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Use completing the square rather than computing the eigenvalues of the symmetric matrix. Successively removing the mixed terms givesThus one suitable invertible linear change of coordinates isThe coordinate matrix is upper triangular with determinant one, so the change is an invertible linear map. Its inverse is $z=Z$, $y=Y+4Z/7$, $x=X-2Y+5Z/14$. Hence the diagonal coefficients can be taken as $2,-7,-3/14$. This is a matrix congruence of a quadratic form; an orthogonal transformation was not required.

Over the complex numbers, every finite-dimensional representation of a solvable Lie algebra has a basis in which every representing matrix is upper triangular. The Lie theorem is often stated first as the existence of a common eigenvector in every nonzero finite-dimensional Lie algebra representation of a solvable Lie algebra. Applying that assertion successively to quotient representations gives an invariant complete flag, and hence the upper triangular form. The same proof works over any algebraically closed field of characteristic zero.

We prove the common eigenvector assertion by induction on $\dim \mathfrak{g}$, writing the action as $av$. The zero Lie algebra is immediate. If $\mathfrak{g}\ne 0$ is solvable, its derived series of a Lie algebra shows that $[\mathfrak{g},\mathfrak{g}]\ne \mathfrak{g}$. Choose a codimension-one ideal of a Lie algebra $\mathfrak{a}$ containing $[\mathfrak{g},\mathfrak{g}]$, and choose $x\notin \mathfrak{a}$. By induction there are $v\ne 0$ and a linear functional $\lambda :\mathfrak{a}\to \mathbb{C}$ such that $av=\lambda (a)v$ for all $a\in \mathfrak{a}$.

Let $W$ be the span of $v,xv,x^{2}v,\dots$. If $m=\dim W$, the first $m$ of these vectors form a basis, and $W$ is $x$-invariant. We claim that for each $a\in \mathfrak{a}$,For $j=0$ this is the definition of $v$. For the induction step, use $ax=xa+[a,x]$ and $[a,x]\in \mathfrak{a}$. Applying the induction hypothesis to both $a$ and $[a,x]$ proves the claim. Consequently $W$ is $\mathfrak{a}$-invariant, and every $a\in \mathfrak{a}$ acts on $W$ by an upper triangular matrix with all diagonal entries $\lambda (a)$.

Since both $x$ and $a$ preserve $W$, the matrix trace of their commutator on $W$ is zero. The claim applied to $[x,a]\in \mathfrak{a}$ givesHere characteristic zero is essential: $m\ne 0$ in the field, so $\lambda ([x,a])=0$. Now the common weight spaceis nonzero and $x$-invariant. Indeed, for $w\in E_{\lambda }$,An endomorphism of a nonzero finite-dimensional complex vector space has an eigenvector, so choose an eigenvector of $x$ in $E_{\lambda }$. It is a common eigenvector for $\mathfrak{g}=\mathfrak{a}+\mathbb{C}x$. This proves the Lie theorem.

For the printed matrices in characteristic $p$, $[x,y]=x$, and there is no common eigenvector. Index the standard basis by $e_0,\dots ,e_{p-1}$. The cyclic entry in the PDF givesThe $p$ diagonal eigenvalues of $y$ are distinct in $k$. Thus every eigenvector of $y$ is a scalar multiple of a single $e_j$. Since $p\ge 2$, $x{e}_j$ is never a scalar multiple of $e_j$, proving the assertion even when $k$ is not an algebraically closed field.

For $1\le j\le p-1$,At the cyclic boundary,Hence $[x,y]=x$. The two-dimensional Lie subalgebra $kx+ky$ has derived algebra $kx$, whose own derived algebra is zero, so it is solvable. It nevertheless has no common eigenvector, including after extending $k$ to its algebraic closure. This is a failure of Lie theorem in positive characteristic. In the proof above, the obstruction is precisely that $\dim W=p$ can vanish as a scalar in $k$.

