Let $T^{Z(G)}$ be the common fixed subtree of the centre, which is nonempty by hypothesis and is $G$-invariant because $Z(G)$ is central. Restrict the action to this subtree. There $x$ and $y$ act pointwise trivially, so the action factors throughIn particular, the induced tree isometries satisfy

Assume, as usual for a combinatorial tree action, that edge inversions have been removed by barycentric subdivision. The finite-order elements $a,b,c$ are then elliptic. Moreoverso each pairwise product is elliptic. Serre lemma for tree actions implies that the fixed subtrees of each pair intersect. Convex subtrees of a tree have the Helly property, soSince $a,b,c$ generate $G/⟨x,y⟩$, $G$ fixes a vertex of $T^{Z(G)}$. Thus the action of $G$ on $T$ is trivial in the tree-action sense.

A hyperbolic isometry of a tree has a unique invariant axis of a tree isometry, on which it acts by a nonzero translation. Let $L$ be the axis of $x$. By part (c), $x$ is central, so for every $g\in G$,Thus $gL$ is another axis of $x$. Uniqueness gives $gL=L$, and hence the whole group $G$ preserves the line $L$.

The equality $x=a^2$ and the relator $[b,x]=1$ show that $x$ commutes with both $a$ and $b$. Since $c=(ab{)}^{-1}$, it also commutes with $c$, and then with $y=b^3$. Hence $x\in Z(G)$. Similarly, $y=b^3$ commutes with $b$ and, by $[a,y]=1$, with $a$; it therefore commutes with $c$ and $x$. Thus

Quotienting by $⟨x,y⟩$ givesA non-elementary Fuchsian group has trivial center: two hyperbolic elements with different pairs of boundary fixed points have only the identity in their common centralizer in ${\mathrm{PSL}}_2(\mathbb{R})$. Therefore the image in the quotient of every element of $Z(G)$ is trivial. It follows that

Letbe the orientation-preserving $(2,3,7)$ hyperbolic triangle group. DefineEvery defining relator of $G$ maps to the identity, and $A,B,C$ lie in the image, so this is a surjective group homomorphism. Since$\Delta (2,3,7)$ is a non-elementary Fuchsian group.

Write the generators additively in the abelianization. The relators giveandAfter substitution, the last relation becomesThe commutator relators disappear automatically, sobecause $\gcd (5,4)=1$. Explicitly, an isomorphism to $\mathbb{Z}$ sends

Use the amalgam from part (c), with edge groupAn odd power of $a$ belongs to $A\setminus C$, because $C$ is the index-two translation subgroup of the Klein bottle group $A$. Similarly, an odd power of $c$ belongs to $B\setminus C$. Thusis a reduced alternating word whose syllables lie in $A\setminus C$ and $B\setminus C$. The normal form theorem for an amalgamated free product says that every nonempty reduced alternating word is nonidentity. The displayed element is therefore nontrivial for every $n\ge 1$.

Introduce $d=c^2$ and $e=a^2$. The two vertex groupsare Klein bottle groups. In $A$, the subgroup $⟨a^2,d⟩$ is ${\mathbb{Z}}^2$ of index two; in $B$, the subgroup $⟨e,c^2⟩$ is also ${\mathbb{Z}}^2$ of index two. IdentifyinggivesEliminating $d,e$ from this amalgamated free product recovers exactly the two given relators.

The Bass-Serre tree is bipartite with vertex sets $G/A$ and $G/B$ and edge set $G/⟨a^2,c^2⟩$. Both edge-group inclusions have index two, so every vertex has degree two. The tree is therefore a bi-infinite line, with $A$- and $B$-type vertices alternating.

The relation isso $K$ is the Klein bottle group. Let it act on ${\mathbb{R}}^2$ byThese are Euclidean isometries and satisfy $ab{a}^{-1}=b^{-1}$. Every element has a normal form $b^m{a}^n$. The orbit of $(0,0)$ is discrete, and a rectangle of finite size meets every orbit, so the action is proper and cocompact.

The squareis a vertical translation, while $b$ is a horizontal translation. They commute, andas an isometry only when $m=n=0$. Hence $⟨a^2,b⟩\cong {\mathbb{Z}}^2$. The normal form shows that every element lies in either $⟨a^2,b⟩$ or $a⟨a^2,b⟩$, so this subgroup has index two in $K$.

Choose a finite generating set $h_1,\dots ,h_r$ for $H$. Under the standard classification of connected covering spaces, each $h_i$ is represented by a based combinatorial loop ${\gamma }_i$ in $Y$. Let $Z$ be the union of the images of these finitely many finite edge paths. Then $Z$ is a finite connected subgraph containing $y_0$.

Part (b) makes the inclusion-induced mapinjective. Its image contains every $h_i$, because every ${\gamma }_i$ lies in $Z$, and therefore contains the subgroup they generate, namely all of $H$. The map is consequently an isomorphism.

Whenever a combinatorial loop $\alpha$ traverses an oriented edge and immediately traverses the same edge backwards, delete that backtracking pair. Each deletion is a homotopy relative to endpoints inside $Z$ and reduces the edge length by two, so the process terminates at a reduced, hence locally injective, combinatorial loop $\beta$.

