Levi-Civita Connection

Abstract

The Levi-Civita connection of a surface $f\colon M \to \mathbb {R}^3$ provides a geometrically meaningful way to take directional derivatives of a vector field Y  on M. Based on the Levi-Civita connection we derive two important equations that are satisfied by the curvature of a surface in $\mathbb {R}^3$: the Gauss equation and the Codazzi equation.

The Levi-Civita connection of a surface $f\colon M \to \mathbb {R}^3$ provides a geometrically meaningful way to take directional derivatives of a vector field Y  on M. Based on the Levi-Civita connection we derive two important equations that are satisfied by the curvature of a surface in $\mathbb {R}^3$: the Gauss equation and the Codazzi equation.

1 Derivatives of Vector Fields

Let $f\colon M \to \mathbb {R}^3$ be a regular surface in $\mathbb {R}^3$ with unit normal $N\colon M\to \mathbb {R}^3$. Let $Y\in \Gamma (TM)$ be a smooth vector field on M, $p\in M$ and $X\in T_pM$. Then, differentiating $0=\langle N,df(Y)\rangle$ in the direction of X, we obtain

$$\displaystyle \begin{aligned} 0=d_X\langle N, df(Y)\rangle = \langle df(AX),df(Y)\rangle + \langle N,d_Xdf(Y)\rangle.\end{aligned} $$

Fixing a point $p\in M$, we can decompose every vector $\mathbf {v}\in \mathbb {R}^n$ uniquely as

$$\displaystyle \begin{aligned} \mathbf{v}=\lambda N +df(Z)\end{aligned}$$

for some $\lambda \in \mathbb {R}$ and some $Z\in T_pM$. The vector $\lambda N$ is called the normal part of $\mathbf {v}$ and $df(Z)$ is called the tangential part of $\mathbf {v}$. In our case, the normal part of $d_Xdf(Y)$ equals $-\langle AX,Y\rangle N$. The tangential part is of the form $df(Z)$ for some vector $Z\in T_pM$ that we denote by $(\nabla Y)(X)$ or also by $\nabla _XY$. This gives us

$$\displaystyle \begin{aligned} d_Xdf(Y)=-\langle AX,Y\rangle N +df(\nabla_X Y).\end{aligned}$$

We leave it to the reader to show that the map $Y\mapsto \nabla Y$ is linear.

Definition 9.1

The linear map $\nabla \colon \Gamma (TM) \to \Gamma (\mathrm {End}\,TM)$ that assigns to a vector field Y  the endomorphism field $\nabla Y$ is called the Levi-Civita connection of f.

$\nabla _XY$ can be interpreted as the directional derivative of the vector field Y  in the direction of X (Fig. 9.1).

Fig. 9.1

Two vector fields $X,Y\in \Gamma (TM)$

Here are some useful properties of the Levi-Civita connection:

Theorem 9.2

Let$X, Y,Z$be vector fields on M and$\lambda \colon M\to \mathbb {R}$a smooth function. Then

• (i) $\nabla _X (\lambda Y) = (d_X \lambda ) Y + \lambda \nabla _X Y$

• (ii) $d_X \langle Y,Z\rangle = \langle \nabla _X Y,Z\rangle +\langle Y,\nabla _X Z\rangle$

Proof

Equation (i) is left as an exercise. For equation (ii) we have

$$\displaystyle \begin{aligned} d_X \langle Y,Z\rangle &= d_X \langle df(Y),df(Z)\rangle \\ &= \langle d_X df(Y), df(Z)\rangle + \langle df(Y),d_X df(Z)\rangle \\&= \langle df(\nabla_XY),df(Z)\rangle +\langle df(Y), df(\nabla_XZ)\rangle \\&= \langle \nabla_XY,Z\rangle + \langle Y,\nabla_XZ\rangle.\end{aligned} $$

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We can also use $\nabla$ to define directional derivatives of endomorphism fields (as defined in Definition 6.19):

Theorem 9.3

If B is a smooth endomorphism field on M and$X\in \Gamma (TM)$is a vector field, then there is a unique smooth endomorphism field$(\nabla _X B)$on M such that for all$Y\in \Gamma (TM)$the following Leibniz rule holds:

$$\displaystyle \begin{aligned} \nabla_X(BY)=(\nabla_X B)Y+B(\nabla_XY).\end{aligned}$$

Proof

Define C as the unique endomorphism field on M for which

$$\displaystyle \begin{aligned} CU&=\nabla_X(BU)-B\nabla_XU \\ CV&=\nabla_X(BV)-B\nabla_XV.\end{aligned} $$

