Critical dimensions D for "smooth manifolds iff triangulable manifolds"

I am aware that at least for lower dimensions,

"smooth manifolds iff triangulable manifolds"

at least for dimensions below a certain critical dimensions D.

My question is that for

• For orientable manifolds, in which lower dimensions $$\leq$$ D, such that "smooth manifolds iff triangulable manifolds" is true, but above that dimensions > D is false? (So that "smooth manifolds may not be triangulable manifolds.")

Consider:

1. orientable SO-manifolds with SO(D) co/bordism structure.

2. orientable Spin-manifolds with Spin(D) co/bordism structure.

• For non-orientable manifolds, in which lower dimensions $$\leq$$ D, such that "smooth manifolds iff triangulable manifolds" is true, but above that dimensions > D is false? (So that "smooth manifolds may not be triangulable manifolds.")

Consider:

1. non-orientable O-manifolds with O(D) co/bordism structure.

What are their critical dimensions D?

In this post, we learn:

"All orientable 5-dimensional manifolds are triangulable." "In 6 dimensions, there are non-triangulable orientable manifolds."

Are these referred to topological manifolds? Or smooth manifolds?

However, I heard an alternative view from a geometry topologist that (maybe standard viewpoint to some of you, but I apologize for my background ignorance):

"All smooth manifolds are uniquely triangulable. No critical dimensions D constraint or orientability constraints."

p.s. Your correct answers are more important. References (the proofs) are secondary but very helpful.

Many thanks! Really appreciate your holiday time answer and inspiration!

1 Answer

All smooth manifolds are triangulable, as you say. This follows from Morse theory, which dictates that you only need to know how to triangulate (PL) handle-attachments, which one can do by hand. The result, though not phrased then in terms of Morse theory, has been known since the 30s. So in my post I meant "topological manifold" when I said "non-triangulable manifold", because that is the only thing that could have made sense!

For dimensions $$D \leq 4$$ any triangulable manifold is smooth with no further requirement.

For any dimension $$D \geq 5$$ there are spin manifolds which are triangulable, but do not admit any smooth structure.

So the answer to your title question is $$D \leq 4$$, and no reasonable assumption on $$M$$ for $$D \geq 5$$ is likely to give you a similar result in higher dimensions. Certainly spin is not enough.

As for "triangulable iff smooth". The standard statement is that the notion of PL and smooth structures are equivalent in dimensions up to 6 (in dimension 7, every PL structure may be smoothed, but not uniquely). See here for a good discussion.

A PL manifold can be thought of as a topological manifold $$X$$ equipped with a(n equivalence class of) triangulation so that the link of any vertex is PL-homeomorphic to $$S^{\dim X - 1}$$.

In dimension 4 it is already difficult to prove that "triangulable and PL-able" are equivalent. In any dimension, it is not hard to show that $$\text{lk}(\sigma)$$ is a homology manifold using the cone-homeomorphism from $$\text{lk}(\sigma) \times \Bbb R$$ to an open subset of a topological manifold; in particular, $$\text{lk}(\sigma) \times \Bbb R$$ is a topological manifold. It is similarly not hard to show that $$\text{lk}(\sigma)$$ must have the same homotopy type as $$S^{\dim M - 1}$$. In dimensions $$\leq 2$$, homology manifolds are in fact manifolds (Theorem 16.32 in Bredon's sheaf theory) and topological manifolds admit unique PL structures, so we see that there is a PL homeomorphism $$\text{lk}(\sigma) \cong S^{\dim M - 1}$$. In particular, any triangulation of a manifold of dimension at most $$3$$ equipped is automatically PL.

In dimension 4 the same is true but one needs to do more work: now you need to use the fact that $$\text{lk}(\sigma)$$ is itself a simplicial complex. A theorem of Edwards (theorem 3.5 here) then indicates that $$\text{lk}(\sigma)$$ is a manifold, and hence by the Poincare conjecture that it is homeomorphic to $$S^3$$.

What you want now is an example, in each dimension $$n \geq 5$$, of a triangulable manifold $$X_n$$ which is not PL (and hence not smooth). This is discussed on MO here, citing Rudyak: if $$X_4$$ is Freedman's E8 manifold, then $$X_{4+k} = X_4 \times T^k$$ is triangulable, but not PL. This is proved by a sort of dimensional reduction for both parts: in Rudyak's Theorem 7.4, he argues that none possess a PL structure by passing to the universal cover $$\widetilde{X}_{4+k} = X_4 \times \Bbb R^k$$. The Kirby-Siebenmann product theorem (relating PL structure sets on $$M$$ and $$M \times \Bbb R$$) states that this carries a PL structure for any $$k \geq 1$$ if and only if $$\widetilde X_5$$ does. Because PL 5-manifolds carry smooth structures, $$\widetilde X_5$$ is smoothable; one then argues by constructing a bordism between $$X_4$$ and a smooth spin manifold with the same signature, which is impossible by Rokhlin's theorem. Therefore no $$\widetilde X_n$$ is PL for $$n \geq 4$$, and hence neither is any $$X_n$$ for $$n \geq 4$$.

