Are $K(\pi_1,1)$ tangentially homotopy equivalent?

Question

Is it known whether any two smooth, compact manifolds $X \simeq K(\pi_1,1) \simeq Y$ are tangentially homotopy equivalent, i.e. the pullback of the tangent bundle of $Y$ along some smooth homotopy equivalence $X \rightarrow Y$ is isomorphic to the tangent bundle of $X$? I suspect this may be difficult, it does not appear stronger or weaker than the Borel conjecture, because even if the Borel conjecture were true, we could have multiple smooth structures which are not tangentially homotopy equivalent.

Answer 1

I think the answer is no: there exists a pair of aspherical closed smooth manifolds which are homotopy equivalent but not tangentially homotopy equivalent.

Claim: Let $X$ be a smooth closed oriented 9-manifold such that $p_2(TX) = 0 \in H^8(X;\mathbb{Z}) = H_1(X;\mathbb{Z})$. For any $v \in H_1(X;\mathbb{Z})$ with $7 v = 0$, there exists a smooth manifold $Y$ and a PL homeomorphism $f: X \to Y$, such that $f^*(p_2(TY)) = v$.

If $v \neq 0$, there can then be no tangential homotopy equivalence $X \to Y$, since it would have to take $p_2(TY) \neq 0$ to $p_2(TX) = 0$. To get a concrete example we can take $X$ to be the product of $(S^1)^6$ and a closed aspherical 3-manifold with non-trivial 7-torsion in $H_1$. Even more concretely, the 3-manifold can be taken as the mapping torus of the diffeomorphism of $S^1 \times S^1$ corresponding to the matrix $\begin{bmatrix}1 & 7\\0 & 1\end{bmatrix}.$

Proof of claim: The 7-torsion in $H^8(X;\mathbb{Z})$ agrees with the 7-torsion in $H^8(X;\mathbb{Z}_{(7)})$, and by smoothing theory it suffices to see that $(0,v)$ is in the image of the homomorphism $$[X,PL/O] \to [X,BO] \xrightarrow{(p_1,p_2)} H^4(X;\mathbb{Z}_{(7)}) \times H^8(X;\mathbb{Z}_{(7)}).$$ But the second map factors through an isomorphism from $[X,BO] \otimes \mathbb{Z}_{(7)}$, and in the domain we may therefore factor over $[X,PL/O] \otimes \mathbb{Z}_{(7)}$. But by the Kervaire-Milnor calculation of exotic spheres there is a map $PL/O \to K(\mathbb{Z}/7\mathbb{Z},7)$ inducing an isomorphism on homotopy groups in a large range (far beyond $9 = \dim(X)$) after tensoring with $\mathbb{Z}_{(7)}$. Furthermore, the connecting map $$H^7(X;\mathbb{Z}/7\mathbb{Z}) \xleftarrow{\cong} [X,PL/O] \otimes \mathbb{Z}_{(7)} \to [X,BO] \otimes \mathbb{Z}_{(7)} \xrightarrow{p_2} H^8(X;\mathbb{Z}_{(7)})$$ may be identified with the Bockstein homomorphism $\beta: H^7(X;\mathbb{Z}/7\mathbb{Z}) \to H^8(X;\mathbb{Z}_{(7)})$, which may in turn be identified with $\beta: H_2(X;\mathbb{Z}/7\mathbb{Z}) \to H_1(X;\mathbb{Z}_{(7)})$. But the image of that is precisely the kernel of multiplication by 7, i.e. the 7-torsion elements. $\Box$