# Definition of $\text{Vect}_0^n(S^k)$ in Hatcher's K-Theory

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 $$\text{Vect}^n_0(S^k)$$?

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 $$x_0$$.
• 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 $\text{Vect}_0^n$. – Jahan Claes May 17 at 19:33
• 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? – Jahan Claes May 18 at 19:19
• 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? – Jahan Claes May 18 at 19:21
• 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 $h'$. – Paul Frost May 23 at 15:36