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I am working through Milnor's Characteristic classes and am currently working problems on the topic of oriented bundles and euler class. I am having trouble computing the euler class of the tangent bundle of a sphere. (The actual question is in the third paragraph)

The first part which worked out fine for me was to show that the total space of the tangent bundle is equal to $S^{n} \times S^{n} \setminus A$ where $A$ is the anti-diagonal subset of the product space and furthermore that $H^{*}(E,E_{0};Z) \cong H^{*}(S^{n} \times S^{n},A;Z)$ (Here $(E,E_{0})$ is the pair (total space, total space - zero section), as used in the book). This follows by excision and arguing that $S^{n}\times S^{n} \setminus (Diagonal) \sim A$.

The part I am having trouble with is showing that, in the $n$ even case, the euler class corresponds to twice a generator of $H^{n}(S^{n};Z)$. It seems obvious to me that the fundamental class ,$u$, is a generator of $Z = H^{n}(E,E_{0})$. Next, since $-\cup u$ is an isomorphism, $u \cup u$ is a generator of $Z = H^{2n}(E,E_{0})$ and finally, since the Thom isomorphism is in fact an isomorphism this seems to suggest that the euler class should in fact be a generator of $H^{n}(S^{n};Z)$.

For one thing, I haven't used the fact that $n$ is even, though I don't see where that would be relevant. I also am not 100% sure about my computation of the fundamental class or the cohomology groups. For reference I will state my computation of the cohomology below.

To start, $H^{i}(S^{n}\times S^{n};Z) = (Z, i = 0);(Z \oplus Z, i = n);(Z, i = 2n)$ as proved with the Kunneth formula. Next, using the long exact sequence of a triple $(S^{n}\times S^{n},A,\not{0})$, we get that $H^{i}(S^{n} \times S^{n},A;Z) = (Z,i = 0,n,2n); 0\,else$.

GabrielB
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2 Answers2

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Almost by definition, the Euler class is the self-intersection number of the zero section of the tangent bundle. And since the zero section is homotopic to any other section, it is also equal to the intersection number of the zero section with any other section. The easiest thing is to pick a section which is transverse to the zero section, with isolated intersection points, and then to add up the local intersection numbers over all the intersection points.

This is just the Poincare-Hopf index formula: a section of the tangent bundle is just a vector field; the intersection points with the zero section are just the zeroes of the vector field; and the local intersection number at each zero is just the index of the vector field at that zero.

So, all you have to do is pick a vector field on $S^n$ with isolated zeroes whose indices are easy to compute. The easiest is the vector field which has a "source" at the south pole and a "sink" at the north pole. The index of a source equals $+1$, and the index of a sink equals $(-1)^n$.

Lee Mosher
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this is in Hatcher's Vector Bundles and K-Theory Book, Page 92. Alternatively you can use the fact that $\chi(M) = <e(M), [M]>$ to conclude that $e(M)$ must correspond to the cohomology class $2$. Hope this helps!

Also, your sphere must be even for otherwise you will have an odd-dimensional bundle which will produce a $2$-torsion in $H^n(S^n; \mathbb{Z}) = \mathbb{Z}$ which is clearly impossible

Elden Elmanto
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