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At the Gaussian fixed point, $[{\varphi }_i]=(d-2)/2$, so all three quartic couplings have engineering dimension $d-4[{\varphi }_i]=4-d$. ThusTake $A>0$ and $ϵ>0$, as in the perturbative epsilon expansion. Define $x=A{g}_1/ϵ$, $y=A{g}_2/ϵ$, $z=A\lambda /ϵ$. The renormalization-group fixed point equations becomeIf $z=0$, each of $x,y$ is independently $0$ or $1/36$, giving four renormalization-group fixed points. If $z\ne 0$, subtracting the first two equations givesThe branch $x+y=1/36$ forces $z=1/12$; inserting this in the first equation yields a repeated root $x=y=1/72$. Hence every nonzero-$z$ solution has $x=y$. Then $z=1/8-3x$, andhas roots $x=1/40$ and $x=1/72$. The complete list of six coupled Ising fixed points near four dimensions, in coordinates $(g_1,g_2,\lambda )$, isThe internal symmetries of these renormalization-group fixed points are:

The last point displays the field-rotation equivalence of decoupled Ising theories. With the orthogonal transformation ${\psi }_{\pm }=({\varphi }_1\pm {\varphi }_2)/\sqrt{2}$,It is the same pair of decoupled Ising models written in fields rotated by $45^{\circ }$, but remains a distinct coordinate solution of the stated renormalization-group beta functions. Here $D_4$ denotes the dihedral group of order eight; an alternative convention calls it $D_8$. We have described linear internal transformations preserving the gradient energy; free massless sectors also have constant scalar-field shift symmetries. At $ϵ=0$, the six coordinates coalesce at the Gaussian fixed point.

Write the dihedral group as $D_{2n}=⟨r,s:r^n=s^2=e,srs=r^{-1}⟩$. Every reflection has order two, so an element of order three must be a rotation. Such rotations exist exactly when $3\mid n$, and then they are $r^{n/3}$ and $r^{2n/3}$. The answer is therefore

The actions on the ordered pair $(x,y)$ satisfyThus every element of $G$ has the form ${\sigma }^i$ or $\tau {\sigma }^i$ for $0\le i<4$, so $|G|\le 8$. Their actions areBecause $x,y$ are algebraically independent elements over $\mathbb{Q}$, these eight substitutions are distinct. Henceand the presentation identifies $G$ with the dihedral group $D_8$.

PutBoth generators $\sigma$ and $\tau$ fix $u$ and $v$, so for $F=\mathbb{Q}(u,v)$,The elements $x^2$ and $y^2$ are the two roots ofso $[F(x^2,y^2):F]\le 2$. Adjoining first $x$ and then $y$ takes at most two quadratic extensions, and the tower law givesPart (b) gives $8=|G|\le [K:K^G]$. Since $F\subseteq K^G\subseteq K$, another use of the tower law givesAll inequalities are equalities and $[K^G:F]=1$. Thereforethe dihedral fixed field of a two-variable rational function field.

For the rotation group $C_5$, the identity fixes all $n^5$ colorings. Because five is prime, each of the four nonidentity rotations is a single five-cycle and fixes exactly the $n$ constant colorings. Burnside lemma therefore gives the number of coloring necklaces

For the dihedral group $D_{10}$, the five reflections each fix one vertex and interchange two pairs of vertices, so each fixes $n^3$ colorings. Adding these to the identity and rotation contributions gives the number of coloring bracelets

In local coordinates centred at $p$ and $f(p)$, a nonconstant rational map has the formThe point $p$ is a ramification point of a holomorphic map when its ramification index of a holomorphic map satisfies $e_p>1$. A branch value of a holomorphic map, also called a branch point in the target, is a value $f(p)$ of some ramification point.

Let $B$ be the finite set of branch values and put $R=f^{-1}(B)$. For every $w\notin B$, all points of $f^{-1}(w)$ have local degree one, so the holomorphic inverse function theorem supplies disjoint neighbourhoods on which $f$ is biholomorphic. Compactness of the fibre lets their target neighbourhoods be intersected to one evenly covered neighbourhood of $w$. Henceis an unramified covering map. Here the displayed restriction requires $R=f^{-1}(B)$; if “ramification point” is reserved only for points with $e_p>1$, then the source deletion must be written $f^{-1}(B)$.

