# euler class of the normal bundle and self intersection number [duplicate]

Let $S$ be a compact submanifold of $X$ smooth manifold. I know that $T_X|_S=T_S\oplus N_{S/X}$ where $N_{S/X}$ is the normal bundle. I have read that the euler class $e(N_{S/X})$ corresponds (via integration over S, i suppose) to the self intersection number $S\cdot S$. I've thought about it, but i don't know how to prove it, also i can't find the proof in any book. Do you know something about it?

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## marked as duplicate by Ben Webster♦Mar 5 '14 at 15:51

To start: what is your definition of $e(N_{S/X})$? –  J. Martel Mar 8 '13 at 2:14
IMO you are depriving yourself of some important suffering. –  J. Martel Mar 8 '13 at 2:20
Your statement that $T_X|_S = T_S \oplus N_{S/X}$ is not true. There is an exact sequence relating these things, but it generally does not split. –  Jack Huizenga Mar 8 '13 at 2:49
@Jack: words like "compact", "manifold", and "integration" suggest to me that Konrad is working in the differentiable category, where he may split to his heart's content. –  Allen Knutson Mar 8 '13 at 2:59
@Konrad: How do you define the self-intersection "number" unless $\dim(X)=2\dim(S)$? –  Mark Grant Mar 8 '13 at 3:07

The result you want is contained in Thm. 4.7. Again, you need $2\dim S=\dim X$.
yes, i forgot the hypotesis $2dim S=dim X$, in particular $S$ is a complex curve in $X$. $S\cdot S=\int_X \eta_S\wedge \eta_S$. If $s_0:S\rightarrow N$ is the zero section of the normal bundle, to use thm 4.7 i guess you are impling that $\int_X \eta_S \wedge \eta_S =\int_N \eta_{s_0(S)}\wedge \eta_{s_0(S)}$.. but why is that? –  Konrad Mar 12 '13 at 15:04