Scalar curvature of the unit round four-sphere

On the unit sphere let $n(p)=p$ be the outward unit normal, and let $D$ be the flat ambient Euclidean connection. For tangent fields $X,Y$, differentiating $⟨Y,n⟩=0$ and using $D_{X}n=X$ gives $⟨D_{X}Y,n⟩=-⟨X,Y⟩$. Thus the tangential projection is ${\mathrm{\nabla }}_{X}Y=D_{X}Y+⟨X,Y⟩n$. It is metric compatible and torsion free, as follows by projecting the corresponding ambient identities, so it is the Levi–Civita connection of the induced metric. In $D_X{D}_{Y}Z$, the tangential projection is ${\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-⟨Y,Z⟩X$; the analogous expression with $X,Y$ exchanged has the same form. Project the ambient flat identity $D_X{D}_{Y}Z-D_Y{D}_{X}Z-D_{[X,Y]}Z=0$. In the convention $R(X,Y)Z={\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-{\mathrm{\nabla }}_{[X,Y]}Z$, this gives $R(X,Y)Z=⟨Y,Z⟩X-⟨X,Z⟩Y$. Hence for orthonormal $X,Y$, $⟨R(X,Y)Y,X⟩=1$, proving the required sectional curvature rather than assuming it.

In dimension four an orthonormal frame has $\left(\genfrac{}{}{0px}{}{4}{2}\right)=6$ unordered two-planes. By step 1.1 each has sectional curvature $1$, so their sum is $6$.

The scalar-curvature trace sums over ordered pairs $(a,b)$; diagonal terms vanish and every unordered pair occurs twice. Therefore $R=2{\sum }_{1\le a<b\le 4}K(e_a,e_b)=2\cdot 6=12$.