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For a locally compact Hausdorff Abelian group, each continuous unitary character defines an algebra character by $f\mapsto \int f(x)\chi (x)dx$. Conversely, if $\Phi$ is a nonzero algebra character and $\Phi (g)\ne 0$, the ratio $\Phi (T_{x}g)/\Phi (g)$ is independent of $g$, continuous and multiplicative. Translation isometries bound all its integer powers, so its modulus is one. The convolution formula $f\ast g=\int f(x)T_{x}gdx$ recovers $\Phi (f)=\int f\chi$. The Gelfand topology corresponds to the compact-open topology on the Pontryagin dual group; conjugating the characters gives the alternative Fourier sign convention.

Evaluation $(x,\chi )\mapsto \chi (x)$ is jointly continuous for the compact-open topology on the dual of a locally compact group. One proof uses $\overline{\chi (x)}=\widehat{T_{x}a}(\chi )/\hat{a}(\chi )$ near any character where the denominator is nonzero, together with norm-continuity of translations. A finite neighbourhood cover of a compact family $C$ then gives the displayed uniform convergence. This justifies compact-set-and-integrable-tail estimates even when the group is not metrizable and approximation is indexed by a net.

The function $\varphi (t)=t/(1+t)$ increases for $t\ge 0$, andThus $a\le b+c$ implies $\boxed{\varphi (a)\le \varphi (b)+\varphi (c)}$. Applying this to the triangle inequality for a metric proves the triangle inequality for the bounded metric transform $d_b=\varphi \circ d$. Symmetry and nonnegativity are immediate, and $d_b(x,y)=0$ exactly when $d(x,y)=0$, so $\boxed{d_b\text{ is a metric}}$.

Each $K_n$ is compact and lies inside $D$, so the supremum norm $\Vert f-g{\Vert }_n$ is finite for continuous $f,g$. Every summand of $\sigma$ lies between zero and $2^{-n}$, giving absolute convergence and $0\le \sigma \le 1/2$. Symmetry is inherited from these norms. If $\sigma (f,g)=0$, all the nonnegative summands vanish, hence $f=g$ on every $K_n$; their union is $D$, so $f=g$ everywhere. Finallyand the subadditivity already proved give the triangle inequality term by term and then after summation. Therefore $\boxed{\sigma \text{ is a metric on }\mathcal{F}}$. This is a weighted metric for local uniform convergence, representing the compact-open topology.

The geometric-series sum is $s(z)=1/(1-z)$ and its remainder is $s-s_k=z^{k+1}/(1-z)$. Writing $r_n=1-1/n$, the triangle inequality givesEquality occurs at the positive real point $z=r_n$. ThusFor every integer $N\ge 2$, split the defining sum into a finite head and its tail. Since $\varphi (t)\le t$ and $\varphi (t)\le 1$,Given $\epsilon >0$, choose $N$ with $2^{-N}<\epsilon /2$. For that fixed $N$, each of the finitely many displayed norms tends to zero, so their sum is below $\epsilon /2$ for sufficiently large $k$. Hence $\boxed{\sigma (s_k,s)\to 0}$, even though convergence is not uniform on the whole open disk.

A map $p:E\to B$ is a Serre fibration if every homotopy $H:D^k\times I\to B$, $k\ge 0$, with a lift of its initial slice has a lift on the whole cylinder extending that initial lift. By attaching cells, this gives the relative homotopy lifting property for CW pairs: a specified lift on the initial slice and on the subcomplex cylinder can be extended.

Give $PY$ the compact-open topology. Suppose the initial lifting is a family of paths ${\gamma }_x$ starting at $y_0$, with ${\gamma }_x(1)=H(x,0)$. DefineThe two expressions coincide on their common boundary, the initial path at $s=0$ is exactly ${\gamma }_x$, and the endpoint is $H(x,s)$. The adjoint map on $D^k\times I\times I$ is continuous by pasting; compactness of the path parameter gives continuity into the path space. Thus endpoint evaluation $PY\to Y$ is a Serre fibration. The variable subdivision avoids changing the initial lift by an unwanted reparametrization.

For a based pair $(E,F,e_0)$, the long exact sequence of relative homotopy groups isending in ${\pi }_1(E)\to {\pi }_1(E,F)\to {\pi }_0(F)\to {\pi }_0(E)$. The low-dimensional end is an exact sequence of pointed sets with its usual actions, rather than an entirely abelian sequence.

The boundary sends a relative disk map to its restriction on the boundary sphere. An absolute class has constant boundary, so $\partial j_{\ast }=0$. Conversely, if a relative map has null-homotopic boundary in $F$, choose a based null-homotopy there. Attach this null-homotopy as a collar to the disk, leaving the original map on a smaller inner disk. The resulting boundary is constant at $e_0$, so it defines an absolute class in ${\pi }_n(E)$; inserting the collar is a relative homotopy, so its image is the original relative class. This proves $\mathrm{im}{j}_{\ast }=\ker \partial$. At $n=1$, append a path in $F$ from the terminal point to $e_0$, giving the corresponding based loop and pointed-set exactness.

