# Can triangulations (or some related combinatorial structure) distinguish smooth structures on $RP^4$?

There are exotic versions of $$RP^4$$, constructed by Cappell-Shaneson, which are homeomorphic but not diffeomorphic to the standard $$RP^4$$. One way to distinguish them is via the $$\eta$$ invariant of $$Pin^+$$ Dirac operators on them, c.f. the article "Exotic structures on 4-manifolds detected by spectral invariants" by Stolz, Invent. math. 94, 147-162 (1988) (pdf here).

I was wondering if there was a known combinatorial way to distinguish the smooth structures, e.g. in the following senses:

1. Can one construct triangulations of $$RP^4$$ (e.g. via Morse theory) that must 'correspond' to one of the smooth structures?

2. If a triangulation by itself can't distinguish smooth structures, is there some additional combinatorial data that one can put on top of the triangulation to distinguish them, like the branching structure on the triangulation?

The motivation for this question is based on some papers (https://arxiv.org/abs/1610.07628, https://arxiv.org/abs/1810.05833) that construct topological invariants via state-sums on triangulations (generalizing the Crane-Yetter sum) that speculate whether exotic structures can be detected via the state-sum. So it's natural to ask whether such manifolds can even be distinguished combinatorially. And something like this could seem plausible because in 4 dimensions, every manifold is smooth iff it is triangulable.

(If low-brow answers exist, that would be nice since I don't know much about exotic manifolds.)

• In dimensions less than 7, the categories DIFF and PL are equivalent, so, yes, one can. Jun 8, 2020 at 2:43
• And the answer to this question is positive as I said in my comment: Take any two inequivalent smooth structures on $RP^4$. Triangulate these. The resulting triangulated manifolds will be PL-inequivalent. The reason is that in dimension 4 DIFF=PL. Jun 8, 2020 at 14:50
• @SamHopkins: It is semidecidable for obvious reason; however, PL non-isomorphism in dimension 4, in general, is undecidable. It does not mean that the existing semi-algorithm is practical. Jun 8, 2020 at 15:43
• Is the piecewise linear category sensitive to branching structures or just the unordered triangulation? I.e. can two PL-inequivalent manifolds share a common triangulation, but only differ in the branching structure on the triangulation?
– Joe
Jun 9, 2020 at 4:46
• @Joe: I do not know what the last question means but you should ask a new question if you have one, just make sure you explain what your terminology means (e.g. "branching structure"). Before asking, I suggest you first take a look at some standard sources on PL manifolds such as Rourke and Sanderson ("Introduction to Piecewise-Linear Topology"), so you avoid making up your own terminology in lieu of the standard notions. Jun 9, 2020 at 14:57


Given a general $$n$$-dimensional pin+ TFT $$Z'\colon\mathsf{Bord}_n(\mathrm{Pin}^+)\to\cC$$, and for a nice choice of target category $$\cC$$, there is expected to be an $$n$$-dimensional unoriented TFT $$Z\colon\mathsf{Bord}_n\to\cC$$ obtained by “summing over pin+ structures,” akin to the finite path integral in Dijkgraaf-Witten theory. For example, if $$M$$ is a closed, unoriented $$n$$-manifold and $$P^+(M)$$ denotes its set of pin+ structures,

$$Z(M) = \sum_{\mathfrak p\in P^+(M)} \frac{Z'(M, \mathfrak p)}{\#\mathrm{Aut}(\mathfrak p)}.$$

If $$Z'$$ is fully extended, and $$\cC$$ is chosen appropriately, it should be possible to define $$Z$$ as a fully extended TFT as well. At present, though, I think this has only been shown up to category number 2 (once-extended TFTs).

Moreover, it's believed that fully extended TFTs (again, for certain choices of target category $$\cC$$) can all be constructed using state sums, with input data of a triangulation. There is work of Kevin Walker on implementing this, though I don't know exactly what assumptions (e.g. choice of $$\cC$$) he works with.

