[Tu Differential Geometry] 6. Affine Connections

This article is one of Manifold, Differential Geometry, Fibre Bundle.

 

Consider a smooth vector field $Y$ on a manifold $M$ and a tangent vector $X_p\in T_pM$ at a point $p$ in $M$. To define the directional derivative of $Y$ in the direction $X_p$, it is necessary to compare the values of $Y$ in a neighborhood of $p$. If $q$ is a point near $p$, in general it is not possible to compare the vectors $Y_q$ and $Y_p$ by taking the difference $Y_q-Y_p$, since they are in distinct tangent spaces. For this reason, the directional derivative of a vector field on an arbitrary manifold $M$ cannot be defined in the same way as in Section 4. Instead, we extract from the directional derivative in $\mathbb{R}^n$ certain key properties and call any map $D:\mathfrak{X}(M)\times \mathfrak{X}(M)\rightarrow \mathfrak{X}(M)$ with these properties an affine connection. Intuitively, an affine connection on a manifold is simply a way of differentiating vector fields on the manifold.

Mimicking the directional derivative in $\mathbb{R}^n$, we define the torsion and curvature of an affine connection on a manifold $M$. Then both torsion and curvature are linear over $C^\infty$ functions in every argument.

We will see in a later section that there are infinitely many affine connections on any manifold. On a Riemannian manifold, however, there is a unique forsion-free affine connection compatible with the metric, called the Riemannian of Levi-Civita connection. As an example, we describle the Riemannian connection on a surface in $\mathbb{R}^3$.

6.1 Affine Connections

On an arbitrary manifold $M$, which is not necessarily embedded in a Euclidean space, we can define the directional derivative of a $C^\infty$ function $f$ in the direction $X_p\in T_pM$ in the same way as before: $$\nabla_{X_p}f=X_p f.$$ However, there is no longer a canonical way to define the directional derivative of a vector field $Y$. 

Whatever definition of directional derivative of a vector field one adopts, it should satisfy the properties in Proposition 4.2. This motivates the following definition.

Definition 6.1. An affine connection on a manifold $M$ is an $\mathbb{R}$-bilinear map $$\nabla:\mathfrak{X}(M)\times \mathfrak{X}(M)\rightarrow \mathfrak{X}(M),$$ written $\nabla_X Y$ for $\nabla(X,Y)$, satisfying the two properties below: if $\mathcal{F}$ is the ring $C^\infty(M)$ of $C^\infty$ functions on $M$, then for all $X, Y\in \mathbb{X}(M)$, 

(i) $\nabla_XY$ is $\mathcal{F}$-linear in $X$,

(ii) (Leibniz rule) $\nabla_XY$ satisfies the Leibniz rule in $Y$: for $f\in \mathcal{F}$, $$\nabla_X(fY)=(Xf)Y+f\nabla_XY.$$

6.2 Torsion and Curvature

Given an affine connection $\nabla$ on a manifold $M$, one might ask whether it satisfies the same properties as Proposition 4.3 for the Euclidean connection on $\mathbb{R}^n$. For $X,Y\in \mathfrak{F}(M)$, set $$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]\in \mathfrak{X}(M),\\ R(X,Y)=[\nabla_X,\nabla_Y]-\nabla_{[X,Y]}=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}\in \mbox{End}(\mathfrak{X}(M)).$$ We call $T$ the torsion and $R$ the curvature of the connection. There does not seem to be a good reason for calling $T(X,Y)$ the torsion, but as we shall see in Section 8, $R(X,Y)$ is intimately related to the Gaussian curvature of a surface.

An affine connection $\nabla$ on a manifold $M$ gives rise to a linear map $$\mathfrak{X}(M)\rightarrow \mbox{End}_\mathbb{R}(\mathfrak{X}(M)),\quad X\mapsto \nabla_X.$$ Here both vector spaces $\mathfrak{X}(M)$ and $\mbox{End}_\mathbb{R}(\mathfrak{X}(M))$ are Lie algebras. The curvature measures the deviation of the map $X\mapsto \nabla_X$ from being a Lie algebra homomorphism.

Recall that $\mathcal{F}$ is the ring $C^\infty(M)$ of $C^\infty$ functions on the manifold $M$. Although $\nabla_XY$ is not $\mathcal{F}$-linear in $Y$, it truns out, amazingly, that both torsion and curvature are $\mathcal{F}$-linear in all their arguments.

Proposition 6.3 (Property of Torsion and Curvature). Let $X,Y,Z$ be smooth vector fields on a manifold $M$ with affine connection $\nabla$.

(i) The torsion $T(X,Y)$ is $\mathcal{F}$-linear in $X$ and in $Y$.

(ii) The curvature $R(X,Y)Z$ is $\mathcal{F}$-linear in $X$, $Y$, and $Z$.

