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Every finite group of odd order is a solvable group. In particular, a nonabelian simple group has even order and therefore has an involution by Cauchy's theorem for finite groups. This is a deep theorem, not an elementary consequence of the Sylow theorems.

Let $W_g$ be a connected sum of oriented manifolds consisting of $g$ copies of $S^r\times S^r$ with odd $r\ge 1$. Suppose an orientation-preserving smooth involution $f$ has finitely many fixed points, all nondegenerate with positive determinant $\det (I-D{f}_x)$. The Lefschetz-Hopf fixed-point theorem and the cohomology ring of a connected sum of odd-dimensional sphere products giveThis middle group has dimension $2g$ and its Poincare duality pairing is alternating. The induced map preserves the pairing and squares to the identity. By eigenspaces of a symplectic involution, its trace is $2g-4b$ for an integer $b$. Therefore $\#\mathrm{Fix}(f)\equiv 2-2g(\mathrm{mod}4)$.

Double a compact orientable genus-zero topological surface with $g+1$ boundary circles. The resulting closed surface has genus $g$. The involution exchanging the halves fixes the seam circles and has local collar form $(s,t)\mapsto (s,-t)$. Compose it with a flow rotating each seam coordinate by a small nonzero amount, with speed supported in its collar. Equality with the initial point would force $t=0$, where the rotation is nonzero; off the collars the original reflection already has no fixed points. The new map is fixed-point-free and homotopic to the reflection through the flow. This gives a surface example in every genus where having fixed points is not a homotopy invariant.

Up to group isomorphism, the groups of order eight are $C_8$, $C_4\times C_2$, $C_2^3$, $D_8$ and $Q_8$, with $D_8$ here denoting the eight-element dihedral group. An element of order eight gives the first case. If every nonidentity element has order two the group is an elementary abelian group. Otherwise, after excluding elements of order eight, a cyclic subgroup of order four has index two; an element outside it either commutes with its generator or inverts it. In the latter case its square is either the identity or the central involution, giving the dihedral group or quaternion group.

A nonidentity involution in the affine group over a finite field ${\mathbb{F}}_{\ell }$, for odd prime $\ell$, has linear coefficient $-1$: a translation has order $\ell$, while $u^2=1$ forces $u=\pm 1$. Its unique fixed point is $b/2$, and its remaining orbits have size two. This orbit count computes primes in an intermediate field as double cosets for an order-two decomposition group acting on the radical roots.

// Target: number-theory.bigb

For the dihedral group of a regular $2n$-gon, $n\ge 2$, use $r^{2n}=s^2=1$ and $srs=r^{-1}$. Its involutions are the central half-turn $r^n$ and all reflections $s{r}^j$. The half-turn has singleton conjugacy class and normal closure $\{1,r^n\}$. Conjugating a reflection changes its exponent by an even integer or negates it, so its conjugacy class is $\{s{r}^{j+2k}:0\le k<n\}$. Products of consecutive reflections generate $r^2$, giving normal closure $⟨r^2,s{r}^j⟩$, of order $2n$ and index two. The two reflection classes are distinguished by exponent parity.

Suppose an involution $R$ of a Markov chain satisfies $P(Rx,Ry)=P(x,y)$. Start one chain at $x$, set the other equal to its image under $R$ until the first hitting time of the fixed-point set of $R$, and subsequently use identical transitions. This is a coalescing coupling, because the symmetry preserves each marginal transition matrix and both chains coincide at that hitting time.

Every nontrivial quotient group of a group generated by two involutions is generated by two involutions, allowing repeated generators. The two generator images have squares equal to the identity. If one image becomes trivial, the other generates $C_2$; if both become trivial, the quotient is trivial and is excluded. Thus the two-involution characterization of a dihedral group is preserved by nontrivial quotients.

