# Connections, The Kähler Condition, and Curvature – VI. Relation of Curvature Tensor with Gaussian Curvature

We want to link our definition to the classical definition of Gaussian curvature. Suppose we have a surface ${S}$ in ${\mathbb{R}^3}$ given by

$\displaystyle \vec{r} = \vec{r}(u, v).$

We give ${S}$ the metric induced from the Euclidean metric of ${\mathbb{R}^3}$. Then the metric in terms of the coordinates ${u}$ and ${v}$ is given by

$\displaystyle d\vec{r} \cdot d\vec{r} = E\,du^2 + 2F\,du\,dv + G\,dv^2,$

where

$\displaystyle E = \vec{r}_u \cdot \vec{r}_u,\text{ }F = \vec{r}_u \cdot \vec{r}_v,\text{ }G = \vec{r}_v \cdot \vec{r}_v,$

and the subscripts ${u}$ and ${v}$ mean partial differentiation with respect to ${u}$ and ${v}$.

Let ${\vec{n}}$ be the unit normal to the surface ${S}$. Choose a point ${P}$ in the surface ${S}$, and let ${C}$ be a curve cut out from ${S}$ by a plane ${\Pi}$ through ${P}$ containing ${\vec{n}(P)}$. Let

$\displaystyle u = u(t),\text{ }v = v(t)$

be the equations defining ${C}$ with the parameter ${t}$ equal to the arc length of ${C}$. We now compute the curvature of the curve ${C}$. Let ${\vec{r}'}$ and ${\vec{r}''}$ denote respectively the first and second order derivatives of ${\vec{r}}$ with respect to ${t}$. Since ${\vec{r}'}$ is a unit vector, the curvature of ${C}$ is given by the length of ${\vec{r}''}$. Since ${C}$ is a plane curve on the same plane as ${\vec{n}(P)}$, the vector ${\vec{r}''(P)}$ is parallel to ${\vec{n}(P)}$. Hence the curvature ${\kappa}$ of ${C}$ at ${P}$ is given by ${\vec{r}'' \cdot \vec{n}}$, which by the chain rule is equal to

$\displaystyle (\vec{r}_{uu} \cdot \vec{n})u'^2 + 2(\vec{r}_{uv} \cdot \vec{n})u'v' + (\vec{r}_{vv} \cdot \vec{n})v'^2,$

where the primes for ${u}$ and ${v}$ mean differentiation with respect to ${t}$. Let

$\displaystyle D = \vec{n} \cdot \vec{r}_{uu},\text{ }D' = \vec{n} \cdot \vec{r}_{uv},\text{ }D'' = \vec{n} \cdot \vec{r}_{vv}.$

Since ${t}$ is the arc length of ${C}$, we can write

$\displaystyle \kappa = {{D\,du^2 + 2D'\,du\,dv + D''\,dv^2}\over{E\,du^2 + 2F\,du\,dv + F\,dv^2}}.$

The expression makes ${\kappa}$ the quotient of two quadratic forms whose variables are the components of the tangent vector at ${P}$ which belongs to ${\Pi}$. As we change this tangent vector, or equivalent, as we change ${\Pi}$, the value of ${\kappa}$ changes, and we have two extremal values. The product of these two extremal values is known as the Gaussian curvature. We let

$\displaystyle \zeta = {{du}\over{dt}},\text{ }\eta = {{dv}\over{dt}}.$

We have

$\displaystyle \kappa(\zeta, \eta)\left(E\,\zeta^2 + 2F\,\zeta\,\eta + G\,\eta^2\right) = D\,\zeta^2 + 2D'\,\zeta\,\eta + D''\,\eta^2.$

By differentiating with respect to ${\zeta}$ and ${\eta}$, we conclude that the extremal values of ${\kappa}$ satisfy the two equations

$\displaystyle (D - \kappa E)\zeta + (D' - \kappa F)\eta = 0,\text{ }(D' - \kappa F)\zeta + (D'' - \kappa G)\eta = 0.$

The vanishing of the determinant of the coefficients of ${\zeta}$ and ${\eta}$ yields a quadratic equation in ${\kappa}$. The product of the two roots of ${\kappa}$ in this equation can be read off from the coefficients of the equation. So we conclude the Gaussian curvature is given by

$\displaystyle {{DD'' - D'^2}\over{EG - F^2}}.$

Another way to see it is that the Gaussian curvature is the product of the eigenvalues of the matrix

$\displaystyle \begin{pmatrix} D & D' \\ D' & D'' \end{pmatrix}$

with respect to the matrix

$\displaystyle \begin{pmatrix} E & F \\ G & H \end{pmatrix},$

and so is equal to the quotient of the determiannts of the two matrices.

