[proofplan] We compute the curvature by applying the connection twice to an arbitrary local section of the tautological bundle. The identities needed come from differentiating the projection equation $P^2=P$ and the section equation $Ps=s$. The first identity shows that $dP$ has no tangent-to-tangent component, and the second identifies the normal part of $ds$ with $dP\,s$. Substituting this normal component into the direct curvature computation gives $F_\nabla s=P\,dP\wedge dP\,P s$ for every section $s$. [/proofplan]

[step:Record the projection identities obtained from $P^2=P$]Let $U\subset \operatorname{Gr}(k,n)$ be an [open set](/page/Open%20Set), and write \begin{align*} P_U:U\to \operatorname{End}_{\mathbb K}(\mathbb K^n) \end{align*} for the restriction of the projection map $P$. To keep notation light, write $P$ for $P_U$ in this proof. Thus $P$ is a smooth endomorphism-valued $0$-form on $U$, and \begin{align*} dP\in \Omega^1(U;\operatorname{End}_{\mathbb K}(\mathbb K^n)). \end{align*} Here $\Omega^1(U;\operatorname{End}_{\mathbb K}(\mathbb K^n))$ denotes the space of smooth $1$-forms on $U$ with values in the $\mathbb K$-linear endomorphisms of $\mathbb K^n$. Since every point of $\operatorname{Gr}(k,n)$ is represented by an idempotent projection, we have \begin{align*} P^2=P. \end{align*} Applying the [exterior derivative](/theorems/1525) entrywise and using the Leibniz rule for the product of an endomorphism-valued $0$-form with an endomorphism-valued $0$-form gives \begin{align*} dP\,P+P\,dP=dP. \end{align*} Multiplying this identity on the left by $P$ gives \begin{align*} P\,dP\,P+P^2\,dP=P\,dP. \end{align*} Using $P^2=P$, this becomes \begin{align*} P\,dP\,P+P\,dP=P\,dP. \end{align*} Subtracting $P\,dP$ from both sides yields \begin{align*} P\,dP\,P=0. \end{align*}[/step]

[guided]We first isolate the algebraic fact about a moving projection that will remove the unwanted component of $ds$ later. On an open set $U\subset \operatorname{Gr}(k,n)$, the projection model gives a smooth map \begin{align*} P:U\to \operatorname{End}_{\mathbb K}(\mathbb K^n). \end{align*} This is a $0$-form with values in endomorphisms of the fixed [vector space](/page/Vector%20Space) $\mathbb K^n$. Its exterior derivative is the endomorphism-valued $1$-form \begin{align*} dP\in \Omega^1(U;\operatorname{End}_{\mathbb K}(\mathbb K^n)). \end{align*} Here $\Omega^1(U;\operatorname{End}_{\mathbb K}(\mathbb K^n))$ means smooth $1$-forms on $U$ whose coefficients are $\mathbb K$-linear endomorphisms of the fixed vector space $\mathbb K^n$. At every point of $U$, the endomorphism $P$ is a projection, so it satisfies \begin{align*} P^2=P. \end{align*} Differentiating this equation entrywise gives the product rule \begin{align*} d(P^2)=dP. \end{align*} Because $P$ has form degree $0$, there is no sign change in the graded Leibniz rule, and therefore \begin{align*} d(P^2)=dP\,P+P\,dP. \end{align*} Hence \begin{align*} dP\,P+P\,dP=dP. \end{align*} The important consequence is that $dP$ has no component that starts and ends inside the image of $P$. To see this algebraically, multiply the identity on the left by $P$: \begin{align*} P\,dP\,P+P^2\,dP=P\,dP. \end{align*} Since $P^2=P$, this is \begin{align*} P\,dP\,P+P\,dP=P\,dP. \end{align*} Subtracting the common term $P\,dP$ gives \begin{align*} P\,dP\,P=0. \end{align*} This identity is the mechanism behind the phrase “only the normal component of $ds$ contributes” in the curvature computation.[/guided]

[step:Differentiate the tautological section equation] Let \begin{align*} s:U\to \mathbb K^n \end{align*} be a smooth local section of $S$, viewed through the inclusion of $S$ into the trivial bundle $U\times \mathbb K^n$. Since $s$ takes values in the image of the projection $P$ at each point, it satisfies \begin{align*} P s=s. \end{align*} Differentiating this vector-valued identity gives \begin{align*} dP\,s+P\,ds=ds. \end{align*} Let \begin{align*} I:\mathbb K^n\to \mathbb K^n \end{align*} denote the identity endomorphism. Rearranging the last identity gives \begin{align*} (I-P)ds=dP\,s. \end{align*} This identifies the component of $ds$ normal to the image of $P$ with $dP\,s$. [/step]

