[proofplan]
We prove that the Euler class of the tangent bundle of a closed oriented manifold evaluates on the fundamental class to give the Euler characteristic: $e(TM)[M] = \chi(M)$. The proof uses two key facts: (1) the Euler class $e(TM)$ is the Thom class of the normal bundle of the diagonal $\Delta: M \hookrightarrow M \times M$, and (2) the self-intersection number of the diagonal equals the Lefschetz number of the identity, which is the Euler characteristic. As a corollary, we deduce that $TS^{2n}$ admits no proper direct sum decomposition.
[/proofplan]
[step:Identify the normal bundle of the diagonal with the tangent bundle]
The diagonal embedding $\Delta: M \hookrightarrow M \times M$ sends $x$ to $(x, x)$. At any point $x \in M$, the differential of $\Delta$ is
\begin{align*}
d\Delta_x: T_x M \to T_{(x,x)}(M \times M) \cong T_x M \oplus T_x M, \quad v \mapsto (v, v).
\end{align*}
The image of $d\Delta_x$ is the diagonal subspace $\{(v, v) : v \in T_x M\}$. The normal bundle $\nu_{\Delta(M) \subseteq M \times M}$ is the quotient
\begin{align*}
\nu_{\Delta(M)} = \frac{T_{(x,x)}(M \times M)}{d\Delta_x(T_x M)} = \frac{T_x M \oplus T_x M}{\{(v,v)\}}.
\end{align*}
The map $T_x M \to (T_x M \oplus T_x M)/\{(v,v)\}$ defined by $w \mapsto [(w, 0)] = [(0, -w)]$ is a linear isomorphism: it is injective because $(w, 0) \in \{(v,v)\}$ implies $w = 0$, and surjective by dimension count ($\dim(\nu_{\Delta(M)}) = 2d - d = d = \dim(T_x M)$). Therefore
\begin{align*}
\nu_{\Delta(M) \subseteq M \times M} \cong TM
\end{align*}
via the isomorphism $(w, 0) \bmod \{(v,v)\} \leftrightarrow w$.
[guided]
Why does the normal bundle of the diagonal equal the tangent bundle? Geometrically, the diagonal $\Delta(M) \subseteq M \times M$ parametrizes "points that are equal." Moving away from the diagonal in $M \times M$ means making the two coordinates differ — the "difference direction" $(w, -w)$ (or equivalently $(w, 0)$ modulo the diagonal) is a tangent direction in $M$. So the space of directions transverse to the diagonal is naturally identified with $T_x M$ itself.
This identification is canonical: it uses no Riemannian metric or choice of complement, only the linear algebra of the diagonal embedding.
[/guided]
[/step]
[step:Relate the Euler class to the self-intersection of the diagonal]
By the Thom isomorphism, the Euler class $e(E)$ of an oriented rank-$d$ vector bundle $E \to M$ is the pullback of the Thom class $u \in H^d(E, E^\#)$ along the zero section $s_0: M \to E$:
\begin{align*}
e(E) = s_0^*(u) \in H^d(M).
\end{align*}
For $E = TM \cong \nu_{\Delta(M)}$, the Thom class $u_\Delta \in H^d(M \times M, M \times M \setminus \Delta(M))$ restricts to the Euler class via $\Delta^*(u_\Delta) = e(TM)$.
The self-intersection number of the diagonal is defined as
\begin{align*}
\Delta \cdot \Delta = \Delta^*(u_\Delta)[M] = e(TM)[M],
\end{align*}
where $[M] \in H_d(M)$ is the fundamental class. So $e(TM)[M]$ equals the self-intersection number of the diagonal in $M \times M$.
[/step]
[step:Compute the self-intersection of the diagonal as the Lefschetz number of the identity]
The self-intersection of the diagonal $\Delta \cdot \Delta$ counts (with signs) the intersection of $\Delta(M)$ with a small perturbation of itself. By the Lefschetz fixed point theorem, this count equals the Lefschetz number of the identity map $\operatorname{id}_M$:
\begin{align*}
L(\operatorname{id}_M) = \sum_{n=0}^{d} (-1)^n \operatorname{tr}\bigl((\operatorname{id}_M)_*: H_n(M; \mathbb{Q}) \to H_n(M; \mathbb{Q})\bigr) = \sum_{n=0}^{d} (-1)^n \dim_{\mathbb{Q}} H_n(M; \mathbb{Q}).
