I'm trying to piece together a proof of the projective bundle formula from several incomplete sources. Here's the statement I'd like to prove:

Projective bundle formula: Let $\pi: E \to X$ be a vector bundle of rank $r$ over a compact space $X$. Let $\mathbb{P}(E)$ be the projective bundle of $E$ with distinguished line bundle $H$. Then we have the following formula: $K(\mathbb{P}(E))=K(X)[H]/( \sum^r_{k=0}(-1)^k [H]^k[\bigwedge^k E] )$.

At this point I feel like I have all the ingredients for the proof but I can't seem to piece them together. Here's what I have:

There's a canonical Kozul complex over $E$:

$$0 \to \mathbb{C} \to E \to \bigwedge^2 E \to \dots \to \bigwedge^r E \to 0$$

Over each point $e \in E$ the maps are given by wedge multiplication $(-) \wedge e$. Therefore this complex is exact outside of the zero section and gives us the thom class in $\lambda_E \in K(E,E-0)=\tilde{K}(Th(E))$. What's a neat way of proving that $K(E,E-0)$ is a free rank 1 module generated by $\lambda_E$?

Suppose wer'e past that. Observe that there's a cofiber sequence

$$\mathbb{P}(E) \to \mathbb{P}(E \oplus \mathbb{C}) \to Th(E)$$

This suggests that the pullback of $\lambda_E$ to $\mathbb{P}(E \oplus \mathbb{C})$ might tell me the relavant information. Two things are not clear to me here.

  1. Why is the pullback of $\lambda_E$ to $\mathbb{P}(E \oplus \mathbb{C})$ equal to $\sum^r_{k=0}(-1)^k [H]^k[\bigwedge^k E]$? (Twisting the kozul complex by powers of $[H]$ must have some geometric interpratation... no?)
  2. It seems rather implausible that for every cofiber sequence $X \to Y \to C$ one has $K(X) = K(Y)/Im(K(C)\to K(Y))$. So can we really justify and prove the formula using these ingredients? Maybe there's an inductive step here that might help simplify the statement?

I apologize if there's a well known source that discusses this in detail. The very reason I'm here is I didn't find one.

  • $\begingroup$ I know it is not the proof you're trying to get, but the usual proof uses the Atiyah-Hirzebruch-Serre spectral sequence for K-theory (or, if you prefer, the Leray-Hirsch theorem) $\endgroup$ – Denis Nardin Apr 14 '16 at 22:12
  • $\begingroup$ @DenisNardin Thanks for the suggestion. Do you think the kind of proof i'm aiming at is possible? $\endgroup$ – Saal Hardali Apr 14 '16 at 22:14
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    $\begingroup$ I think it is likely, although I cannot see all the details (btw for point 2 you might want to use the long exact sequence in K-theory from the cofiber sequence together with the surjectivity of some map). Also the best way to prove the Thom isomorphism is, in my opinion, the proof in Milnor-Stasheff. It's simple short and easy to remember $\endgroup$ – Denis Nardin Apr 14 '16 at 22:20
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    $\begingroup$ Over $\mathbb P(E)$ your Koszul complex starts with $\mathbb C\hookrightarrow E\otimes H$, that's why you have powers of $[H]$. $\endgroup$ – Marc Hoyois Apr 14 '16 at 22:51
  • $\begingroup$ @MarcHoyois what does this map do over a point $l_x \in P(E)$? (Line in $E_x$) $\endgroup$ – Saal Hardali Apr 14 '16 at 22:59

First, for any bundle $V$ of dimension $d$ over $Y$ put $$ \lambda(V)(t) = \sum (-1)^k[\Lambda^k(V)]t^{d-k} \in K^0(Y)[t]. $$ This is a monic polynomial of degree $d$ over $K^0(Y)$. It satisfies $\lambda(L)(t)=t-[L]$ if $L$ is a line bundle, and $\lambda(A\oplus B)(t)=\lambda(A)(t)\lambda(B)(t)$. Thus, if $L$ is isomorphic to a subbundle of $V$ then $V\simeq L\oplus W$ for some $W$ and we find that $\lambda(V)([L])=0$.