[step:Compute the curvature action in the ambient trivial bundle] By definition of the universal connection, \begin{align*} \nabla s=P\,ds. \end{align*} For a $\mathbb K^n$-valued differential form $\alpha$ on $U$ whose values lie in $S$, the exterior covariant derivative induced by the projected connection is represented in the ambient trivial bundle by \begin{align*} d^\nabla\alpha=P\,d\alpha. \end{align*} Thus, to compute the curvature action on $s$, view $\nabla s$ as a $\mathbb K^n$-valued $1$-form and apply this extension rule. Therefore \begin{align*} F_\nabla s=d^\nabla(\nabla s)=P\,d(P\,ds). \end{align*} Using the graded Leibniz rule with $P$ of degree $0$ and $ds$ of degree $1$, we obtain \begin{align*} d(P\,ds)=dP\wedge ds+P\,d(ds). \end{align*} Since $s$ is a smooth $\mathbb K^n$-valued function on $U$, the identity $d^2s=0$ in the trivial bundle gives \begin{align*} d(ds)=0. \end{align*} Thus \begin{align*} F_\nabla s=P\,dP\wedge ds. \end{align*} [/step]

[step:Replace $ds$ by its normal component]Decompose the $\mathbb K^n$-valued $1$-form $ds$ using the complementary projections $P$ and $I-P$: \begin{align*} ds=P\,ds+(I-P)ds. \end{align*} Substituting this into the curvature expression gives \begin{align*} F_\nabla s=P\,dP\wedge P\,ds+P\,dP\wedge (I-P)ds. \end{align*} The first term vanishes because $P\,dP\,P=0$. More explicitly, let $\mathfrak X(U)$ denote the space of smooth vector fields on $U$, and let $X,Y\in \mathfrak X(U)$. The value of $P\,dP\wedge P\,ds$ is \begin{align*} (P\,dP)_X(P\,ds)_Y-(P\,dP)_Y(P\,ds)_X, \end{align*} and each summand contains the factor $P\,dP\,P$. Therefore \begin{align*} P\,dP\wedge P\,ds=0. \end{align*} Using the identity $(I-P)ds=dP\,s$ from the tautological section equation, we get \begin{align*} F_\nabla s=P\,dP\wedge dP\,s. \end{align*}[/step]

[guided]At this point we have reduced the curvature computation to \begin{align*} F_\nabla s=P\,dP\wedge ds. \end{align*} The remaining question is which part of $ds$ survives after left multiplication by $P\,dP$. Since $P$ is a projection, the endomorphism $I-P$ is the complementary projection on the ambient trivial bundle. Hence every $\mathbb K^n$-valued $1$-form decomposes as \begin{align*} ds=P\,ds+(I-P)ds. \end{align*} Substituting this decomposition gives \begin{align*} F_\nabla s=P\,dP\wedge P\,ds+P\,dP\wedge (I-P)ds. \end{align*} We now prove that the first term is zero. Let $X,Y\in \mathfrak X(U)$ be smooth vector fields. Evaluating the vector-valued $2$-form $P\,dP\wedge P\,ds$ on $(X,Y)$ gives \begin{align*} (P\,dP\wedge P\,ds)(X,Y)=(P\,dP)(X)(P\,ds)(Y)-(P\,dP)(Y)(P\,ds)(X). \end{align*} Since $(P\,ds)(Y)=P(ds(Y))$ and $(P\,ds)(X)=P(ds(X))$, the two summands contain the endomorphism products \begin{align*} P\,dP(X)\,P \end{align*} and \begin{align*} P\,dP(Y)\,P. \end{align*} The projection identity proved earlier says exactly that $P\,dP\,P=0$ as an endomorphism-valued $1$-form. Therefore both summands vanish, and \begin{align*} P\,dP\wedge P\,ds=0. \end{align*} Thus only the complementary component remains: \begin{align*} F_\nabla s=P\,dP\wedge (I-P)ds. \end{align*} We now rederive the identity for the complementary component inside this guided argument. The section $s$ is tautological, so $Ps=s$. Differentiating this vector-valued identity gives \begin{align*} dP\,s+P\,ds=ds. \end{align*} Rearranging gives \begin{align*} (I-P)ds=dP\,s. \end{align*} Substituting it gives \begin{align*} F_\nabla s=P\,dP\wedge dP\,s. \end{align*} This is the key geometric point: the derivative of a section splits into a tangential part and a normal part, and the curvature is controlled by how the projection changes into the normal direction.[/guided]

[step:Insert the tautological projection on the right] Since $s$ is a section of $S$, we have $s=Ps$. Therefore \begin{align*} F_\nabla s=P\,dP\wedge dP\,P s. \end{align*} Because the computation holds on every open set $U\subset \operatorname{Gr}(k,n)$ and for every local smooth section $s$ of $S$, the curvature of the universal connection is the endomorphism-valued $2$-form \begin{align*} F_\nabla=P\,dP\wedge dP\,P \end{align*} acting on $S$. This is the desired formula. [/step]