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Choose an orthonormal basis $(e_1,e_2)$ of $T_{p}S$ and write $e(\theta )=\cos \theta e_1+\sin \theta e_2$. On a punctured disc on which the exponential map is a diffeomorphism, the mapdefines geodesic polar coordinates centred at $p$. The curves $r\mapsto X(r,\theta )$ are unit-speed geodesics, so the first coefficient of the first fundamental form isThe Gauss lemma says that radial and angular coordinate vectors are orthogonal, hence

The angular variation $J=X_{\theta }$ is a Jacobi field along each radial geodesic, with initial dataSince $G=⟨X_{\theta },X_{\theta }⟩=|J{|}^2$, smooth dependence of the geodesic flow givesConsequentlyas required.

Suppose $S$ lies in the closed ball of radius $R$ centred at $c$. By compactness, the continuous function $q\mapsto |q-c|$ attains its maximum $r\le R$ at some $p\in S$. The sphere of radius $r$ about $c$ is a supporting sphere tangent to $S$ at $p$.

Put $N=(p-c)/r$. For any unit vector $v\in T_{p}S$, choose a surface curve $\alpha$ with $\alpha (0)=p$ and ${\alpha }^{\prime }(0)=v$. Since $|\alpha (t)-c{|}^2$ has a local maximum at zero,The scalar $⟨{\alpha }^{\prime \prime }(0),N⟩$ is the normal curvature in direction $v$, so every normal curvature is at most $-1/r$ with this choice of unit normal. Applying this to principal directions shows that both principal curvatures satisfy $k_i\le -1/r$. ThereforeThis proves both assertions, including the existence of an elliptic point on every compact regular surface.

Choose a point $p\in C$ and project from $p$ to a line not containing $p$. Every line through $p$ meets the quartic in four points counted with intersection multiplicity by Bézout theorem; removing the fixed intersection at $p$ leaves a generic fibre of three points. This projection from a point on a plane curve therefore defines a nonconstant morphismof degree three, so the gonality is at most three.

It cannot be one, because a degree-one morphism between smooth projective curves is an isomorphism of algebraic varieties, whereas $C$ has genus three and the projective line has genus zero. It cannot be two either: that would make $C$ a hyperelliptic curve. The canonical map of a hyperelliptic curve is not a closed embedding, but for a smooth plane quartic the canonical map of a smooth plane curve is exactly its given embedding $C\hookrightarrow {\mathbb{P}}^2$, since $K_C\cong {\mathcal{O}}_C(1)$. This contradiction proves

The intersection with a line $\ell$ is a hyperplane section:The canonical bundle of a smooth projective hypersurface formula, obtained from the adjunction formula, gives for a smooth degree-four curve in ${\mathbb{P}}^2$Thus ${\mathcal{O}}_C(D)\cong K_C$, so the divisor class of $D$ is the canonical divisor class. Therefore $D$ is a canonical divisor. Its degree is also $4=2g-2$, consistently with $g=3$.

PutBy the Jacobian criterion, a projective plane curve is singular only at a common projective zero of $F$ and all its partial derivatives. HereTheir simultaneous vanishing forces $X_0=X_1=X_2=0$, which is not a point of projective space. Hence the Fermat quartic $C$ is nonsingular. In particular it is a smooth plane quartic and has geometric genus $(4-1)(4-2)/2=3$.

Let the smooth projective curve be the projective completion of $f(x,y)=0$, of degree $d$, and suppose after interchanging $x$ and $y$ if necessary that $f_y=\partial f/\partial y$ is nonzero. A basis of its holomorphic differential forms isEquivalently one may use $-x^i{y}^{j}dy/f_x$. These expressions agree on their common domain because differentiating $f(x,y)=0$ gives $f_{x}dx+f_{y}dy=0$. The allowed monomials are precisely the homogeneous polynomials of degree at most $d-3$, in accordance with the adjunction formula $K_C\cong {\mathcal{O}}_C(d-3)$.

On $C_c^0({\mathbb{R}}^n)$ equipped with the $L^{\infty }$ norm, point evaluation is linear and satisfieswith equality for a compactly supported function equal to one near zero. Thus $\Vert \Lambda \Vert =1$. The Hahn-Banach theorem extends it, without increasing its norm, to a continuous functional $\tilde{\Lambda }$ on $L^{\infty }({\mathbb{R}}^n)$.

