[proofplan] We first define local index by the degree of the normalised coordinate representative on a small sphere around an isolated zero. We then compute one model case using a Morse function: after choosing the gradient convention correctly, each critical point of Morse index $k$ contributes $(-1)^k$, and the [Morse cell decomposition theorem](/page/Morse%20Cell%20Decomposition%20Theorem) gives the Euler characteristic. Finally, instead of invoking a global index theorem for vector fields, we identify the sum of local indices with the Euler number of $TM$ by the [Thom class](/page/Thom%20Class) definition of the [Euler class](/page/Euler%20Class); applying the same identity to the Morse model gives the desired equality for the original field. [/proofplan]

[step:Define the local and global indices through degree]For each zero $p \in Z(X)$, choose an oriented coordinate chart $(U_p,\varphi_p)$ with $p \in U_p$ and choose $r_p>0$ such that $\overline{B}(\varphi_p(p),r_p) \subset \varphi_p(U_p)$ and $p$ is the only zero of $X$ in $\varphi_p^{-1}(\overline{B}(\varphi_p(p),r_p))$. Define the coordinate representative \begin{align*} Y_p: \overline{B}(\varphi_p(p),r_p) &\to \mathbb{R}^n \\ y &\mapsto d(\varphi_p)_{\varphi_p^{-1}(y)}\bigl(X_{\varphi_p^{-1}(y)}\bigr). \end{align*} Since $Y_p(y) \ne 0$ for $y \in \partial B(\varphi_p(p),r_p)$, define \begin{align*} \nu_p: \partial B(\varphi_p(p),r_p) &\to S^{n-1} \\ y &\mapsto \frac{Y_p(y)}{|Y_p(y)|}. \end{align*} The local index is $\operatorname{ind}_p(X)=\deg(\nu_p)$. This definition is independent of the oriented chart and radius by the [homotopy invariance of degree](/page/Homotopy%20Invariance%20of%20Degree) applied to nested spheres and orientation-preserving coordinate changes. Define the global index of $X$ by \begin{align*} I(X) := \sum_{p\in Z(X)} \operatorname{ind}_p(X). \end{align*} For each $q\in M$, let $0_q\in T_qM$ be the zero vector, and let \begin{align*} 0_M:M&\to TM\\ q&\mapsto 0_q \end{align*} be the zero section. The set $Z(X)$ is finite because it is a discrete closed subset of the [compact space](/page/Compact%20Space) $M$: isolatedness makes it discrete, smoothness makes $Z(X)=\{q\in M:X(q)=0_M(q)\}$ closed as the equalizer of the two continuous sections $X,0_M:M\to TM$, and closedness of $M$ means compactness.[/step]

[guided]The first task is to make the quantity in the theorem precise enough that we can compare it across different vector fields. Let $p \in Z(X)$ be a zero. Because the zero is isolated, there is an oriented coordinate chart $(U_p,\varphi_p)$ around $p$ and a radius $r_p>0$ such that the closed coordinate ball $\overline{B}(\varphi_p(p),r_p)$ lies inside $\varphi_p(U_p)$ and contains no zero of $X$ except $p$. In these coordinates the vector field becomes the map \begin{align*} Y_p: \overline{B}(\varphi_p(p),r_p) &\to \mathbb{R}^n \\ y &\mapsto d(\varphi_p)_{\varphi_p^{-1}(y)}\bigl(X_{\varphi_p^{-1}(y)}\bigr). \end{align*} This is a genuine map into $\mathbb{R}^n$ because $d(\varphi_p)_q:T_qM\to \mathbb{R}^n$ identifies the tangent space at $q$ with the coordinate [vector space](/page/Vector%20Space). On the boundary sphere $\partial B(\varphi_p(p),r_p)$ the vector $Y_p(y)$ is nonzero, so we can normalise it and obtain \begin{align*} \nu_p: \partial B(\varphi_p(p),r_p) &\to S^{n-1} \\ y &\mapsto \frac{Y_p(y)}{|Y_p(y)|}. \end{align*} The local index is the [topological degree](/page/Topological%20Degree) $\operatorname{ind}_p(X)=\deg(\nu_p)$. The choice of oriented chart does not change the degree because an orientation-preserving change of coordinates has degree $1$ on small boundary spheres. The choice of radius does not change the degree because two admissible radii