Connection

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Connections and Covariant Differentiation

Using the Lie derivative, we can differentiate vector fields or differential forms, but it is impossible to extend this concept to sections \(\Gamma(E)\) of an arbitrary vector bundle \(\pi:E\rightarrow M\). On the tangent bundle \(TM\), given two points \(p,q\) on an integral flow \(\phi\), there exists a natural isomorphism \(d\phi^{-t}\) connecting the two tangent spaces \(T_pM\) and \(T_qM\), but no such map exists between two fibers \(E_p\) and \(E_q\) of an arbitrary vector bundle \(E\). Therefore, we additionally define a connection joining these fibers.

Definition 1 Given a vector bundle \(E\rightarrow M\) defined over a manifold \(M\), a connection \(\nabla:\mathfrak{X}(M)\times\Gamma(E)\rightarrow\Gamma(E)\) defined on \(E\) is a map satisfying the following conditions.

1. (Tensoriality) \(\nabla_XY\) is \(C^\infty\)-linear in the first argument.
2. (Linearity) \(\nabla_XY\) is \(\mathbb{R}\)-linear in the second argument.

3. For any \(f\in C^\infty(M)\), \(\nabla\) satisfies the following Leibniz rule:

   \[\nabla_X(fY)=f\nabla_XY+(Xf)Y\]

In this case, \(\nabla_XY\) is also called the covariant derivative of \(Y\) in the direction of \(X\). The following proposition shows that to compute \((\nabla_XY)_p\), it suffices to know \(X\) and \(Y\) in a neighborhood of \(p\).

Proposition 2 Let a manifold \(M\) be given, and let \(X\in\mathfrak{X}(M)\) and \(Y\in\Gamma(E)\). For any point \(p\in M\), \((\nabla_XY)_p\) depends only on

1. the value \(X_p\) of the vector field \(X\) at the point \(p\),
2. the vector field \(Y\vert_U\) in an open neighborhood \(U\) of the point \(p\).

Proof

First, we show that \((\nabla_XY)_p\) depends only on the vector field in an open neighborhood \(U\) of \(p\). To show that \((\nabla_XY_1)_p=(\nabla_XY_2)_p\) whenever two vector fields \(Y_1,Y_2\) agree on an open neighborhood \(U\) of \(p\), it suffices to show that if a vector field \(Y\) is identically zero on all points of an open neighborhood \(U\), then \((\nabla_XY)_p=0\). Let \(\varphi\) be a bump function satisfying \(\supp(\varphi)\subseteq U\) and \(\varphi(p)=1\); then the vector field \(\varphi Y\) is identically zero on all of \(M\). Therefore, by the second condition of Definition 1, \(\nabla_X(\varphi Y)=0\). On the other hand, by the Leibniz rule,

\[0=\nabla_X(\varphi Y)=\varphi\nabla_XY+(X\varphi)Y\]

and evaluating the right-hand side at the point \(p\) yields

\[\varphi(p)(\nabla_XY)_p+(X\varphi)(p)Y_p=(\nabla_XY)_p\]

so \((\nabla_XY)_p=0\).

Now we show that \(\nabla_XY\) depends only on the vector \(X_p\) at the point \(p\). As above, it suffices to assume \(X_p=0\) and show that \((\nabla_XY)_p=0\). Take a coordinate chart \((U,(x^i))\) in a neighborhood of \(p\), and write \(X\) with respect to this coordinate system as

\[X=X^1\frac{\partial}{\partial x^1}+\cdots+X^n\frac{\partial}{\partial x^n}\]

Then

\[(\nabla_XY)_p=(\nabla_{\sum X^i\frac{\partial}{\partial x^i}}Y)_p=\left(\sum_{i=1}^n X^i\nabla_{\partial/\partial x^i} Y\right)_p\]

and since \(X^i(p)=0\) for all \(i\), we obtain the desired result.

Covariant Differentiation on the Tangent Bundle

In particular, let us examine a connection defined on the tangent bundle \(TM\rightarrow M\). Then \(\nabla\) is a map from \(\mathfrak{X}(M)\times\mathfrak{X}(M)\) to \(\mathfrak{X}(M)\). It should be noted that although the Lie derivative and the connection are both concepts for thinking about differentiation, they produce different results. For instance, for the Lie derivative

\[\mathcal{L}_{fX}Y=[fX,Y]=fX(Y)-Y(fX)=fX(Y)-(Yf)X-fY(X)=f[X,Y]-(Yf)X=f\mathcal{L}_XY-(Yf)X\]

but, by definition,

\[\nabla_{fX}Y=f\nabla_XY\]

so it is always possible to choose \(Y\) and \(f\) appropriately such that \(\nabla_{fX}Y\neq\mathcal{L}_{fX}Y\).

