Over ${\mathrm{Gr}}_k({\mathbb{C}}^N)$ this rank-k complex vector bundle has fibre W at the point W. In a chart of planes that are graphs of maps from a fixed k-plane, $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The direct-limit bundle over the infinite Grassmannian classifies rank-k bundles over compact Hausdorff spaces.

Tensoring a Tate module with the P-adic numbers gives its rational Tate module. For a CM elliptic curve with base-defined multiplication by $K$, it is free of rank one over $K\otimes {\mathbb{Q}}_{\ell }$, including when this algebra is a product of two fields. In the split case the two distinct roots of a nonrational CM endomorphism each have a one-dimensional eigenspace.

Let $R=K\cap {\mathrm{End}}_k(E)$. If $\alpha \in R$ maps the Tate module into ${\ell }^n{T}_{\ell }(E)$, it kills $E[{\ell }^n]$ and factors through the quotient isogeny $[{\ell }^n]$. The resulting endomorphism is $\alpha /{\ell }^n$, belongs to $K$ and is defined over $k$, hence belongs to $R$. Approximation in $R\otimes {\mathbb{Z}}_{\ell }$ now identifies the displayed stabilizer. This does not assume that the integral Tate module is free over a possibly nonmaximal CM order.

For a Tarski monster group, any nonidentity element generates a proper subgroup of order $p$. Adjoining an element outside that subgroup generates the whole group. A proper nontrivial normal subgroup would have order $p$; conjugation into its finite automorphism group forces it to be central. Combining it with another order-$p$ cyclic subgroup would then give a proper subgroup of order $p^2$, a contradiction.

// Target: algebra.bigb

A Conway mutation cuts a link along a sphere meeting it in four points, rotates the enclosed two-string tangle through $180^{\circ }$, and reglues it.

Objects are finite words of strand orientations. Morphisms are boundary-fixed ambient-isotopy classes of oriented tangles in a slab, composed by stacking compatible boundary words; horizontal juxtaposition is the tensor product. Cups, caps, identity strands and crossings generate the category. Ordinary unframed isotopy includes all three Reidemeister moves; framed variants impose different first-move relations.

A coordinate chart $x$ trivializes the tangent bundle by expressing $v\in T_{p}M$ in the coordinate basis. For $y=f(x)$, the chain rule gives the displayed transition. Smoothness and fibrewise linear invertibility of the Jacobian matrix construct the natural smooth vector bundle structure. The chain rule also supplies the transition cocycle.

For the real tautological bundle $\gamma$ over Real projective space, the tangent space at a line is $\mathrm{Hom}(\gamma ,{\gamma }^{\perp })$. Adding $\mathrm{Hom}(\gamma ,\gamma )\cong {\epsilon }^1$ gives $\mathrm{Hom}(\gamma ,{\epsilon }^{m+1})\cong (m+1)\gamma$, using a metric to identify a real line with its dual. Hence its total Stiefel–Whitney class is $(1+a{)}^{m+1}$, with $a=w_1(\gamma )$.

For a finite Galois extension of non-Archimedean local fields, this is a homomorphism from the inertia group to $k_L^{\times }$ with kernel the wild inertia group. Inertia acts trivially on residue units, which proves multiplicativity and independence of the chosen uniformizer. Passing to the maximal unramified subextension reduces the kernel calculation to a totally ramified extension. Hence it embeds the tame inertia quotient into the cyclic multiplicative group of the residue field.

For a Finite Galois extension of $K(T)$ with algebraically closed characteristic-zero constants, the local decomposition and inertia groups coincide and are cyclic. A fixed compatible choice of primitive roots of unity selects a generator through its uniformizer character of inertia. Changing the prime above a base point conjugates this generator, and changing the uniformizer does not change its character. Without the roots-of-unity convention, only the inertia-subgroup conjugacy class is intrinsic. Over $\mathbb{C}$, positive local orientation gives the standard convention.

On $\{0,1{\}}^{\mathbb{N}}$, the event of infinitely many heads isIt says that after every index there is another head. For each $n$, the same intersection can start at $m=n$, so $A\in \sigma (X_n,X_{n+1},\dots )$; therefore $A$ is a tail event. This verifies measurability directly using countable unions and intersections.

Assume now that the coordinates are fair independent Bernoulli trials. For fixed $m$, let $B_{m,r}$ be the event of at least one head among $X_m,\dots ,X_{m+r-1}$. Its complement requires $r$ tails, soThe last equality follows from continuity of probability. Now $A={\bigcap }_{m\ge 1}{B}_m$, a countable intersection of almost sure events, and hence $\mathbb{P}(A)=1$. This determines which of the two possibilities in Kolmogorov's zero–one law occurs. Equivalently, the Borel-Cantelli second lemma applies to the independent events $\{X_j=1\}$.