enclose an annulus on which the normalised vector field gives a homotopy between the two boundary maps, and [homotopy invariance of degree](/page/Homotopy%20Invariance%20of%20Degree) preserves the degree. Thus the definition is intrinsic. Finally define \begin{align*} I(X) := \sum_{p\in Z(X)} \operatorname{ind}_p(X). \end{align*} This is a finite sum. For each $q\in M$, let $0_q\in T_qM$ be the zero vector, and let $0_M:M\to TM$ be the zero section $q\mapsto 0_q$. Then $Z(X)=\{q\in M:X(q)=0_M(q)\}$ is closed because it is the equalizer of the continuous sections $X$ and $0_M$, and it is discrete because all zeros are isolated. A closed discrete subset of the compact manifold $M$ is finite.[/guided]

[step:Compute the index of a Morse gradient field]Choose a Riemannian metric $g$ on $M$ and a Morse function $f:M\to \mathbb{R}$. The [Morse existence theorem](/page/Morse%20Existence%20Theorem) applies because $M$ is a closed smooth manifold, and it gives a smooth function whose critical points are nondegenerate. Let \begin{align*} G: M &\to TM \\ q &\mapsto -\nabla f(q) \end{align*} be the negative gradient vector field with respect to $g$. The zeros of $G$ are precisely the critical points of $f$, hence are isolated because nondegenerate critical points are isolated. Let $p$ be a critical point of $f$ with Morse index $k$. By the [Morse lemma](/page/Morse%20Lemma), there is an oriented coordinate chart $(U,\varphi)$ with coordinates $x=(x_1,\dots,x_n)$, centred at $p$, in which \begin{align*} f\circ \varphi^{-1}(x)=f(p)-x_1^2-\cdots-x_k^2+x_{k+1}^2+\cdots+x_n^2. \end{align*} Let \begin{align*} D_k: \mathbb{R}^n &\to \mathbb{R}^n \\ x &\mapsto (-2x_1,\dots,-2x_k,2x_{k+1},\dots,2x_n) \end{align*} be the Hessian [linear map](/page/Linear%20Map) of $f$ in these coordinates, and let $G_p$ be the positive definite matrix representing the metric $g_p$ at $p$. The derivative at $p$ of the local representative of the negative gradient is \begin{align*} L_k:=-G_p^{-1}D_k. \end{align*} Thus, for some $\rho>0$, the local representative of $G$ on $B(0,\rho)$ has the form $L_kx+R(x)$, where the remainder map $R:B(0,\rho)\to \mathbb{R}^n$ satisfies \begin{align*} \frac{|R(x)|}{|x|}\to 0 \quad\text{as }x\to 0. \end{align*} Since $G_p$ is positive definite, $\det(G_p)>0$, and therefore \begin{align*} \operatorname{sgn}(\det L_k)=\operatorname{sgn}(\det(-D_k))=(-1)^{n-k}. \end{align*} Choose $0<r<\rho$ such that $|R(x)|<|L_kx|$ for every $x\in\partial B(0,r)$. Then \begin{align*} H:[0,1]\times \partial B(0,r)&\to \mathbb{R}^n\setminus\{0\}\\ (t,x)&\mapsto L_kx+tR(x) \end{align*} is a homotopy through nonvanishing boundary fields, because on $\partial B(0,r)$ one has \begin{align*} |L_kx+tR(x)|\ge |L_kx|-|R(x)|>0. \end{align*} The normalised boundary map $x\mapsto L_kx/|L_kx|$ has degree $\operatorname{sgn}(\det L_k)=(-1)^{n-k}$. If instead one uses the positive gradient, the sign is $(-1)^k$; with the negative gradient convention above, the local index is $(-1)^{n-k}$. The [Morse cell decomposition theorem](/page/Morse%20Cell%20Decomposition%20Theorem) applies to the Morse function $f$ on the closed smooth manifold $M$: the sublevel sets of $f$ give a finite CW decomposition with exactly one $k$-cell for each critical point of Morse index $k$. Therefore \begin{align*} \chi(M)=\sum_{k=0}^{n}(-1)^k c_k, \end{align*} where $c_k$ is the number of critical points of Morse index $k$. Define the model vector field \begin{align*} V:M&\to TM\\ q&\mapsto \nabla f(q), \end{align*} or equivalently replace $f$ by $-f$ and use the negative gradient convention. The preceding determinant computation for the positive gradient gives local contribution $(-1)^k$ at each critical point of Morse index $k$. Hence $V$ has isolated zeros and \begin{align*} I(V)=\chi(M). \end{align*}[/step]