The Monodromy theorem says that analytic continuations along endpoint-fixed homotopic paths have the same terminal germ. Equivalently here, the path lifting theorem lifts a loop $\gamma$ based at $w\notin B$ from each point of $f^{-1}(w)$; taking the endpoint of each lift permutes that fibre. The permutation depends only on $[\gamma ]\in {\pi }_1({\mathbb{C}}_{\infty }\setminus B,w)$, giving the monodromy group of a covering.

For the stated function, make the Möbius change of coordinateThenThe critical points are $t=0,\pm 1/\sqrt{2},\infty$, with branch valuesA loop around $0$ interchanges the two roots born from $t=0$, so its branch cycle is a transposition. A loop around $-1/4$ simultaneously interchanges the two pairs that collide there, so its branch cycle is a product of two disjoint transpositions. With a suitable labelling these areThey generate a transitive group of order eight; their product is a four-cycle. Therefore the full monodromy group isthe dihedral group in its action on the four vertices of a square.

The Poincare half-plane model iswith the orientation inherited from the complex plane. Forone hasThese identities show directly that ${\gamma }^{\ast }d{s}^2=d{s}^2$. The map is holomorphic with nonzero derivative, so it preserves orientation. The matrices $I$ and $-I$ induce the same map, giving the action of PSL2(R).

Conversely, let $F$ be an orientation-preserving isometry. The PSL2(R) action is transitive on $\mathbb{H}$, so compose $F$ with an element taking $F(i)$ back to $i$. The resulting isometry fixes $i$ and acts on $T_i\mathbb{H}$ by an orientation-preserving orthogonal map, hence a rotation. The stabilizer of $i$ in PSL2(R),realizes every such tangent rotation. An isometry is determined by its value and differential at one point because it preserves geodesics and the exponential map. Thus the composed isometry belongs to PSL2(R), and so does $F$.

The map $\tau (z)=-\overline{z}$ is an orientation-reversing isometry. Composing any orientation-reversing isometry with $\tau$ gives an orientation-preserving one, so

A hyperbolic line is a vertical Euclidean line or a semicircle orthogonal to the real axis. Its hyperbolic reflection ${\sigma }_{\ell }$ is the unique orientation-reversing isometry that fixes every point of $\ell$. If $\ell ,{\ell }^{\prime }$ meet at angle $\theta$, thenis the hyperbolic rotation about $A$ through angle $2\theta$. The generators satisfyIf $\theta /\pi =p/q$ in lowest terms, then $\rho$ has order $q$ and the generated group is the finite dihedral group of order $2q$. If $\theta /\pi$ is irrational, $\rho$ has infinite order and the generated group is the infinite dihedral group. In the degenerate case $\ell ={\ell }^{\prime }$, the group has order two.

Direct multiplication givesHenceThis is the defining conjugation relation between a rotation and reflection in a dihedral group.

The relationsimply $RM=M{R}^{-1}$. Moving every occurrence of $M$ to the left and reducing exponents therefore puts every word into one of the normal formsExplicitly, with $\theta =2\pi /n$,andThe $n$ rotations are distinct. The $n$ reflected matrices are also distinct, and no reflected matrix equals a rotation because their determinants are $-1$ and $1$, respectively. Hence there are exactlymatrices, forming the dihedral group.

Irreducible degrees divide the order of a finite $2$-group and satisfyThe number of linear characters is $|G/G^{\prime }|$. For a nontrivial finite $2$-group this is at least $4$ unless the group is abelian, and every nonlinear degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, thenThus the only degree collections arewith respectively $r=16,10,7$.

For $1^{16}$ take the cyclic group $G=C_{16}=⟨g⟩$. Every conjugacy class is a singleton, and its character table iswhere ${\zeta }_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ and
$C_2=⟨z⟩$. On the five classesthe four linear characters of $D_8$, indexed by
$\epsilon ,\delta \in \{\pm 1\}$, and its degree-two character areThe ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists ofwhere $\chi$ runs over the above five rows and
$\eta \in \{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the dihedral groupOrder its seven classes asThe four linear rows areThe three degree-two rows, for $k=1,2,3$, areThe row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.

Let $\zeta =e^{2\pi i/n}$ and define Möbius transformationsThey satisfyand the $2n$ maps $r^k$ and $s{r}^k$ are distinct. They therefore give a faithful action of the dihedral group $D_{2n}$ on the Riemann sphere.

