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mme
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In high dimensions Galewski and Stern (and independently Matsumoto) proved that a manifold $M$ is triangulable iff $\beta \Delta(M)=0$. Here $\Delta \in H^4(M;\Bbb Z/2)$, and $\beta$ is the Bockstein corresponding to the short exact sequence $\ker \mu \to \Theta_{\Bbb Z} \to \Bbb Z/2$. The middle term is the homology cobordism group of 3-manifolds and the last map is the Rokhlin invariant of 3-manifolds.

These invariants are additive under connected sum. Therefore if $\beta \Delta(N)$ is nonzero, so is $\beta \Delta(M \# N)$ as long as the dimension is at least 6, and so $M \# N$ remains non-triangulable.

The only leftover mystery cases are dimensions 4 and 5. In dimension 4 triangulable manifolds are automatically PL and therefore smooth. So we see that $M \# N$ is smoothable only if $M$ and $N$ are not both definite manifolds of the same signature, but otherwise we are out of luck. In dimension 5, $M \# N$ is not triangulable if ($\beta \Delta(M \#N) = \beta \Delta(M) + \beta \Delta(N) = \beta \Delta(N) \neq 0$$M$ is but $N$ is not: $$\beta \Delta(M \#N) = \beta \Delta(M) + \beta \Delta(N) = \beta \Delta(N) \neq 0.$$ But eg $M \# M$ is always triangulable.)

In high dimensions Galewski and Stern (and independently Matsumoto) proved that a manifold $M$ is triangulable iff $\beta \Delta(M)=0$. Here $\Delta \in H^4(M;\Bbb Z/2)$, and $\beta$ is the Bockstein corresponding to the short exact sequence $\ker \mu \to \Theta_{\Bbb Z} \to \Bbb Z/2$. The middle term is the homology cobordism group of 3-manifolds and the last map is the Rokhlin invariant of 3-manifolds.

These invariants are additive under connected sum. Therefore if $\beta \Delta(N)$ is nonzero, so is $\beta \Delta(M \# N)$ as long as the dimension is at least 6, and so $M \# N$ remains non-triangulable.

The only leftover mystery cases are dimensions 4 and 5. In dimension 4 triangulable manifolds are automatically PL and therefore smooth. So we see that $M \# N$ is smoothable only if $M$ and $N$ are not both definite manifolds of the same signature, but otherwise we are out of luck. In dimension 5, $M \# N$ is not triangulable ($\beta \Delta(M \#N) = \beta \Delta(M) + \beta \Delta(N) = \beta \Delta(N) \neq 0$.)

In high dimensions Galewski and Stern (and independently Matsumoto) proved that a manifold $M$ is triangulable iff $\beta \Delta(M)=0$. Here $\Delta \in H^4(M;\Bbb Z/2)$, and $\beta$ is the Bockstein corresponding to the short exact sequence $\ker \mu \to \Theta_{\Bbb Z} \to \Bbb Z/2$. The middle term is the homology cobordism group of 3-manifolds and the last map is the Rokhlin invariant of 3-manifolds.

These invariants are additive under connected sum. Therefore if $\beta \Delta(N)$ is nonzero, so is $\beta \Delta(M \# N)$ as long as the dimension is at least 6, and so $M \# N$ remains non-triangulable.

The only leftover mystery cases are dimensions 4 and 5. In dimension 4 triangulable manifolds are automatically PL and therefore smooth. So we see that $M \# N$ is smoothable only if $M$ and $N$ are not both definite manifolds of the same signature, but otherwise we are out of luck. In dimension 5, $M \# N$ is not triangulable if $M$ is but $N$ is not: $$\beta \Delta(M \#N) = \beta \Delta(M) + \beta \Delta(N) = \beta \Delta(N) \neq 0.$$ But eg $M \# M$ is always triangulable.

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mme
mme
  • 9.6k
  • 5
  • 48
  • 73

In high dimensions Galewski and Stern (and independently Matsumoto) proved that a manifold $M$ is triangulable iff $\beta \Delta(M)=0$. Here $\Delta \in H^4(M;\Bbb Z/2)$, and $\beta$ is the Bockstein corresponding to the short exact sequence $\ker \mu \to \Theta_{\Bbb Z} \to \Bbb Z/2$. The middle term is the homology cobordism group of 3-manifolds and the last map is the Rokhlin invariant of 3-manifolds.

These invariants are additive under connected sum. Therefore if $\beta \Delta(N)$ is nonzero, so is $\beta \Delta(M \# N)$ as long as the dimension is at least 6, and so $M \# N$ remains non-triangulable.

The only leftover mystery cases are dimensions 4 and 5. In dimension 4 triangulable manifolds are automatically PL and therefore smooth. So we see that $M \# N$ is smoothable only if $M$ and $N$ are not both definite manifolds of the same signature, but otherwise we are out of luck. In dimension 5, $M \# N$ is not triangulable ($\beta \Delta(M \#N) = \beta \Delta(M) + \beta \Delta(N) = \beta \Delta(N) \neq 0$.)