[guided]We now build one vector field for which the theorem can be computed directly. Choose a Riemannian metric $g$ on $M$ and a Morse function $f:M\to \mathbb{R}$. The [Morse existence theorem](/page/Morse%20Existence%20Theorem) applies because $M$ is a closed smooth manifold, and it gives a smooth function whose critical points are nondegenerate. Define the negative gradient vector field \begin{align*} G: M &\to TM \\ q &\mapsto -\nabla f(q). \end{align*} The zeros of $G$ are exactly the critical points of $f$, because $\nabla f(q)=0$ if and only if $df_q=0$. Since the critical points of a Morse function are nondegenerate, they are isolated, so $G$ has isolated zeros. Let $p$ be a critical point of Morse index $k$. The [Morse lemma](/page/Morse%20Lemma) gives local coordinates $x=(x_1,\dots,x_n)$ centred at $p$ in which \begin{align*} f\circ \varphi^{-1}(x)=f(p)-x_1^2-\cdots-x_k^2+x_{k+1}^2+\cdots+x_n^2. \end{align*} In these coordinates the Hessian linear map of $f$ is \begin{align*} D_k: \mathbb{R}^n &\to \mathbb{R}^n \\ x &\mapsto (-2x_1,\dots,-2x_k,2x_{k+1},\dots,2x_n). \end{align*} Let $G_p$ be the positive definite matrix representing $g_p$ in these coordinates. The leading linear part of the negative gradient is \begin{align*} L_k:=-G_p^{-1}D_k. \end{align*} We must justify that the nonlinear gradient field has the same local index as this linear model. For some $\rho>0$, the local coordinate representative on $B(0,\rho)$ has the form $L_kx+R(x)$, where $R:B(0,\rho)\to\mathbb{R}^n$ satisfies \begin{align*} \frac{|R(x)|}{|x|}\to 0 \quad\text{as }x\to 0. \end{align*} Since $L_k$ is invertible, there is a constant $m_k>0$ such that \begin{align*} |L_kx|\ge m_k|x| \end{align*} for all $x\in\mathbb{R}^n$. Choose $0<r<\rho$ so small that \begin{align*} |R(x)|<m_k|x| \end{align*} for $x\in\partial B(0,r)$. Then \begin{align*} H:[0,1]\times\partial B(0,r)&\to\mathbb{R}^n\setminus\{0\}\\ (t,x)&\mapsto L_kx+tR(x) \end{align*} is a homotopy through nonzero boundary fields, because \begin{align*} |H(t,x)|\ge |L_kx|-|R(x)|>0. \end{align*} Therefore the actual gradient field and its linear part have the same local index by [homotopy invariance of degree](/page/Homotopy%20Invariance%20of%20Degree). The normalised map $x\mapsto L_kx/|L_kx|$ is the sphere map induced by an invertible linear transformation, and its degree is the sign of the determinant. Since $\det(G_p)>0$ and $D_k$ has $k$ negative eigenvalues and $n-k$ positive eigenvalues, \begin{align*} \deg\left(x\mapsto \frac{L_kx}{|L_kx|}\right)=\operatorname{sgn}(\det L_k)=(-1)^{n-k}. \end{align*} Thus the negative gradient contributes $(-1)^{n-k}$ at a critical point of Morse index $k$. If we use the positive gradient, or equivalently replace $f$ by $-f$, the contribution becomes $(-1)^k$. The [Morse cell decomposition theorem](/page/Morse%20Cell%20Decomposition%20Theorem) applies to this Morse function on the closed smooth manifold $M$. It says that the sublevel sets of $f$ determine a finite CW decomposition of $M$ with exactly one $k$-cell for each critical point of Morse index $k$. Hence, if $c_k$ denotes the number of critical points of index $k$, then \begin{align*} \chi(M)=\sum_{k=0}^{n}(-1)^k c_k. \end{align*} Choose the model vector field explicitly as \begin{align*} V:M&\to TM\\ q&\mapsto \nabla f(q), \end{align*} or, equivalently, replace $f$ by $-f$ and use the negative gradient convention. With this choice the local contribution at Morse index $k$ is $(-1)^k$, so $V$ has isolated zeros satisfying \begin{align*} I(V)=\sum_{p\in Z(V)}\operatorname{ind}_p(V)=\chi(M). \end{align*} This supplies the model case for the theorem.[/guided]