Now suppose $F=p^{-1}(b_0)$. Use a cube model in which the bottom and side faces are based and the top face maps into $F$. A based $n$-cube in $B$ lifts from its constant bottom face, with sides held at $e_0$ by relative lifting. Its top face lies in $F$, proving surjectivity of the relative homotopy of a Serre fibration comparison $p_{\ast }:{\pi }_n(E,F)\to {\pi }_n(B,b_0)$.

For injectivity, lift a based homotopy of two projected cubes from the first relative lift; this ends at some lift of the second base cube. Any two lifts of that same base cube are relatively homotopic: insert an extra parameter between them, prescribe the two liftings on its two ends and the constant lift on the bottom, and apply relative lifting in the remaining cube direction. The top face stays in $F$ because its projection is $b_0$. Combining these homotopies proves injectivity. ThereforeAt $n=1$ the same construction gives a bijection of pointed sets, and one can transport the base fundamental-group law across it. An arbitrary pair's relative ${\pi }_1$ has no general group law. The printed target index $a$ is a typographical error: the map preserves degree $n$.

The last-column map $U(n)\to S^{2n-1}$ is surjective because a unit vector can be completed to a unitary basis. Its stabilizer is the block subgroup $\mathrm{diag}(U(n-1),1)$, so it induces a continuous bijection $U(n)/U(n-1)\to S^{2n-1}$. Unitary matrices form a compact set in their matrix-space topology; the quotient is compact and the sphere Hausdorff. HenceThis quotient is a fiber bundle. Near the last standard unit vector, project the other standard vectors by $I-v{v}^{\ast }$ onto $v^{\perp }$ and apply Gram-Schmidt orthogonalization; their independence persists in a small neighborhood. The resulting orthonormal columns together with $v$ give a continuous local section. Translating these sections covers the sphere and trivializes the projection, so it is a Serre fibration.

Apply this to $U(n)\to U(n+1)\to S^{2n+1}$. Its homotopy sequence containsFor $i\le 2n-1$, both outer groups vanish by cellular approximation theorem. Thus ${\pi }_i(U(n))\cong {\pi }_i(U(n+1))$ in the stated range. The component assertion also holds: diagonalizing a unitary matrix and varying its eigenvalue phases continuously connects it to the identity. This establishes the unitary sphere quotient and stability range.

Give $\mathcal{O}(U)$ the compact-open topology. Choose a finite polygonal cycle $\Gamma$ in $U\setminus {\sigma }_A(x)$ whose winding number is one on the Banach algebra spectrum and zero outside $U$. One can take the oriented boundary of a finite union of small squares surrounding the compact Banach algebra spectrum and contained in $U$. DefineThe integral is an algebra-norm integral of a continuous function on finitely many segments. Cauchy's theorem, applied after any bounded linear functional on $A$, makes it independent of the cycle. Moreoverso it is continuous for the compact-open topology.

Here is a direct verification of the algebra properties and uniqueness using the allowed Runge theorem. The version required is that every holomorphic function on an open plane set is a locally uniform limit of rational functions with poles outside that set, allowing a polynomial part. A rational function $r=p/q$ of this kind has the unambiguous value $r(x)=p(x)q(x{)}^{-1}$, since the roots of $q$ avoid the Banach algebra spectrum. These rational values form a unital algebra homomorphism.

They agree with the contour integral. First $\frac{1}{2\pi i}{\int }_{\Gamma }(z1-x{)}^{-1}dz=1$, by replacing the cycle with a large circle and integrating its Neumann expansion. The identity $z^{j}1-x^j=(z1-x){\sum }_{k=0}^{j-1}{z}^{j-1-k}{x}^k$ then proves the assertion for polynomials. For $a\notin U$, the resolvent identity givesThe cycle has winding number zero at $a$, so integration gives $(x-a1{)}^{-1}$. This identity holds on a neighborhood of $a$ disjoint from the cycle and the Banach algebra spectrum. Differentiating with respect to $a$ gives the analogous result for every power $(z-a{)}^{-k}$, and partial fractions give it for all the rational functions under consideration.

If $r_n\to f$ and $s_n\to g$ locally uniformly, the contour estimate makes their evaluated sequences converge. Since $r_n{s}_n\to fg$ locally uniformly and multiplication in $A$ is continuous, their limits satisfy ${\Theta }_x(fg)={\Theta }_x(f){\Theta }_x(g)$. Also ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. Conversely any continuous unital complex-algebra homomorphism sending $Z$ to $x$ must send $Z-a$ to $x-a1$ and its reciprocal to the inverse. It therefore agrees on rational functions and then, by Runge density and continuity, on all of $\mathcal{O}(U)$. This proves existence and uniqueness of the holomorphic functional calculus.

