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.

1$\begingroup$ What happens for tori? There it is known that the Borel conjecture holds, but is it clear that every smooth manifold homeomorphic to $T^n$ is parallelizable? $\endgroup$ – Jens Reinhold Nov 28 '20 at 11:03

3$\begingroup$ As Wall says in his book (section "Fake tori") all PL homotopy tori are parallelizable. Another (perhaps more detailed) reference is HsiangShaneson "Fake tori". $\endgroup$ – Igor Belegradek Nov 28 '20 at 14:40

$\begingroup$ @IgorBelegradek Does this somehow imply they are parallelizable in the usual sense? $\endgroup$ – Connor Malin Nov 28 '20 at 14:42

5$\begingroup$ @ConnorMalin: yes, I think this can be deduced from (i) T^n splits stably as a wedge of spheres, and (ii) the map $BO \to BPL$ induces an injection on homotopy groups. $\endgroup$ – user169545 Nov 28 '20 at 14:54
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 9manifold 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 3manifold with nontrivial 7torsion in $H_1$. Even more concretely, the 3manifold 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 7torsion in $H^8(X;\mathbb{Z})$ agrees with the 7torsion 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 KervaireMilnor 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 7torsion elements. $\Box$

$\begingroup$ Thanks for the response, I see that the calculation of the exotic spheres gives that $\pi_* (PL/O) \otimes \mathbb{Z}_{(7)}$ agrees with $\pi_*(K(\mathbb{Z}/7,7)) \otimes \mathbb{Z}_{(7)}$ in a range, but I can't see why this is realized by a map $PL/O \rightarrow K(\mathbb{Z}/7,7)$ before localizing at 7. Could you say what this map is? $\endgroup$ – Connor Malin Nov 28 '20 at 15:51

2$\begingroup$ Yes, it is the composition of the localization map $PL/O \to (PL/O)_{(7)}$ and the Postnikov truncation $(PL/O)_{(7)} \to \tau_{\leq 7} (PL/O)_{(7)}$. The fact that $\tau_{\leq 7} (PL/O)_{(7)}$ has only one nonzero homotopy group $\pi_7 \cong \mathbb{Z}/7\mathbb{Z}$ implies that it is weakly equivalent to the EilenbergMacLane space, finally compose with such a weak equivalence. $\endgroup$ – user169545 Nov 28 '20 at 15:55

3$\begingroup$ Another omitted detail is why the composition is $\beta$ and not zero (these are the only options, up to rechoosing the identification of $\pi_7(PL/O)$). I think that's well known, but otherwise it can be deduced from the fact that an almostparallelizable closed 8manifold must have signature divisible by 7 if it is smoothable, but can be nonzero mod 7 in general. $\endgroup$ – user169545 Nov 28 '20 at 16:50