Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 115 (Differential Geometry)

Let $\overline{\nabla }$ be the Euclidean connection and split the ambient tangent bundle along the embedded submanifold as $T{\mathbb{R}}^{n+m}{|}_M=TM\oplus NM$. The second fundamental form is the normal-bundle-valued bilinear formIt is symmetric because $\overline{\nabla }$ is torsion-free and the Lie bracket of tangent vector fields remains tangent:This is the symmetry of the second fundamental form.

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With the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Gauss equation for a Euclidean embedded submanifold isThe Codazzi equation iswhere the derivative uses the Levi-Civita connection on tangent arguments and the normal connection on the value of $A$.

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On the unit sphere $S^n\subseteq {\mathbb{R}}^{n+1}$ choose the outward unit normal $N(x)=x$. For tangent vector fields $X,Y$,soIf $X,Y$ are orthonormal, the Gauss equation givesThus the round unit sphere has sectional curvature one. Tracing over an orthonormal basis gives its scalar curvature

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The metric and orientation determine the Riemannian volume form: in a positively oriented coordinate chart $(x^1,\dots ,x^n)$,Writing $|g|=\det (g_{ij})$ and $(g^{ij})=(g_{ij}{)}^{-1}$, the Dirichlet energy on a Riemannian manifold with source $f$ is

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For an arbitrary smooth variation $u_t=u+t\phi$, differentiation under the integral givesSince $M$ is compact without boundary, integration by parts turns this intoThe fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationor equivalently $-{\Delta }_{g}u=f$ for the Laplace-Beltrami operator.

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Integrating the Euler-Lagrange equation and applying the divergence theorem on the compact boundaryless manifold givesEquivalently, if this integral were nonzero, replacing $u$ by $u+C$ would leave the gradient term unchanged and make the energy unbounded below in one direction, so no minimizer could exist.

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WriteThe identity belongs to $G$. If $A,B\in G$, thenTaking determinants in $A^T\Omega A=\Omega$ shows that $A$ is invertible, and multiplying that identity by $A^{-T}$ and $A^{-1}$ gives $(A^{-1}{)}^T\Omega A^{-1}=\Omega$. Thus $G$ is a group.

Let ${\mathrm{Skew}}_{2n}(\mathbb{R})$ denote the vector space of skew-symmetric $2n\times 2n$ matrices and defineThen $G=F^{-1}(\Omega )$. At $A\in G$,Given any skew-symmetric matrix $S$, set $H=AB$ with $B=-\frac{1}{2}\Omega S$. Since $A^T\Omega A=\Omega$, a direct calculation givesThe derivative is therefore surjective at every point of $F^{-1}(\Omega )$. The regular level set theorem proves that the symplectic group is an embedded submanifold of $M_{2n}(\mathbb{R})$.

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The ambient matrix space has dimension $4n^2$, while the space of skew-symmetric $2n\times 2n$ matrices has dimensionBecause $\Omega$ is a regular value, the codimension of its level set is the dimension of the target. Hence

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Ambient matrix multiplicationis bilinear and therefore smooth. Its restriction to the embedded submanifold $G\times G$ is smooth, and part a shows that its image lies in $G$. By the defining smooth structure on an embedded submanifold, this restriction is a smooth map $G\times G\to G$.

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Use the conventionAn embedded submanifold is minimal when its mean curvature vector $\mathbf{H}$ vanishes identically. For a variation $M_t$ with velocity field $V$, the first variation of area iswhere $\eta$ is the outward unit conormal. For a boundaryless $M$, only the first integral remains. This sign convention is consistent with the outward variation of a round sphere increasing its area, because its $\mathbf{H}$ points inward.

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Let $\overline{\nabla }f$ and $\overline{\mathrm{Hess}}f$ be the ambient gradient and Hessian. The Laplacian of a restricted ambient function isTo prove it, choose a local orthonormal tangent frame $e_1,\dots ,e_n$ with ${\nabla }_{e_i}^M{e}_j=0$ at the point under consideration. There,Both sides are intrinsic scalars, so the pointwise calculation proves the formula everywhere.

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Suppose that a positive-dimensional compact boundaryless minimal submanifold $M^n\subseteq {\mathbb{R}}^{n+m}$ existed. Apply part b to the squared ambient distance $f(x)=|x{|}^2$. Its ambient Hessian is $2$ times the Euclidean metric, its gradient is $2x$, and $\mathbf{H}=0$, soCompactness makes $|x{|}^2$ attain a maximum. The Laplacian at a local maximum is nonpositive, contradicting $2n>0$. Hence no such compact Euclidean minimal submanifold exists.

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