For its spectral mapping property, first note compatibility on a smaller open spectral neighborhood $W\subseteq U$: composing restriction to $W$ with the calculus there gives the calculus on $U$ by uniqueness. If $\mu \notin g({\sigma }_A(x))$, choose such a $W$ where $\mu -g$ never vanishes. Thenwith the right side evaluated on $W$. Conversely, for $\lambda \in {\sigma }_A(x)$, write $g(z)-g(\lambda )=(z-\lambda )q(z)$ with $q$ holomorphic on $U$. The evaluated factors commute. If their product were invertible, both factors would be invertible, contradicting $\lambda \in {\sigma }_A(x)$. ThusPullback $h\mapsto h\circ g$ is continuous from $\mathcal{O}(V)$ to $\mathcal{O}(U)$ because the image under $g$ of a compact set is compact. The composite $h\mapsto {\Theta }_x(h\circ g)$ is a continuous unital homomorphism taking the coordinate function to $y={\Theta }_x(g)$. Uniqueness of the calculus at $y$ provesFor the exponential conclusion, let $K={\sigma }_A(x)$. There is a simple polygonal arc from zero through its unbounded complementary component to outside a disk containing $K$, continued by a ray to infinity and avoiding $K$. Loops in an initial polygonal path can be removed to make the arc simple. The complement of this slit is a simply connected open neighborhood of $K$ avoiding zero, and hence admits a holomorphic logarithm $\ell$. Set $y={\Theta }_x(\ell )$. The composition rule givesThe calculus of the entire exponential agrees with the usual algebra exponential because its power-series partial sums converge locally uniformly. Finally an invertible matrix over the complex numbers has a finite Banach algebra spectrum avoiding zero. Removing finitely many plane points leaves a connected set, so the hypothesis always holds: every invertible matrix over the complex numbers is an exponential of a matrix over the complex numbers.

Let $G$ be locally compact, Abelian and Hausdorff, and fix a Haar measure. Write $T_{x}f(y)=f(x^{-1}y)$ and use convolutionThe L1 group algebra is a commutative Banach algebra. Here a group character means a continuous homomorphism $\chi :G\to \mathbb{T}$, not the trace of an arbitrary group representation. A Gelfand character is a nonzero multiplicative complex-linear functional on the convolution algebra. We use the conventionWith the alternative standard Fourier convention involving $\overline{\chi }$, one simply conjugates every group character in this identification.

For such $\chi$, the functional is bounded by $\Vert f{\Vert }_1$, and Fubini's theorem, the substitution $y=xz$, and the homomorphism law giveIt is nonzero: if $u\ge 0$ is nonzero in $C_c(G)$, then $f=\overline{\chi }u$ has ${\Phi }_{\chi }(f)=\int u>0$.

Conversely let $\Phi$ be an algebra character. It is automatically bounded, with $|\Phi (f)|\le \Vert f{\Vert }_1$. Indeed, extend it to the unitization by $\tilde{\Phi }(f+\lambda 1)=\Phi (f)+\lambda$. If $|\Phi (f)|>\Vert f{\Vert }_1$, the Neumann series makes $f-\Phi (f)1$ invertible, whereas its character value is zero, contradicting multiplicativity of an invertible element and its inverse.

Choose $g$ with $\Phi (g)\ne 0$ and put $\chi (x)=\Phi (T_{x}g)/\Phi (g)$. Commutativity of the group gives $(T_{x}f)\ast g=f\ast (T_{x}g)$; applying $\Phi$ yieldsThus the ratio is independent of the chosen $g$. Applying the relation successively to $T_x{T}_{y}g=T_{xy}g$ gives $\chi (xy)=\chi (x)\chi (y)$, and $\chi (e)=1$. In particular it never vanishes.

The translation map $x\mapsto T_{x}g$ is continuous in $L^1$. For $g\in C_c(G)$, translated supports near a fixed $x$ lie in one compact set; continuity on that set and a finite cover give uniform convergence of the translated values, hence $L^1$ convergence because the compact set has finite Haar measure. For general $g$, approximate by $C_c(G)$ and use $\Vert T_{x}g{\Vert }_1=\Vert g{\Vert }_1$. Therefore $\chi$ is continuous. For all integers $n$,Positive powers exclude modulus greater than one; negative powers exclude modulus less than one. Hence $|\chi (x)|=1$.

Finally, the $L^1$-valued convolution formula $f\ast g=\int f(x)T_{x}gdx$ and boundedness of $\Phi$ giveCancel $\Phi (g)$ to obtain the displayed formula for every $f$. Different characters give different functionals: if their continuous difference is nonzero at a point, integration against a suitably phased compactly supported test function in a small neighbourhood detects that difference. Thus the correspondence is bijective.