The derived algebra of a complex solvable Lie algebra is nilpotent. First suppose $\mathfrak{g}\subseteq \mathfrak{gl}(V)$. By the Lie theorem, put every element of $\mathfrak{g}$ in upper triangular form. The diagonal of a commutator of upper triangular matrices is zero, so $\mathfrak{d}=[\mathfrak{g},\mathfrak{g}]$ consists of strictly upper triangular matrices. The Lie algebra of all such matrices is a nilpotent Lie algebra: if $F_r$ consists of matrices whose entries vanish whenever $j-i<r$, thenTherefore the lower central series of a Lie algebra of $\mathfrak{d}$ reaches zero. Alternatively, every element of $\mathfrak{d}$ is a nilpotent linear map, and the Engel theorem states that a finite-dimensional Lie subalgebra consisting of nilpotent linear maps is a nilpotent Lie algebra.

For an abstract complex solvable Lie algebra $\mathfrak{g}$, apply the preceding result to its Adjoint representation. The Lie algebra $\mathrm{ad}\mathfrak{d}=[\mathrm{ad}\mathfrak{g},\mathrm{ad}\mathfrak{g}]$ is nilpotent. Since $\ker (\mathrm{ad}{|}_{\mathfrak{d}})=\mathfrak{d}\cap Z(\mathfrak{g})$ is central in $\mathfrak{d}$, some term of the lower central series of a Lie algebra of $\mathfrak{d}$ lies in that central ideal; the next term is zero. Thus $\mathfrak{d}$ itself is nilpotent.

Conversely, every nilpotent Lie algebra is solvable, since its derived series of a Lie algebra is contained term by term in its lower central series of a Lie algebra. If $\mathfrak{d}$ is nilpotent, it is therefore solvable, and $\mathfrak{g}/\mathfrak{d}$ is Abelian. More directly, the derived series of a Lie algebra of $\mathfrak{g}$, after its first term, is the derived series of a Lie algebra of $\mathfrak{d}$. Consequently the derived algebra nilpotence criterion is

The Hopf boundary point lemma requires an interior sphere condition, strictness of the boundary maximum, and quantitative control of the uniformly elliptic operator near the contact point. Assume that $y\in \partial \Omega$, that $B_R(z)\subset \Omega$ and $\overline{B_R(z)}\setminus \{y\}\subset \Omega$, and that $|y-z|=R$. After shrinking the tangent open ball, we may suppose that the coefficients are bounded there and that, for constants $0<\lambda \le \Lambda$ and $B<\infty$,The coefficient matrix may be replaced by its symmetric matrix part because the Hessian matrix is symmetric. Assume also that $u(x)<u(y)$ throughout the tangent open ball. In the usual statement, one assumes $u(x)<u(y)$ throughout $\Omega$; when $\Omega$ is connected, the strong maximum principle for elliptic operators supplies this strictness from a nonconstant solution attaining its global maximum at $y$.

Put $\mathbf{n}=(y-z)/R$, the outward unit normal of the tangent open ball. With only the given continuity at $y$, the precise conclusion isIf the outward normal derivative exists, it is strictly positive. In particular, if $u$ is continuously differentiable up to $y$ and the domain has this outward unit normal, then ${\partial }_{\mathbf{n}}u(y)>0$. The stated $C^0$ boundary regularity alone does not assert existence of a normal derivative.

For the proof, use the annulus $A=\{R/2<|x-z|<R\}$ and the barrier for the Dirichlet problemWriting $q=x-z$, direct differentiation givesChoose $k>2(n\Lambda +BR)/(\lambda R^2)$, so that $Lv>0$ on $A$. On the inner sphere, compactness and the strict maximum give $m={\min }_{|x-z|=R/2}(u(y)-u(x))>0$. Choose $\epsilon >0$ with $\epsilon {\max }_{|x-z|=R/2}v\le m$. On the outer sphere $v=0$ and $u-u(y)\le 0$, including at $y$ by continuity. Consequently $w=u-u(y)+\epsilon v$ has $Lw\ge 0$ and nonpositive boundary values. The allowed comparison principle, or the weak maximum principle for elliptic operators, yields $w\le 0$ in $A$.

Along the inward radius this givesThis proves the Hopf boundary point lemma. There is no sign restriction on $u(y)$ here, because the operator has no zeroth-order term.