Let $p\colon \mathbb{P}(E)\to X$ be the obvious projection. For any $U\subseteq X$, put \begin{align*} A^*(U) &= K^*(U)[t]/\lambda(E|_U)(t) \\ B^*(U) &= K^*(p^{-1}(U)) \end{align*} The bundle $H$ over $\mathbb{P}(E)$ is tautologically a subbundle of $p^*E$, and using this we obtain a ring map $A^*(X)\to B^*(X)$ sending $t$ to $[H]$. Essentially the same construction gives maps $\phi_U\colon A^*(U)\to B^*(U)$ for all $U$. Recall also that $\lambda(E)(t)$ is a monic polynomial of degree $d$. It follows that the set $T=\{1,t,\dotsc,t^{d-1}\}$ is a basis for $A^*(U)$ over $K^0(U)$.

Now suppose we have open subsets $U$ and $V$. There is a Mayer-Vietoris sequence relating the $K^*$ groups of $U$, $V$, $U\cup V$ and $U\cap V$. Using the basis $T$, we obtain a long exact sequence relating the $A^*$ groups of $U$, $V$, $U\cup V$ and $U\cap V$. The Mayer-Vietoris sequence for $p^{-1}(U)$ and $p^{-1}(V)$ gives another long exact sequence relating the $B^*$ groups of $U$, $V$, $U\cup V$ and $U\cap V$. The maps $\phi$ link these two long exact sequences. They are obviously compatible with ther restriction maps $A^*(U)\to A^*(U\cap V)$ and so on, but a little work is needed to check that they are also compatible with the connecting morphisms.

Now put $\mathcal{U}=\{U\;|\;\phi_U \text{ is iso } \}$. If $U$ is contractible then $E|_U\simeq U\times\mathbb{C}^d$ and the standard calculation of $K^*(\mathbb{C}P^{d-1})$ shows that $U\in\mathcal{U}$. If $U$, $V$ and $U\cap V$ lie in $\mathcal{U}$ then the Mayer-Vietoris sequences together with the five lemma show that $U\cup V\in\mathcal{U}$. Now suppose that $X$ can be covered by open sets $U_1,\dotsc,U_n$ such that all intersections $U_{i_1}\cap\dotsb\cap U_{i_r}$ are empty or contractible. Then one can check by induction on $n$ that $\phi_X$ is iso. If $X$ is a finite simplicial complex then the open stars of vertices provide a cover of the required type, so $\phi_X$ is iso. The case of a completely general $X$ follows by homotopy invariance and a limit argument.

  • $\begingroup$ That's amazing. So much thanks!. This method seems to suggest that $A^*$ and $B^*$ are complexes of sheaves while the mayer vietoris sequence is the one you obtain from the gluing condition. Now $\phi$ is just a map of complexes of sheaves. After this is established we can conclude that $\phi$ is isomorphism by observing that it's an isomorphism on stalks (By local contractibility for a suitably nice space). Is this a valid way of looking at this? Or am I missing some technical point? $\endgroup$ – Saal Hardali Apr 15 '16 at 10:36
  • $\begingroup$ @SaalHardali I don't think that $A^*$ and $B^*$ are sheaves (rather, they are homotopy groups of sheaves of spectra), so if you don't want to throw spectra in the mix you need to work with the M-V sequence $\endgroup$ – Denis Nardin Apr 15 '16 at 12:10
  • $\begingroup$ @DenisNardin My problem here is this seems to only prove that $A^*$ and $B^*$ are degree wise isomorphic as graded groups not as rings. Does it follow that the ring structure is the same? By the way It seems to me that the method indicated implies that $A^*$ and $B^*$ are complexes of sheaves $\endgroup$ – Saal Hardali Apr 15 '16 at 12:34
  • $\begingroup$ @Saal Well, the map $A^*(U)\to B^*(U)$ is a map of rings by definition, so if it is an isomorphism of graded groups it is also an isomorphism of rings $\endgroup$ – Denis Nardin Apr 15 '16 at 12:35
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    $\begingroup$ @Saal I don't see any such differential. If you want to discuss this further drop by the homotopy theory chat since the software is scowling at me as we speak :). $\endgroup$ – Denis Nardin Apr 15 '16 at 12:41

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