There is a separate problem with the printed linear-span assertion, when $Z_{((n-1),1)}$ denotes the usual centralizer of a subalgebra. Let $A_{n-1}$ be the sum of all transpositions in $H$. Direct expansion in the group algebra givesThus $Y=X_n^2-2A_{n-1}$ commutes with $H$ and satisfies ${\Pi }_n(Y)=(n-1)1$. For $n\ge 3$ it contains nonzero coefficients on three-cycles involving $n$, whereas every element of $\mathrm{span}\{1,X_n\}$ is supported only on the identity and transpositions. Therefore $Y$ is an explicit counterexample to the asserted equality.

One correct description of the entire preimage is obtained by putting $C=C_{\mathbb{C}{S}_n}(\mathbb{C}H)$. Equivariance gives ${\Pi }_n(C)\subseteq Z(\mathbb{C}H)$, and ${\Pi }_n$ is the identity on that center. It follows thatA useful corrected two-dimensional statement restricts the support to the identity and the transpositions $(in)$: invariance under conjugation by $H$ then forces their coefficients to be equal, yielding exactly $\mathrm{span}\{1,X_n\}$. This additional support restriction is absent from the PDF.

The printed uniqueness assertion has an exceptional case at $n=4$. To locate it, use the two double cosetsAn $H$-bimodule-equivariant set map is determined by $u=\Phi (1)$ and $v=\Phi (t)$. Equivariance requires $u\in Z(H)$ andbecause $H\cap tHt=S_{n-2}$ and $t$ commutes with this subgroup. These conditions are also sufficient: define $\Phi (h)=hu$ and $\Phi (ht{h}^{\prime })=hv{h}^{\prime }$. Changing the double-coset expression only inserts an element of $S_{n-2}$, which commutes with $v$.

For $n\ge 5$, the center of $H$ and this centralizer are both trivial. For the latter, a commuting permutation must fix the unique point outside $\{1,\dots ,n-2\}$ and lie in the center of $S_{n-2}$; that center is trivial when $n-2\ge 3$. Therefore $u=v=1$, giving deletion. For $n=4$, however, $C_{S_3}(S_2)=\{1,(12)\}$. The choiceis well defined and satisfies all the printed equivariance conditions, but sends $(34)$ to $(12)$ rather than to the identity. Uniqueness holds for $n\ge 5$; it is false at $n=4$ as printed. Even requiring the map to restrict to the identity on $H$ does not remove this counterexample.

The deletion projection for a symmetric group removes $n$ from the cycle containing it. A cycle $(a_1\cdots a_{r}n)$ becomes $(a_1\cdots a_r)$, with a singleton understood as a fixed point; all other cycles remain unchanged. This describes a genuine permutation of $\{1,\dots ,n-1\}$ and immediately proves the two identity assertions.

Put $H=S_{n-1}$. For $\sigma ,\theta \in H$ and $k<n$, if $\pi (\theta (k))\ne n$, deletion gives $\sigma \pi (\theta (k))$. If $\pi (\theta (k))=n$, it gives $(\sigma \pi \theta )(n)=\sigma \pi (n)$. These are exactly the two cases in $\sigma {\pi }_n\theta (k)$. Thus

For the second generation assertion use the permitted Olshanskii centralizer lemma, in the precise formIn particular $Z_r\subseteq \mathrm{alg}(Z_{r-1},X_r)$. Induction puts every $Z_r$ in $\mathrm{alg}(1,X_1,\dots ,X_r)$. Conversely, if $A_r$ is the sum of all transpositions in $S_r$, then $A_r\in Z_r$ and $X_r=A_r-A_{r-1}\in G{Z}_n$. ThereforeHere generation is unital; the zero generator $X_1$ causes no problem.

The primitive idempotent products used above show that the Gelfand–Tsetlin algebra is generated by the centers at all levels: those products give every diagonal matrix unit, whereas each center acts diagonally. Hence

For $a,b<0$, ${\mu }_s<1$. The local attracting nonzero segments have $x<-b/(2a)$ or $x>0$, again subject to $M>-1$; the segment between the saddle-node and zero repels. The crossings at ${\mu }_s$ and one are saddle-node and transcritical. The quadratic-cubic flip criticality at the origin requires additional information: writing $\mu =-1+ϵ$ givesIt is supercritical if $a+b^2>0$ and subcritical if $a+b^2<0$, with a degenerate flip if $a+b^2=0$. The sign assumptions in case (iii) do not determine this criticality. The original “no further bifurcations in the region sketched” qualification is local; it does not remove this possible degeneracy or the additional $M=-1$ crossings farther along fixed branches.

For $a,b>0$, ${\mu }_s>1$. The negative branch segment $-b/(2a)<x<0$ has $M<1$ and is attracting wherever $M>-1$; all other nonzero segments repel or undergo further flips according to the full multiplier criterion. Zero and the crossing branch exchange stability at one, and the two negative branches meet at the saddle-node. The origin's flip at minus one is supercritical.

