diff --git a/content/notes/math/differential_geometry.mdx b/content/notes/math/differential_geometry.mdx
index 57d5bc8..c617c76 100644
--- a/content/notes/math/differential_geometry.mdx
+++ b/content/notes/math/differential_geometry.mdx
@@ -1256,7 +1256,7 @@ $$
 The Riemannian connection on a smooth, not necessarily orientable surface $M$ in $\R^3$ is given by $\nabla_X Y = \operatorname{pr}(\mathrm{D}_X Y)$.
 </MathBox>
 
-## Gauss's Theorema Egregium
+## Gauss's Theorema Egregium for $\R^3$
 
 Suppose $M$ is a regular submanifold of $\R^n$. For $X$ a tangent vector field on $M$ and $Z$ a vector field along $M$ in $\R^n$, the directional derivative $\mathrm{D}_X Z$ is a vector field along $M$ in $\R^n$. Since the Riemannian connection $\nabla$ of a surface $M$ in $\R^3$ is defined in terms of the directional derivative $\mathrm{D}$ on $M$, it is easy to compare the curvature tensors of $\nabla$ and $\mathrm{D}$. This will lead to a formula for the curvature $R$ of $\nabla$, called the *Gauss curvature equation*.
 
@@ -1468,6 +1468,210 @@ $$
 
 ## Generalizations to hypersurfaces in $\R^{n+1}$
 
+### The shape operator of a hypersurface
+
+A *hypersurface* $M$ in $\R^{n+1}$ is a submanifold of codimension $1$. Assume there is a smooth unit normal vector field $N$ on $M$; note that this is always possible locally on any hypersurface. Recall that the directional derivative $\mathrm{D}$ on $\R^{n+1}$ is the Riemannian connection of $\R^{n+1}$.
+
+For any point $\mathbf{p}\in M$ and tangent vector $X_\mathbf{p}\in T_\mathbf{p}M$, since $\langle N,N\rangle\equiv 1$
+
+$$
+  0 = X_\mathbf{p}\langle N,N\rangle = 2\langle\mathrm{D}_{X_\mathbf{p} N, N}\rangle
+$$
+
+Therefore, $\mathrm{D}_{X_\mathbf{p}}N$ is tangent to $M$. The *shape operator* $L_\mathbf{p}: T_\mathbf{p}M \to T_\mathbf{p}M$ is defined to be
+
+$$
+  L_\mathbf{p}(X_\mathbf{p}) = -\mathrm{D}_{X_\mathbf{p}}N,\; X_\mathbf{p}\in T_\mathbf{p}M
+$$
+
+As the point $\mathbf{p}$ varies over $M$, the shape operator globalizes to an $\mathcal{F}$-linear map $L:\mathfrak{X}(M)\to\mathfrak{X}(M)$ given by $L(X)_\mathbf{p} = L_\mathbf{p}(X_\mathbf{p})$.
+
+<MathBox title='Properties of the shape operator on hypersurfaces in $\R^{n+1}$' boxType='definition'>
+Let $X,Y\in\mathfrak{X}(M)$ be smooth vector fields on an oriented hypersurface $M$ in $\R^{n+1}$. Then
+1. $\langle L(X), Y\rangle = \langle\mathrm{D}_X Y, N\rangle$
+2. the shape operator is self-adjoint with respect to the Euclidean metric inherited from $\R^{n+1}$
+$$
+  \langle L(X),Y\rangle = \langle X,L(Y)\rangle
+$$
+
+<details>
+<summary>Proof</summary>
+
+Since $\langle Y, N\rangle = 0$, by the compatibility of $\mathrm{D}$ with the metric
+
+$$
+  0 = X\langle Y,N\rangle = \langle \mathrm{D}_X Y, N\rangle + \langle Y, \mathrm{D}_X N\rangle
+$$
+
+Hence
+
+$$
+  \langle \mathrm{D}_X Y, N\rangle = \langle Y, -\mathrm{D}_X N\rangle = \langle Y, L(X)\rangle
+$$
+
+Reversing the roles of $X$ and $Y$ gives
+
+$$
+  \langle \mathrm{D}_Y X, N\rangle = \langle X, L(Y)\rangle
+$$
+
