[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*}
[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]
[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*}
[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]
[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]
