Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / Section II

Identify the Möbius group with ${\mathrm{PSL}}_2(\mathbb{C})$, the orientation-preserving isometry group of three-dimensional hyperbolic space. Here a properly discontinuous group action permits finite stabilizers: for each compact set $K$, only finitely many $g$ satisfy $gK\cap K\ne \varnothing$.

Fix $o\in {\mathbb{H}}^3$. The set of isometries with $d(o,go)\le R$ is compact. To see this, an orientation-preserving isometry is uniquely determined by the image of an oriented orthonormal frame at $o$. The image point lies in a compact closed hyperbolic ball and its frame lies in a compact rotation-group fiber. These choices parametrize the indicated set of isometries continuously.

A discrete subgroup of a topological group is closed: otherwise distinct subgroup elements accumulating at the same point would have quotients tending to the identity, contradicting isolation of the identity. Its intersection with a compact set is therefore finite. If $K\subseteq B(o,R)$ and $gK\cap K\ne \varnothing$, choosing $x,gx\in K$ gives $d(o,go)\le d(o,gx)+d(gx,go)\le 2R$. Hence discreteness implies a properly discontinuous group action. Conversely, if the subgroup is not discrete, distinct $g_n$ tend to the identity. For a compact ball $K$ with $o$ in its interior, eventually $g_{n}o\in K$, so infinitely many $g_{n}K$ meet $K$, contradicting proper discontinuity. This proves the equivalence.

Matrices of determinant one over the Gaussian integers are closed under multiplication and inversion, with inverse $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. Quotienting by $\{\pm I\}$ gives the specified group of Möbius transformations. The Gaussian integers form a discrete set in the complex plane, so these matrices form a discrete subgroup of ${\mathrm{SL}}_2(\mathbb{C})$, and the finite quotient remains discrete. Thus this group acts properly discontinuously on ${\mathbb{H}}^3$.

Explicit representatives of all four types areThe first is $z\mapsto -1/z$ and has order two; the second is the nonidentity translation $z\mapsto z+1$. The third has positive real reciprocal eigenvalues $(3\pm \sqrt{5})/2$, giving translation along an axis without rotation. For the last, a nonreal trace excludes both unit-modulus and real eigenvalues. Its multiplier therefore has modulus different from one and a nonzero rotation angle: it is a loxodromic Möbius transformation.

Use the square-root encryption convention called the Rabin-Williams scheme in the Cambridge course notes. Choose distinct large primes $p,q\equiv 3(\mathrm{mod}4)$, keep them secret, and publish $N=pq$. A message block $m\in \{0,\dots ,N-1\}$ is encrypted asUsually require $\gcd (m,N)=1$; otherwise the message itself reveals a factor. For a quadratic residue $c$ modulo $p$, Euler criterion gives $c^{(p-1)/2}=1$, so $c^{(p+1)/4}$ is a square root. Compute both signs modulo each prime, then combine the four sign choices by the Chinese remainder theorem. For a unit ciphertext there are four roots modulo $N$. Message redundancy identifies the intended root; squaring alone does not identify it uniquely.

To reduce integer factorization to recovery of square roots, choose $x$ uniformly from $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$, compute $c=x^2$, and ask the alleged decoder for any root $y$ of $c$. Conditional on $c$, the secret $x$ is uniform among its four roots, even if the decoder deterministically chooses $y$. With probability $1/2$, $y\not\equiv \pm x(\mathrm{mod}N)$. Then the signs agree modulo one prime and disagree modulo the other. Sincethe greatest common divisor factors $N$. Independent trials require an expected two successful oracle calls. If a trial's chosen $x$ is not a unit, $\gcd (x,N)$ already factors $N$. This establishes the computational equivalence between factorization and inversion of the squaring map; it does not claim that learning any partial information about a formatted message is equivalent to full inversion.