To see that $$X_n$$ is triangulable for $$n \geq 5$$, see Rudyak's Theorem 21.5: that every compact orientable 5-manifold is triangulable is a theorem of Siebenmann. Then $$X_{5 + k} = X_5 \times T^k$$ is also triangulable, being a product of triangulable manifolds.

Here is a reasonable approach one might try to see that any orientable 5-manifold is triangulable, but I think it is circular, or at least circuitous.

That any orientable 5-manifold is triangulable might also follow (but see below) from Galewski-Stern. One of the relevant theorems is that if $$0 \to \text{ker}(\mu) \to \Theta \xrightarrow{\mu} \Bbb Z/2 \to 0$$ is the short exact sequence with $$\Theta$$ the 3-dimensional homology cobordism group and $$\mu$$ the Rokhlin homomorphism (take a spin manifold $$X$$ that a homology 3-sphere $$\Sigma$$ bounds; then $$\mu([\Sigma]) = \sigma(X)/8 \mod 2$$), then if $$\Delta(M) \in \Bbb H^4(M; \Bbb Z/2)$$ is the Kirby-Siebenmann class and $$\beta_\Theta: H^*(M;\Bbb Z/2) \to H^{*+1}(M; \text{ker}(\mu))$$ the associated Bockstein map, then a closed manifold $$M$$ of dimension $$n \geq 5$$ is triangulable if and only if $$\beta_\Theta \Delta(M) = 0 \in H^5(M;\text{ker}(\mu))$$.

(Manolescu's [2013 contribution] was essentially that $$\beta_\Theta$$ is not identically zero, which is equivalent to the group theoretic statement that $$\mu: \Theta \to \Bbb Z/2$$ has no section.)

Any short exact sequence $$0 \to H \to G \to K \to 0$$ gives rise to a long exact sequence on cohomology (with boundary map the Bockstein). If $$M$$ is an oriented closed manifold of dimension $$n$$, then the end of this sequence is precisely $$H^{n-1}(M;K) \xrightarrow{\beta} H^n(M; H) \to H^n(M; G) \to H^n(M; K) \to 0;$$ because $$H^n(M; A) \cong A$$ naturally for an oriented closed $$n$$-manifold, the end of this sequence is precisely our original short exact sequence $$H \to G \to K \to 0$$; by exactness, we see that $$\beta: H^{n-1}(M; K) \to H^n(M; H)$$ is identically zero.

In particular, if $$M$$ is a closed oriented 5-manifold, we must have $$\beta_\Theta \Delta(M) = 0 \in H^5(M; \text{ker} (\mu))$$. Therefore, $$M$$ is triangulable.

But... the Galewski-Stern paper relies on the Siebenmann paper in which compact oriented 5-manifolds are shown more quickly to be triangulable.

• thanks +1, for the excellent answers + examples! Dec 28, 2018 at 1:20
• This is great. Is there perhaps a typo that every compact orientiable 5-manifold is triangulable, not orientiable? Dec 28, 2018 at 2:28
• @GabeK Indeed, I have corrected it. That every compact orientable 5-manifold is orientable was surely known long before Siebenmann :)
– mme
Dec 28, 2018 at 2:49
• No. The iff is only true for $D \leq 4$, which requires no orientability constraint. The final part of my answer details the construction of triangulable manifolds of any dimension $D \geq 5$, which are not PL and hence not smoothable. And one direction is true: every smooth manifold of any dimension is triangulable.
– mme
Dec 28, 2018 at 22:59
• @wonderich If it clears up confusion: triangulable and PL are not the same. The latter comes with a triangulation with special properties (the link of every vertex is a sphere). But you can have non-PL triangulations. The most famous is $\Sigma^2 P$ for $P$ an integral homology sphere; the result is homeomorphic to a sphere with a non-PL triangulation. The final portion of the answer discusses manifolds which are triangulable, but so that every triangulation must have a vertex whose link is not a sphere (in fact, not a manifold).
– mme
Dec 28, 2018 at 23:44