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According to the last answer to this question. It seems there is a way, using Differential Geometry, to proof that Euler Characteristic is multiplicative on fiber bundles i.e.: $$\chi(E)=\chi(F)\cdot \chi(B)$$ where $p\colon E\to B$ is a smooth fiber bundle with fiber $F$.

My question: Could you give me a reference (article or book) for a proof of the multiplicativity of Euler characteristic on smooth fiber bundles which uses Riemanninan Geometry or Differential Geometry.

Remark: I am aware of the proof using spectral sequences and the combinatorial one.

Thanks in advance!

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  • $\begingroup$ Trying to solve your question, I ended up searching for relations between holomogy/cohomology groups in the fiber bundle context since $$\chi(X) =\sum_{i=0}^\infty (-1)^i H_i(X).$$ Do you know anything in this area? I am aware of the injection between $H_i(E)$ and $H_i(B)$, but this does not seem to help. $\endgroup$
    – alerouxlapierre
    Jun 29, 2017 at 19:30
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    $\begingroup$ As a possible route, have you thought about how you'd use the Poincaré-Hopf theorem to prove this in the case of a trivial bundle (i.e., product of two manifolds)? $\endgroup$
    – Ted Shifrin
    Jun 29, 2017 at 21:03
  • $\begingroup$ @AlexisLeroux-Lapierre this paper by Spanier along with the Kunneth Formula is enough to obtain the result using (co)homology and betti numbers but OP wants Dif Geo proof $\endgroup$
    – Birch Bryant
    Jun 30, 2017 at 1:34
  • $\begingroup$ @TedShifrin It seems that the Poincaré-Hopf theorem could be used in the non-trivial case too, don't you think? $\endgroup$
    – Amitai Yuval
    Jun 30, 2017 at 1:40
  • $\begingroup$ Both @AlexisLeroux-Lapierre and @ user253929 thanks for pointing out that argument. I was not aware of it. $\endgroup$
    – D1811994
    Jun 30, 2017 at 12:20

2 Answers 2

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If you know about Morse Bott functions, the proof is simple. If a $f:B\to \bf R$, is a Morse function, then $F=f\circ \pi : E\to \bf R$ is a Morse-Bott function whose critical submanifold are exactly the fibers over critical point and the same indices. Thus $\chi (E)= \sum _c ind(c) \chi (F_c)= \chi (F)\times \chi (B)$.

The proof of Morse inequalities (and equalities) are exactly the same for Morse-Bott and Morse functions (see the book of Milnor eg).

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Here is a self-contained proof using Poincare-Hopf theorem (as suggested by Ted). Let $\nabla$ be a connection on the bundle $p: E\to B$. Take a nondegenerate vector field $X$ on $B$ and lift it to a vector field $Y$ on $E$ using $\nabla$. Let $b_1,...,b_n$ denote the singular points of $X$; set $F_i:= p^{-1}(b_i), i=1,...,n$. Take small (pairwise disjoint) tubular neighborhoods $p^{-1}(U_i)$ of the fibers $F_i$, $i=1,...,n$. These neighborhoods admit a product structure $F_i\times U_i$ consistent with the fibration $p$. Pick a nondegenerate vector field $Z$ on $F\cong F_i, i=1,...,n$, extend it to each $U_i$ (using the product decomposition) and then multiply by pull-backs (via $p$) of suitable bump-functions, supported on $U_i$'s. Then extend the resulting vector field $W$ to the rest of $E$. Lastly, take the vector field $V=Y+W$. The set $Sing(V)$ of singular points of $V$ is the disjoint union of copies of $Sing(Z)$ in $F_1,...,F_n$. At each point $q\in F_i\cap Sing(V)$, $$ index(V,q)= index(Z,q) \times index(X, b_i)$$ (this follows from the fact that the determinant of a square block-diagonal matrix with blocks $A_1, A_2$ equals $det(A_1)\times det(A_2)$).

Now, what's left is just to count the number of singular points of $V$ (with sign): Each fiber $F_i$ contributes $$ index(X,b_i)\times index(Z)= index(X,b_i)\times \chi(F).$$ Summing up over all points $b_i$ we obtain $$ index(V)= index(X)\times \chi(F)= \chi(B)\times \chi(F). $$ qed

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