The rational functionis invariant under both $r$ and $s$, so it is constant on every orbit. Conversely, for nonzero finite $z_1,z_2$, put $u=z_1^n$ and $v=z_2^n$. Equality of the function values givesIf $u=v$, then $z_2={\zeta }^k{z}_1=r^k{z}_1$ for some $k$. If $uv=1$, then $z_2={\zeta }^k/z_1=r^{k}s{z}_1$ for some $k$. Finally, zero and infinity both map to infinity and are interchanged by $s$. HenceThis is the orbit-separating invariant for the standard dihedral action on the Riemann sphere.

For unit vectors, the defining inequality for $Y$ is equivalent to $v\cdot w\ge 0$. WriteThen $t=(1-\Vert u{\Vert }^2{)}^{1/2}$, sois a homeomorphism $D(T{S}^{2n})\to Y$. Replacing $v$ by $-v$ gives the same description of $Y^{\prime }$. Their intersection isthe unit tangent bundle, equivalently a Stiefel manifold. Since $e(T{S}^{2n})$ evaluates to $\chi (S^{2n})=2$, its Gysin sequence gives the cohomology of the unit tangent bundle of an even-dimensional sphere:Every product of positive-degree classes vanishes.

Let $G$ consist of the eight signed permutations of the two factors:It is the group of signed permutation matrices in dimension two, hence isomorphic to the dihedral group $D_8$, and every element preserves the equation $v\cdot w=0$. If $a,b$ generate $H^{2n}(X;\mathbb{Z})$ from the two sphere factors, these maps act by the corresponding signed permutation matrix, because the antipodal map of the even-dimensional sphere has degree $-1$. The eight actions are distinct, so only the identity can be homotopic to $1_X$.

Every element of $G$ acts trivially on $H^{\ast }(Z;\mathbb{Z})$. The degree-$2n$ group is $\mathbb{Z}/2$, so its automorphism is forced to be the identity. On the top class, regard $Z$ as the Stiefel manifold of two-frames. The signed permutations are the right action of $O(2)$. The determinant-one component is connected, while a determinant-minus-one element is homotopic within that component to $(v,w)\mapsto (v,-w)$, the antipodal map on the $S^{2n-1}$ fibers. That map has degree $+1$, so the top class is also fixed.

The subgroup preserving $Y$ iswhere $A(v,w)=(-v,w)$, $B(v,w)=(v,-w)$, and $S(v,w)=(w,v)$. It consists exactly of the elements with ${\epsilon }_1{\epsilon }_2=1$. Under the deformation retraction $Y\simeq S^{2n}$ onto the diagonal, $1$ and $S$ act as the identity, while $AB$ and $ABS$ act as the antipodal map. They therefore act on $H^{2n}(Y;\mathbb{Z})$ by $+1,+1,-1,-1$, respectively.

The Thom isomorphism theorem identifies $H^{\ast }(Y,Z)$ with a degree-$2n$ shift of $H^{\ast }(S^{2n})$. On the Thom class in degree $2n$, the same four elements act by $+1,+1,-1,-1$: the factor swap reverses each normal vector, which preserves the orientation because the normal rank $2n$ is even, while the simultaneous antipodal map reverses the oriented tangent fiber. On the degree-$4n$ relative class every element acts by $+1$, since the base and fiber signs for the last two elements cancel.

When $n=1$, $Z=V_2({\mathbb{R}}^3)\cong SO(3)$. Every signed permutation of $(v,w)$ extends, after sending the third frame vector $v\times w$ to the required sign, to right multiplication by an element of $SO(3)$. Since $SO(3)$ is path-connected, every right translation is homotopic to the identity. Therefore

Put $\alpha =\sqrt[4]{7}$. The roots of $T^4-7$ are $\alpha ,i\alpha ,-\alpha ,-i\alpha$, so its splitting field isThe polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $7$, and $\mathbb{Q}(\alpha )\subset \mathbb{R}$ does not contain $i$. The tower law for field extensions therefore gives $[L:\mathbb{Q}]=8$.