[step:Identify the local index sum with the Euler number]Let $E:=TM$ be the oriented rank-$n$ tangent bundle over $M$, let $\pi:E\to M$ be its bundle projection, and let $0_E:M\to E$ be the zero section. Let $\tau\in H^n(E,E\setminus 0_E(M);\mathbb{Z})$ denote the [Thom class](/page/Thom%20Class) compatible with the orientation of $E$. The [Euler class](/page/Euler%20Class) is \begin{align*} e(E):=0_E^*\tau\in H^n(M;\mathbb{Z}). \end{align*} Let $[M]\in H_n(M;\mathbb{Z})$ denote the fundamental class determined by the given orientation of $M$. Put $Z:=Z(X)$. Since $X$ is nonzero on $M\setminus Z$, the section $X$ is a map of pairs \begin{align*} X_Z:(M,M\setminus Z)&\to (E,E\setminus 0_E(M)). \end{align*} The pullback $X_Z^*\tau\in H^n(M,M\setminus Z;\mathbb{Z})$ maps to $e(E)$ under the natural map to $H^n(M;\mathbb{Z})$. Indeed, after forgetting the relative condition, $X$ and $0_E$ are homotopic through the fibrewise linear homotopy $(t,q)\mapsto tX(q)$, so [homotopy invariance of cohomology](/page/Homotopy%20Invariance%20of%20Cohomology) makes their ordinary pullbacks of the Thom class agree. Let \begin{align*} [M,M\setminus Z]\in H_n(M,M\setminus Z;\mathbb{Z}) \end{align*} be the relative fundamental class obtained from the oriented fundamental class $[M]$ under the natural map $H_n(M;\mathbb{Z})\to H_n(M,M\setminus Z;\mathbb{Z})$. Choose pairwise disjoint closed oriented coordinate balls $B_p\subset M$, one around each $p\in Z(X)$, containing no other zero, and put $N:=\bigcup_{p\in Z(X)}B_p$. Since $X$ is nonzero on $M\setminus N$, [excision](/page/Excision) gives an isomorphism \begin{align*} H^n(M,M\setminus Z;\mathbb{Z}) \cong H^n(N,N\setminus Z;\mathbb{Z}). \end{align*} Under this isomorphism, additivity of relative fundamental classes identifies $[M,M\setminus Z]$ with the sum of the classes $[B_p,B_p\setminus\{p\}]$. The punctured ball $B_p\setminus\{p\}$ [deformation retracts](/page/Deformation%20Retract) onto $\partial B_p$, so homotopy invariance for pairs allows the pairing to be computed on $(B_p,\partial B_p)$. Therefore \begin{align*} \langle e(TM),[M]\rangle &=\langle X_Z^*\tau,[M,M\setminus Z]\rangle\\ &=\sum_{p\in Z(X)}\langle X_Z^*\tau,[B_p,\partial B_p]\rangle. \end{align*} Under the oriented trivialisation of $TM$ over $B_p$ induced by the chosen chart, the Thom class restricts, by the local form of the [Thom isomorphism](/page/Thom%20Isomorphism), to the standard generator of $H^n(\mathbb{R}^n,\mathbb{R}^n\setminus\{0\};\mathbb{Z})$. The [relative degree](/page/Relative%20Degree) interpretation of this generator identifies \begin{align*} \langle X_Z^*\tau,[B_p,\partial B_p]\rangle=\deg(\nu_p)=\operatorname{ind}_p(X). \end{align*} Therefore \begin{align*} I(X)=\sum_{p\in Z(X)}\operatorname{ind}_p(X)=\langle e(TM),[M]\rangle. \end{align*} Apply the same identity to the Morse model vector field $V$ constructed above. The previous step gives $I(V)=\chi(M)$, hence \begin{align*} \langle e(TM),[M]\rangle=I(V)=\chi(M). \end{align*} Combining the two displayed identities gives \begin{align*} \sum_{p\in Z(X)}\operatorname{ind}_p(X)=I(X)=\langle e(TM),[M]\rangle=\chi(M), \end{align*} which is the desired identity.[/step]

