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.

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