New answers tagged

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 ...


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

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


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 ...


0

What your first observation amounts to is that $H$ is (isomorphic to) a subgroup of $S_3 \times S_2$, where the first factor acts on $\{1,2,3\}$ and the second factor acts on $\{4,5\}$. This only has $12$ elements, which is a drastic reduction from the $120$ elements of $S_5$ we would have to test by "brute force". Moreover, it can be seen that any element ...


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 ...


0

What you want to read is Serre's book on Linear Representation of Finite Groups. Section 8.2 of chapter 8 covers the case of a semi-direct product by an abelian group. For the more general case, you want to use the theory of induced characters in a more subtle way. This is not as trivial as it sounds though ; make sure you have the mathematical background to ...


0

Actually, $(1,4,2,3)K$ is one of the elements of the quotient group. Where you're getting is confused is the fact that it is the very same element as $(1,3,2,4)K$ – we're just picking different representatives.


0

Elements of $S_A$ can be looked at as bijections $p:\{1,\dots,n\}\to A$ where $n$ denotes the cardinality of $A$. Likewise elements of $S_B$ can be looked at as bijections $p:\{1,\dots,n\}\to B$. Every pair $\langle p_A,p_B\rangle$ induces a bijection $A\to B$ prescribed by $p_A(i)\to p_B(i)$ for $i=1,\dots,n$. It seems you want $Y\subset X$ to be a set ...


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

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

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

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).


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 ...


0

First: from $f \circ \cdots \circ f (x) = x$, are you able to conclude that $f$ is bijective (therefore a "permutation" of $A$)? Second: Are you familiar with the decomposition of permutations into a composite of disjoint cycles? Third: If you are, do you know the equation for the order of a permutation (the smallest "power" of that permutation, greater or ...


0

$K \trianglelefteq G$ is easy to be seen because each $L \trianglelefteq S_3$ as $L$ has index 2 in $S_3$. For $H \trianglelefteq K$, we can use the fact: If $H$ has a prime index $p$ in $G$ and there is no prime divisor of $|G|$ less than $p$, then $H \trianglelefteq G$. $|K:H| = 3$ and there is no smaller prime dividing $|K|=9$, so we see that $H ...


0

Observe that $|L| = 3$ and $|S_3| = 6$, so the index of $L$ in $S_3$ is $2$. Therefore, $L \lhd S_3$. Consequently, $K = L \times L \lhd S_3 \times S_3 = G$. Then, observe that $K$ is abelian, because it is a direct product of abelian groups, so all of its subgroups are normal. Therefore $H \lhd K$. To see that $H$ is not normal in $G$, define $$\tau = ...


0

Using generators, there won't be much computation needed. $K$ is generated by $(1,p)$ and $(p,1)$. If $H$ is invariant under conjugation by the generators of $K$, then $H$ is normal in $K$. Lagrange's theorem proves that $S_3$ is generated by $p$ and any transposition, e.g. $t= \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 ...


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 ...


0

It is difficult to predict the order of the permutation $\sigma\rho$ just from the order of the permutations $\sigma$ and $\rho$. In this case, we may use $\sigma = (1 5)(2 4)$ and $\rho = (5 2)(4 3)$. Both clearly have order $2$ (being the product of two disjoint $2$-cycles), but we have $$ \sigma\rho = (1 5)(2 4)(5 2)(4 3) = (12345) $$ when read ...


0

The fact that $\sigma$ is of order $5$ doesn't imply that it can't be the product of two permutations of order $2$. That would only be the case if those permutations commute. You can find two such permutations by trial and error. A permutation of order $2$ only has cycles of length $1$ (fixed points) and $2$ (transpositions). There are only two different ...


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 - ...


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 ...


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 ...


0

By the time I typed up the question, I realized the answer. We typically have so much freedom in choosing $S$ that it's no problem to meet either condition. Note that the following are equivalent: $\pi$ is the discrete partition $G_\pi$ is trivial $s=n!$ and conversely, the following are equivalent: $\pi$ is non discrete; it has an equivalence class ...


0

$N_2$ is a subgroup. In that case, $\pi \circ \pi_2$ will have uniform distribution on each coset, while the probability of each coset is the same for both $\pi \circ \pi_2$ and $\pi \circ \pi_1$. Note that $\pi_1$ does not need to be uniform. Also, by translation invariance, the result holds equally well if $N_2$ is a coset. (The same idea can be applied ...



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