For coin flipping by telephone, Alice publishes $N$ with its factorization secret. Bob chooses uniformly a unit $x$ with $0<x<N/2$ and sends $c=x^2\mathrm{mod}N$. Alice computes the two roots $r_1,r_2$ in this half interval, selects a binary digit in which they differ, and announces a guess for that digit of Bob's $x$. Bob reveals $x$; Alice verifies its range and square, and their common bit records whether her guess was correct. Alice then reveals $p,q$, so Bob verifies the primes, product, and differing digit.

Conditional on $c$, Bob's original root is equally likely to be $r_1$ or $r_2$, so Alice's guess succeeds with probability $1/2$. Bob cannot change his answer to the other half-interval root without finding a nontrivial second square root, which would factor $N$ by the argument above. Thus, assuming factorization is hard and the prescribed sampling and completed exchange are followed,The revealed factors make the completed transcript checkable. The protocol does not prevent a party from aborting communication.

Use a normal linear modelThe distance and climb are treated as fixed predictors; the assumptions are a linear conditional mean, independent errors with common variance, and normal errors for the exact tests. Fit by ordinary least squares using

The overall null hypothesis is ${\beta }_1={\beta }_2=0$, against at least one nonzero slope. The intercept-only and full models have respectively 34 and 32 residual degrees of freedom. Under the null, the F-test statistic isThe displayed residual sums are rounded; use the reported statistic for the test. Its upper-tail p-value is below $2.2\times 10^{-16}$, so reject the no-relationship hypothesis overwhelmingly. At least one predictor adds information beyond a constant mean.

In the coefficient table, Estimate gives one of the ordinary least squares estimators, Std. Error is the estimated standard deviation of that estimator, t value is the estimate divided by its standard error, and the last column is the two-sided p-value for a zero coefficient. Under the corresponding null, the statistic has a Student's t-distribution with 32 degrees of freedom. The fitted mean isAt fixed climb, an extra mile predicts about 6.22 extra minutes; at fixed distance, an extra foot of climb predicts about 0.01105 extra minutes. Both slopes have strong evidence of being positive. Specifically the climb test uses $t=5.387$ and $p=6.45\times 10^{-6}$, so climb remains significant after controlling for distance. The intercept differs from zero at the 5% level, but extrapolating to a zero-distance, zero-climb race has no useful physical interpretation.

The regression diagnostics show a very large positive residual for Knock Hill and a smaller positive outlier for Bens of Jura. The Q-Q plot has a pronounced upper tail, contradicting the assumed normal-error pattern. These observations can distort both the fitted coefficients and the estimated error variance. Check the source records, units, transcription and any unusual race conditions first, especially for Knock Hill; correct a demonstrated data error and refit. If the records are valid, consider omitted terrain effects, nonlinear distance/climb effects, a suitable response transformation or robust regression, and compare fits with and without influential observations. A diagnostic outlier alone is not a reason to silently delete a valid record.

Extend $q$ by zero to $x<0$ and apply the assumed Fourier inversion theorem. For sufficiently regular, decaying data and $x>0$ this gives $q(x)=(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{ikx}\hat{q}(k)dk$, with $\hat{q}(k)={\int }_0^{\infty }{e}^{-ikx}q(x)dx$. If $\mathrm{Im}k\le 0$, the exponential in this last integral is bounded in magnitude by one; it defines a holomorphic transform in the lower half-plane, with its boundary values on the real axis.

The PDF orients $L$ from $i\infty$ down to zero, then from zero to $+\infty$. The function $e^{ikx}\hat{q}(-k)$ is holomorphic in the upper half-plane, and for $x>0$ its exponential decays there. Close the two rays through a large first-quadrant arc. Cauchy integral theorem and Jordan lemma giveFor example, if $q,q^{\prime }$ are integrable, integration by parts gives $\hat{q}(-k)=O(1/k)$ in that quadrant, which suffices for the arc estimate. Standard limiting arguments extend the identity to the admissible Fourier-inversion class. Adding any constant multiple of this zero integral to ordinary inversion provesThe extra integral is a reflected, zero-supported contribution, not an extra datum.