\end{align*}
The trace of the identity on a vector space of dimension $b_n$ is $b_n$, so
\begin{align*}
L(\operatorname{id}_M) = \sum_{n=0}^{d} (-1)^n b_n = \chi(M),
\end{align*}
where $b_n = \dim_{\mathbb{Q}} H_n(M; \mathbb{Q})$ are the Betti numbers and $\chi(M)$ is the Euler characteristic.
More directly, the diagonal class $\delta \in H^d(M \times M; \mathbb{Q})$ — the Poincare dual of $\Delta_*[M]$ — satisfies
\begin{align*}
\Delta^*(\delta)[M] = \sum_{n=0}^{d} (-1)^n \operatorname{tr}(\operatorname{id}_{H_n(M;\mathbb{Q})}) = \chi(M),
\end{align*}
by the diagonal class computation from the [Diagonal Class](/theorems/2296) theorem, which expresses $\delta = \sum_{i,j} (-1)^{|a_i|} (a_i \times a_j^*)$ in terms of a basis $\{a_i\}$ of $H^*(M; \mathbb{Q})$ and its Poincare dual basis $\{a_j^*\}$.
[guided]
The connection between the Euler class and the Euler characteristic passes through three identifications:
1. $e(TM)[M] = \Delta \cdot \Delta$ (the Euler class evaluates to the self-intersection number of the diagonal, by the Thom class interpretation).
2. $\Delta \cdot \Delta = L(\operatorname{id}_M)$ (the self-intersection of the diagonal counts fixed points of the identity — every point is fixed — with the correct algebraic signs given by the Lefschetz trace formula).
3. $L(\operatorname{id}_M) = \chi(M)$ (the Lefschetz number of the identity is the alternating sum of Betti numbers).
Each identification uses a different piece of the theory: the first uses the Thom isomorphism, the second uses the Lefschetz fixed point theorem, and the third is a definition.
[/guided]
[/step]
[step:Deduce that $TS^{2n}$ is irreducible as a direct sum]
For even-dimensional spheres, $\chi(S^{2n}) = 1 + (-1)^{2n} = 2 \neq 0$, so $e(TS^{2n})[S^{2n}] = 2 \neq 0$, and in particular $e(TS^{2n}) \neq 0$ in $H^{2n}(S^{2n})$.
Suppose for contradiction that $TS^{2n} = E \oplus F$ for subbundles $E$ and $F$ of ranks $k$ and $2n - k$ respectively, with $0 < k < 2n$. The Whitney product formula for the Euler class gives
\begin{align*}
e(TS^{2n}) = e(E) \smile e(F) \in H^{2n}(S^{2n}).
\end{align*}
Since $e(E) \in H^k(S^{2n})$ and $e(F) \in H^{2n-k}(S^{2n})$, and $H^j(S^{2n}) = 0$ for $0 < j < 2n$, both $e(E) = 0$ and $e(F) = 0$. Therefore $e(TS^{2n}) = 0 \smile 0 = 0$, contradicting $e(TS^{2n}) \neq 0$.
This contradiction shows that $TS^{2n}$ cannot be decomposed as a direct sum of two proper subbundles — the tangent bundle of $S^{2n}$ is irreducible.
[guided]
This argument is a striking application of characteristic classes: a purely topological invariant (the Euler class) obstructs a geometric decomposition (splitting the tangent bundle). The key inputs are:
- The Euler characteristic of $S^{2n}$ is non-zero ($\chi = 2$), which by the theorem gives $e(TS^{2n}) \neq 0$.
- The cohomology of $S^{2n}$ is concentrated in degrees $0$ and $2n$, so any class in an intermediate degree must vanish.
- The Whitney product formula converts a bundle decomposition into a factorization of the Euler class as a cup product of classes in intermediate degrees.
For odd-dimensional spheres, $\chi(S^{2n+1}) = 0$, so $e(TS^{2n+1}) = 0$ and this obstruction vanishes. Indeed, $S^{2n+1}$ admits a nowhere-zero vector field (given by multiplication by $i$ when $S^{2n+1}$ is viewed as the unit sphere in $\mathbb{C}^{n+1}$), which trivializes a rank-1 subbundle of $TS^{2n+1}$.
[/guided]
[/step]