The universal cover $\tilde{Y}$ of a connected graph is a tree. The lift $\tilde{\beta }$ is also locally injective because a covering map is a local graph isomorphism. A locally injective edge path in a tree cannot repeat a vertex: the segment between two successive visits would be a nonempty reduced closed path, whereas every closed path in a tree backtracks. Thus $\tilde{\beta }$ is injective unless $\beta$ is constant.

Now let a loop in $Z$ become null-homotopic in $Y$. Its reduced representative $\beta$ lifts to a closed path in $\tilde{Y}$. The preceding injectivity forces that lift, and hence $\beta$, to be constant. The original loop is null-homotopic in $Z$, proving thatis an injective group homomorphism.

Suppose for a contradiction that $Y\times \mathbb{R}$ carries a complete Ricci-flat metric. Since $Y$ is closed, the two subsets $Y\times (A,\infty )$ and $Y\times (-\infty ,-A)$ are different unbounded components outside the compact set $Y\times [-A,A]$. Thus $Y\times \mathbb{R}$ is disconnected at infinity and, by part (a), contains a line in a Riemannian manifold.

Its Ricci curvature is zero, so the Cheeger-Gromoll splitting theorem gives an isometryThe product Ricci tensor shows that $N$ is a complete three-dimensional Ricci-flat manifold. By the allowed fact, $N$ is flat, and hence so is $Y\times \mathbb{R}$.

The universal cover of a complete flat manifold is complete, simply connected, and has zero sectional curvature. The Hadamard-Cartan theorem therefore identifies it diffeomorphically with ${\mathbb{R}}^4$, so it is contractible. On the other hand, the universal cover of the product iswhich deformation retracts onto $\tilde{Y}$. It is contractible only if $\tilde{Y}$ is contractible, contrary to the hypothesis. Hence no such complete Ricci-flat metric exists.

The Cheeger-Gromoll splitting theorem states that a complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line in a Riemannian manifold is isometric to a Riemannian product

The Hadamard-Cartan theorem states that if a complete simply connected Riemannian manifold has nonpositive sectional curvature, then for every point $p$ its exponential mapis a diffeomorphism. In particular, the manifold is diffeomorphic to Euclidean space and is contractible.

A line in a Riemannian manifold is a unit-speed geodesic $\gamma :\mathbb{R}\to M$ that minimizes globally:for all $s,t\in \mathbb{R}$. A connected noncompact manifold is disconnected at infinity if some compact set $K$ has a complement with at least two unbounded connected components.

Choose points $p_j$ and $q_j$ in two such components withThe Hopf-Rinow theorem supplies a length-minimizing geodesic ${\gamma }_j$ from $p_j$ to $q_j$. Its image must meet $K$, since otherwise it would connect the two different components of $M\setminus K$. Reparametrize so that ${\gamma }_j(0)=x_j\in K$. After taking a subsequence, compactness gives $x_j\to x\in K$ and the unit tangent vectors ${\dot{\gamma }}_j(0)$ converge to some unit $v\in T_{x}M$.

Both endpoint parameters tend to infinity because their distances from $K$ do. Smooth dependence of geodesics on initial data therefore makes ${\gamma }_j$ converge on every compact parameter interval to the complete geodesicEvery finite segment of every ${\gamma }_j$ minimizes length. Passing to the limit gives $d_g(\gamma (s),\gamma (t))=|s-t|$, so $\gamma$ is a line.

On $p$-forms in dimension $n$, the defining identity for the Hodge star operator givesFor $n=4$ and $p=2$, therefore, ${\ast }^2=1$. For every $\alpha \in {\Omega }^2(N)$ defineThen $\ast {\alpha }_{+}={\alpha }_{+}$, $\ast {\alpha }_{-}=-{\alpha }_{-}$, and $\alpha ={\alpha }_{+}+{\alpha }_{-}$. The two eigenspaces of the involution $\ast$ have zero intersection, which proves uniqueness. They are respectively the spaces of self-dual and anti-self-dual two-forms.

Now suppose $N$ is compact and let $\beta$ be an exact three-form, say $\beta =d\theta$. Apply the Hodge decomposition theorem to the two-form $\theta$:Set $a=\delta \psi$. Then $da=d\theta =\beta$ and $\delta a={\delta }^2\psi =0$. For a two-form in dimension four, $\delta =-\ast d\ast$, so $d\ast a=0$. The self-dual formsatisfiesThus every exact three-form is the exterior derivative of a self-dual two-form.

Yes to both questions. Since ${\omega }_g$ has top degree, $d{\omega }_g=0$. Moreover $\ast {\omega }_g=1$, so the formula for the codifferential givesThereforeand the Riemannian volume form is a harmonic differential form.

The Levi-Civita connection preserves both the Riemannian metric and its chosen orientation. At any point, extend a positively oriented orthonormal basis to a local frame whose covariant derivatives vanish at that point. Differentiating ${\omega }_g(e_1,\dots ,e_n)=1$ there gives $\nabla {\omega }_g=0$. Hence ${\omega }_g$ is a parallel differential form.