\rightarrow Aut(V)$ for some vector space $V$; an (invertible) matrix is an automorphism of $V$ (once you choose a basis), and $\phi^A$ is the representation $x\mapsto A\phi(x)A^{-1}$ (or replace $A$ with $A^{-1}$ depending on the author's conventions). In other words, it's just the ...



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