This extension cannot be represented by an $L^1$ function. If $h\in L^1$ satisfied $\tilde{\Lambda }(f)=\int fh$, choose continuous functions $0\le f_k\le 1$ with $f_k(0)=1$ and support in the ball of radius $1/k$. Thenby the absolute continuity of the Lebesgue integral, a contradiction. Therefore $(L^{\infty }{)}^{\prime }$ strictly contains the functionals represented by $L^1$; this is a singular functional on L infinity.

Finally, $L^p({\mathbb{R}}^n)$ is a reflexive Banach space exactly whenThe endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive.

The continuous dual space $X^{\prime }$ consists of all bounded linear maps $\Lambda :X\to \mathbb{C}$, withIf $Y=L^p({\mathbb{R}}^n)$ with $1\le p<\infty$ and $q$ is the conjugate exponent, then duality of Lp spaces identifies $Y^{\prime }$ isometrically with $L^q({\mathbb{R}}^n)$, including $q=\infty$ when $p=1$. Every functional has the formand $\Vert {\Lambda }_h\Vert =\Vert h{\Vert }_q$.

Write $A=T-\lambda I$. We first prove that $\mathrm{im}A$ is dense. Otherwise some nonzero $y$ lies in its orthogonal complement, soBy part (b), choose finite-dimensional spaces $E_j$ containing $y$ such that $S_j=P_{E_j}T\to T$ in operator norm. Sincethe operator $S_j^{\ast }-\overline{\lambda }I$ is not injective. Therefore $S_j-\lambda I$ is not surjective; the contrapositive of part (c) makes $\lambda$ an eigenvalue of $S_j$. Choose unit vectors $x_j$ with $S_j{x}_j=\lambda x_j$. ThenCompactness gives a subsequence for which $T{x}_j$ converges. Since $\lambda \ne 0$, the displayed relation makes $x_j$ converge to a unit vector $x$, and continuity gives $Tx=\lambda x$, contradicting the hypothesis. Thus $\mathrm{im}A$ is dense.

Next suppose $A$ were not bounded below, or equivalently that its lower norm were zero. There would be unit vectors $x_j$ with $A{x}_j\to 0$, henceAgain a convergent subsequence of $T{x}_j$ forces the corresponding $x_j$ to converge to a unit vector $x$ satisfying $Ax=0$, another contradiction. Therefore some $c>0$ satisfies

This lower bound makes $\mathrm{im}A$ closed: if $A{x}_j$ converges, then $x_j$ is Cauchy and its limit maps to the same limit. Since the image is both dense and closed, it is all of $H$. We have proved the Fredholm alternative for a compact operator: every nonzero spectral value of a compact operator is an eigenvalue.

Put $E=\mathrm{im}S$, a finite-dimensional subspace, and decompose $H=E\oplus E^{\perp }$. The restrictionis injective, because a nonzero vector in its kernel would be a $\lambda$-eigenvector of $S$. By finite-dimensionality it is therefore surjective.

Given $y=y_E+y_{\perp }$, first put $x_{\perp }=-y_{\perp }/\lambda$. Then $S{x}_{\perp }\in E$, so choose $x_E\in E$ such thatFor $x=x_E+x_{\perp }$ we obtain $(S-\lambda I)x=y$. Hence $S-\lambda I$ is surjective. This is the finite-rank Fredholm alternative.

Compactness gives a stronger statement than the one requested. Since $T(B_H)$ has compact closure, finitely many open balls of radius $\delta$ cover it. Choose their centres from $T(B_H)$ and let $E$ be their finite-dimensional linear span. ThenFor the orthogonal projection $P_E$, best approximation in a Hilbert space givesIn particular this holds for every vector $e_n$ in the given orthonormal basis.

The operator $P_{E}T$ has image in $E$ and hence has finite rank, whileThus every compact operator is an operator norm limit of finite-rank operators. This is the finite-rank approximation theorem for compact operators on a Hilbert space.

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

Define a simplicial map $j:K\to L$ on vertices byThe image of every simplex is a simplex of $L$, and $j\circ i={\mathrm{id}}_L$. The maps $i\circ j$ and ${\mathrm{id}}_K$ are contiguous simplicial maps, since the union of their images on every simplex is contained in a simplex of $K$. Their realizations are therefore homotopic, so $|j|$ is a homotopy inverse of $|i|$.

Here is an explicit chain homotopy. Put $f=i_{\bullet }{j}_{\bullet }$ and use the orientations from part (b). DefineWe verifyIt is immediate on every vertex except $v_3$, where $\partial h_0(v_3)=v_3-v_1=v_3-f(v_3)$. It is immediate on the fixed edges. For the remaining edges,because $f(e_{23})=[v_2{v}_1]=-e_{12}$. Finally,since $f(t)$ is degenerate and hence zero. Thus $h$ is the required chain homotopy.