It also identifies the character space of an Abelian L1 group algebra topologically with the Pontryagin dual group. For a net converging uniformly on compact subsets in $\hat{G}$, convergence of the functionals follows first for compactly supported $f$, then for all $L^1$ functions using the common norm bound one. Conversely suppose the functionals converge in the Gelfand topology. Choose $g$ whose limiting character value is nonzero; its values are eventually bounded away from zero. For compact $K\subset G$, the set $\{T_{x}g:x\in K\}$ is compact in $L^1$. Convergence of uniformly bounded linear functionals is uniform on this compact set: a finite norm-net reduces it to convergence at finitely many functions, with the common bound controlling the approximation error. The ratio formula then gives uniform convergence of the group characters on $K$. This proves equality with the compact-open topology, without assuming first countability of $G$.

For later use, these transforms form a self-adjoint subalgebra dense in $C_0(\hat{G})$. The set of algebra characters together with the zero functional is weak-star compact by Banach-Alaoglu theorem and the closed multiplicativity equations. Removing zero gives a locally compact Hausdorff character space; each evaluation $\Phi \mapsto \Phi (f)$ vanishes at infinity, since its modulus-at-least-$ϵ$ set is compact and avoids zero. Transforms separate characters, and none vanishes identically at a given character. They are closed under conjugation because $f^{\ast }(x)=\overline{f(x^{-1})}$ satisfies ${\Phi }_{\chi }(f^{\ast })=\overline{{\Phi }_{\chi }(f)}$. The locally compact Stone-Weierstrass theorem therefore gives the claimed density.

For a connected smooth Riemannian manifold $(M,g)$ without boundary, let $I(M)$ be the group of all global diffeomorphisms $f$ with $f^{\ast }g=g$, endowed with its compact-open topology. Equivalently these maps preserve the intrinsic distance; the Myers-Steenrod theorem supplies smoothness of distance-preserving bijections and makes $I(M)$ a finite-dimensional Lie group acting smoothly. The group is metric-dependent: a topological manifold can carry a very symmetric metric or one with no continuous symmetries. Connectedness matters here. With infinitely many identical disconnected components, permutations of those components can produce a group which is not a finite-dimensional Lie group.

An isometry is determined by its value and derivative at one point. To prove this, suppose $f(p)=p$ and $d{f}_p=I$. Preservation of the Levi-Civita connection giveswhere the exponential map is defined locally. Thus $f$ is the identity in a normal neighborhood of $p$. The set where its value and derivative agree with the identity is both closed and, by the same argument, open. Connectedness makes it all of $M$. No completeness is needed for this determination argument.

The stabilizer $I(M{)}_p$ consequently acts faithfully and orthogonally on $T_{p}M$, so embeds in $O(n)$. More generally an isometry takes a chosen orthonormal frame at $p$ to one at its image point, providing an injective map into the orthonormal frame bundle. It is a smooth embedding, and the full isometry action is proper. The orbit-stabilizer dimension formula therefore givesIsotropy is compact. If $M$ is compact then $I(M)$ itself is compact, by the same frame description or Arzela-Ascoli applied to isometries and their inverses. Properness also means that the quotient by isometries has well-controlled local orbit structure, even though orbit dimensions may vary.

The infinitesimal equation is the Killing equation. Differentiating $f_t^{\ast }g=g$ at zero givesConversely a vector field satisfying this equation has local flows preserving the metric. A global one-parameter subgroup requires the field to be complete. On a geodesically complete manifold every Killing field is complete: its length is constant along each of its own trajectories, so a finite-time trajectory has bounded speed and stays in a closed bounded ball. The Hopf-Rinow theorem makes that ball compact, and the ordinary differential equation can then be continued. The caveat is real: on the Euclidean interval $(0,1)$, the field ${\partial }_x$ is Killing but its translations are not globally defined for all time. This illustrates Killing fields on incomplete manifolds need not generate global isometries.

The Killing transport identities determine a Killing field from $X(p)$ and the skew-adjoint map $(\nabla X{)}_p$. Along a geodesic the field is a Jacobi field, so uniqueness for its second-order differential equation propagates these initial data. They have $n+n(n-1)/2$ components, reproducing the dimension bound infinitesimally. The important distinction on an incomplete manifold is between all local Killing solutions and the Lie algebra of complete global isometry flows.

In dimension three the bound is six, and rotational isotropy gives stronger information. The Lie subalgebras of $\mathfrak{so}(3)$ have dimensions zero, one or three: identifying the bracket with the cross product, a two-plane is not closed under brackets. Suppose $\dim I(M)=5$. Since an orbit has dimension at most three, its stabilizer would have dimension at least two, and hence exactly three. Its connected isotropy acts as all of $SO(3)$ on the tangent space. But the orbit tangent space is invariant under that action and would have dimension $5-3=2$, impossible because the standard rotational representation is irreducible. Therefore dimension five cannot occur.