Clearly, if the endomorphism field $\nabla _XB$ exists, it has to be equal to C. This proves the uniqueness part of the theorem. To prove existence, we show that C has the property we claim for $\nabla _XB$. Let us write Y  as a linear combination of U and V :

$$\displaystyle \begin{aligned} Y=aU+bV.\end{aligned}$$

Then, by Theorem 9.2,

$$\displaystyle \begin{aligned} \nabla_X(BY)&= \nabla_X(aBU+bBV)\\&=(d_Xa)BU +a\nabla_X (BU) +(d_Xb)BV+b\nabla_X(BV)\\&=(d_Xa)BU +a(CU+B\nabla_XU)+(d_Xb)BV+b(CV+B\nabla_XV)\\&=CY+B\nabla_X Y.\end{aligned} $$

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A smooth endomorphism fieldB for which $\nabla B=0$ is said to be parallel.

Theorem 9.4

Let$X, Y,Z$be vector fields on M and$\lambda \colon M\to \mathbb {R}$a smooth function. Then

• (i) $\nabla _X (JY) = J\nabla _X Y$

• (ii) $d_X \det ( Y,Z) = \det (\nabla _X Y,Z)+\det ( Y,\nabla _X Z).$

Proof

For Equation (i) we choose a positively oriented orthonormal vector fields $X,Y\in \Gamma (TM)$. Then $JX=Y$ and $JY=-X$ and any vector field $W\in \Gamma (TM)$ can be written as $W=\beta X + \delta Y$. Then for $Z\in \Gamma (TM)$

$$\displaystyle \begin{aligned} \nabla_Z (JW) = \nabla_Z(\beta Y - \delta X) = (d_Z\beta) Y + \beta\nabla_ZY - (d_Z\delta) X - \delta \nabla_ZX\, \end{aligned} $$

where we use equation (ii) of Theorem 9.2 for the last equality. Moreover,

$$\displaystyle \begin{aligned} J\nabla_ZW &= J\left( (d_Z\beta)X + \beta\nabla_ZX + (d_Z\delta)Y + \delta\nabla_ZY \right)\\ &= (d_Z\beta)Y + \beta J\nabla_ZX - (d_Z\delta)X + \delta J\nabla_ZY\,. \end{aligned} $$

We define $\omega (Z):= \langle \nabla _ZX,Y\rangle$ and conclude that

$$\displaystyle \begin{aligned} \nabla_Z (JW) = (d_Z\beta)Y - \beta\,\omega(Z)X - (d_Z\delta)X - \delta\,\omega(Z)Y = J\nabla_ZW\,. \end{aligned} $$

In physics, the 1-form $\omega$ is known as the angular velocity of the orthonormal basis $X_p,Y_p\in T_pM$ in the direction $Z_p\in T_pM$.

Now equation (ii) follows:

$$\displaystyle \begin{aligned} d_X \det(Y,Z)&= d_X \langle JY,Z \rangle \\&= \langle \nabla_X(JY),Z\rangle + \langle JY, \nabla_X Z\rangle \\&= \langle J(\nabla_XY),Z\rangle + \langle JY, \nabla_X Z\rangle \\&= \det (\nabla_X Y,Z)+\det( Y,\nabla_X Z).\end{aligned} $$

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Remark 9.5

Note that the equations from 9.2 and the second equation from 9.4 have the flavor of a Leibniz rule.

Theorem 9.6

For the coordinate vector fields we have

$$\displaystyle \begin{aligned} \nabla_UV=\nabla_VU.\end{aligned}$$

Proof

This can be seen by looking at the tangential component of

$$\displaystyle \begin{aligned} -\langle AU,V\rangle N+df(\nabla_UV)=f_{vu}=f_{uv}=-\langle AV,U\rangle N+df(\nabla_VU).\end{aligned}$$

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2 Equations of Gauss and Codazzi

The derivative $\nabla A$ (defined in Theorem 9.3) of the shape operator A of a surface $f\colon M\to \mathbb {R}$ has an important symmetry property:

Theorem 9.7

For all vector fields$X,Y\in \Gamma (TM)$theCodazzi equationholds:

$$\displaystyle \begin{aligned} (\nabla_XA)Y=(\nabla_YA)X.\end{aligned}$$

Proof

We can write X and Y  as linear combinations (with functions as coefficients) of U and V . If we expand both sides of the equation in question accordingly, we see that it is sufficient to consider the special case $X=U$ and $Y=V$. In this case, our claim follows from the fact that partial derivatives of N commute: Equality for the normal part of

$$\displaystyle \begin{aligned} -\langle AU,AV\rangle N +df(\nabla_U(AV)) &=d_U df(AV)\\ &=d_Ud_V(N) \\ &= d_Vd_U(N) \\&=d_V df(AU) \\&= -\langle AV,AU\rangle N +df(\nabla_V(AU))\end{aligned} $$

is automatically satisfied, while the tangential part gives us what we want to prove. □

