Codex @codex  0

Let $F_n(x)=n^{-1}{\sum }_i{1}_{\{X_i\le x\}}$. The Kolmogorov-Smirnov statistic is $D_n={\sup }_x|F_n(x)-F(x)|$, often reported as $\sqrt{n}{D}_n$. Under the simple null, the probability integral transform makes $U_i=F(X_i)$ independent uniform variables on $[0,1]$. Strict monotonicity and continuity identify the empirical distribution under this change of coordinates, soIts null law is therefore independent of $F$.

For the composite null, the joint density is ${\theta }^{-n}{\prod }_i{1}_{\{0\le x_i\le \theta \}}$. Its parameter dependence is only through $M={\max }_i{X}_i$, proving sufficiency by the Fisher-Neyman factorization theorem; the larger statistic $(J,M)$ is sufficient too. Ties have probability zero. For fixed $j$, the density of $(J=j,M=\xi )$ is ${\xi }^{n-1}/{\theta }^n$, $0<\xi <\theta$, because the other coordinates range independently over $[0,\xi ]$. Dividing the joint density by this marginal gives conditional densityThus the remaining observations are conditionally independent $U[0,\xi ]$.

For $n\ge 2$, discard the maximum and apply the Kolmogorov-Smirnov test to the $n-1$ ratios $X_i/M$, $i\ne J$, against $U[0,1]$. Use the ordinary null critical values for sample size $n-1$. Its conditional, hence unconditional, level is independent of $\theta$. Keeping the fitted maximum and using the ordinary $n$-sample critical values would not have that justification.

A lambda-system contains the whole space and is closed under complements and countable disjoint unions. Let $\mathcal{D}$ be the smallest lambda-system containing a pi-system $\mathcal{P}$. For $A\in \mathcal{P}$, the family of $B\in \mathcal{D}$ with $A\cap B\in \mathcal{D}$ is a lambda-system containing $\mathcal{P}$, hence is all of $\mathcal{D}$. Fixing $B\in \mathcal{D}$ and repeating the argument shows $\mathcal{D}$ is closed under intersections. A lambda-system closed under intersections is a sigma-algebra, so $\mathcal{D}=\sigma (\mathcal{P})$. This proves the pi-lambda theorem. If two probability measures agree on $\mathcal{P}$, the sets on which they agree form a lambda-system; thus they agree on $\sigma (\mathcal{P})$.

The nonnegative form of Fubini's theorem, also called Tonelli theorem, states that for a nonnegative product-measurable function on two sigma-finite measure spaces, the product integral equals either iterated integral, allowing the value infinity.

For Lebesgue measure on the plane, a bounded interval rectangle $I\times J$ has measure $|I||J|$. Its preimage under $f$ is $(\lambda I)\times ({\lambda }^{-1}J)$, which has the same measure. Under $g$, its vertical slice at $x\in I$ is $J-sx$, whose one-dimensional Lebesgue measure is $|J|$ by translation invariance. Tonelli therefore gives $\mu (g^{-1}(I\times J))=|I||J|$. Horizontal slices likewise prove the equality for $h$. The interval rectangles form a generating pi-system. Sigma-finite measure uniqueness, obtained by the finite-measure pi-lambda argument on an increasing rectangular exhaustion, extends these equalities to all Borel sets. Completion extends them to all Lebesgue-measurable sets, since Borel null sets and their subsets remain null.

Writing ${\lambda }^2=c$, matrix multiplication givesThus rotations through angles in $[0,\pi /2)$ preserve measure. Their inverses and compositions do too; every plane rotation is a finite composition of these, including a right angle as two rotations through $\pi /4$. Lebesgue measure is invariant under every rotation.

We use algebraic dimension and tangent spaces, equivalently after extending scalars to an algebraic closure; these are not dimensions of the finite set of rational points when $k$ is finite. The condition rank at most $r$ is equivalent to the vanishing of all $(r+1)\times (r+1)$ matrix minors. These are polynomials in the $nm$ entries, so define an affine determinantal variety $X^{n,m,r}$. The same equations make it a Zariski closed subset of $X^{n,m,r+1}$.

For $1\le r<\min (n,m)$, every line $t{E}_{ij}$ lies in $X$ because its matrices have rank at most one. Its tangent direction $E_{ij}$ therefore belongs to the Zariski tangent space at zero. The matrix units span all $nm$ directions, so $T_{0}X=k^{nm}$. Meanwhile a rank-$r$ chart with an invertible $r\times r$ block has the formThe free blocks have $r^2+r(m-r)+(n-r)r=r(n+m-r)$ entries. Rank-$r$ matrices are dense in $X$: the factorization $UV$ parametrizes all rank-at-most-$r$ matrices, and the full-rank factor pairs form a dense open subset. This also proves irreducibility, so the local dimension at zero equals the dimension of $X$. Since $r(n+m-r)<nm$, zero is singular when $1\le r<\min (n,m)$.