Set $\Theta =e^{-ikx+i{k}^{2}t}$. Expanding the proposed local relation givesThus its vanishing is exactly $i{q}_t+q_{xx}=0$. Integrating over $x>0$, using decay at infinity, yieldsIntegrate in time and use the initial data to obtain the global relationHere ${\tilde{g}}_j(z,t)={\int }_0^t{e}^{iz\tau }{g}_j(\tau )d\tau$, with $g_0=q(0,\cdot )$ and $g_1=q_x(0,\cdot )$. In particular these finite-time transforms are entire in $z$.

Write $F(k)=e^{ikx-i{k}^{2}t}$ and $B_{\pm }(k)=k{\tilde{g}}_0(k^2,t)\pm i{\tilde{g}}_1(k^2,t)$. Substitution into part (i), including replacing $k$ by $-k$ in the reflected transform, givesFor the boundary integrands, rotate the negative real ray to the positive imaginary ray through the second quadrant. There are no singularities, and for $k=a+ib$ with $a\le 0,b\ge 0$,After writing $F{B}_{+}$ as its finite time integral, this estimate and the usual smooth-data arc bounds justify the rotation. The positive real ray stays fixed and the new imaginary ray is directed downwards, givingThe same argument applies separately to the ${\tilde{g}}_0$ and ${\tilde{g}}_1$ terms. Thus the net boundary integral in the inversion formula isFor Dirichlet boundary conditions choose $c=-1$, eliminating the unknown normal derivative. For Neumann boundary conditions choose $c=+1$, eliminating the unknown boundary value. The resulting unified transform representations areEach contains only the initial transform and the prescribed boundary transform. The signs depend on the PDF's orientation of $L$, checked above.

Differentiate the Friedmann equation and use cosmological energy conservation:For an expanding universe $H>0$, division by $2H$ and substitution of the Friedmann equation give the Friedmann acceleration equationThe strong energy condition makes $\ddot{a}\le 0$. Since $aH=\dot{a}$ and $\dot{a}>0$,Thus the quantity is nonincreasing, with strict decrease when the energy condition is strict. Dividing the Friedmann equation by $H^2$ yieldsFor fixed nonzero curvature, the magnitude of $\Omega -1$ grows as expansion decelerates. The observed closeness to one therefore requires much tighter early-time closeness to one in the standard decelerating model: this is the flatness problem. Accelerated expansion during inflation reverses this evolution and suppresses the curvature contribution.

Normalize the scale factor by $a(t_0)=1$. Separate conservation for a constant equation-of-state parameter $w$ gives $d\log {\rho }_i=-3(1+w_i)d\log a$. Thus radiation has ${\rho }_R={\rho }_{R0}{a}^{-4}$ while the dark energy component with $w=-1$ has constant density. Flatness gives ${\Omega }_{R0}+{\Omega }_{\Lambda 0}=1$, soHere $\rho$ has mass-density units in the stated Friedmann equation; the corresponding energy density is $\rho c^2$.

Assume both density parameters are positive and choose the Big Bang at $t=0$. The expanding branch satisfies $\dot{a}=H_0\sqrt{{\Omega }_{R0}{a}^{-2}+{\Omega }_{\Lambda 0}{a}^2}$. Set $u=a^2$ and integrate:ConsequentlyThe factor $H_0$ is required because $\beta$ has inverse-time units. As $t\downarrow 0$, $a(t)\sim (2H_0\sqrt{{\Omega }_{R0}}t{)}^{1/2}$, the expected radiation-dominated power law. As $t\to \infty$, $a(t)\sim \alpha e^{H_0\sqrt{{\Omega }_{\Lambda 0}}t}/\sqrt{2}$, the exponential expansion driven by a cosmological constant.

