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Put $T={\alpha }^{\prime }(s)$. Since $s$ is arc length, $|T|=1$, so $T^{\prime }\perp T$. The curvature of a space curve isWhere $k(s)\ne 0$, define the principal normal and binormal byand define the torsion of a space curve byThus torsion requires nonzero curvature, so that $N$ and $B$ are defined.

Differentiating the orthonormality relations for the frame $(T,N,B)$ shows that its derivative matrix is skew-symmetric. Since $T^{\prime }=kN$, differentiating $B=T\times N$ and using the definition of $\tau$ gives the Frenet-Serret formulas

Let $P(x)={\prod }_{i=1}^r(x-{\lambda }_i)$. Since both $x$ and $y$ are invertible on $X$, the formuladefines a regular map $F:X\to {\mathbb{A}}^1$.

Fix $t\in k$. To solve $F(x,y)=t$, setThe curve equation becomesThis is a nonconstant polynomial, so it has a root because $k$ is algebraically closed. At a root, neither $x$ nor $t-x$ can vanish, and the equation also forces $P(x)\ne 0$. The displayed value of $y$ is therefore defined and nonzero, satisfies $y^2=P(x)$, and obeys $F(x,y)=t$. Thus $F$ is surjective.

A derivation of $A$ centred at the $k$-point $p:A\to k$ is a $k$-linear map $D:A\to k$ satisfying the twisted Leibniz ruleThe Zariski tangent space isSince $p(f)\ne 0$, $p$ extends uniquely to the localization $A[f^{-1}]$ by $p(a/f^n)=p(a)/p(f{)}^n$. Restriction gives a linear mapConversely, every $D\in T_{p}A$ has the unique extensionThe quotient rule verifies the Leibniz identity, and the assumption that $f$ is not a zero divisor makes the localization representation compatible with equality of fractions. Extension and restriction are inverse, so the natural map is an isomorphism.

View infinity as the puncture of the Riemann sphere and consider $g(1/z)$ near $z=0$. If this map had an essential singularity, the Great Picard theorem in a coordinate chart would make almost every nearby value occur infinitely often, contradicting the injectivity of $g$. The singularity is therefore removable as a map into the compact Riemann surface $R$, and $g$ extends to a nonconstant holomorphic mapThe extension is injective. A collision between $\tilde{g}(\infty )$ and $g(z_0)$, or local degree greater than one at infinity, would give nearby values two distinct preimages by the local normal form of a nonconstant holomorphic map, again contradicting injectivity on $\mathbb{C}$. Thus $\tilde{g}$ is a holomorphic bijection between compact Riemann surfaces and hence a biholomorphism. Therefore $R\cong {\mathbb{P}}^1$, whose genus is zero.

The map $f$ is bounded by one near the puncture, so the Riemann removable singularity theorem extends it holomorphically to a map $\tilde{f}:\Delta \to \mathbb{C}$. The maximum modulus principle shows that its image still lies in $\Delta$: if $|\tilde{f}(0)|=1$, the extension would be constant, contradicting its values in $\Delta \setminus \{0\}$ unless that constant also had modulus less than one.

If $f$ is biholomorphic, its inverse $g:\Delta \setminus \{0\}\to \Delta \setminus \{0\}$ also extends holomorphically. Were $a=\tilde{f}(0)\ne 0$, continuity of $\tilde{g}\circ \tilde{f}$ and the identity $g(f(z))=z$ on the punctured disk would give $g(a)=0$. This is impossible because $a\in \Delta \setminus \{0\}$ and $g$ takes values in the punctured disk. Hence

For $|z|<1$, sum the geometric series and its derivative:ThereforeThis rational continuation is holomorphic across every point of the unit circle except $z=-1$, where it has a double pole. The set of singular points is consequently

Here “covering map” uses the course convention of a local homeomorphism; “regular” adds the evenly-covered product condition. The derivative of $p(z)=z^4$ is $4z^3$, which never vanishes on the upper half-plane $\mathbb{H}$. The holomorphic inverse function theorem therefore makes $p$ a local biholomorphism and hence a covering map in this sense.

It is not regular. The point $1$ has only the preimage $i$ in $\mathbb{H}$, whereas $e^{i\epsilon }$ for small positive $\epsilon$ has the two preimages with arguments $\epsilon /4$ and $\pi /2+\epsilon /4$. Thus the fibre cardinality is not locally constant near $1$, so no neighbourhood of $1$ has a preimage of the form $U\times \Delta$ for a fixed discrete set $\Delta$.