[guided]The remaining task is to compare $X$ with the Morse model without assuming the theorem we are proving. We do this through the [Euler class](/page/Euler%20Class) of the tangent bundle. Let $E:=TM$ be the oriented rank-$n$ tangent bundle over $M$, let $\pi:E\to M$ be the projection, and let $0_E:M\to E$ be the zero section. Let $\tau\in H^n(E,E\setminus 0_E(M);\mathbb{Z})$ be the [Thom class](/page/Thom%20Class) determined by the orientation of $E$. Let $[M]\in H_n(M;\mathbb{Z})$ be the fundamental class determined by the orientation of $M$. By definition, the Euler class of $E$ is \begin{align*} e(E):=0_E^*\tau\in H^n(M;\mathbb{Z}). \end{align*} Why may we use the section $X$ to localise the Euler class? Put $Z:=Z(X)$. Since $X$ is nonzero on $M\setminus Z$, it defines a map of pairs \begin{align*} X_Z:(M,M\setminus Z)&\to (E,E\setminus 0_E(M)). \end{align*} The relative class $X_Z^*\tau\in H^n(M,M\setminus Z;\mathbb{Z})$ maps to $e(E)$ in ordinary cohomology. The reason is that, after forgetting the relative condition, the fibrewise linear homotopy $(t,q)\mapsto tX(q)$ connects $0_E$ to $X$, so [homotopy invariance of cohomology](/page/Homotopy%20Invariance%20of%20Cohomology) gives the same ordinary Euler class. The relative information is concentrated at the points where $X$ meets the zero section, namely the zeros of $X$. Let \begin{align*} [M,M\setminus Z]\in H_n(M,M\setminus Z;\mathbb{Z}) \end{align*} be the relative fundamental class obtained from the oriented fundamental class $[M]$ by the natural map from absolute to relative homology. Now choose pairwise disjoint closed oriented coordinate balls $B_p\subset M$, one for each $p\in Z(X)$, such that $p$ is the only zero of $X$ in $B_p$. This is possible because $Z(X)$ is finite. Let \begin{align*} N:=\bigcup_{p\in Z(X)}B_p. \end{align*} On $M\setminus N$ the section $X$ avoids the zero section, so it contributes nothing to the relative Thom pairing there. [Excision](/page/Excision) gives \begin{align*} H^n(M,M\setminus Z;\mathbb{Z}) \cong H^n(N,N\setminus Z;\mathbb{Z}). \end{align*} Under this isomorphism the relative fundamental class becomes the sum of the local classes $[B_p,B_p\setminus\{p\}]$. Since $B_p\setminus\{p\}$ [deformation retracts](/page/Deformation%20Retract) onto $\partial B_p$, homotopy invariance for pairs lets the same local pairing be written on $(B_p,\partial B_p)$. Hence \begin{align*} \langle e(TM),[M]\rangle &=\langle X_Z^*\tau,[M,M\setminus Z]\rangle\\ &=\sum_{p\in Z(X)}\langle X_Z^*\tau,[B_p,\partial B_p]\rangle. \end{align*} It remains to identify each summand with the local degree already used to define $\operatorname{ind}_p(X)$. Use the oriented coordinate chart on $B_p$ to trivialise $TM|_{B_p}$ as $B_p\times\mathbb{R}^n$. Under this oriented trivialisation, the [Thom class](/page/Thom%20Class) becomes, by the local form of the [Thom isomorphism](/page/Thom%20Isomorphism), the standard generator of \begin{align*} H^n(\mathbb{R}^n,\mathbb{R}^n\setminus\{0\};\mathbb{Z}). \end{align*} Pulling this generator back by the coordinate representative of $X$ computes the [relative degree](/page/Relative%20Degree) of the map from the ball to $\mathbb{R}^n$ that sends the boundary sphere away from $0$. The boundary version of that relative degree is exactly the degree of the normalised map \begin{align*} \nu_p:\partial B_p&\to S^{n-1}, \end{align*} so \begin{align*} \langle X_Z^*\tau,[B_p,\partial B_p]\rangle=\deg(\nu_p)=\operatorname{ind}_p(X). \end{align*} Thus \begin{align*} I(X)=\sum_{p\in Z(X)}\operatorname{ind}_p(X)=\langle e(TM),[M]\rangle. \end{align*} This identity is not a homotopy assertion between $X$ and the Morse model; it is a local computation of the Euler class around the zeros of a single section. Applying the same computation to the Morse model vector field $V$ gives \begin{align*} I(V)=\langle e(TM),[M]\rangle. \end{align*} The Morse calculation from the previous step gives $I(V)=\chi(M)$. Therefore \begin{align*} \sum_{p\in Z(X)}\operatorname{ind}_p(X)=I(X)=\langle e(TM),[M]\rangle=I(V)=\chi(M). \end{align*} This proves the Poincaré-Hopf identity for the original vector field $X$ without invoking the Poincaré-Hopf theorem as an input.[/guided]