Choose a finite symmetric generating set for , a base point , and put . If the word length of is at most , the triangle inequality and invariance of Riemannian distance under isometries give .
A properly discontinuous group action has finite point stabilizer and a discrete orbit. More explicitly, properness on compact sets shows that only finitely many elements move into any bounded ball, since closed bounded balls are compact by the Hopf-Rinow theorem. Thus there is such that distinct orbit points have distance more than : choose it using the finitely many orbit points within a fixed small ball of , and translate the resulting separation by isometries. If the whole orbit is one point, the group itself is finite and the conclusion is immediate.
Let be the number of distinct orbit points represented by the word metric ball of radius . Their radius- balls are disjoint, have equal positive volume , and lie inside . The allowed Bishop-Gromov inequality for nonnegative Ricci curvature states that is nonincreasing for . ThereforeEach orbit point has at most preimages inside the word ball, since its full fiber is a coset of . For the growth function of a finitely generated group, this givesHence has polynomial growth of a group of degree at most . The finite stabilizer factor makes the proof valid even when the properly discontinuous action is not free. This is polynomial growth of properly discontinuous isometry groups.
Let . A compact connected smooth manifold has finite CW complex homotopy type, hence a finitely generated fundamental group. Its universal cover with the lifted Riemannian metric is complete by the lifted-geodesic argument in part 2(b) and has nonnegative Ricci curvature. The deck transformations act freely and properly discontinuously by isometries. Part (a) givesThe degree-one Hurewicz theorem identifies with the abelianization of . This finitely generated abelian group has free rank , so there is a surjection . Choose lifts of a basis and let bound their word lengths. For every with , the product has word length at most and maps to that vector. Distinct vectors therefore give distinct group elements. The integer cube shows, for , thatfor some and all sufficiently large . Comparing with the degree- upper bound forces ; the case is immediate. ThusThis is the first Betti-number bound from polynomial fundamental-group growth.
The standard connectedness convention for a Riemannian manifold is needed here. If disconnected manifolds are allowed, the bound applies to each connected component; a disjoint union of two flat -tori has first Betti number and contradicts a global bound by .
Consider the orientation-preserving affine group of the real line, realized asThe coordinates identify it smoothly with , so this Lie group is connected. Choose the Lie algebra element and the group element :Direct matrix multiplication gives .
If a bi-invariant Riemannian metric existed, every conjugation map would be an isometry fixing the identity. Its differential there, the Adjoint representation of a Lie group, would preserve the positive-definite inner product at the identity. Thusa contradiction since . Therefore this connected Lie group admits no bi-invariant Riemannian metric. This is the adjoint dilation obstruction to a bi-invariant Riemannian metric, an elementary alternative permitted by the question. The positive definite Riemannian metric interpretation matters; no claim about indefinite metrics is needed.
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