An odd involution $A$ reverses fermion parity and is its own inverse, so it identifies the bosonic and fermionic subspaces. They have equal dimension in a finite-dimensional representation. For a supercharge $q$ with $q^2=0$ and $\{q,q^{†}\}=cI$, $c>0$, use $A=(q+q^{†})/\sqrt{c}$. This gives a direct proof of boson-fermion degeneracy in a supermultiplet at positive energy. A zero-energy supersymmetric vacuum need not have an accompanying fermionic vacuum.

At fixed nonzero four-momentum, take the finite physical supermultiplet and write $P_F=(-1{)}^F$ for fermion parity. Each supercharge is odd, so $P_F{Q}_{\alpha }=-Q_{\alpha }{P}_F$ and the same holds for its adjoint. By the cyclic property of the trace,Consequently $\mathrm{Tr}[P_F\{Q_{\alpha },Q_{\alpha }^{†}\}]=0$. Sum the two spinor components of the $\mathcal{N}=1$ Super-Poincaré algebra. Since ${\mathrm{Tr}}_{2\times 2}{\sigma }^i=0$, their sum is $4H$. On a fixed-energy representation, $H=EI$, yieldingFor a nonzero physical one-particle momentum, $E>0$, including the massless case. Hence the bosonic and fermionic physical state counts agree:Equivalently a nonzero normalized supercharge and its adjoint define an odd involution pairing the two subspaces. This is supertrace pairing at positive energy; it counts spin polarizations, not merely distinct names of particles.

The phrase “every supermultiplet” requires this positive-energy qualification. A zero-energy supersymmetric vacuum with all supercharges acting as zero is a one-dimensional bosonic representation with no fermionic partner. It is a counterexample to the unrestricted statement, because division by $E$ is then impossible. It does not contradict the requested equality for particle supermultiplets.

By part (b), $M=S^n/\Gamma$, where the group of deck transformations acts freely by round isometries. Every round isometry is the restriction of an orthogonal transformation of ${\mathbb{R}}^{n+1}$: it preserves the inner product $⟨x,y⟩=\cos d(x,y)$, so its images of an orthonormal basis determine its linear extension.

Suppose $n$ is even. Every $A\in SO(n+1)$ has eigenvalue one. Indeed, the nonreal eigenvalues come in conjugate pairs with product one, while the remaining real eigenvalues are $\pm 1$. Their number is odd and their product is one, forcing at least one $+1$. Such a transformation fixes a unit vector. Freeness of the deck transformation action consequently implies that the determinant map $\Gamma \to \{\pm 1\}$ has trivial kernel. Thus $|\Gamma |\le 2$.

If $\Gamma$ is nontrivial with generator $A$, then $A^2=I$. Its eigenvalues are $\pm 1$, and freeness excludes the eigenvalue $+1$. Hence $A=-I$, the antipodal involution. We obtainThis proves the even-dimensional part of the classification of complete positive constant-curvature manifolds.

The conclusion fails in odd dimensions. For $n=2m-1\ge 3$, identify $S^{2m-1}$ with the unit sphere in ${\mathbb{C}}^m$. Scalar multiplication by $\zeta =e^{2\pi i/r}$, with $r\ge 3$, generates a freely acting cyclic group of isometries: ${\zeta }^{j}z=z$ for nonzero $z$ forces ${\zeta }^j=1$. The quotient is complete with constant sectional curvature one and has fundamental group $C_r$, so it is neither the simply connected sphere nor Real projective space with fundamental group $C_2$. In particular $S^3/C_3$ is a lens space counterexample.

Let $P$ be a sphere with $g+1$ disjoint open disks removed, and form its double $S$ by gluing two copies along the boundary. Then $\chi (S)=2\chi (P)=2(1-g)$, so $S$ is an oriented closed surface of genus $g$. The involution $\iota$ swapping the two halves fixes precisely the $g+1$ seam circles.