+Since $[X,Y]$ is tangent to $M$
+
+$$
+  \langle [X,Y], N\rangle = 0
+$$
+
+By torsion-freeness, subtracting $\langle D_Y X, N\rangle$ and $\langle [X,Y], N\rangle$ from $\langle \mathrm{D}_X Y, N\rangle$ gives
+
+$$
+\begin{align*}
+  0 =& \langle \mathrm{D}_X Y - \mathrm{D}_Y X - [X,Y], N\rangle \\
+  =& \langle Y, L(X)\rangle - \langle X,L(Y)\rangle
+\end{align*}
+$$
+</details>
+</MathBox>
+
+<MathBox title='Principal, mean and Gaussian curvature of hypersurfaces in $\R^{n+1}$' boxType='definition'>
+Let $M$ be a hypersurface in $\R^{n+1}$ and $L_\mathbf{p}: T_\mathbf{p}M \to T_\mathbf{p} M$ be the shape operator at a point $\mathbf{p}\in M$. The following curvature measures can be defined on $M$ at $\mathbf{p}$:
+1. The *principal curvatures* are the eigenvalues $\lambda_i$ of $L_\mathbf{p}$.
+2. The *mean curvature* is the average of the principal curvatures, or the trace of $L_\mathbf{p}$
+$$
+  H(\mathbf{p}) = \frac{1}{n}\sum_{i=1}^n \lambda_i = \frac{1}{n}\operatorname{tr}(L_\mathbf{p})
+$$
+3. The *Gaussian curvature* is the product of the principal curvatures, or the determinant of $L_\mathbf{p}$
+$$
+  K(\mathbf{p}) = \prod_{i=1}^n \lambda_i = \det(L_\mathbf{p})
+$$
+</MathBox>
+
+### The Levi-Civita connection on a hypersurface
+
+<MathBox title='' boxType='theorem'>
+Let $M$ be a hypersurface in $\R^{n+1}$ and $\mathrm{D}$ the directional derivative on $\R^{n+1}$. For $X,Y\in\mathfrak{X}(M)$ the tangential component of $\mathrm{D}_X Y$ defines the Riemannian connection $\nabla$ of $M$
+
+<details>
+<summary>Proof</summary>
+
+Since it is evident that $\nabla$ satisfies the two defining properties of a connection it suffices to check that $\nabla_X Y$ is torsion free and compatible with the metric.
+
+**Torsion-freeness:** Let $T_\mathrm{D}$ and $T_\nabla$ be the torsions of $\mathrm{D}$ and $\nabla$, respectively. By definition, for $X,Y\in\mathfrak{X}(M)$
+
+$$
+\begin{align*}
+  \mathrm{D}_X Y =& \nabla_X Y + (\mathrm{D}_X Y)_{\mathrm{nor}} \\
+  \mathrm{D}_Y X =& \nabla_Y X + (\mathrm{D}_Y X)_{\mathrm{nor}}
+\end{align*}
+$$
+
+Since $\mathrm{D}$ is torsion-free, $\mathrm{D}_X Y - \mathrm{D}_Y X = [X,Y]$. Equating the normal components of both sides, we get $(\mathrm{D}_X Y)_\mathrm{nor} - (\mathrm{D}_Y X)_\mathrm{nor} = 0$. Thus,
+
+$$
+\begin{align*}
+  0 =& T_\mathrm{D}(X,Y) \\
+  =& \mathrm{D}_X Y - \mathrm{D}_Y X - [X,Y] \\
+  =& (\nabla_X Y - \nabla_Y X - [X,Y]) + (\mathrm{D}_X Y)_\mathrm{nor} - (\mathrm{D}_Y X)_\mathrm{nor} \\
+  =& \nabla_X Y - \nabla_Y X - [X,Y] = T_\nabla (X,Y)
+\end{align*}
+$$
+
+**Compatibility with the metric:** For $X,Y,Z\in\mathfrak{X}(M)$
+
+$$
+\begin{align*}
+  X\langle Y, Z\rangle =& \langle\mathrm{D}_X Y, Z\rangle + \langle Y, \mathrm{D}_X Z\rangle \\
+  =& \langle\nabla_X Y + (\mathrm{D}_X Y)_\mathrm{nor} Z \rangle + \langle Y, \nabla_X Z + (\nabla_X Y)_\mathrm{nor}\rangle \\
+  =& \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle
+\end{align*}
+$$
+</details>
+</MathBox>
+
+### Fundamental forms
+