Proof. We will first check tha $\mathcal{F}$-linearity of $R(X,Y)Z$ in $X$. For this, it is useful to recall the following formula from the theory of manifolds: if $f,g\in \mathcal{F}=C^infty(M)$, then $$\begin{align}[fX,gY]=fg[X,Y]+f(Xg)Y-g(Yf)X.\end{align}$$

By the definition of curvature, $$\begin{align}R(fX,Y)Z=\nabla_{fX}\nabla_YZ-\nabla_Y\nabla_{fX}Z-\nabla_{[fX,Y]}Z.\end{align}$$ By the $\mathcal{F}$-linearity of $\nabla$ in $X$, the first term in (2) is $f\nabla_X\nabla_YZ$, and by the Leibniz rule, the second term is $$\nabla_Y(f\nabla_XZ)=(Yf)\nabla_XZ+f\nabla_Y\nabla_XZ.$$ Applying (1), the last term in (2) is $$\nabla_{[fX,Y]}Z=\nabla_{f[X,Y]-(Yf)X}Z=f\nabla_{[X,Y]}Z-(Yf)\nabla_XZ,$$ since $X(1)=0$. Combining the three terms gives $$R(fX,Y)Z=f(\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z)=fR(X,Y)Z.$$

Because $R(X,Y)$ is skew-symmetric in $X$ and $Y$, $\mathcal{F}$-linearity in $Y$ follows from $\mathcal{F}$-linearity in the first argument: $$R(X,fY)=-R(fY,X)=-fR(Y,X)=fR(X,Y).$$

6.3 The Riemannian Connection

To narrow down the number of affine connections on a manifold, we impose additional conditions on a connection. On any manifold, we say that a connection is torsion-free if its torsion is zero. On a Riemannian manifold, we say that a connection is compatible with the metric if for all $X,Y,Z\in \mathfrak{X}(M)$, $$Z\left\langle X,Y\right\rangle=\left\langle \nabla_ZX,Y\right\rangle+\left\langle X,\nabla_ZY\right\rangle.$$ It turns out the these two additional conditions are enouth to determine a connection uniquely on a Riemannian manifold.

Definition 6.4. On a Riemannian manifold a Riemannian connection, sometimes called a Levi-Civita connection, is an affine connection that is torsion-free and compatible with the metric.

Lemma 6.5 (Uniqueness of Vector Field). A $C^\infty$ vector field $X$ on a Riemannian manifold $(M,\left\langle\ ,\ \right\rangle )$ is uniquely determined by the values $\left\langle X,Z\right\rangle$ for all $Z\in \mathfrak{X}(M)$.

Proof. We need to show that if $X'\in \mathfrak{X}(M)$ and $\left\langle X,Z\right\rangle=\left\langle X',Z\right\rangle$, for all $Z\in \mathfrak{X}(M)$, the n$X=X'$. With $Y=X-X'$, this is equivalent to showing that if $\left\langle Y,Z\right\rangle=0$ for all $Z\in \mathfrak{X}(M)$, then $Y=0$. Take $Z=Y$. By the positive-definiteness of the inner product at each point $p\in M$, $$\left\langle Y,Y\right\rangle =0\ \Rightarrow \left\langle Y_p,Y_p\right\rangle =0\mbox{ for all } p\in M\\ \Rightarrow Y_p=0\mbox{ for all }p\in M \Rightarrow Y=0.$$

Theorem 6.6 (Uniqueness of Riemannian Connection). On a Riemannian manifold there is a unique Riemannian connection.

Proof. Suppose $\nabla$ is a Riemannian connection on $M$. By Lemma 6.5, to specify $\nabla_XY$, it suffices to know $\left\langle \nabla_XY,Z\right\rangle$ for every vector field $Z\in \mathfrak{X}(M)$. So we will try to find a formula for $\left\langle \nabla_XY,Z\right\rangle$ involving only the Riemannian metric and canonical operations on vector fields such as the Lie bracket.