The inductive definition of ordinal addition is $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ for a nonzero limit ordinal $\lambda$. The synthetic definition is the order type of the order sum obtained by putting a copy of $\beta$ after a disjoint copy of $\alpha$.

These agree by transfinite induction on the second argument. Appending no points gives $\alpha$; appending one new greatest point gives a successor. At a limit stage, the order sum is exhausted by the increasing initial segments obtained by appending each $\beta <\lambda$, so its order type is their supremum. Thus the synthetic construction satisfies all three recursion clauses, and uniqueness of transfinite recursion proves equivalence.

False. By the successor and limit clauses for ordinal multiplication,Multiplication is not commutative because it counts ordered copies, rather than merely the cardinality of their union.

Always true. In the order-sum interpretation of ordinal multiplication, $\alpha (\beta +\gamma )$ consists of copies of $\alpha$ indexed first by $\beta$ and then by $\gamma$. The first portion has order type $\alpha \beta$ and the second $\alpha \gamma$, proving $\alpha (\beta +\gamma )=\alpha \beta +\alpha \gamma$. Equivalently, transfinite induction on $\gamma$ uses the successor rule, associativity of ordinal addition, and continuity of both operations in the right argument at limit ordinals.

False. Take $\alpha ={\omega }^2+1$, which satisfies the proposed lower bound. Associativity of ordinal addition and $1+{\omega }^2={\omega }^2$ giveThe final successor survives only in the second expression.

False. Take $\alpha ={\omega }_1+1$. Continuity of ordinal multiplication in the right argument gives $({\omega }_1+1){\omega }_1={\omega }_1^2$. For a precise bound, for each countable ordinal $\beta$ monotonicity givesBoth bounding expressions have supremum ${\omega }_1^2$ as $\beta$ ranges below ${\omega }_1$, because $2\beta$ is still countable and these indices are cofinal. But the successor rule gives

A matching in a graph is a set of edges with no shared endpoints. A matching from $X$ to $Y$ here covers every vertex of $X$. Necessity of the Hall marriage theorem condition is immediate: the partners of vertices of $A\subseteq X$ are distinct members of its neighbour set $\Gamma (A)$.

For sufficiency, take a matching of maximum size. If it leaves $x\in X$ unmatched, follow alternating unmatched and matched edges from $x$. Let $A\subseteq X$ and $B\subseteq Y$ be the reachable vertices. No vertex of $B$ can be unmatched, because the resulting augmenting path in a matching would increase the matching. Every neighbour of $A$ is in $B$, and the matching pairs $B$ bijectively with $A\setminus \{x\}$. Thus $|\Gamma (A)|=|B|=|A|-1$, contradicting Hall's condition. The maximum matching therefore covers $X$.

For the deficiency bound, adjoin $d$ new vertices to $Y$, each adjacent to all of $X$. Every nonempty $A$ then has $|\Gamma (A)|+d\ge |A|$ neighbours. Apply the Hall marriage theorem and remove the at most $d$ edges using new vertices. This leaves at least $|X|-d$ independent edges. The lower bound is vacuous if $d>|X|$.

Write the adjacency matrix of a graph in bipartite block form $\left(\begin{array}{cc}0& C\\ C^T& 0\end{array}\right)$. If Hall's condition failed for $A\subseteq X$, the linear map from vectors supported on $A$ to their neighbour coordinates in $Y$ would have fewer output than input coordinates. It therefore has a nonzero kernel, giving a nonzero vector $v$ supported on $A$ with $C^{T}v=0$. The vector $(v,0)$ would be a zero eigenvector of the adjacency matrix. Consequently absence of the zero eigenvalue implies a matching covering $X$.

The converse is false, even with equal part sizes. The four-cycle $K_{2,2}$ has a perfect matching, but its matrix $C=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ is singular. Its adjacency matrix has eigenvalues $2,-2,0,0$.