If the dimension is six, each orbit has dimension three and each stabilizer dimension three. All orbits are open, so connectedness makes the action transitive. At each point the isotropy is transitive on tangent two-planes, forcing sectional curvature to be independent of the plane; transitivity makes its value independent of the point. Thus a maximally symmetric connected three-manifold has constant sectional curvature. Homogeneous Riemannian manifolds are complete, so its simply connected cover is the round sphere, Euclidean space or hyperbolic space after normalization. Conversely these three simply connected metrics attain dimension six:Here $O^{+}(3,1)$ preserves the chosen sheet of the hyperboloid; it includes orientation-reversing hyperbolic isometries. Constant curvature alone does not imply that a quotient retains six-dimensional global symmetry.

Dimension four likewise forces transitivity: stabilizer dimension three would give a one-dimensional invariant orbit tangent space, again impossible for full rotational isotropy. The stabilizer thus has dimension one and the orbits are three-dimensional. Typical four-dimensional isometry groups occur for $S^2\times \mathbb{R}$ and ${\mathbb{H}}^2\times \mathbb{R}$, where the factor groups have dimensions three and one. The usual Nil metric has three translation symmetries and one rotational isotropy symmetry. In centred Heisenberg coordinates it iswhich exhibits the horizontal rotations. Its distinct vertical and horizontal Ricci eigenvalues bound continuous isotropy by that rotation group, so its full isometry dimension is four. Standard metrics on $\widetilde{S{L}_2(\mathbb{R})}$ and nonround Berger spheres also have dimension four. The standard Sol geometry has a three-dimensional translation group and only finite isotropy, giving dimension three. These examples show how a preferred vertical direction reduces rotational symmetry.

The isometry dimensions in dimension three are precisely $0,1,2,3,4,6$, all realizable. A compact hyperbolic three-manifold gives zero; the product of a circle and a closed hyperbolic surface gives one. For dimension two take $T^2\times (0,1)$ withIts scalar curvature is the constant $-14$, but its Ricci eigenvalues are the distinct constants $-3,-6,-5$ in the two horizontal directions and the vertical direction. Hence connected isotropy is trivial and possible isometries preserve these line distributions. A local isometry preserving them has $t^{\prime }=t+c$ and would require constant dilations $e^{-c}$ and $e^{-2c}$ of the two circle coordinates. A global circle diffeomorphism with such a constant dilation has degree $\pm 1$, forcing $c=0$. The identity component is therefore exactly the two circle translations. A flat three-torus gives dimension three; $S^2\times \mathbb{R}$ gives four; a simply connected constant-curvature model gives six. This construction distinguishes local homogeneity from the smaller global isometry group.

For quotients, let a complete simply connected Riemannian manifold $X$ cover $M=X/\Gamma$. Every isometry of $M$ lifts to an isometry of $X$ which normalizes the deck transformation group. Conversely, every normalizing isometry descends. Two lifts differ by a deck transformation, giving the isometry group of a Riemannian quotient formulaThus one should not replace the isometry group of a quotient by that of its model geometry. For a flat torus, for example,The second factor is finite, whereas translations give the three-dimensional identity component.

There is a useful curvature test for compact manifolds. Contracting and differentiating the Killing equation gives ${\nabla }^{\ast }\nabla X=\mathrm{Ric}(X)$, with the nonnegative rough-Laplacian convention. Integration yields the integrated Bochner identity for Killing fieldsIf Ricci curvature is negative definite, no nonzero Killing field exists. The compact isometry group then has dimension zero and is finite. This explains why closed hyperbolic manifolds have finite global symmetry, despite the six-dimensional isometry group of their universal cover. Compactness is essential to this conclusion.

For a compact flat Riemannian manifold, the same identity makes every Killing field parallel. Parallel fields are exactly the vectors fixed by linear holonomy. ThereforeApplied to the six flat types of Question 1, this gives dimensions three for the torus, one for each of the four nontrivial cyclic types, and zero for the Hantzsche-Wendt manifold. These are identity-component dimensions; disconnected finite symmetries can still be present. Local curvature symmetry, global deck-group constraints, and completeness thus play different roles in determining the isometry group.

Give $\mathcal{O}(U)$ the compact-open topology, that is, uniform convergence on every compact subset of $U$. Choose a bounded finite union of polygonal domains $D$ whose closure lies in $U$, whose interior contains $\sigma (x)$ and whose boundary avoids the spectrum. Orient its boundary $\Gamma$ positively, with holes oriented negatively. It has winding number one on the spectrum and zero outside $U$. For the resolvent of an element $R(\zeta ,x)=(\zeta 1-x{)}^{-1}$ define the holomorphic functional calculus byThis is an integral of a continuous Banach space-valued function. Its value is independent of the admissible contour: the integrand is holomorphic away from the spectrum, and the difference of two such cycles has winding number zero there, so the Cauchy integral theorem applies. The vector-valued theorem follows by applying bounded linear functionals and then the Hahn-Banach theorem.

Linearity is immediate. To prove multiplicativity, choose two admissible contours with the inner domain compactly contained in the outer one, and use the resolvent identityIn the double integral for ${\Theta }_x(f){\Theta }_x(g)$, integrate the first term in the outer variable $\eta$. The Cauchy integral formula gives $g(\zeta )$. For the second term, integration of $f(\zeta )/(\eta -\zeta )$ over the inner contour is zero, since $\eta$ lies outside that inner domain. Therefore the double integral equals ${\Theta }_x(fg)$. The continuous integrands on disjoint compact contours justify exchanging the integrals.