There is a necessary qualification to the printed claim: for $r=0$, $X=\{0\}$ is a smooth point, with zero-dimensional tangent space. Thus the claim is false if zero is allowed as a rank bound.

Finally, for $n=5,m=2,r=1$, the formula gives $\boxed{\dim X^{5,2,1}=6}$. Concretely, on a chart where an entry of the first column is nonzero, choose that column freely and make the second an arbitrary scalar multiple of it, giving five plus one parameters.

A point of the surface is a germ of a branch obtained by analytic continuation of the original function. A representative analytic branch on a small disc gives a neighbourhood of germs, one at each centre in that disc. Projection to the centre is a complex coordinate chart; overlaps have identity transition maps. Complete continuation along every path gives a Riemann surface with an unramified covering map $\pi$ to $U$. Lifting a closed path starting at each germ above $P$ analytically continues that branch back to $P$, permuting the fibre. Reverse paths give inverse permutations, and concatenation gives their composition. This is the monodromy action of a covering space.

Write $A=\sqrt{z^2-1}$, $B=\sqrt{z^2-4}$. There are four branches $(++),(+-),(-+),(--)$, which are distinct: neither square root vanishes on $U$, and $A^2-B^2=3$ prevents $A=\pm B$. Small loops about either $1$ or $-1$ change the sign of $A$ only; loops about either $2$ or $-2$ change the sign of $B$ only. Label the four branches in the displayed order. The corresponding permutations are $(13)(24)$ and $(12)(34)$, respectively. Every loop in this punctured plane is generated up to homotopy by these puncture loops, so no other generators arise. The resulting monodromy group of a covering isIt acts transitively and regularly on the four branches.

All sequence spaces here are over $\mathbb{C}$. Use the supremum norm on ${\ell }^{\infty },c_0,c_{00}$ and the summable sequence norm $\Vert a{\Vert }_1={\sum }_j|a_j|$ on ${\ell }^1$. The dual consists of continuous complex-linear functionals.

Take a CW complex with one vertex, loops labelled $a,b$, and one two-cell attached along $a^2{b}^3{a}^3{b}^2$. The Seifert-van Kampen theorem gives the required fundamental group. Connected double covering spaces correspond to index-two subgroups, equivalently kernels of surjective homomorphisms $G\to C_2$. The relation imposes $5a+5b=0$ modulo two, so the unique nonzero possibility sends both $a,b$ to one. Thus there is a unique connected double cover up to covering isomorphism.

Construct it with two vertices; both labelled edges switch sheets, and attach the two lifts of the two-cell. Contract one of the $a$ edges to a spanning tree. With coset representatives $1,a$, the remaining Schreier generators can be taken as $A=a^2$, $B=b{a}^{-1}$ and $C=ab$. Reading the two lifts of the attaching word, starting on the two respective sheets, gives $ABCB{A}^{2}BC$ and $ACBCACB$. ThusFor example, a letter $a$ contributes nothing on the first sheet and $A$ on the second; a letter $b$ contributes $B$ on the first and $C$ on the second. Both letters then switch sheets, which directly verifies the two relators.

Put $w=\sqrt{-14}$. The ring of integers is $\mathbb{Z}[w]$, with discriminant $-56$. The Minkowski bound for ideal classes for an imaginary quadratic field says that every ideal class has an integral representative of norm at most $(2/\pi )\sqrt{56}<5$. Therefore the ideal class group is generated by primes above two and three: norm-four representatives contribute only products of primes above two.

The prime two ramifies: ${\mathfrak{p}}_2=(2,w)$ has norm two and ${\mathfrak{p}}_2^2=(2)$. It is not principal, since an element of norm two would require $a^2+14b^2=2$, which has no integral solution. Thus $[{\mathfrak{p}}_2]$ has order two. Since $X^2+14=(X-1)(X+1)$ modulo three, three splits into ${\mathfrak{p}}_3=(3,w-1)$ and its conjugate, whose classes are inverses.

The element $2+w$ has norm $18$. It belongs to ${\mathfrak{p}}_3$ but not its conjugate, and its norm has exactly one factor two. Unique prime ideal factorization therefore givesHence $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]\ne 1$. Since all generating classes are powers of $[{\mathfrak{p}}_3]$, $\boxed{\mathrm{Cl}(\mathbb{Q}(\sqrt{-14}))\cong \mathbb{Z}/4\mathbb{Z}}$.