There is another important relation between the shape operator A and the Levi-Civita connection $\nabla$, the so-called Gauss equation:

If $h\colon M\to \mathbb {R}^k$ is a smooth function, then the partial derivatives of h commute, i.e.

$$\displaystyle \begin{aligned} d_Ud_Vh-d_Vd_Uh=0.\end{aligned}$$

For vector fields this is not true, and the failure of “partial derivatives” of vector fields to commute is determined by the Gaussian curvature of f:

Theorem 9.8

For any vector field$Z\in \Gamma (TM)$theGauss equationholds:

$$\displaystyle \begin{aligned} \nabla_U \nabla_V Z -\nabla_V \nabla_U Z = -K \det(U,V)JZ\end{aligned}$$

where$K=\det A$is the Gaussian curvature of f.

Proof

Collecting only the terms that are orthogonal to N in

$$\displaystyle \begin{aligned} d_U (-\langle AV,Z\rangle N +df(\nabla_V Z))&=d_U d_V df(Z)\\&=d_V d_U df(Z)\\&=d_V(-\langle AU,Z\rangle N+df(\nabla_U Z))\end{aligned} $$

we obtain

$$\displaystyle \begin{aligned} -\langle AV,Z\rangle AU +\nabla_U\nabla_V Z = -\langle AU,Z\rangle AV+\nabla_V\nabla_U Z.\end{aligned}$$

Substituting in Theorem 6.24AU for X, AV  for Y  and using

$$\displaystyle \begin{aligned} \det(AU,AV)=\det A \det(U,V).\end{aligned}$$

we arrive at the equality this we wanted to prove. □

3 Theorema Egregium

The following theorem is due to Gauss. He called it the “Theorema Egregium”, which means “most excellent theorem”.

Theorem 9.9 (Theorema Egregium)

Suppose that the surfaces$f,\tilde {f}\colon M\to \mathbb {R}^3$induce the same Riemannian metric on M. Then f and$\tilde {f}$have the same Gaussian curvature$K\colon M\to \mathbb {R}$.

Proof

By the Gauss equation (Theorem 9.8), it is sufficient to prove that if f and $\tilde {f}$ induce the same Riemannian metric on M, they also induce the same Levi-Civita connection. This in turn follows from Theorem 9.10. □

Theorem 9.10

Suppose that the surfaces$f,\tilde {f}\colon M\to \mathbb {R}^3$induce the same Riemannian metric on M. Then the Levi-Civita connections of f and$\tilde {f}$are identical.

Proof

We show that the Levi-Civita connection $\nabla$ induced on M by f is already completely determined by the induced metric $\langle \,,\rangle _{f}$. By Theorem 9.6 and the second equation of Theorem 9.2

$$\displaystyle \begin{aligned} \langle\nabla_UU,U\rangle &= \frac{1}{2}d_U\langle U,U\rangle \\ \langle \nabla_UU,V\rangle &= d_U\langle U,V\rangle -\langle U,\nabla_UV\rangle \\&= d_U\langle U,V\rangle -\langle U,\nabla_VU\rangle \\&= d_U\langle U,V\rangle -\frac{1}{2}d_V\langle U,U\rangle\\ \langle \nabla_UV,U\rangle &= \langle\nabla_VU,U\rangle \\&=\frac{1}{2}d_V\langle U,U\rangle \\ \langle \nabla_UV,V\rangle &=\frac{1}{2}d_U\langle V,V\rangle.\end{aligned} $$

Hence $\nabla _UU$ and $\nabla _UV=\nabla _VU$ are completely determined, as well as (by a similar calculation) $\nabla _VV$. Therefore, $\nabla U$ and $\nabla V$ are completely determined by the knowledge of $\langle \,,\rangle _{f}$ alone. By the first equation of Theorem 9.2, then also $\nabla Y$ is determined for an arbitrary vector field $Y=b_1U+b_2V$. □

Figure 9.2 reveals the reason why the leather patch on the smoothed dodecahedron in Fig. 6.6 was stuck.

Fig. 9.2

By the Theorema Egregium, any isometric motion of the patch has to preserve Gaussian curvature (indicated by color), so the patch cannot slide