For $f=1$ and $f=Z$, deform to a large circle and use the Neumann series $R(\zeta ,x)={\sum }_{m\ge 0}{x}^m{\zeta }^{-m-1}$. Its relevant coefficients give ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. The estimateproves continuity in the stated topology.

For uniqueness, the Runge approximation theorem in the required form says that rational functions with poles outside $U$, allowing a pole at infinity, are dense in $\mathcal{O}(U)$ for uniform convergence on compact subsets. Every complex-linear unital homomorphism taking $Z$ to $x$ has prescribed values on polynomials and on $(Z-a{)}^{-1}$ for $a\notin U$: the latter value must be $(x-a1{)}^{-1}$. Hence its rational-function values are prescribed. Continuity and Runge approximation force its values on every holomorphic function. This proves continuity and uniqueness of holomorphic functional calculus, including disconnected $U$.

To prove the spectral mapping theorem, first suppose $\mu \notin f(\sigma (x))$. On a smaller open neighborhood $U^{\prime }$ of the spectrum, $f-\mu$ has no zeros, so $h=1/(f-\mu )$ is holomorphic there. The contour definition is compatible with this restriction. Multiplicativity gives ${\Theta }_x(f)-\mu 1$ the inverse ${\Theta }_x(h)$, excluding $\mu$ from its spectrum.

Conversely, if $\mu =f(\lambda )$ for $\lambda \in \sigma (x)$, write $f(z)-\mu =(z-\lambda )g(z)$, defining $g(\lambda )=f^{\prime }(\lambda )$ at its removable singularity. ThenThese two factors commute. An invertible product of commuting elements makes each factor invertible, by multiplying the inverse of the product by the other factor. That would contradict $\lambda \in \sigma (x)$. Thus

Write $K={\sigma }_A(x)$ and equip $\mathcal{O}(U)$ with the compact-open topology. All algebra homomorphisms here are complex linear. Choose a finite positively oriented boundary cycle $\Gamma$ in $U\setminus K$ with winding number one on a neighborhood of $K$ and zero outside $U$. Such a cycle can be obtained as the boundary of a finite union of sufficiently small squares whose interior contains $K$ and whose closure is contained in $U$; holes are included with negative boundary orientation. Define the holomorphic functional calculus by the norm-convergent contour integralThe resolvent identity proves that $R$ is a Banach-space-valued holomorphic function off $K$. The Cauchy integral theorem, applied after each bounded linear functional, makes the integral independent of the chosen boundary cycle. The Hahn-Banach theorem then gives equality of the algebra-valued integrals. Moreover,which proves complex linearity and continuity for the compact-open topology.

For the constant function $1$ and the coordinate function $Z$, the contour can be deformed to a large circle because their integrands are holomorphic everywhere outside $K$. There the Neumann series gives $R(z)={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$. Integrating term by term showsTo prove multiplicativity, take nested cycles ${\Gamma }_o,{\Gamma }_i$ around $K$, with the outer cycle winding once on the inner cycle and the inner winding zero on the outer cycle. Use the outer cycle for $f$ and the inner for $g$. The resolvent identity readsIn the resulting double integral, the term containing $R(w)$ gives $f(w)g(w)R(w)$ after the $z$ integral, by the Cauchy integral formula. The term containing $R(z)$ vanishes after the $w$ integral, because $w\mapsto g(w)/(z-w)$ is holomorphic on and inside the inner cycle's region. ThereforeThe integrals are over finite compact curves and have continuous integrands, so exchanging their order is legitimate. This proves the existence of the required continuous unital algebra homomorphism.

For uniqueness, the Runge approximation theorem gives density in $\mathcal{O}(U)$, for the compact-open topology, of rational functions with poles outside $U$. More explicitly, exhaust $U$ by compact sets that are holomorphically convex relative to $U$, and approximate on those sets by rational functions whose poles lie in the complement of $U$; this gives convergence on every compact subset. Any unital algebra homomorphism taking $Z$ to $x$ must take $(Z-\alpha {)}^{-1}$, for $\alpha \notin U$, to $(x-\alpha 1{)}^{-1}$, since it preserves inverse equations. Partial fractions and polynomial algebra therefore fix its value on every such rational function. Continuity fixes its value on their limits. Thus the continuous unital homomorphism with the prescribed coordinate value is unique. Restricting a function to a smaller neighborhood of $K$ gives the same value, either by the contour definition or this uniqueness argument.