Define automorphismsThey satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct, so they exhaust the Galois group. Hencethe dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, isFor example, $\sqrt{7}+i$ has stabilizer $⟨r^2⟩$, and $(1+i)\alpha$ has stabilizer $⟨rs⟩$; the other entries follow similarly. Also $\alpha +i$ has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion $L=\mathbb{Q}(\alpha +i)$ holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $⟨r⟩$, the two index-two Klein four-groups, $⟨r^2⟩$, and $\{1\}$, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.

A complex representation of a finite group $G$ is a homomorphismfor a finite-dimensional complex vector space $V$. It is an irreducible representation when its only invariant subspaces are $0$ and $V$, and its degree is $\dim V$. Two representations are isomorphic when there is an invertible linear map intertwining their $G$-actions.

Write the dihedral group asIn an irreducible representation choose an eigenvector $0\ne v\in V$ of $\rho (r)$, which is possible because $r^n=1$. If its eigenvalue is $\lambda$, thenThus $W=\mathrm{span}\{v,\rho (s)v\}$ is invariant under both $r$ and $s$. Irreducibility forces $W=V$, proving $\dim V\le 2$ as in irreducible complex representations of a finite dihedral group.

Let $\omega =e^{2\pi i/n}$. Besides the one-dimensional representations described below, define for each indicated $j$These matrices satisfy the defining relations. Since ${\omega }^j\ne {\omega }^{-j}$, the only one-dimensional $r$-invariant subspaces are the two coordinate axes, and $s$ interchanges them. Hence ${\rho }_j$ is irreducible. Their characters satisfywhose values are distinct in the ranges below, so the ${\rho }_j$ are pairwise nonisomorphic. This is the family of two-dimensional representations of a finite dihedral group.

If $n$ is odd, a one-dimensional character must send $r$ to a scalar $\epsilon$ satisfying ${\epsilon }^n=1$ and $\epsilon ={\epsilon }^{-1}$. Thus $\epsilon =1$, while $s$ may independently map to $1$ or $-1$. These give two one-dimensional irreducibles. Takinggives $(n-1)/2$ two-dimensional irreducibles, for a total of

If $n$ is even, $r$ may map to either $1$ or $-1$, and again $s$ may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Takinggives $n/2-1$ two-dimensional irreducibles, for a total ofDifferent dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.

On $P=\mathrm{span}\{\alpha ,\beta \}$, each $s_{\alpha }$ is reflection in the line ${\alpha }^{\perp }\cap P$. The product of two plane reflections is a rotation. If $\theta$ is the oriented angle from $\beta$ to $\alpha$, thenrotates through $2\theta$ modulo $2\pi$.

If this rotation has order $m$, conjugation by either reflection inverts it. Henceis a dihedral group, with rotational subgroup $⟨r⟩=⟨s_{\alpha }{s}_{\beta }⟩$.

For the simple-root angles from part ii, the rotation orders and Weyl groups areThese are the rank-two Weyl groups.

The Coxeter number is the order of a Coxeter element. In the stated families the values areHere $D_{2n}$ is the dihedral group of order $2n$, $S_n$ is the symmetric group, $S{S}_n$ is the signed symmetric group, and $ES{S}_n$ is the even signed symmetric group.

The six-element dihedral group generated by the displayed rotations already forces every invariant second-rank tensor to have the same form as an axially invariant second-rank tensor. The axial threefold rotation eliminates mixed entries and leaves only a planar scalar plus a planar antisymmetric part; the transverse half-turn eliminates the latter. No smaller finite subgroup suffices: cyclic rotation groups retain an axial antisymmetric tensor, while a four-element noncyclic rotation group retains arbitrary diagonal tensors along its three mutually perpendicular half-turn axes.

Put $\alpha =\sqrt[4]{7}$ and $L=\mathbb{Q}(\alpha ,i)$. The dihedral group $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ acts byThe ten subgroups and their fixed fields areThe normal subgroups are $D_8$, $⟨r⟩$, the two displayed Klein four-groups, $⟨r^2⟩$, and $\{1\}$.

A group generated by two involutions is dihedral in the extended convention allowing $C_2$ and the infinite dihedral group. If $a\ne b$, put $x=ab$, $y=b$; then $yx{y}^{-1}=x^{-1}$. Conversely, from the inversion relation, $y$ and $xy$ square to the identity and generate. If $xy=e$, the group is $C_2$ and the same nonidentity element is used twice. Exact order two, rather than merely a square equal to the identity, requires this exceptional case. This convention extends the usual finite dihedral group definition.