Use disjoint collars with coordinates $(s,t)\in \mathbb{R}/\mathbb{Z}\times (-\eta ,\eta )$, where $\iota (s,t)=(s,-t)$. Choose a smooth even bump function $\chi (t)$ supported inside the collar, with $\chi (0)=1$, and choose $0<ϵ<1$. Define an orientation-preserving diffeomorphism $R$ on each collar by $R(s,t)=(s+ϵ\chi (t),t)$ and as the identity outside the collars. Scaling $ϵ$ from zero gives an isotopy from the identity to $R$, so $F=R\circ \iota$ is homotopic to $\iota$.

A fixed point of $F$ in a collar would satisfy $-t=t$, hence $t=0$. Its other coordinate would then require $s+ϵ=s$ modulo one, impossible for the chosen $ϵ$. Outside the collars, $F=\iota$ exchanges the interiors of the halves and has no fixed points. Therefore $\iota$ has fixed circles but is homotopic to the fixed-point-free map $F$. This is the fixed-point-free perturbation of a doubled-surface reflection; the construction includes genus zero, where the double is a sphere.

Take the double cover of $S^2$ branched at $2g+2$ points, with deck involution $\iota$. One explicit construction cuts the sphere along $g+1$ disjoint arcs pairing the branch points and cross-glues two copies along the cuts. Its Euler characteristic is $2\chi (S^2)-(2g+2)=2-2g$, so it is an oriented closed surface of genus $g$. Locally at a branch point the projection is $z\mapsto z^2$ and the involution is $z\mapsto -z$; thus the involution is nonidentity and orientation-preserving, also when $g=0$.

The invariant rational cohomology of a finite surface action identifies with the cohomology of its quotient. For this action, triangulate the quotient with the branch points as vertices and lift the triangulation. Quotient cochains identify with invariant cochains, and averaging over the two deck transformations makes taking invariants exact over $\mathbb{Q}$. Hence the invariant part of $H^1(S;\mathbb{Q})$ is $H^1(S^2;\mathbb{Q})=0$. Since ${\iota }^2=1$, its eigenvalues are $\pm 1$ and it is therefore $-I$ on the $2g$-dimensional $H^1$. Its actions on $H^0$ and $H^2$ are both the identity. ThusHomotopy invariance and the result just proved show that every map homotopic to this involution has fixed points, for every genus.

Let $h=h^{\ast }$. The exponential $u_t=e^{ith}$ is defined by its norm-convergent power series. Continuity and conjugate linearity of the involution give $u_t^{\ast }=e^{-ith}$, and multiplication of the commuting exponential series gives $u_t^{\ast }{u}_t=u_t{u}_t^{\ast }=1$. Thus $u_t$ is a Unitary element of a C-star algebra and, by the C-star identity, $\Vert u_t\Vert =1$.

A character of an algebra is automatically continuous with $|\phi (a)|\le \Vert a\Vert$, as proved in Question 5. Passing it through the exponential series givesThe left side is $e^{-t\Im \phi (h)}$. Considering both signs of $t$ forces $\Im \phi (h)=0$. So characters take real values on self-adjoint C-star elements.

Now write $x=h+ik$ withboth self-adjoint C-star elements. ThenThis proves that characters of a C-star algebra respect the involution. The argument itself does not require commutativity, although a noncommutative algebra need not have any characters.

First the C-star identity and submultiplicativity give $\Vert a{\Vert }^2=\Vert a^{\ast }a\Vert \le \Vert a^{\ast }\Vert \Vert a\Vert$, hence $\Vert a\Vert \le \Vert a^{\ast }\Vert$ when $a\ne 0$. Applying this to $a^{\ast }$ gives equality, so the involution is isometric and continuous. Also $\Vert 1\Vert =\Vert 1{\Vert }^2$ and $1\ne 0$, yielding $\Vert 1\Vert =1$.