To prove the holomorphic spectral mapping theorem, first let $\mu \notin f(K)$. There is an open neighborhood $V\subseteq U$ of $K$ on which $f-\mu$ has no zeros. The holomorphic function $h=1/(f-\mu )$ on $V$ supplies a two-sided inverse throughThus ${\sigma }_A({\Theta }_x(f))\subseteq f(K)$. Conversely, if $\lambda \in K$ and $\mu =f(\lambda )$, the divided differenceis holomorphic on $U$, and multiplicativity givesThe two factors commute. A commuting product $uv$ can be invertible only if both factors are invertible: its inverse commutes with each factor, and $v(uv{)}^{-1}$ is a two-sided inverse for $u$. Invertibility of the displayed product would therefore contradict $\lambda \in K$. This proves

For the root, let $D=\mathbb{C}\setminus (-\infty ,0]$ and $q(z)=\exp (\mathrm{Log}z/3)$, using the principal complex logarithm. Define $y={\Theta }_x(q)$. Since $q(z{)}^3=z$, multiplicativity and the holomorphic spectral mapping theorem giveThis is the principal cube root of a Banach-algebra element.

For uniqueness, suppose $v^3=x$ and ${\sigma }_A(v)\subset S$. Cubing maps the whole open sector $S$ into $D$, sois a continuous unital algebra homomorphism: pullback by cubing is continuous for the compact-open topology, since images of compact sets are compact. It takes the coordinate function to $v^3=x$, and uniqueness already proved gives $F={\Theta }_x$. For $z\in S$, the scalar principal cube root satisfies $q(z^3)=z$, because $3\arg z\in (-\pi ,\pi )$. HenceThis proves uniqueness with the stated spectral sector, without assuming the ambient Banach algebra is commutative or that algebra characters separate its elements.

Use additive notation and fix Haar measure $m$ on $G$. A continuous unitary character is a continuous homomorphism $\chi :G\to \mathbb{T}$. With the conventionthe Fubini theorem and the character identity imply $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Also $|\hat{f}(\chi )|\le \Vert f{\Vert }_1$, and choosing $f=\chi a$ for a nonzero nonnegative $a\in C_c(G)$ shows that evaluation at $\chi$ is a nonzero functional. Thus each character gives a character of an algebra on the L1 convolution algebra. Here a multiplicative linear functional means a nonzero multiplicative linear functional; the zero functional must be excluded for the claimed bijection.

Conversely, let $\phi$ be such a functional. It is automatically bounded with $|\phi (f)|\le \Vert f{\Vert }_1$: extend it to the unitization of an algebra by $\tilde{\phi }(f+z1)=\phi (f)+z$. If $|z|>\Vert f{\Vert }_1$, the Neumann series makes $z1-f$ invertible, so its image under the unital homomorphism cannot be zero. Hence $\phi (f)$ cannot lie outside the norm disc.

Choose $a\in L^1(G)$ with $\phi (a)\ne 0$, and let $T_{x}a(t)=a(t-x)$. The Abelian translation identities $(T_{x}a)\ast b=a\ast (T_{x}b)$ giveTherefore, defining $c(x)=\phi (T_{x}a)/\phi (a)$, one has $\phi (T_{x}b)=c(x)\phi (b)$ for every $b$. Applying this to two successive translations gives $c(x+y)=c(x)c(y)$ and $c(0)=1$. By translation continuity in Lp on a locally compact group, $c$ is continuous. It is uniformly bounded by $\Vert a{\Vert }_1/|\phi (a)|$; applying this bound to $c(nx)=c(x{)}^n$ for positive and negative integers forces $|c(x)|=1$. Thus $\chi (x)=\overline{c(x)}$ is a continuous unitary character.

The Bochner integral identity $a\ast b={\int }_{G}b(y)T_{y}adm(y)$ holds in $L^1$, since $\Vert T_{y}a{\Vert }_1=\Vert a{\Vert }_1$. Applying the bounded functional yieldsCancel $\phi (a)\ne 0$ to obtain the required correspondenceThe recovery formula $\overline{\chi (x)}=\phi (T_{x}a)/\phi (a)$ proves uniqueness. These functionals have norm exactly one, since $b=\chi a$ with $a\ge 0$ and $\int a=1$ has $\Vert b{\Vert }_1=\phi (b)=1$.

For the topology, identify the Pontryagin dual group with these functionals and give it the Gelfand topology, namely pointwise convergence of $\hat{f}(\chi )$ for every $f\in L^1$. If a net ${\chi }_{\lambda }$ converges to $\gamma$ uniformly on every compact subset of $G$, then for $f\in C_c(G)$,Density of $C_c$ in $L^1$ and the uniform functional norm bound one extend this convergence to every $f\in L^1$.