Use the orientationsThe simplicial chain complex haswithThe map ${\partial }_2$ is injective, so $H_2(K)=0$. The complex is connected, so $H_0(K)\cong \mathbb{Z}$ and $\mathrm{rank}{\partial }_1=3$. Hence $\ker {\partial }_1$ has rank $5-3=2$; quotienting by the rank-one primitive subgroup $\mathrm{im}{\partial }_2$ givesThere are no higher simplices. ThereforeThe result agrees with the Euler characteristic $4-5+1=0$.

The geometric realization of a simplicial complex consists of the filled triangle $⟨v_1,v_2,v_3⟩$ together with the two extra edges $⟨v_1,v_4⟩$ and $⟨v_2,v_4⟩$. Thus the triangle on $v_1,v_2,v_4$ has only its boundary:

The eleventh cyclotomic field $L=\mathbb{Q}({\xi }_{11})$ hasThe cyclic group $C_{10}$ has exactly one subgroup of each order $1,2,5,10$. The Galois correspondence therefore gives exactly four subfields:The Quadratic Gaussian period in the eleventh cyclotomic field satisfies $\eta =(-1+\sqrt{-11})/2$, while $\theta$ generates the maximal real subfield of the eleventh cyclotomic field. Every subgroup of the abelian Galois group is normal, so all four subfields are Galois over $\mathbb{Q}$.

Each nontrivial proper subfield has prime degree, two or five. If $a\in M\setminus \mathbb{Q}$, then $\mathbb{Q}⊊\mathbb{Q}(a)\subseteq M$, and the tower law forces $\mathbb{Q}(a)=M$. Thus every such element is a primitive element. The displayed $\eta$ and $\theta$ are the requested examples. Finally, ${\xi }_{11}$ itself is a primitive element for $L/\mathbb{Q}$.

Because $L/K$ is a finite Galois extension,

Suppose the irreducible quadratic $f$ became reducible over $L$. It would then have a root $\alpha \in L$, and $K(\alpha )/K$ would be a quadratic intermediate extension. By the Fundamental theorem of Galois theory, $\mathrm{Gal}(L/K(\alpha ))$ would be a subgroup of index two in $A_{10}$ and hence a normal subgroup by the index-two subgroup is normal result. But the Simplicity of alternating groups says that $A_{10}$ is simple, while this subgroup would be proper and nontrivial. This contradiction proves that $f$ remains irreducible in $L[x]$.

If the edge $uv$ lies in no triangle, no vertex can be adjacent to both endpoints; hence the neighbourhoods satisfy $\Gamma (u)\cap \Gamma (v)=\varnothing$.

Suppose $|\Gamma (u)|+|\Gamma (v)|=2n$. The two disjoint neighbourhoods then partition $V(G)$. If both were independent, every edge would run between these two parts, so $G$ would be bipartite andcontrary to $e(G)=n^2+1$. Thus at least one neighbourhood contains an edge, and that edge forms a triangle with $u$ or $v$.

We now prove by induction on $n\ge 2$ the Triangle lower bound one edge above the Mantel threshold: every graph on $2n$ vertices with at least $n^2+1$ edges has at least $n$ triangles. Part (ii) is the base case. For $n\ge 3$, if every edge lies in a triangle, part (iii) givesOtherwise choose an edge $uv$ lying in no triangle and put $s=d(u)+d(v)\le 2n$. Deleting $u,v$ removes $s-1$ edges, so the graph $G^{\prime }=G-u-v$ hasIf $s=2n$, induction gives at least $n-1$ triangles in $G^{\prime }$, while the nonindependence proved above supplies a further triangle through $u$ or $v$.

If $s\le 2n-1$, then $e(G^{\prime })\ge (n-1{)}^2+2$. Induction first supplies a triangle $\Delta$ in $G^{\prime }$. Delete one edge of $\Delta$; the remaining graph still has at least $(n-1{)}^2+1$ edges, so induction supplies $n-1$ triangles avoiding that edge. Together with $\Delta$, these give at least $n$ distinct triangles in $G$.

Let $t(G)$ be the number of triangles in a graph. Count incidences $(e,\Delta )$ in which the edge $e$ lies in the triangle $\Delta$. Every edge contributes at least one incidence by hypothesis, while every triangle contributes exactly three. ThereforeThis is the edge-triangle incidence bound.