For $F_1$, the functions satisfy $\Vert \sin (\pi nx)/n{\Vert }_{\infty }\le 1/n$, so the sequence converges uniformly to zero. Its closure is the sequence together with its limit and is compact. The same argument applies to $F_2$, since $\Vert \sin (\pi nx)/\sqrt{n}{\Vert }_{\infty }\le n^{-1/2}$. Thus $F_1$ and $F_2$ are relatively compact.

The family $F_3$ is not equicontinuous. At $x=0$ its members vanish, but for $x_n=1/(2n)\to 0$,By the Arzelà-Ascoli theorem, a relatively compact subset of $C[0,1]$ must be equicontinuous. Hence $F_3$ is not relatively compact.

Every function with the stated two linear pieces can be writtenfor suitable real constants $a,b,c$. SincedefineThis is a polynomial. The argument of $P_i$ lies in $[0,1/4]\subseteq [0,1]$, so

The real Stone-Weierstrass theorem states that if $X$ is a compact Hausdorff space and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points of $X$, then $A$ is dense in $C(X,\mathbb{R})$ in the uniform norm.

The Tietze extension theorem states that if $X$ is a normal topological space, $F\subseteq X$ is closed, and $f:F\to \mathbb{R}$ is continuous, then $f$ extends to a continuous function on $X$. In its bounded form, an $f$ taking values in $[-M,M]$ has an extension with the same range bound. The underlying separation step is the Urysohn lemma.

Writing the square coordinates modulo one, the curve has the parametrizationIt is therefore the $(2,3)$ torus knot, or trefoil, on the common boundary of the two solid tori. Applying the Seifert-van Kampen theorem to the complements on the two sides gives one generator $a$ for the meridian of one solid torus and one generator $b$ for the other; the boundary slope identifies two turns of the first with three turns of the second. HenceThis group is nonabelian: imposing the additional relation $a^2=b^3=1$ gives a quotientwhich maps onto the nonabelian group $S_3$ by sending $a$ to a transposition and $b$ to a three-cycle. An abelian group cannot have a nonabelian quotient.

Regard the 3-sphere asIt is the union of the two closed solid toriExplicit homeomorphisms from the two copies in the question areandOn their boundaries both maps send $(u,v)\in S^1\times S^1$ to $(u,v)/\sqrt{2}$. They therefore descend through the quotient to a continuous bijection $S\to S^3$. Since $S$ is compact and $S^3$ is Hausdorff, this bijection is a homeomorphism.

Here $n=2$, $r_2=1$, and $d_K=-47$, so the Minkowski bound for ideal classes isEvery ideal class therefore has a representative of norm at most four. Such classes are generated by the prime ideals above $2$ and $3$. Putin additive notation. Since conjugate prime ideals represent inverse classes, the two principal-ideal factorizations from part (b) giveThus $y=-2x$ and $5x=0$, so the class group is generated by $x$ and has order dividing five.

It is not trivial. If ${\mathfrak{p}}_2$ were principal, there would be an element of norm two, butcannot equal two for integers $a,b$. Therefore $x\ne 0$. Since five is prime,

The element $\omega$ satisfiesModulo both $2$ and $3$, this polynomial factors as $X(X-1)$. Henceandwhere the respective ideal norms are $2,2,3,3$.

Now $N(\omega )=12$, and $\omega$ lies in ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ but not in their conjugate primes. Similarly $N(2+\omega )=18$, while $2+\omega$ lies in ${\mathfrak{p}}_2$ and ${\mathfrak{q}}_3$. Comparing norms gives

If $L$ has degree $n=r_1+2r_2$ and discriminant $d_L$, then every class in the ideal class group contains a nonzero integral ideal $\mathfrak{a}$ satisfying the Minkowski bound for ideal classes

Let ${\chi }_{\mathrm{reg}}$ be the character of the regular representation. It has ${\chi }_{\mathrm{reg}}(1)=|G|$ and vanishes away from the identity. The multiplicity of an irreducible character ${\chi }_j$ in it isThus the regular representation decomposes as ${⨁}_j{n}_j{\rho }_j$. Comparing dimensions gives