A field homomorphism fixing $K$ is injective, since its kernel is an ideal of a field and cannot contain one. It is also $K$-linear. Equal finite field extension degrees then make it a bijection, hence a $K$-isomorphism.

Equal degrees alone do not ensure that any such homomorphism exists. For example $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ both have degree two, but a $\mathbb{Q}$-isomorphism would require a square root of two in $\mathbb{Q}(\sqrt{3})$. Squaring $a+b\sqrt{3}$ with rational $a,b$ would give $2ab=0$ and $a^2+3b^2=2$. Neither $a^2=2$ nor $b^2=2/3$ has a rational solution. Thus the two degree-two extensions are not isomorphic over $\mathbb{Q}$.

An isomorphism fixing $L$ fixes its subfield $K$, so an $L$-isomorphism is a $K$-isomorphism. The converse is false because an isomorphism over $K$ may act nontrivially on $L$.

Take $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2})$, $\alpha =2^{1/4}>0$, $F=L(\alpha )$ and $F^{\prime }=L(i\alpha )$. Both are degree-two extensions of $L$. The polynomial $X^4-2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion, and both $\alpha$ and $i\alpha$ are roots. Sending $\alpha$ to $i\alpha$ therefore gives a $\mathbb{Q}$-isomorphism $F\to F^{\prime }$; on $L$ it sends $\sqrt{2}$ to $-\sqrt{2}$.

There is no $L$-isomorphism. The two quadratic extensions adjoin square roots of $\sqrt{2}$ and $-\sqrt{2}$. Their square classes would agree only if $-1$ were a square in $L$. To check this criterion directly, a square root of $\sqrt{2}$ in $L(i\alpha )$ would have form $a+b(i\alpha )$; its square forces $ab=0$. The case $b=0$ would put $\alpha$ in $L$, while $a=0$ would require $b^2=-1$. Both are impossible, the latter since $L\subseteq \mathbb{R}$. Thus the extensions are isomorphic over $K$ but not over $L$.

If $g$ belongs to the centre of a group, then $\rho (g)$ commutes with $\rho (h)$ for every $h\in G$. The Schur lemma says that an endomorphism of a finite-dimensional complex irreducible representation commuting with the representation is scalar. ThereforeSince $g$ has finite order, ${\lambda }_g$ is a root of unity and has modulus one.

Let $\chi$ be the character of a representation of $\rho$. The character orthogonality identity for an irreducible character is ${\sum }_{g\in G}|\chi (g){|}^2=|G|=n$. For $g\in Z$, the scalar formula from part (i) gives $|\chi (g)|=d$. Keeping these $m$ nonnegative summands yields

If $\rho$ is a faithful representation, $g\mapsto {\lambda }_g$ embeds $Z$ into the multiplicative group of complex numbers. Its image is a finite group of roots of unity, hence lies in the cyclic group of $n$th roots of unity and is itself cyclic. Thus $Z$ is cyclic.

For the order-18 group, the index-two subgroup $P$ is normal and $G=⟨P,t⟩$. Restrict a complex irreducible representation $V$ to the finite abelian group $P$. Simultaneous diagonalization decomposes it into character spaces $V_{\chi }$. The inversion relation makes $t$ exchange $V_{\chi }$ and $V_{{\chi }^{-1}}$. The sum of spaces belonging to any such character orbit is $G$-invariant, so irreducibility leaves only one orbit.

If the character is trivial, all of $P$ is in the representation's kernel. Otherwise $\chi$ has order three and $\chi \ne {\chi }^{-1}$. The only spaces are $V_{\chi }$ and $V_{{\chi }^{-1}}$, and both have the same kernel in $P$. A nontrivial character of $P\cong C_3\times C_3$ maps onto $C_3$, so that kernel has order three. It acts trivially on all of $V$. Therefore the group has no faithful irreducible complex representation. In fact choosing nonzero $v\in V_{\chi }$ makes $\mathrm{span}(v,tv)$ invariant, so these nontrivial-character irreducibles have degree two.

