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The ordinary Thom isomorphism says $H^{*+n}(E,E_{0}) \simeq H^{*}(X)$, where $E$ is a vector bundle over $X$ and $E_{0}$ is $E$ minus the zero section. Now assume that $S$ is a non vanishing section for the vector bundle $E$. In each fiber $E_{x}$ we remove two points $0_{x}$ and $S(x)$. Then we put $E_{0,1}$for the union of all 2-points punctured fibers.

Motivating by the ordinary Thom isomorphism, my question is

What should be a relevant right side of the following equality(equivalency):

\begin{equation} H^{*+n}(E,E_{0,1}) \simeq \;? \end{equation}

What should be a generalized Thom class?

Does this right side depend on choosing a particular non vanishing section $S$?

It is obvious that we can generalize the main question to multi- point punctured fibers. That is, assume we have m sections $S_{1},\ldots ,S_{m}$ such that we have m distinct vector $S_{1}(x),\ldots,S_{m}(x)$. We remove these m ponts from each fiber $E_{x}$we denote the resulting total space by $E_{1,2,\ldots m}$. We search for a relevant right side for:

\begin{equation} H^{*+n}(E,E_{1,2, \ldots, m})=? \end{equation}

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2 Answers 2

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You get two copies of $H^*(X)$. In general ($k$ pairwise disjoint sections), by excision it's just like disjoint union of $k$ copies of the original bundle, hence $k$ copies of $H^*(X)$. (An extra observation is that, for one section, the result does not depend on its choice, as any section is homotopic to $0$.)

There's no generalized Thom class: the usual one would do. You can regard it as a class in $H^n(E,E\setminus D)$, where $D$ is a disk bundle containing all the sections. This class restricts to $H^n(E,E_{\text{whatever}})$ in your notation.

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  • $\begingroup$ Could you please more explain. What you are considering for excision? $\endgroup$
    – Ali Taghavi
    Commented Mar 6, 2014 at 23:07
  • $\begingroup$ "For one section", is not necessary that we care about the following situation:Assume $s$ and $t$are two non vanishing section.Is there a homotopy of non vanishing sections which connects $s$ to $t$? $\endgroup$
    – Ali Taghavi
    Commented Mar 6, 2014 at 23:21
  • $\begingroup$ Sorry, I was away. "Excision" means the excision theorem/axiom; the standard wording is $H_*(X,A)=H_*(X\setminus B,A\setminus B)$ if $\operatorname{closure}(B)\subset\operatorname{interior}(A)$. Any section is homotopic to the zero section: $h(x,t)=ts(x)$; ergo, any two sections are homotopic. $\endgroup$
    – Alex Degtyarev
    Commented Mar 6, 2014 at 23:28
  • $\begingroup$ Thanks, My question was how you apply the excision axiom in the case of my question, that is what is your $B$. Morover could you please answer to last my previous comment? $\endgroup$
    – Ali Taghavi
    Commented Mar 6, 2014 at 23:32
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    $\begingroup$ OK, it is not true that you can homotope nonvanishing sections through nonvanishing sections. But it is not needed. By excision, in your notation, $H(E,E_0)=H(D,D_0)$, where $D$ is a "tubular neighborhood" (disk bundle) about the zero section. The same applies to any section; moreover, you can choose these neighborhoods $D_i\supset s_i$ disjoint, so that $H^*$ splits into direct sum. Thus, it remains to discuss one section only (or, rather, one at a time), and then it doesn't matter if it is nonvanishing: just homotope it to zero. $\endgroup$
    – Alex Degtyarev
    Commented Mar 6, 2014 at 23:39
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You can see from the special case when $X$ is a point that what you proposed cannot work.

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  • $\begingroup$ Yes the particular case X=point shows that the "right side" can not depend only on $X$, but it depend also on the number of deleted points. So i am trying to underestand the first answer to my question.In fact my question: Does my propose leads to triviality? $\endgroup$
    – Ali Taghavi
    Commented Mar 6, 2014 at 23:10
  • $\begingroup$ This question goes nowhere. Have a look at the book by Bott and Tu Differential forms in algebraic topology. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 6, 2014 at 23:22
  • $\begingroup$ at this time the book is not available to me. do you remember a particular statment or proposition in the book related to my question? $\endgroup$
    – Ali Taghavi
    Commented Mar 7, 2014 at 0:05
  • $\begingroup$ Then read csrefully any proof of the Thom isomorphism theorem to understand why your question is ill posed. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 7, 2014 at 11:55
  • $\begingroup$ @LiviuNicolaescu: Could you please explain your objection to the question? As far as I see it, Alex Degtyarev's answer describes what happens (even when $X$ is a point). The question has an easy answer, but one which I can imagine being useful to know, and it seems to be a genuine generalization of the Thom isomorphism. $\endgroup$
    – Mark Grant
    Commented Mar 9, 2014 at 9:40

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