Nevertheless, no finite prefix certifies membership in $A$: every finite record can be extended by all tails or by alternating heads and tails. Under the fair independent model, conditioning on any exact finite prefix leaves $\mathbb{P}(A)=1$, since $A$ is independent of the initial sigma-algebra. The excluded all-tails continuation has probability zero, but remains a possible member of the sample space. Logical determination by a finite record and a conclusion holding almost surely are different statements.

Let $1\le t\le r\le n$, and choose an integer $i$ withThe Frankl family ${\mathcal{A}}_i$ consists of the $r$-element subsets of $[n]$ containing at least $t+i$ points of $[t+2i]$. Two members share at least $2(t+i)-(t+2i)=t$ points in this distinguished prefix. Hence it is a t-intersecting family. For $i=0$, it is the uniform set family of all $r$-element subsets containing a fixed $t$-element set.

For consecutive admissible indices $i,i+1$, write $N=n-t-2i-2$ and $a=r-t-i-1$. The members gained by enlarging the prefix contain both new points and exactly $t+i-1$ old prefix points; the lost members contain neither new point and exactly $t+i$ old prefix points. ThusWhen $n>2r-t$, the relevant binomial coefficients are positive and their ratio shows that the difference is positive, zero or negative according as $n$ is below, equal to or aboveThis comparison explains the consecutive-candidate transitions in the Ahlswede-Khachatrian theorem; the prefix range ensures both new coordinates actually exist.

Complementing all members preserves cardinality and gives $|A^c\cap B^c|=n-2r+|A\cap B|$. Thus complementation bijects t-intersecting families of $r$-sets with $(n-2r+t)$-intersecting families of $(n-r)$-sets. It swaps left and right coordinate shifts of a set family. When the transformed intersection parameter is positive, extremal families and their generating-support lemmas can therefore be compared in opposite compression directions.

The maximum size of a t-intersecting family of $r$-sets on $[n]$ is the largest complete-intersection candidate family. In strict nontrivial intervals $n_{k+1}<n<n_k$, the maximizing index is $k$, where $n_k=(r-t+1)(2+(t-1)/k)$ and $n_0=\infty$. At admissible interior boundaries the consecutive candidates tie. If $n\le 2r-t$, the whole level is optimal. A proof uses left compression, the small-support generating lemma for extremal intersecting families for the original and complement families, and the compatibility bound for complementary generating families. The latter forces one generator family to have the candidate threshold size, placing the original family inside a candidate.

If $|A\cap B|\ge t$, then $|A△B|=|A\cup B|-|A\cap B|\le n-t$. The Kleitman diametric theorem therefore gives the sharp nonuniform intersection bound. For $n-t=2q$, all sets of size at least $n-q$ attain it. For $n-t=2q+1$, take all sets of size at least $n-q$, together with all $(n-q-1)$-sets avoiding one fixed point.

For two-intersecting families on four points, all supersets of a fixed pair form a maximal family of size four: any set missing part of that pair fails its intersection test with the pair itself. The maximum instead has size five, attained by all sets of size at least three. Thus maximality and maximum cardinality diverge for intersection parameter two.

On eight points, the rank-four Frankl family consisting of sets containing at least three of $1,2,3,4$ has seventeen members and is two-intersecting. The first fifteen rank-four sets in lexicographic order contain $1,2$, while the sixteenth is $1345$. The earlier set $1278$ meets it only at one. This disproves preservation of two-intersection by lexicographic replacement even when the rank is half the ground-set size.

For aggregate monetary loss $L$ and monetary attachment/exhaustion levels $A<D$, this is the loss allocated to the tranche. It is zero until attachment, rises with losses between $A$ and $D$, and is capped at $D-A$. Expected tranche losses and surviving tranche notional determine protection and premium legs. For basket products whose payoff occurs at a first default, the joint default-time distribution matters as well as the final loss distribution.

For a rank-$(N-k)$ parity-check matrix of a linear code $C$, two words have equal syndromes exactly when their difference lies in $C$. Thus attainable syndromes label cosets bijectively. For a received word $c+e$, its syndrome is that of the error $e$, so minimum-weight representatives provide minimum-distance decoding.

// Target: algebra.bigb

For a linear code with parity-check matrix $H$, write the received word as $y=c+e$, where $c$ is a codeword and $e$ an error pattern. Its syndrome depends only on the error because $H{c}^T=0$. A syndrome decoder chooses a representative error $\hat{e}$ with that syndrome and returns $y-\hat{e}$. Choosing a minimum-Hamming weight representative gives minimum-distance decoding; within the guaranteed correction radius it recovers the unique transmitted codeword. Several representatives may tie outside that radius.