Now we verify that the Gaussian curvature agrees with the notion of curvature defined as a measure of the failure of the commutativity of partial covariant differentiation. We define covariant differentiation of a vector field on ${S}$ as the orthogonal projection of the usual differentiation onto the tangent planes of ${S}$. First, let us check that this is a torsion-free connection compatible with the metric. Suppose we have a vector field ${\vec{a}}$ along a curve ${\gamma}$ of ${S}$ parametrized by ${t}$ and assume that the covariant derivative of ${\vec{a}}$ along the curve vanishes. Then ${(d/dt)\vec{a}}$ is proportional to ${\vec{n}}$ and

$\displaystyle {d\over{dt}}(\vec{a} \cdot \vec{a}) = 2\left({d\over{dt}} \vec{a}\right) \cdot \vec{a} = 0$

and ${\vec{a} \cdot \vec{a}}$ is constant. Next, we have to verify that the Christoffel symbols with respect to a local coordinate system are symmetric in the two covariant indices. The vectors ${\partial/\partial u}$ and ${\partial/\partial v}$ are respectively the vectors ${\vec{r}_u}$ and ${\vec{r}_v}$. The total covariant derivative of ${\partial/\partial u}$ is the projection of ${d\vec{r}_u}$ onto the tangent plane of ${S}$, i.e. ${d\vec{r}_u - (d\vec{r}_u \cdot \vec{n})\vec{n}}$. Relabel the two variables ${u, v}$ respectively as ${x^1, x^2}$. Then

$\displaystyle \vec{r}_{x^jx^k} - (\vec{r}_{x^jx^k} \cdot \vec{n})\vec{n} = \Gamma_{jk}^i \vec{r}_{x^i}.$

Clearly the Christoffel symbols ${\Gamma_{jk}^i}$ are symmetric in ${k}$ and ${j}$.

Let us now compute the curvature from the connection. We have seen that ${\nabla_j(\partial/\partial x^k)}$ corresponds to ${\vec{r}_{x^jx^k} - (\vec{r}_{x^jx^k} \cdot \vec{n})\vec{n}}$. Then ${\nabla_\ell \nabla_j(\partial/\partial x^k)}$ corresponds to the orthogonal projection ${(\partial/\partial x^\ell)(\vec{r}_{x^j x^k} - (\vec{r}_{x^j x^k} \cdot \vec{n})\vec{n})}$ onto the tangent plane of ${S}$. Now

$\displaystyle {\partial\over{\partial x^\ell}}(\vec{r}_{x^j x^k} - (\vec{r}_{x^j x^k} \cdot \vec{n})\vec{n}) = \vec{r}_{x^j x^k x^\ell} - (\vec{r}_{x^j x^k} \cdot \vec{n})\vec{n}_{x^\ell} \text{ mod }\vec{n}.$

So ${\partial_\ell \partial_j (\partial/\partial x^k) - \partial_j \partial _\ell(\partial/\partial x^k)}$ corresponds to the orthogonal projection of ${(\vec{r}_{x^\ell x^k} \cdot \vec{n})\vec{n}_{x^j} - (\vec{r}_{x^j x^k} \cdot \vec{n}) \vec{n}_{x^\ell}}$ onto the tangent plane of ${S}$. Since ${\vec{n}}$ is a unit vector, it is always perpendicular to its derivatives. So ${(\vec{r}_{x^\ell x^k} \cdot \vec{n}) \vec{n}_{x^j} - (\vec{r}_{x^j x^k} \cdot \vec{n})\vec{n}_{x^\ell}}$ equals its own orthogonal projection onto the tangent plane of ${S}$. We assume without loss of generality that at ${P}$, the two vectors ${\vec{r}_u}$ and ${\vec{r}_v}$ are orthonormal. Then ${R_{kij\ell}}$ is the inner product of ${(\vec{r}_{x^\ell x^k} \cdot \vec{n})\vec{n}_{x^j} - (\vec{r}_{x^j x^k} \cdot \vec{n}) \vec{n}_{x^\ell}}$ with ${\vec{r}_{x^i}}$. So

$\displaystyle R_{kij\ell} = (\vec{r}_{x^\ell x^k} \cdot \vec{n})(\vec{n}_{x^j} \cdot \vec{r}_{x^i}) - (\vec{r}_{x^j x^k} \cdot \vec{n})(\vec{n}_{x^\ell} \cdot \vec{r}_{x^i}),$

which is equal to

$\displaystyle -(\vec{r}_{x^\ell x^k} \cdot \vec{n})(\vec{r}_{x^i x^j} \cdot \vec{n}) + (\vec{r}_{x^j x^k} \cdot \vec{n})(\vec{r}_{x^\ell x^i} \cdot \vec{n})$

from differentiating

$\displaystyle \vec{r}_{x^i} \cdot \vec{n} = 0$

with respect to ${x^j}$. In particular, ${R_{1212}}$ with respect to an orthonormal frame is

$\displaystyle -(\vec{r}_{uv} \cdot \vec{n})^2 + (\vec{r}_{uu} \cdot \vec{n})(\vec{r}_{vv} \cdot \vec{n}) = DD'' - D'^2 = {{DD'' - D'^2}\over{EG - F^2}},$

which is the Gaussian curvature. Incidentally, this verification proves the theorem of Gauss that the Gaussian curvature of a surface in ${\mathbb{R}^3}$ depends only on the first fundamental form.