+At each point $\mathbf{p}$ of an oriented hypersurface $M\subseteq\R^{n+1}$, there is a sequence of symmetric bilinear maps on $T_\mathbf{p}M$ called the *first*, *second* and *third fundamental forms* and so on
+
+$$
+\begin{align*}
+  \matrm{I}(X_\mathbf{p},Y_\mathbf{p}) =& \langle X_\mathbf{p}, Y_\mathbf{p}\rangle \\
+  \mathrm{II}(X_\mathbf{p},Y_\mathbf{p}) =& \langle L(X_\mathbf{p}), Y_\mathbf{p}\rangle = \langle X_\mathbf{p}, L(Y_\mathbf{p})\rangle \\
+  \mathrm{III}(X_\mathbf{p},Y_\mathbf{p}) =& \langle L^2 (X_\mathbf{p}), Y_\mathbf{p}\rangle = \langle L(X_\mathbf{p}),L(Y_\mathbf{p})\rangle = \langle X_\mathbf{p}, L^2 (Y_\mathbf{p})\rangle, \dots
+\end{align*}
+$$
+
+For $X,Y\in\mathfrak{X}(M)$, the directional derivative $\mathrm{D}_X Y$ decomposes into the sum of a tangential and normal component. The tangential component is the Levi-Civita connection on $M$; the normal component is essentially the second fundamental form.
+
+<MathBox title='' boxType='theorem'>
+If $N$ is a smooth unit normal vector field on the hypersurface $M$ in $\R^{n+1}$ and $X,Y\in\mathfrak{X}(M)$ are smooth vector fields on $M$, then
+
+$$
+  (\mathrm{D}_X Y)_\mathrm{nor} = \mathrm{II}(X,Y)N
+$$
+
+<details>
+<summary>Proof</summary>
+
+The normal component $(\mathrm{D}_X Y)_\mathrm{nor}$ is a multiple of $N$, so
+
+$$
+  \mathrm{D}_X Y =& (\mathrm{D}_X Y)_\mathrm{tan} + (\mathrm{D}_X Y)_\mathrm{nor} \\
+  =& lambda_X Y + \lambda N
+$$
+
+for some $\lambda\in\R$. Taking the inner product of both sides with $N$ gives
+
+$$
+  \langle \mathrm{D}_X,N\rangle = \lambda\langle N,N\rangle = \lambda
+$$
+
+Hence
+
+$$
+\begin{align*}
+  (\mathrm{D}_X Y)_\mathrm{nor} =& \langle\mathrm{D}_X Y,N\rangle N \\
+  =& \langle L(X), Y\rangle N \\
+  =& \mathrm{II}(X,Y)N
+\end{align*}
+$$
+</details>
+</MathBox>
+
+### The Gauss curvature and Codazzi-Mainardi equations
+
+Let $M$ be an oriented, smooth hypersurface in $\R^{n+1}$, with a smooth unit normal vector field $N$. The relation between the Levi-Civita connection $\mathrm{D}$ on $\R^{n+1}$ and the Levi-Civita connection on $M$ implies a relation between the curvature $R_\mathrm{D}(X,Y)Z = 0$ on $\R^{n+1}$ and the curvature $R(X,Y)Z$ on $M$. The tangential component of this relation gives the Gauss curvature equation
+
+$$
+  R(X,Y) = \langle L(Y), Z\rangle L(X) - \langle L(X), Z\rangle L(Y)
+$$
+
+and the normal component gives the Codazzi-Mainardi equation
+
+$$
+  \nabla_X L(Y) - \nabla_Y L(X) - L([X,Y]) = 0
+$$
+
+<MathBox title='' boxType='corollary'>
+If $M$ is a smooth oriented hypersurface in $\R^{n+1}$ with curvature $R$ and $X,Y\in\mathfrak{X}(M)$, then
+
+$$
+  \langle R(X,Y)Y, X\rangle = \mathrm{II}(X,X)\mathrm{II}(Y,Y) - \mathrm{II}(X,Y)^2
+$$
+
+<details>
+<summary>Proof</summary>
+
+By the Gauss curvature equation
+
+$$
+\begin{align*}
+  \langle R(X,Y)Y, X\rangle =& \langle L(Y), Y\rangle \langle L(X),X\rangle - \langle L(X), Y\rangle\langle L(Y),X\rangle \\
+  =& \mathrm{II}(X,X)\mathrm{II}(Y,Y) - \mathrm{II}(X,Y)^2
+\end{align*}
+$$
+</details>
+</MathBox>
+
+
 # Curvature and differential forms
 
 # Geodesics