Assume $u\in W^{1,2}(B_2(0))$ is nonnegative almost everywhere, and that the real coefficient matrix $A=(a^{ij})$ is measurable and satisfies, for almost every $x$ and every $\xi \in {\mathbb{R}}^n$,The operator norm bound controls all coefficients, including any antisymmetric part. For a symmetric matrix, these assumptions are exactly the usual lower and upper quadratic-form bounds. The weak solution condition isThen the Harnack inequality for uniformly elliptic divergence-form equations isA nonnegative weak solution has its supremum controlled by its infimum on the smaller ball. The De Giorgi-Nash-Moser theorem supplies a Hölder continuous function representative, for which ordinary supremum and infimum can replace the essential supremum and essential infimum. Nonnegativity and the larger ball are essential hypotheses; there is no differentiability requirement on the coefficients.

Use a radial geometric Kähler potential. Write $s=|z_1{|}^2+|z_2{|}^2$ and $y(s)={\phi }^{\prime }(s)$. The coefficient Hermitian matrix of $i\partial \bar{\partial }\phi (s)$ isIts eigenvalues are $y$ in the complex tangential direction and $y+s{y}^{\prime }$ in the complex radial direction. A Kähler metric is positive precisely when both are positive. The standard form has coefficient matrix $\frac{1}{2}I$, so equality of the Riemannian volume forms is the conditionIndeed, the volume form is ${\omega }^2/2$ and its coefficient is proportional to $\det G$. Multiplying the determinant equation by $2s$ gives $(s^2{y}^2{)}^{\prime }=s/2$, so we can takeAn explicit antiderivative isIt is smooth for $s>0$. Its two eigenvalues areThe form is real and closed because it is $i\partial \bar{\partial }$ of a real function, so it is a Kähler metric on the punctured space. Since $a>0$, neither eigenvalue equals $1/2$, and the metric differs from the Euclidean one. The radial and tangential stretches compensate exactly, preserving the volume form:This is a radial Kähler metric with Euclidean volume in complex dimension two.

An open migration process is a continuous-time Markov chain recording the populations $n=(n_1,\dots ,n_J)$ of finitely many colonies. Independent external Poisson processes bring individuals to colony $j$ at rate ${\nu }_j$. When that colony contains $k>0$ individuals, a departure opportunity occurs at total rate ${\varphi }_j(k)>0$, with ${\varphi }_j(0)=0$. The departing individual moves to colony $l$ with probability $p_{jl}$, or leaves the system with probability $p_{j0}=1-{\sum }_l{p}_{jl}$. Routing choices and the clocks are independent. A return to the same colony does not change the state.

The word open means that the routing matrix $P=(p_{jl})$ is transient: every individual eventually leaves, or equivalently the spectral radius of $P$ is less than one. The traffic equations of an open migration process, with row-vector convention, give the total arrival rates:These rates include migrations as well as external arrivals. On the active colonies assume ${\alpha }_j>0$ and the usual irreducible Markov chain condition. PutThe product form and its normalization condition areTo establish the product-form stationary distribution of an open migration process, first use its unnormalized weight $b(n)={\prod }_j{b}_j(n_j)$. Its ratios satisfy $b_j(k){\varphi }_j(k)={\alpha }_j{b}_j(k-1)$. The incoming external arrivals into $n$, divided by $b(n)$, contribute ${\sum }_j{\nu }_j{\varphi }_j(n_j)/{\alpha }_j$. Incoming exits contribute ${\sum }_i{\alpha }_i{p}_{i0}$. Incoming migrations from $i$ to $j\ne i$ contribute ${\alpha }_i{p}_{ij}{\varphi }_j(n_j)/{\alpha }_j$. Terms involving an empty destination are zero. Thus the full incoming rate divided by $b(n)$ isHere the traffic equations of an open migration process give the second equality, including ${\sum }_i{\alpha }_i{p}_{i0}={\sum }_j{\nu }_j$. The right side is precisely the total outgoing rate. This proves global balance for a continuous-time Markov chain; pairwise detailed balance is not required for general routing.