Put $s=\sqrt{-17}$. The ring of integers of a number field is $\mathbb{Z}[s]$ and the field discriminant is $-68$. The Minkowski bound for ideal classes says that each ideal class has an integral ideal representative of norm at most $(2/\pi )\sqrt{68}<6$.

The prime two ramifies as $(2)={\mathfrak{p}}_2^2$, where ${\mathfrak{p}}_2=(2,1+s)$. Three splits as $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, choosing ${\mathfrak{p}}_3=(3,1+s)$, whereas five is inert because $-17\equiv 3(\mathrm{mod}5)$ is not a square. The norm bound therefore leaves only the classes of $1$, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\overline{\mathfrak{p}}}_3$; a norm-four ideal is ${\mathfrak{p}}_2^2$ and is principal.

The ideal ${\mathfrak{p}}_2$ is not principal, since an element of norm two would solve $a^2+17b^2=2$. On the other hand $N(1+s)=18$, and $1+s$ is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$. Unique prime ideal factorization and the norm give $(1+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Thus $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2]$, which has order two, and

For the Diophantine equation, $x$ must be positive. If $x$ were even, $y^2\equiv 15(\mathrm{mod}32)$, impossible for a square. Thus $x$ is odd and $y$ even. If $17\mid x$, then $17\mid y$, but $y^2+17$ would have 17-adic valuation one while $x^5$ has valuation divisible by five, also impossible.

The ideals $(y+s)$ and $(y-s)$ are consequently coprime. A common prime divides $2s$ and hence lies over two or seventeen, both excluded above. Their product is $(x{)}^5$, so unique prime ideal factorization gives $(y+s)={\mathfrak{a}}^5$. Since the ideal class group has order four, $[\mathfrak{a}{]}^5=1$ forces $[\mathfrak{a}]=1$. Write $\mathfrak{a}=(a+bs)$ with integers $a,b$. The only units are $\pm 1$ by their norm, and the sign can be absorbed into the fifth power. Hence $y+s=(a+bs{)}^5$.

Comparing the coefficient of $s$ givesFor $b=1$, the bracket would be one, but it is four modulo five. For $b=-1$, the bracket would be minus one, but it is one modulo three for either possible square residue of $a$. Both cases are impossible. Therefore there are no integer solutions.

True. A path $u:x\to y$ gives the basepoint-change isomorphism of fundamental groups $[\gamma ]\mapsto [\overline{u}\ast \gamma \ast u]$. Concatenating paths is compatible with homotopy and loop multiplication. Using the reversed path gives its inverse, because $u\ast \overline{u}$ and $\overline{u}\ast u$ contract relative to endpoints.

False. Changing the connecting path changes the isomorphism by an inner automorphism. Take $x=y$, one path constant and the other representing $a$ in the Klein bottle group $⟨a,b:ab{a}^{-1}=b^{-1}⟩$. The first induced isomorphism is the identity, whereas the second sends $b$ to $b^{-1}\ne b$. The element $b$ has infinite order, as is also witnessed by mapping it to a translation and $a$ to a reflection in the infinite dihedral group. A fixed path to a different endpoint can be appended if distinct basepoints are desired.

True. Coordinate projections define ${\pi }_1(X\times Y,(x,y))\to {\pi }_1(X,x)\times {\pi }_1(Y,y)$. A pair of based loops gives the inverse by $t\mapsto ({\gamma }_X(t),{\gamma }_Y(t))$. Homotopies combine and project coordinatewise, and loop concatenation does too. These maps are well-defined inverse group homomorphisms, requiring no local connectedness assumptions.

False. Take a bouquet of two circles and the two-dimensional torus. Both are connected polyhedra and have first homology group ${\mathbb{Z}}^2$. Their fundamental groups are respectively the nonabelian free group $F_2$ and the abelian group ${\mathbb{Z}}^2$, so they are not isomorphic. First homology only remembers the abelianization of the fundamental group.

