0
$\begingroup$

I'm reading Hatcher's book on K-Theory, and on pages 25-26, he talks about real vector bundles over $S^k$. He defines the object $\text{Vect}_0^n(S^k)$ as

the n dimensional vector bundles over $S^k$ with an orientation specified in the fiber over one point $x_0 \in S^{k−1}$, with the equivalence relation of isomorphism preserving the orientation of the fiber over $x_0$.

I'm interpreting this to mean that an element of $\text{Vect}_0^n(S^k)$ is a vector bundle plus a specified orientation of the fiber over some arbitrary point $x_0$ (chosen for convenience to lie on the "equator" $S^{k-1}$ of $S^k$). Another way to say this is, the trivializing functions $\phi_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times\mathbb{R}^n$ must send the oriented basis of $\pi^{-1}(x_0)$ to the standard basis of $\mathbb{R}^n$ for any $\alpha$ with $x_0\in U_\alpha$. And we consider any two vector bundles to be equivalent if they are isomorphic AND have the same specified orientation on $x_0$.

However, I'm worried this interpretation is incorrect, or there is something else I'm missing, because later he says:

The map $\text{Vect}^n_0(S^k)\rightarrow\text{Vect}^n(S^k)$ that forgets the orientation over $x_0$ is a surjection that is two-to-one except on vector bundles that have an automorphism (an isomorphism from the bundle to itself) reversing the orientation of the fiber over $x_0$, where it is one-to-one.

According to my interpretation, this is exactly backwards. I would assume that if a vector bundle has an automorphism that reverses the orientation of $x_0$, there would be TWO elements of $\text{Vect}^n_0(S^k)$ that map to it, one for each orientation of $x_0$, while if a vector bundle does NOT have an automorphism that reverses the orientation of $x_0$, then there would only be one element in $\text{Vect}^n_0(S^k)$ that maps to it.

Can someone point out where I am thinking about this incorrectly, or just provide more details on the object $\text{Vect}^n_0(S^k)$?

$\endgroup$
1
$\begingroup$

The elements of $\text{Vect}^n(S^k)$ are equivalence classes $[E]$ of $n$-dimensional vector bundles $E$ over $S^k$, where $E, E'$ are equivalent iff there exists a bundle isomorphism $\phi : E \to E'$. The elements of $\text{Vect}_0^n(S^k)$ are equivalence classes $[E,\omega]$ of pairs $(E,\omega)$ consisting of an $n$-dimensional vector bundle $E$ over $S^k$ and an orientation $\omega$ of the fiber over $x_0$ (which has two orientations $\omega^\pm$), where $(E,\omega), (E',\omega')$ are equivalent iff there exists a bundle isomorphism $\phi : E \to E'$ such that $\phi(\omega) = \omega'$.

The map $$P : \text{Vect}_0^n(S^k) \to \text{Vect}^n(S^k),\quad P:[E,\omega] \mapsto [E]$$ is obviously a surjection. We certainly have $$P^{-1}([E]) = \{ [E,\omega^+], [E,\omega^-] \} .$$ But $[E,\omega^+] = [E,\omega^-]$ iff there exists a bundle isomorphism $\phi : E \to E$ such that $\phi(\omega^+) = \omega^-$, i.e. a bundle automorphism on $E$ which reverses the orientation of the fiber over $x_0$.

Thus Hatcher is right.

$\endgroup$
  • $\begingroup$ Ok this cleared everything up. I think my issue was I was thinking about there being some intrinsic definition of "positive" orientation, but of course there is no intrinsic definition of positive orientation, there are just two DIFFERENT orientations. So orientation-reversing automorphisms are also automorphisms in $\text{Vect}_0^n$. $\endgroup$ – Jahan Claes May 17 at 19:33
  • $\begingroup$ Ok, after thinking about this a bit, o have a follow-up question. With your definition of $\text{Vect}_0^n$, is it obvious that the elements of this space are in bijective correspondence with the homotopy classes of clutching functions that take $x_0$ to a positive matrix? $\endgroup$ – Jahan Claes May 18 at 19:19
  • $\begingroup$ Naively, it seems like the clutching function described by some $h_\pm$ could be homotopic to the clutching function defined by the orientation-reversed versions of $h_\pm$, is there some reason that can't happen? $\endgroup$ – Jahan Claes May 18 at 19:21
  • $\begingroup$ Clutching fuinctions are not really related to $\text{Vect}_0^n$. Each element of $\text{Vect}^n$ corresponds to a homotopy class of clutching functions, so each element of $\text{Vect}_0^n$ corresponds to a homotopy class of clutching functions + an orientation of the fiber over $x_0$, and these two components are independent. Nevertheless I think you may be right that $[E,\omega^+] = [E,\omega^-]$ iff the clutching function $h$ for $E$ is homotopic to the orientation reversed clutching function $h'$. $\endgroup$ – Paul Frost May 23 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.