For finite rates at each finite population, the continuous-time Markov chain is nonexplosive Markov chain: its total population is bounded over a finite time interval by the initial population plus the finitely many external arrivals, so it visits a finite set on which all jump rates are bounded. Consequently finite $Z_j$ normalize the invariant weight to a stationary distribution. Under the stated irreducible Markov chain assumptions, this is the unique stationary distribution, and it is a positive recurrent Markov chain. Conversely, the unique invariant measure of an irreducible recurrent chain is proportional to $b$, so a positive recurrent Markov chain requires its normalization to be finite. A sufficient condition is ${\mathrm{liminf}}_{k\to \infty }{\varphi }_j(k)>{\alpha }_j$; the series criterion itself is the exact condition. Colonies with ${\alpha }_j=0$ have zero stationary population and can be removed from the active class.

For the single-server M-M-1 queue specialization, ${\varphi }_j(k)={\varphi }_j$ for $k>0$, each stationary colony has the geometric distribution of an M-M-1 queue:The traffic equations of an open migration process do not involve the service rates, so the ${\alpha }_j$ remain fixed during this service-capacity allocation in an open migration process. Write $A={\sum }_j{\alpha }_j$, $B=F-A$ and $b_j={\varphi }_j-{\alpha }_j$. A stationary allocation requires $B>0$, with $b_j>0$ and ${\sum }_j{b}_j=B$. The Cauchy-Schwarz inequality givesEquality holds exactly when $b_j$ is proportional to $\sqrt{{\alpha }_j}$. Hence, for positive arrival rates, the unique optimal service allocation and the minimum expected value of the total population areIf $F\le A$ and some arrival rate is positive, no stable service allocation exists. Zero-arrival colonies need no service in their stationary empty class: allowing zero service there gives the same formula restricted to active colonies. If strictly positive service is demanded even at inactive colonies, the displayed minimum is an infimum approached as that unused allocation tends to zero. If all arrivals vanish, the stationary mean population is zero for every allocation that drains the initial population.

The coefficient matrix in the computational basis isFor a bipartite pure state, the partial traces give ${\psi }_A=M{M}^{†}$ and ${\psi }_B=M^T\overline{M}$. Therefore the requested reduced density matrices areThe columns of $M$ are orthogonal. Their norms are $\sqrt{2/3}$ and $\sqrt{1/3}$, so normalizing them gives the orthonormal setWith $|v_0⟩=|0⟩$ and $|v_1⟩=|1⟩$, a Schmidt decomposition isExpanding this expression reproduces the minus sign in the $|01⟩$ coefficient. The two reduced density matrices both have eigenvalues $2/3$ and $1/3$, consistent with these Schmidt coefficients.

For any normalized pure state on finite-dimensional Hilbert spaces ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$, the Schmidt decomposition givesThe $u_j$ and $v_j$ form orthonormal sets in their respective spaces. The positive numbers $\sqrt{p_j}$ are the Schmidt coefficients, and $r$ is the Schmidt rank. Taking a partial trace gives the reduced density matricesThus the nonzero eigenvalues of the two reduced density matrices agree.

One construction is a singular value decomposition of the coefficient matrix $M$ in $|\psi ⟩={\sum }_{a,b}{M}_{ab}|a⟩|b⟩$. If $M=U\Sigma V^{†}$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the singular values give $\sqrt{p_j}$. This also explains why the Schmidt coefficients are real and nonnegative.

The matrix $N$ is a skew-symmetric matrix, so $x^{T}Nx=0$ for real $x$. Therefore the energy-neutral rotational nonlinearity contributes nothing directly to the energy balance:The nonlinear term rotates the state without changing its Euclidean norm. It can still change the orientation, and thereby affect later energy growth through the linear term. Thus energy cancellation does not mean the nonlinear and linear trajectories have identical energy histories.

Use the Einstein summation convention. Substitute the given basis transformation into the expansion of the vector:By linear independence of the basis vectors, $v_j=R_{ij}{v}_i^{\prime }$. In matrix notation $v=R^T{v}^{\prime }$. A rotation matrix is orthogonal, so multiplication by $R$ givesThe placement of the transpose follows from the specified convention $e_i^{\prime }=R_{ij}{e}_j$. It must not be guessed from a different convention for an active rotation of a vector.