A Riemannian connection satisfies the two formlas $$\begin{align}\nabla_XY-\nabla_YX-[X,Y]=0\end{align}$$ and $$\begin{align}X\left\langle Y,Z\right\rangle=\left\langle \nabla_XY,Z\right\rangle+\left\langle Y,\nabla_XZ\right\rangle.\end{align}$$ Cyclically permuting $X,Y,Z$ in (4) gives two other formulas: $$\begin{align} Y\left\langle Z,X\right\rangle = \left\langle \nabla_YZ,X\right\rangle +\left\langle Z,\nabla_YX\right\rangle,\\ Z\left\langle X,Y\right\rangle=\left\langle \nabla_ZX,Y\right\rangle+\left\langle X,\nabla_ZY\right\rangle.\end{align}$$ Using (3) we can rewrite $\nabla_YX$ in (5) in terms of $\nabla_XY$: $$\begin{align}Y\left\langle Z,X\right\rangle =\left\langle \nabla_YZ,X\right\rangle+\left\langle Z,\nabla_XY\right\rangle-\left\langle Z,[X,Y]\right\rangle.\end{align}$$ Subtracting (6) from (4) and then adding (7) to it will create terms involving $\nabla_XZ-\nabla_ZX$ and $\nabla_YZ-\nabla_ZY$, which are equal to $[X,Z]$ and $[Y,Z]$ by torsion-freeness: $$X\left\langle Y,Z\right\rangle +Y\left\langle Z,X\right\rangle -Z\left\langle X,Y\right\rangle \\ =2\left\langle \nabla_XY,Z\right\rangle + \left\langle Y,\nabla_XZ-\nabla_ZX\right\rangle +\left\langle X,\nabla_YZ-\nabla_ZY\right\rangle-\left\langle Z,[X,Y]\right\rangle\\ =2\left\langle \nabla_XY,Z\right\rangle+\left\langle Y,[X,Z]\right\rangle + \left\langle X,[Y,Z]\right\rangle -\left\langle Z,[X,Y]\right\rangle .$$ Solving for $\left\langle \nabla_XY,Z\right\rangle$, we get $$\begin{align} 2\left\langle \nabla_XY,Z\right\rangle=X\left\langle Y,Z\right\rangle +Y\left\langle Z,X\right\rangle-Z\left\langle X,Y\right\rangle - \left\langle X,[Y,Z]\right\rangle + \left\langle Y,[Z,X]\right\rangle + \left\langle Z, [X,Y]\right\rangle.\end{align}$$ This formula proves that a Riemannian connection, if it exists, is unique.

Define $\nabla_XY$ by the formula (8). It is a straightforward exercise to check that so defined, $\nabla$ is a torsion-free affine connection compatible with the metric. This proves the existence of a Riemannian connection on a Riemanning manifold.

6.4 Orthogonal Projection on a Surface in $\mathbb{R}^3$

Suppose $M$ is a smooth surface in $\mathbb{R}^3$, which we do not assume to be orientable. At each point $p\in M$, let $v_p$ be the normal line through $p$. It is perpendicular to the tangent spaces $T_pM$, so there is an orthogonal decomposition $$T_p\mathbb{R}^3\simeq T_pM\oplus v_p.$$ Let $\mbox{pr}_p:T_p\mathbb{R}^3\rightarrow T_pM$ be the projection to the tangent spacec of $M$ at $p$.

It $X$ is a $C^\infty$ vector field along $M$, then $X_p\in T_p\mathbb{R}^3$. Define a vector field $\mbox{pr}(X)$ on $M$ by $$\mbox{pr}(X)_p=\mbox{pr}_p(X_p)\in T_pM.$$

Proposition 6.8 (Vector field along Surface). If $X$ is a $C^\infty$ vector field along $M$ in $\mathbb{R}^3$, then the vector field $\mbox{pr}(X)$ on $M$ defined above is $C^\infty$.

Proof. For each $p\in M$, there is a neighborhood $U$ of $p$ in $M$ on which there is a $C^\infty$ unit normal vector field $N$. For $q\in U$, $$\mbox{pr}_q(X_q)=X_q-\left\langle X_q,N_q\right\rangle N_q.$$ Note the $\mbox{pr}_q$ does not depend on the choice of the unit normal vector field: $-N_q$ would have given the same answer. As $q$ varies in $U$, $$\mbox{pr}(X)=X-\left\langle X,N\right\rangle N,$$ which shows that $\mbox{pr}(X)$ is $C^\infty$ on $U$. Hence, $\mbox{pr}(X)$ is $C^\infty$ at $p$. Since $p$ is an arbitrary point of $M$, the vector field $\mbox{pr}(X)$ is $C^\infty$ on $M$.

According to this proposition, the projection operator is a map $$\mbox{pr}:\Gamma(T\mathbb{R}^3|_M)\rightarrow \mathfrak{X}(M).$$

6.5 The Riemannian Connection on a Surface in $\mathbb{R}^3$

We continue to consider a smooth, not necessarily orientable surface $M$ in $\mathbb{R}^3$. Given two $C^\infty$ vector fields $X$ and $Y$ on $M$ and a point $p\in M$, the directional derivative $D_{X_p}Y$ is not in general tangent to $M$. Define $$\begin{align}(\nabla_XY)_p:=\nabla_{X_p}Y=\mbox{pr}_p(D_{X_p}Y)\in T_p M,\end{align}$$ where $\mbox{pr}_p:T_p\mathbb{R}^3\rightarrow T_pM$ is the orthogonal projection defined in Section 6.4. As $p$ varies over $M$, $$\nabla_XY=\mbox{pr}(D_XY),$$ which, by Proposition 6.8, shows that $\nabla_XY$ is a $C^\infty$ vector field on $M$. So we have a map $\nabla:\mathfrak{X}(M)\times \mathfrak{X}(M)\rightarrow \mathfrak{X}(M).$$

Proposition 6.9. The Riemannian connection on a smooth, not necessarily orientable surface $M$ in $\mathbb{R}^3$ is given by $\nabla_XY=\mbox{pr}(D_XY)$.