The real Stone-Weierstrass theorem says that a subalgebra of $C(X,\mathbb{R})$ on a compact metric space is uniformly dense if it contains the constants and separates points. The finite sums ${\sum }_i{f}_i(x)g_i(y)$ form such an algebra on $[0,1{]}^2$: multiplication preserves this form, the constant one belongs to it, and the coordinate functions distinguish every pair of different points. Thus they approximate the continuous kernel uniformly.

A finite-rank operator has finite-dimensional range. A compact operator maps the unit ball to a set with compact closure, equivalently sends every bounded sequence to a sequence with a convergent subsequence. The identity on $C[0,1]$ is bounded but not compact: the functions $x^n$ have supremum norm one, and every subsequence has the discontinuous pointwise limit which is zero on $[0,1)$ and one at one. It therefore has no uniformly convergent subsequence.

A bounded finite-rank operator is compact because bounded sets in its finite-dimensional range have compact closure. Suppose $T_n\to T$ in operator norm. For any $\epsilon >0$, choose $n$ with $\Vert T-T_n\Vert <\epsilon /3$ and a finite $\epsilon /3$-net for $T_n$ of the unit ball. The same centres give an $\epsilon$-net for $T$ of that ball. Its closure is totally bounded and complete in the Banach space $Y$, hence compact. Thus an operator-norm limit of compact operators is compact.

For $K_n(x,y)={\sum }_i{f}_i(x)g_i(y)$ uniformly approximating $K$, the integral operator $T_n$ has range in $\mathrm{span}\{f_i\}$ and is bounded. MoreoverConsequently $T_n\to T$ in operator norm, proving that the stated integral operator is compact.

Near an isolated point $P\in A$, choose source and target coordinate charts, using continuity of $\alpha$ to keep a sufficiently small source neighbourhood in the target chart. The coordinate representation is holomorphic on the punctured neighbourhood and is bounded near $P$ by continuity. The Riemann removable singularity theorem extends it holomorphically across $P$, with the given continuous value. Repeating at all points of $A$ proves that $\alpha$ is analytic.

For a nonconstant holomorphic function $f$ on a connected Riemann surface, let $A$ be the set where its coordinate derivative vanishes. This condition is independent of the chart. The derivative cannot vanish identically on an open set, because the identity theorem would then make $f$ constant globally. Its zeros are isolated and closed locally, so $A$ is a closed discrete subset. At every other point the holomorphic inverse function theorem makes $f$ a local coordinate chart.

Now suppose $\alpha$ is a homeomorphism and $g=f\circ \alpha$ is analytic. Outside ${\alpha }^{-1}(A)$, use $f$ as a target coordinate to write locally $\alpha =f^{-1}\circ g$. This is analytic. The preimage ${\alpha }^{-1}(A)$ is discrete because $\alpha$ is a homeomorphism, and the first argument extends analyticity across it. An injective holomorphic map cannot have a zero derivative: a zero of order at least two would give a locally multiple-to-one map. Its local inverse is therefore holomorphic, proving that $\alpha$ is a conformal equivalence.

The plane $\mathbb{C}$ and the unit disk are homeomorphic by $z\mapsto z/(1+|z|)$, with inverse $w\mapsto w/(1-|w|)$. They are not conformally equivalent: a holomorphic equivalence from the plane to the disk would be a bounded nonconstant entire function, contradicting Liouville's theorem.

An affine variety is irreducible when it is nonempty and cannot be expressed as the union of two proper Zariski-closed sets. If a nonempty open subset $U$ were not dense, $X=\overline{U}\cup (X\setminus U)$ would be such a union: the first set is proper by assumption, and the second is proper because $U$ is nonempty. Hence every nonempty open subset of an irreducible variety is dense.