The pushout of groups isIts maps from $G_1,G_2$ agree on $H$, and every other pair of group homomorphisms agreeing on $H$ factors through one unique homomorphism from $G$. This is the universal property. When the $i_j$ are injective, it is the amalgamated free product, and the two factors and the identified copy of $H$ embed in it. Some conventions use this name for the quotient even without injectivity; the embedding assertion requires injectivity.

for an estimator $\delta (X)$ and loss $L$, its risk function is $R(p,\delta )={\mathbb{E}}_{p}L(\delta (X),p)$, where expectation is over repeated data at fixed $p$. Its Bayes risk under prior $\pi$ is $r(\pi ,\delta )=\int R(p,\delta )\pi (dp)$, equivalently the joint expectation under the prior and sampling model.

If an invertible probability measure-preserving system has a finite one-sided generator $\xi$, then its future sigma-algebra ${\mathcal{F}}_1=T^{-1}\mathcal{B}$ equals $\mathcal{B}$ modulo null sets. Hence $H_{\mu }(\xi \mid {\mathcal{F}}_1)=0$, and the infinite-future formula for partition entropy rate and Kolmogorov-Sinai generator theorem give $h_{\mu }(T)=0$. Invertibility matters: a fair binary one-sided Bernoulli shift has entropy $\log 2$ and a finite one-sided generator.

A finite locally free sheaf of ideals on a reduced scheme has rank at most one at every point. On a nonempty affine open subscheme where its rank is $r$, localize its inclusion into the structure sheaf at a minimal prime ideal. The resulting local ring is a field $K$, giving an injection $K^r\hookrightarrow K$, hence $r\le 1$. This works without a Noetherian assumption. The empty scheme is a vacuous exception to claims phrased as nonexistence of a sheaf of a prescribed rank.

Evaluation at a point identifies parallel tensor fields with tensors fixed by the full Riemannian holonomy group. A parallel field is fixed after transport around every loop. Conversely transport a fixed tensor along paths from the basepoint; loop invariance makes the result independent of the path, and local parallel transport shows smoothness and parallelness. Connectedness gives uniqueness.

A nonzero parallel one-form determines a nonzero parallel vector by metric duality. Its value is fixed by the full Riemannian holonomy group, so its span is an invariant line. This contradicts irreducibility when $\dim M\ge 2$. The dimension condition matters: the standard circle has trivial but irreducible one-dimensional holonomy representation and a nonzero parallel one-form.

The locus of nonzero holomorphic one-forms with a fixed list of zero multiplicities $k_1,\dots ,k_r$, summing to $2g-2$. The SL2R action on differentials preserves this list. For example $\mathcal{H}(2)$ and $\mathcal{H}(1,1)$ are distinct genus-two strata.

On a star-shaped holomorphic coordinate neighbourhood, a closed holomorphic differential form of positive degree has a holomorphic primitive. Radial integration gives the homotopy operator. This proves exactness of the holomorphic de Rham complex on stalks.

The sheaf is ${\mathcal{Z}}_M^1=\ker (\partial :{\Omega }_M^1\to {\Omega }_M^2)$. The holomorphic Poincaré lemma gives $0\to \underline{\mathbb{C}}\to {\mathcal{O}}_M\stackrel{\partial }{\to }{\mathcal{Z}}_M^1\to 0$. On a compact complex surface its long exact sequence in sheaf cohomology yields $0\to H^0(M,{\Omega }_M^1)\to H^1(M,\mathbb{C})\to H^1(M,{\mathcal{O}}_M)$.

For a short root $\alpha$ in the G2 root system, restriction of the Adjoint representation to the sl2 subalgebra associated with a root givesHere $V(n)$ is the irreducible sl2 Lie algebra representation of highest weight $n$ and dimension $n+1$. Every nonzero root vector $x\in {\mathfrak{g}}_{\alpha }$ has $\dim Z_{\mathfrak{g}}(x)=6$: the raising operator has a one-dimensional kernel in each of these six irreducible summands.

The area of the flat metric of a quadratic differential is ${\int }_X|q|$. Multiplying $q$ by a scalar $a$ multiplies this area by $|a|$ and lengths by $|a{|}^{1/2}$. It is preserved by the SL2R action on differentials.

For an elliptic curve over ${\mathbb{F}}_p$, put $N=\#E({\mathbb{F}}_p)$ and $a=p+1-N$. The trace of the square of an elliptic-curve endomorphism gives $\mathrm{tr}({\pi }^2)=a^2-2p$ for the Frobenius isogeny. Therefore $\#E({\mathbb{F}}_{p^2})=p^2+1-a^2+2p=N(2p+2-N)$.

For odd $p$ and $k\not\equiv 0(\mathrm{mod}p)$, take $E:y^2=x^3+kx$. If $p\equiv 3(\mathrm{mod}4)$, pairing $x$ with $-x$ cancels the Legendre symbols. If $p\equiv 1(\mathrm{mod}4)$, the coefficient formula for trace of Frobenius modulo p givesThe binomial coefficient is nonzero because every factorial involved is shorter than $p$. Thus the Trace of Frobenius is not zero in the second case.