For a Normal element of a C-star algebra $x$, put $b=x^{\ast }x$, which is a self-adjoint C-star element. Normality gives $(x^{\ast }{)}^2{x}^2=b^2$, and the C-star identity applied to $x^2$ and to $b$ givesThus $\Vert x^2\Vert =\Vert x{\Vert }^2$. Every power of $x$ is normal because $x$ commutes with $x^{\ast }$, so iteration yieldsThe spectral radius formula has a limit along all positive integers, and along this subsequence its root norms are exactly $\Vert x\Vert$. ThereforeThis proves spectral radius norm equality for normal elements directly from the defining norm identity, rather than assuming a spectral representation in advance.

For the unheaded representation request, retain the question's standing unital hypothesis. Let $K$ be the character space of an algebra of $A$, with its weak-star or Gelfand topology. The standard commutative Banach-algebra results allowed here give a nonempty compact Hausdorff character space and ${\sigma }_A(a)=\{\phi (a):\phi \in K\}$. In particular, character compactness follows from the Banach-Alaoglu theorem: the equations $\phi (1)=1$ and $\phi (ab)=\phi (a)\phi (b)$ define a weak-star closed subset of the dual unit ball.

The Gelfand transform is the unital homomorphismEvery element of the commutative C-star algebra is normal. Part i and the character spectrum formula therefore giveThus $\Gamma$ is an isometry, hence injective, and its range is closed because $A$ is complete. Part ii gives $\Gamma (a^{\ast })=\overline{\Gamma (a)}$. The range contains constants and separates points of $K$, since two distinct characters differ on some element of $A$. The complex Stone-Weierstrass theorem states that a unital, conjugation-closed subalgebra of $C(K)$ separating points of a compact Hausdorff space is uniformly dense. Its application here, together with closedness of the range, yieldsThis proves the Commutative Gelfand--Naimark theorem from the preceding parts and the stated approximation theorem.

The word “every” in the printed deduction must be read under the initial unital hypothesis. If nonunital commutative C-star algebras are included, the compact-space statement is false: the nonzero algebra $c_0$ has no multiplicative identity, whereas $C(K)$ for nonempty compact $K$ does. The general nonunital version is $C_0(K)$ on a locally compact Hausdorff space, with vanishing at infinity. This qualification preserves the proved unital result.

For the final real spectrum criterion for a normal C-star element, the ambient algebra may be noncommutative. Let $B$ be the norm-closed unital subalgebra generated by $x$ and $x^{\ast }$. Normality makes these two generators commute. The involution preserves their polynomial algebra and is continuous, so $B$ is a commutative C-star subalgebra.

We first show that $x$ has the same spectrum in $B$ as in $A$, without assuming spectral permanence. Put $\Omega =\mathbb{C}\setminus {\sigma }_A(x)$. A compact subset of the real axis has connected complement: upper and lower half-plane paths join around either end of a bounded interval containing it, and each omitted real point can be joined locally to the upper half-plane. DefineThis set is nonempty because the Neumann series puts the inverse in $B$ for $|\lambda |>\Vert x\Vert$. It is relatively open: from an inverse in $B$, the local inverse formulastays in $B$ for small $h$. It is relatively closed by continuity of the ambient resolvent and norm-closedness of $B$. Connectedness gives $V=\Omega$. Conversely an inverse in $B$ is an inverse in $A$, soThis is the connected-resolvent case of spectrum in a closed unital subalgebra.

Apply the just-proved commutative C-star representation to $B$. For every character $\psi$ of $B$, $\psi (x)$ lies in this real spectrum. Part ii makes $\psi (x^{\ast })=\overline{\psi (x)}=\psi (x)$. The Gelfand transform of $B$ is injective by its isometry, soFor a nonunital ambient C-star algebra, perform the same argument in its standard C-star unitization, using the usual unitization spectrum.

Normality cannot be dropped. In $M_2(\mathbb{C})$, takeSince $N^2=0$, its spectrum is $\{0\}$: for $\lambda \ne 0$, $(\lambda I-N{)}^{-1}={\lambda }^{-1}I+{\lambda }^{-2}N$, while $N$ is not invertible. Yet $N\ne N^{\ast }$ andThus a real spectrum alone does not force self-adjointness; normality is essential.