The coefficients of $\det (TI-A)$ are polynomial functions of the matrix entries. Its polynomial discriminant $\Delta (A)$ is therefore a polynomial in those entries, and it is nonzero exactly when all roots of the characteristic polynomial are distinct. Thus the required set is the principal open subset $D(\Delta )$ of affine space. It is nonempty: an algebraically closed field is infinite, so choose a diagonal matrix with $n$ pairwise distinct entries.

It is also affine as a variety. The map $A\mapsto (A,1/\Delta (A))$ identifies it with the closed hypersurfaceProjection to $A$ is its regular inverse. This proves the requested affineness while retaining the open embedding into matrix space.

Each entry of ${\mathrm{char}}_A(A)$ is a polynomial in the entries of $A$, so their simultaneous zero set $Z$ is Zariski closed. A matrix in $D(\Delta )$ has distinct eigenvalues and is diagonalizable. If $A=S\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)S^{-1}$, thenThus $D(\Delta )\subseteq Z$. Affine space is irreducible because its coordinate ring is an integral domain; by part (i), its nonempty open subset $D(\Delta )$ is dense. The closed set $Z$ containing it must be all of affine space. This proves the Cayley-Hamilton theorem:

A point $x$ is critical for a smooth map $f:X\to Y$ when $D{f}_x:T_{x}X\to T_{f(x)}Y$ is not surjective. A critical value is the image of a critical point; a regular value has no critical point in its fiber, including the case of an empty fiber.

To prove the regular level set theorem, take $x_0\in f^{-1}(y)$ and write $n=\dim X$, $m=\dim Y$. In local coordinates, surjectivity supplies an invertible $m\times m$ derivative minor; relabel source coordinates so it uses the first $m$ columns. The maphas an invertible derivative. The inverse function theorem makes it a local diffeomorphism. In these coordinates the fiber is obtained by fixing its first $m$ coordinates, so it is locally an open subset of ${\mathbb{R}}^{n-m}$. These smooth charts prove that a nonempty regular fiber is a smooth embedded manifold of dimension $n-m$, with tangent space $\ker D{f}_x$.

For the special orthogonal group, use $F:M_n(\mathbb{R})\to {\mathrm{Sym}}_n(\mathbb{R})$ defined by $F(A)=A{A}^T$. Its derivative is $D{F}_A(H)=H{A}^T+A{H}^T$. At an orthogonal $A$, any symmetric matrix $B$ is obtained by taking $H=BA/2$, so $I$ is a regular value. Therefore $O(n)=F^{-1}(I)$ has dimension $n^2-n(n+1)/2=n(n-1)/2$. The determinant takes only the values $\pm 1$ there, so $SO(n)$ is an open and closed part of $O(n)$ and has the same manifold dimension:It is closed in matrix space and bounded because every orthogonal matrix column has norm one. The Heine-Borel theorem proves compactness. Finally the tangent space is the derivative kernel,Thus its translated tangent space is the vector space of skew-symmetric matrices.

The Lp space is the vector space of equivalence classes, modulo equality almost everywhere, of measurable functions $f$ with $\int |f{|}^{p}d\mu <\infty$, equipped with $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. To prove it is a Banach space, let $(f_n)$ be Cauchy in this norm. Choose a subsequence $(f_{n_j})$ such thatTake measurable representatives and put $g_j=|f_{n_{j+1}}-f_{n_j}|$, $G_M={\sum }_{j=1}^M{g}_j$. The Minkowski inequality gives $\Vert G_M{\Vert }_p\le {\sum }_j{2}^{-j}\le 1$. By the monotone convergence theorem, $G={\sum }_{j\ge 1}{g}_j$ satisfies $\int G^{p}d\mu \le 1$. Consequently $G$ is finite almost everywhere, and the telescoping seriesconverges absolutely almost everywhere. Define it arbitrarily on the null exceptional set. The bound $|f|\le |f_{n_1}|+G$ puts $f$ in $L^p$.