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Oct 15, 2017 at 19:47 comment added Ali Taghavi @TylerLawson Thank you for your attention to my question. I try to understand your comment.
Oct 15, 2017 at 19:43 comment added Ali Taghavi @TomGoodwillie I was not aware of the concept "microbundle" I try to understand it and its relation to my question.Thank you for introducing it to me.
Oct 14, 2017 at 0:00 comment added Tom Goodwillie $BHomeo_\ast(\mathbb R^n)$ is a classifying space for microbundles (the Kister-Mazur Theorem).
Oct 13, 2017 at 22:30 comment added Tyler Lawson (And it certainly can't be a trivial based $\Bbb R^n$-bundle if the tangent microbundle is nontrivial.)
Oct 13, 2017 at 22:17 comment added Tyler Lawson Here are my thoughts. The tangent bundle has a zero section, so let's assume that the fibers are based. The tangent bundle as a vector bundle is classified by a map $S^n \to BGL_n(\Bbb R)$, but as a (based) bundle is classified by a map $S^n \to BHomeo_*(\Bbb R^n)$. Asking for it to be equivalent to the trivial torus bundle is asking for a lift $S^n \to \{*\} \to BHomeo_*(\Bbb T^n) \to BHomeo_*(\Bbb R^n)$ (where the second map classifies "fiberwise universal cover"). As a result I think that this is asking when the tangent bundle is a trivial $\Bbb R^n$-bundle.
Oct 13, 2017 at 21:16 comment added Tom Goodwillie Is it true that the spheres with trivial tangent bundle are the only ones with trivial tangent microbundle?
Oct 13, 2017 at 20:08 history edited Ali Taghavi CC BY-SA 3.0
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Oct 12, 2017 at 5:35 comment added Ali Taghavi @MarkGrant Thanks for your attention to my question. May you ellaborate your comment? What is "haut(T^n"?
Oct 12, 2017 at 5:30 history edited Ali Taghavi CC BY-SA 3.0
edited tags; edited title
Oct 12, 2017 at 5:22 history edited Ali Taghavi CC BY-SA 3.0
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Oct 11, 2017 at 15:36 comment added Mark Grant I was thinking about the clutching functions $S^{n-1}\to GL_n(\mathbb{R})$ of the tangent bundles, and somehow this would seem to say that they preserve a lattice. Also, torus bundles clearly have something to do with principal $GL_n(\mathbb{Z})$-bundles, since $\operatorname{haut}(T^n)=GL_n(\mathbb{Z})$. I haven't yet been able to make this argument precise.
Oct 11, 2017 at 14:48 comment added Ali Taghavi @MarkGrant I do not see why this equivalent to such reduction?
Oct 11, 2017 at 12:25 comment added Mark Grant Is this equivalent to reduction of the structure group from $GL_n (\mathbb{R})$ to $GL_n (\mathbb{Z}) $? If so then the answer to the sphere question is yes.
Oct 11, 2017 at 10:07 history edited Ali Taghavi CC BY-SA 3.0
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Oct 11, 2017 at 9:56 history edited Ali Taghavi CC BY-SA 3.0
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Oct 11, 2017 at 9:50 history asked Ali Taghavi CC BY-SA 3.0