Conversely, suppose convergence in the Gelfand topology and choose $a$ with $\hat{a}(\gamma )\ne 0$. For compact $K\subseteq G$, the set $\{T_{x}a:x\in K\}$ is compact in $L^1$ by translation continuity in Lp on a locally compact group. Evaluation convergence is uniform on this compact set: a finite $\delta$-net reduces its supremum to finitely many convergent evaluations plus $2\delta$, using the common functional norm one. ThusNow $\widehat{T_{x}a}(\chi )=\overline{\chi (x)}\hat{a}(\chi )$, and $\hat{a}({\chi }_{\lambda })\to \hat{a}(\gamma )\ne 0$. Dividing this identity proves ${\sup }_{x\in K}|{\chi }_{\lambda }(x)-\gamma (x)|\to 0$. Hence the Gelfand topology is exactly the compact-open topology, with the requested neighbourhood basisFor compact $K$, the continuous difference attains its maximum, so this strict pointwise inequality on $K$ is equivalent to a strict uniform bound. Finite intersections contain another such set by taking the union of the compact sets and the minimum of the tolerances.

Fix Haar measure $m_G$ and use $\hat{f}(\chi )={\int }_{G}f(x)\overline{\chi (x)}d{m}_G(x)$. A positive-definite function $p$ satisfies ${\sum }_{i,j}{z}_i\overline{z_j}p(x_i-x_j)\ge 0$ for every finite choice of points and coefficients. The Bochner theorem states that a continuous positive-definite function on a locally compact Hausdorff Abelian group has a unique representationwhere ${\mu }_p$ is a Radon measure. Conversely, any finite positive Radon measure on the Pontryagin dual group has a continuous positive-definite inverse transform. The sign convention here is the one compatible with the displayed Fourier transform.

First note that the dual is a locally compact Hausdorff topological group. Its group operations are continuous in the compact-open topology proved in question 3. For local compactness, adjoin the zero functional to the character space of an algebra of $L^1(G)$. All its elements lie in the closed unit ball of $L^1(G{)}^{\ast }$, and the equations $\phi (a\ast b)=\phi (a)\phi (b)$ define a closed subset in the weak-star topology. The Banach-Alaoglu theorem makes that subset compact Hausdorff. Removing the zero functional gives an open, hence locally compact, subspace. This provides the setting for constructing the measure on the dual, without assuming a Fourier inversion theorem.

Let $u^{\ast }(x)=\overline{u(-x)}$ and let $\mathcal{P}$ be the finite sums of convolution squares $u\ast u^{\ast }$, $u\in C_c(G)$. Each such square is continuous with compact support, and is positive-definite becauseIts transform is $|\hat{u}{|}^2$, so every $p\in \mathcal{P}$ has $\hat{p}\ge 0$. Also $p\ast q\in \mathcal{P}$: convolution of two squares is $(u\ast v)\ast (u\ast v{)}^{\ast }$, and finite sums distribute. Insert the Bochner representation of $q$ into $p\ast q$ and use the Fubini theorem to obtainThe representing measure on the right is finite and positive. Uniqueness in the Bochner theorem and commutativity now give the Bochner consistency identity for convolution squares

For every compact $K\subseteq \hat{G}$ there is $p\in \mathcal{P}$ with $\hat{p}>0$ throughout $K$. Indeed, at any ${\chi }_0$ choose $u={\chi }_{0}a$, where $a\ge 0$ is a nonzero member of $C_c(G)$. Then $\hat{u}({\chi }_0)=\int a>0$, and continuity supplies a neighbourhood where it stays nonzero. A finite subcover of $K$ and the sum of the corresponding squares give such a $p$.

For $\psi \in C_c(\hat{G})$, choose $p$ positive in transform on its support and defineThe quotient is set to zero away from the support; its denominator is bounded away from zero there. If $q$ is another choice, the consistency identity, multiplied by $\psi /(\hat{p}\hat{q})$ on that compact support, shows that the integral is unchanged. Choosing one $p$ for a union of supports proves linearity; positivity follows from positivity of ${\mu }_p$. On each fixed compact support it is bounded by ${\mu }_p(K)/{\min }_K\hat{p}$ times the uniform norm. Moreover, for every $p\in \mathcal{P}$ and $\psi \in C_c(\hat{G})$, choosing $q$ positive on the support of $\psi$ givesA nonzero square has ${\mu }_p(\hat{G})=p(0)=\Vert u{\Vert }_2^2>0$; a suitable nonnegative compactly supported $\psi$ consequently makes this last expression positive. Thus $J$ is nonzero.

To prove translation invariance, fix $\eta \in \hat{G}$ and modulate $p$ by $p_{\eta }(x)=\eta (x)p(x)$. This again belongs to $\mathcal{P}$, since modulation takes $u\ast u^{\ast }$ to $(\eta u)\ast (\eta u{)}^{\ast }$. Direct calculation and Bochner uniqueness giveHence, for ${\psi }_{\eta }(\chi )=\psi ({\eta }^{-1}\chi )$, using $p_{\eta }$ as the denominator function and changing variables gives $J({\psi }_{\eta })=J(\psi )$. The Riesz representation on compactly supported continuous functions now produces a nonzero translation-invariant positive Radon measure $\nu$ on the dual. It is finite on compact sets, so it is Haar measure. This is the compatible dual Haar measure, denoted $m_{\hat{G}}=\nu$.