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Assume that we have a vector bundle $E$ over $S^n$.

Is there a continuous family of invertible linear maps $T_x:E_x \to E_{-x}$?

Here continuity has the obvious meaning as soon as we have trivialization for the bundle around $x$ and $-x$

There is an obvious affirmative answer for the non orientable line bundle over $S^1$, for the tangent bundle of $S^n$ and for the $2$ dimensional real vector bundle $E \to S^2 \simeq \mathbb{C}P^1$ where $E$ is the tautological complex line bundle over $\mathbb{C}P^1$.

If the answer is yes, what about if we consider $E$ as a Riemannian bundle and we require that every $T_x$ be an isometric linear map?

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There is no such family of complex linear maps for the tautological bundle on $\mathbb CP^1$. The antipodal map is orientation reversing, hence the first Chern class changes its sign. The same argument should work for the positive complex spinor bundle on any even-dimensional sphere.

The same argument also works for a real vector bundle on $S^{4k}$ with non-vanishing $k$-th Pontryagin class. Probably again the positive real spinor bundle is an example.

On the other hand, the answer to the second question is "yes" because $O(r)$ or $U(r)$ is a deformation retract of $GL(r,\mathbb R)$ or $GL(r,\mathbb C)$, respectively.

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  • $\begingroup$ Thank you for your answer. I consider the tautological bundle as a real 2 dimensional bundle over S^2. The antipodal map on $S^2 \simeq \mathbb{C}P^1$, I think, is read as the map $[z,w] \mapsto [-w,z]$. So the desired linear map is $([z,w],(x,y)) \mapsto {[-w,z],(-y,x})$. So am I mistaken to think that the map exist for $S^2$ with two dimensional real vector bundle $E=$ complex line tautological bundle? $\endgroup$
    – Ali Taghavi
    Commented Jun 3, 2017 at 10:10
  • $\begingroup$ My apology for my mistake in previous comment. The antipodal map has no fixed point but my representaion has fixed point on CP^1. But what is representaion of the antipodal map on $S^2$ in terms of coordinate of $\mathbb{C}P^2$? $\endgroup$
    – Ali Taghavi
    Commented Jun 3, 2017 at 10:26
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    $\begingroup$ I think it is $(z:w)\mapsto(-\bar w:\bar z)$. $\endgroup$
    – Sebastian Goette
    Commented Jun 3, 2017 at 10:35
  • $\begingroup$ Ah Yes Thank you. So your answer says that as complex bundle there are no such complex linear maps $T_x$ yes? $\endgroup$
    – Ali Taghavi
    Commented Jun 3, 2017 at 10:37
  • $\begingroup$ Accoring to the first line of your answer I realize that the answer to my question is positive iff and oly if $f*(E) \simeq E$ where $f$ is the antipodal map. But the complex line bundle $0(-1)$ pull back to $O(1)$, yes? $\endgroup$
    – Ali Taghavi
    Commented Jun 3, 2017 at 10:44
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You are asking if a real vector bundle of rank $k$ over $S^n$ is isomorphic to its antipodal image. If $n$ is odd, the answer is always yes because the antipodal map is homotopic to the identity of $S^n$. If $k=2$, the answer is easily seen to be yes, because the oriented $2$-bundles are classified by an integer, the Euler class. If $n$ is even, such a real vector bundle amounts to a free homotopy class of continuous maps $S^n\to BO(k)$. i.e. to a pair of elements $\alpha, \bar\alpha$ of the homotopy group $\pi_n(BSO(k))$, where $BSO(k)$ is the Grassmannian space of the oriented $k$-planes in $R^\infty$, and where $\bar\alpha$ is the image of $\alpha$ by an orientation-reversing isometry. One has $\pi_n(BSO(k))\cong\pi_{n-1}(SO(k))$. Your question amounts to ask that $\alpha=\alpha^{-1}$ or $\alpha=\bar\alpha^{-1}$. I believe that the second always holds, but I'm not sure.

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  • $\begingroup$ Is $\alpha^{-1}$ your notation for the inverse in $\pi_n(-)$? Note that the antipodal map has degree $(-1)^{n+1}$. So $\bar\alpha=\alpha^{(-1)^{n+1}}$ by Hopf's theorem of self maps of $S^n$. $\endgroup$
    – Sebastian Goette
    Commented Jun 4, 2017 at 6:05
  • $\begingroup$ @Sebastian Goette Yes, $\alpha^{-1}$ denotes the inverse in $\pi_n(-)$, and as you mention, for $n$ even, this is the same as $\alpha$ composed at the source by the antipodal map. But $\bar\alpha$ denotes something else: $\alpha$ being the homotopy class of a map $f:S^n\to SO(k)$, by $\bar\alpha$ I mean the class of $\sigma\circ f$, where $\sigma:SO(k)\to SO(k)$ is the conjugation by an element of $O(k)\setminus SO(k)$. $\endgroup$
    – Gael Meigniez
    Commented Jun 4, 2017 at 7:59
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    $\begingroup$ I see, so $\bar\alpha$ is the same bundle, but with opposite orientation? In my answer I named the Pontryagin class as an obstruction. Note that $\alpha$ and $\bar\alpha$ have the same Pontryagin classes, only the Euler class changes sign (and therefore does not give an obstruction). To see that the positive and the negative spinor bundles $S^\pm$ in dimension $4k$ have opposite nontrivial $p_k$, note that the Chern character is a multiple of $p_k$. Then apply the index theorem to the Dirac operator twisted by $S^+$ and $S^-$. $\endgroup$
    – Sebastian Goette
    Commented Jun 4, 2017 at 14:51
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    $\begingroup$ @Sebastian Goette Thank you Sebastian, so, this gives examples where $\alpha^{-1}\neq \alpha$ nor $\bar\alpha$, and where the answer to Ali's question is negative. $\endgroup$
    – Gael Meigniez
    Commented Jun 4, 2017 at 17:31
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    $\begingroup$ @GaelMeigniez Thank you again for your interesting answer. I am sorry that I can not accept two answers simultaneously. $\endgroup$
    – Ali Taghavi
    Commented Jun 6, 2017 at 14:21

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