Let's use this strategy to build a 4d unoriented TFT $$Z$$ which distinguishes $$\RP^4$$ from $$Q$$. Let $$\zeta := e^{i\pi/8}$$ and $$\mu_{16}\subset\C^\times$$ denote the multiplicative group of 16th roots of unity, which is generated by $$\zeta$$. The 4d pin+ $$\eta$$-invariant is a $$\mu_{16}$$-valued invariant of the Dirac operator on a pin+ 4-manifold; for the two pin+ structures on $$\RP^4$$, it takes on the values $$\zeta^{\pm 1}$$, and for the two pin+ structures on $$Q$$, it takes on the values $$\zeta^{\pm 9}$$. This is discussed in Kirby-Taylor, “Pin structures on low-dimensional manifolds”; they also show this $$\eta$$-invariant is a pin+ bordism invariant.

Freed-Hopkins show that any $$\mathrm U_1$$-valued bordism invariant $$\alpha$$ lifts to an invertible TFT $$Z'$$ such that in top dimension, $$Z'(M) = \alpha(M)$$. Such a TFT is expected to be fully extended, but so far has only been constructed down to codimension 2, with target 2-category the Morita category of superalgebras over $$\C$$. In any case, applying this to the $$\eta$$-invariant produces a 4d pin+ TFT, which will be our $$Z'$$. Summing over pin+ structures as above, we obtain a 4d unoriented TFT $$Z$$, with values

$$Z(\RP^4) = \frac{\zeta + \zeta^{-1}}{2},\qquad\quad Z(Q) = \frac{\zeta^9 + \zeta^{-9}}{2}.$$

Thus $$Z(\RP^4)$$ is a positive real number and $$Z(Q)$$ is a negative real number, so we have an (in principle) fully extended 4d unoriented TFT distinguishing $$\RP^4$$ and $$Q$$, hence which should admit a state-sum description.

• What are the physical interpretations of these TFTs? It looks like $Z'$ corresponds to the 3d time reversal invariant ("class DIII") topological superconductor? And $Z$ corresponds to a TFT obtained from it by gauging fermion parity? Nov 16, 2020 at 20:07
• @pianyon I think that is correct, yes! Nov 17, 2020 at 4:28

I'll convert my comment to an answer:

Yes, triangulations can distinguish two non-diffeomorphic smooth structures on any 4-dimensional manifold; in particular, given an exotic $$RP^4$$, there exists an exotic triangulation of topological $$RP^4$$ which is not PL-isomorphic to the standard triangulation. The reason is 2-fold:

a. The easy part is that each smooth manifold $$(M, s)$$ (regardless of its dimension) admits a compatible PL structure: One can find a smooth triangulation $$\tau_s$$ of $$M$$ whose links will be triangulated spheres.

b. The hard part is a theorem due to Kirby and Siebenmann,

Kirby, Robion C.; Siebenmann, Laurence C., Foundational essays on topological manifolds, smoothings and triangulations, Annals of Mathematics Studies, 88. Princeton, N.J.: Princeton University Press and University of Tokyo Press. V, 355 p. hbk: $$24.50; pbk:$$ 10.75 (1977). ZBL0361.57004.

that in dimensions $$\le 6$$, the categories PL and DIFF are equivalent.

In particular, if $$s_1, s_2$$ are non-diffeomorphic smooth structures on a topological manifold $$M$$ of dimension $$\le 6$$, then $$\tau_i=\tau_{s_i}, i=1,2$$, define non-isomorphic PL structures on $$M$$. Concretely, one can say that triangulations given by $$\tau_1, \tau_2$$ do not admit isomorphic subdivisions. (This property fails in dimension 7: Famously, there are 28 non-diffeomorphic smooth structures on $$S^7$$, but all PL structures on $$S^7$$ are PL-isomorphic. The other difference between DIFF and PL categories in dimensions $$\ge 7$$ is that there are PL manifolds of dimension $$\ge 7$$ which do not admit compatible smooth structures.)