The quotient theorem for Cartesian tensors here says: if contracting $T_{ij}$ with every vector $v_j$ gives the components $w_i=T_{ij}{v}_j$ of a vector in every right-handed orthonormal basis, then $T$ obeys the second-order tensor transformation law. Conversely that transformation law guarantees the contraction is a vector. One may equivalently test that $u_i{T}_{ij}{v}_j$ is a scalar for every pair of vectors $u,v$.

By part (a), $v^{\prime }=Rv$ and $w^{\prime }=Rw$. The contraction property in the new basis saysSince this holds for every test vector, $T^{\prime }R=RT$. Using $R^{-1}=R^T$ givesFor the converse, $T^{\prime }{v}^{\prime }=RT{R}^{T}Rv=RTv=Rw$, exactly the vector law. For the equivalent scalar formulation, invariance of $u^{T}Tv$ for every $u,v$ gives $u^T{R}^T{T}^{\prime }Rv=u^{T}Tv$, forcing the same matrix identity.

The quantifier “every” matters: an arbitrary nonzero matrix can annihilate one fixed vector, so a single successful contraction cannot establish tensoriality. The stated bases test proper rotations; transformation under orthogonal reflections would require an additional hypothesis if that were wanted. “Second-order” counts tensor indices, rather than the matrix rank of a particular component matrix.

Centre the cube at the origin with faces having normals $\pm e_1,\pm e_2,\pm e_3$. Every rotational symmetry fixes the centre and permutes these normals, so its rotation matrix is a signed permutation matrix of determinant one. Conversely any such matrix permutes the cube’s coordinates up to signs and preserves the cube. For each of $3!$ coordinate permutations, four of the eight choices of signs give determinant one. Consequently the rotational symmetry group of a cube hasThis counts the rotations directly, without an unverified stabilizer calculation. The TeX’s $2^k$ is damaged transcription: the PDF prints 24.

Let $X$ be the $3^6$ face colourings. A colouring fixed by a rotation must be constant on each permutation cycle of faces, giving $3^c$ fixed colourings if the rotation has $c$ face cycles. The rotation types are:

There are three face axes, each with two quarter-turns and one half-turn; six edge axes, each with one half-turn; and four vertex axes, each with two nonidentity third-turns. These rotations are distinct and, with the identity, total 24, so the table is exhaustive. Face-axis rotations fix their two axial faces; the remaining four faces cycle or pair. Edge half-turns pair all faces, while vertex third-turns form two triples.

Burnside lemma now givesThe three colours are labelled, orthogonal reflections are excluded, and no condition requires every colour to appear.

First prove the hinted Caccioppoli inequality. Testing the weak formulation with $u_k{\phi }^2$ is legitimate by approximation in the zero-boundary Sobolev space. Let ${\Lambda }_{\ast }=\mathrm{esssup}\Vert a(x){\Vert }_{\mathrm{op}}\le n\Lambda$. Thenby Young inequality. HenceNo derivative of the measurable coefficient matrix is taken.

Set $w_k=u_k/{\sup }_{B_1}|u_k|$. The denominator is finite and positive because $u_k$ is continuous on the compact closure and nonzero. Each $w_k$ solves the same linear equation and $|w_k|\le 1$. For every $\theta <1$, choose a cutoff function equal to one on $B_{\theta }$ and supported in a slightly larger interior Euclidean ball. The energy bound gives a uniform $W^{1,2}(B_{\theta })$ bound.

Take radii ${\theta }_j\uparrow 1$. Weak compactness and a diagonal subsequence argument produce one subsequence converging weakly in $W^{1,2}(B_{{\theta }_j})$ for every $j$, with consistent restrictions defining $v\in W_{\mathrm{loc}}^{1,2}(B_1)$. Part (b) and the Arzela-Ascoli theorem allow a further subsequence converging uniformly on $\overline{B_{1/4}}$. Its uniform limit agrees almost everywhere with the weak limit and supplies its continuous representative there.

For every smooth compactly supported test function, choose $j$ containing its support. Since $a^{ij}{D}_i\phi \in L^2$ and $D{w}_k$ converges weakly there,Thus the compactness of normalized weak elliptic solutions givesThe diagonal step is needed to retain the equation beyond the small Euclidean ball of uniform convergence. The limit need not be nonzero: for the Laplacian in dimension at least two, the normalized harmonic functions $\mathrm{Re}(x_1+i{x}_2{)}^k$ tend uniformly to zero on each smaller Euclidean ball.