Here one is working with unordered simplicial complexes. Thus, "branching structures" which one can assign (possibly after a subdivision) to triangulations $$\tau_1, \tau_2$$ are irrelevant.

• I think there is no need to quote Kirby-Siebenmann. The relevant 4-dimensional result is a theorem of Cerf. As Kirby-Siebenmann say on p.39 of their book: "It is of course a telling fault that we do not deal with compatible DIFF structures on PL manifolds of dimension $\le 4$. But the theory is rather joyless there; compatible structures exist and are unique up to concordance". And then they refer to [J.Cerf, Groupes d'automorphismes et groupes de diffeomorphismes des varietes compactes de dimension 3, Bull. Soc. Math. France 87 (1959), 319-329]. Jun 13, 2020 at 20:45
• And in higher dimensions PL/Diff smoothing theory is treated in Hisch-Mazur well before Kirby-Siebenmann. Jun 13, 2020 at 20:45

It is very hard to construct state-sum invariants that distinguish smooth structures in dimension 4, for this simple but crucial fact that is worth mentioning: if $$M$$ and $$N$$ are homeomorphic smooth 4-manifolds, it is often the case (I don't remember what condition is needed here) that $$M \#_h( S^2 \times S^2)$$ and $$N\#_h (S^2 \times S^2)$$ are diffeomorphic for some $$h$$. Therefore any combinatorial invariant where the value on $$M$$ may be deduced from that on $$M \# (S^2 \times S^2)$$ will not work. So for instance if your invariant is multiplicative on connected sums, it should vanish on $$S^2 \times S^2$$.

The most famous state-sum invariant in dimension 3 is the Turaev-Viro one, and it is multiplicative on connected sums and is almost never zero.

• It is often but not always true that homeomorphic $M$ and $N$ are $S^2\times S^2$-stably diffeomorphic. (It is true when $M$ and $N$ are oriented, by a theorem of Gompf.) One counterexample is Capell-Shaneson's fake $\mathbb{RP}^4$ and the standard $\mathbb{RP}^4$, in fact, which can be shown with Kreck's modified surgery theory, reducing the problem to the possible values they can take on in the 4th pin+ bordism group. Jun 17, 2020 at 15:47
• Aha, I didn't know that. What happens if we replace $S^2 \times S^2$ with some other fixed 4-manifold $X$, like for instance the complex projective plane? Are the connected sums $M\#X$ and $N\# X$ always guaranteed to be non-diffeomorphic for these two specific 4-manifolds $M$ and $N$? Jun 18, 2020 at 15:20
• That's a great question, and I don't know what happens in general. The trick that makes this work is that $S^2\times S^2$ is null-bordant, allowing bordism to be used to check stable diffeomorphism. So something different would have to be done for $\mathbb{CP}^2$. Jun 18, 2020 at 15:47
• For these specific $M$ and $N$, I think $X := \mathbb{CP}^2 \# (S^2\times S^2)$ works: for some $i$ and $j$, $M \#_i X$ and $M \#_j X$ are diffeomorphic. This is because $M \# \mathbb{CP}^2$ and $N \#\mathbb{CP}^2$ are $S^2\times S^2$-stably diffeomorphic: one once again uses Kreck's modified surgery theory, but $M \# \mathbb{CP}^2$ and $N \# \mathbb{CP}^2$ aren't pin+ or pin-, so the calculation takes place in the 4th unoriented bordism group, where they are bordant. Jun 18, 2020 at 15:53
• It's actually showed $RP^4 \# CP^2$ and $Q \# CP^2$ ($Q$ being the fake $RP^4$) are diffeomorphic in this article: [S. Akbulut, A fake 4-manifold, 4-Manifold Topology, Contemporary Math. 35 (1984), 75-141.]
– Joe
Jul 28, 2020 at 4:19