Use integer coefficients. The mapping-torus homology from Mayer–Vietoris works for every continuous map $f:X\to X$, without assuming that the mapping torus is a fiber bundle. Project to the time circle and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a mapping cylinder neighborhood with a deformation retraction to its target copy of $X$; the inverse image $V$ of the other arc is a product with a deformation retraction to $X$. Their intersection has two components, each with a deformation retraction to $X$. Label the upper component first. With the convention that the second inclusion in the Mayer–Vietoris sequence has a minus sign, its homology map isIndeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_{\ast }-\mathrm{id})v,u)$. Negating $v$ gives a direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the latter summands in the long exact sequence yieldsIn particular, there is a short exact sequenceNo splitting of this short exact sequence is assumed at this stage.

For the torus, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the basis of $H_1$. Their lifts to ${\mathbb{R}}^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, soThis also follows from the identification of the first homology group with the abelianization of the fundamental group. The Künneth theorem gives $H_2(X)=\mathbb{Z}$, with generator the product orientation class. If $\alpha ,\beta$ are the dual degree-one cohomology classes, then $\alpha ⌣\beta$ evaluates to one on this generator, andThus the degree-two homology map is multiplication by $\det A$, as in homology of an integer torus endomorphism. The claim that the torus map is a homeomorphism requires $A$ to be a unimodular matrix, equivalently an integer inverse or $\det A=\pm 1$; nonsingularity over $\mathbb{R}$ alone would only give a covering map of finite degree. The particular matrix here is unimodular.

For this matrix, $\det A=1$ andIts kernel is zero and its Smith normal form is $\mathrm{diag}(1,2)$, so its cokernel is $\mathbb{Z}/2$. The maps $\mathrm{id}-f_{\ast }$ on $H_0$ and $H_2$ are zero. Substituting into the short exact sequences, and splitting the degree-one extension because its quotient $\mathbb{Z}$ is free, gives

A vector bundle morphism over $M$ is a smooth map $F:E\to E^{\prime }$ with ${\pi }^{\prime }\circ F=\pi$ whose restriction $E_p\to E_p^{\prime }$ is linear for every $p$. A vector bundle isomorphism is such a map with a smooth bundle-morphism inverse. Equivalently, a fiberwise bijective smooth bundle morphism is an isomorphism: in local vector bundle trivializations it is multiplication by an invertible smooth matrix, and its inverse matrix is smooth.

For a diffeomorphism $f:N\to M$, defineThe derivative $d{f}_p$ is a linear isomorphism by the chain rule, since $d(f^{-1}{)}_{f(p)}\circ d{f}_p=\mathrm{id}$. In manifold charts, $F$ is represented by the smooth Jacobian matrix of $f$, so it is a smooth bundle morphism. Its inverse iswhich is also smooth and fiberwise linear. ConsequentlyThis identifies the bundles over the same base $N$; the tangent map by itself is a map from $TN$ to $TM$ covering $f$.

Under the standard smooth manifold hypotheses, the first answer is yes, although the isomorphism requires a choice. A partition of unity combines Euclidean inner products in manifold charts to produce a Riemannian metric $g$. Nonnegative local weights summing to one preserve positivity, and local finiteness preserves smoothness. The musical isomorphism isIt is smooth and fiberwise invertible, with inverse having local matrix $(g_{ij}{)}^{-1}$. It is therefore a vector bundle isomorphism. There is no preferred such map before a metric or equivalent additional structure is chosen.

The second answer is no. On $S^1=\mathbb{R}/2\pi \mathbb{Z}$, compare the trivial vector bundle $S^1\times \mathbb{R}$ with the Möbius line bundleBoth have rank one. A continuous section of a vector bundle of $L$ is represented on $[0,2\pi ]$ by a continuous function $s$ satisfying $s(2\pi )=-s(0)$. If it were nowhere zero, the intermediate value theorem would prevent its endpoint signs from being opposite. Thus $L$ has no nowhere-zero section. The trivial bundle has the section $1$, and a bundle isomorphism would carry that section to a nowhere-zero section of $L$, a contradiction. Hence