Anyway, considering a connection defined on \(TM\), by Proposition 2, the value of \(\nabla_XY\) at a point \(p\) is completely determined by local frames \((E_i)\) in a neighborhood of \(p\). This is because

\[(\nabla_XY)_p=\nabla_{\sum X^i(p)E_i(p)}\left(\sum Y^i(p)E_i(p)\right)\]

holds. On the other hand,

\[\nabla_X\left(\sum_{j=1}^n Y^jE_j\right)=\sum_{j=1}^n \nabla_X(Y^jE_j)=\sum_{j=1}^n\left(Y^j\nabla_XE_j+X(Y^j)E_j\right)=\sum_{i,j=1}^nX^iY^j\nabla_{E_i}E_j+\sum_{j=1}^n X(Y^j)E_j\tag{1}\]

so \(\nabla_XY\) is completely determined by the \(n^2\) vector fields \(\nabla_{E_i}E_j\). Writing the vector field \(\nabla_{E_i}E_j\) again as a linear combination of the \(E_k\)’s,

\[\nabla_{E_i}E_j=\Gamma_{ij}^k E_k\]

there exist \(n^3\) \(C^\infty\) functions \(\Gamma_{ij}^k\) satisfying this. Then equation (1) above can be written as

\[\nabla_XY=\sum_{k=1}^n\left(\sum_{i,j=1}^nX(Y^k)+X^iY^j\Gamma_{ij}^k\right)E_k\]

Definition 3 The \(n^3\) functions \(\Gamma_{ij}^k\) defined above are called the connection coefficients.

On the other hand, the tangent bundle over any manifold \(M\) always admits a connection. To verify this, just as with the Riemannian metric, one can take the connection on Euclidean space

\[\nabla_vY:=v(Y^1)\frac{\partial}{\partial x^i}+\cdots+v(Y^n)\frac{\partial}{\partial x^n}\]

and patch it together via a partition of unity.

Covariant Differentiation on the Cotangent Bundle

We show that a connection \(\nabla\) defined on the tangent bundle \(TM\) extends nicely to any \((r,s)\)-tensor field \(\mathcal{T}^{r,s}(M)\). () To do this, we must first specify how \(\nabla\) extends to the cotangent bundle \(T^\ast M\).

Proposition 4 Let a manifold \(M\) and a connection \(\nabla\) on the tangent bundle \(TM\) be given. Define a map \(\nabla^\ast:\mathfrak{X}(M)\times\Gamma(T^\ast M)\rightarrow\Gamma(T^\ast M)\) by the formula

\[(\nabla_X^\ast\alpha)_p(Y)=X\bigl(\alpha(Y)\bigr)-\alpha_p\bigl(\nabla_XY\bigr)_p\]

Then \(\nabla^\ast\) becomes a connection on \(T^\ast M\).

Proof

First, since we can show that \(\nabla^\ast\alpha\) defined by the right-hand side expression is a \(1\)-form, the codomain of \(\nabla^\ast\) is unproblematic.

That \(\nabla^\ast\) actually satisfies the conditions of a connection is obvious except for the Leibniz rule. In fact, the Leibniz rule also follows obviously from

\[\begin{aligned}(\nabla_X^\ast f\alpha)_pY&=X(f\cdot\alpha(Y))-(f\alpha)_p(\nabla_XY)_p\\&=(Xf)(\alpha(Y))+f(p)\bigl(X(\alpha(Y))-\alpha_p(\nabla_XY)_p\bigr)\\&=\bigl((Xf)\alpha+f\nabla_X\alpha\bigr)Y\end{aligned}\]

By a slight abuse of notation, we write the \(\nabla^\ast\) defined above also as \(\nabla\).

Covariant Differentiation on Tensor Fields

Now we can finally extend the connection on \(TM\) to \((r,s)\)-tensor fields \(\mathcal{T}^{r,s}(M)\). As above, we shall write this connection also as \(\nabla\).

Proposition 5 Let a connection \(\nabla\) defined on the tangent bundle \(TM\rightarrow M\) be given. Then \(\nabla\) can be extended to all tensor fields \(\mathcal{T}^{r,s}(M)\) satisfying the following two conditions

\[\nabla_X(F\otimes G)=(\nabla_X F)\otimes G+F\otimes(\nabla_XG),\qquad\nabla_X(F+G)=\nabla_XF+\nabla_XG\]

and the extension is uniquely determined by the additional condition that \(\nabla_Xf=Xf\) for \(\mathcal{T}^{0,0}M\).

An arbitrary \((r,s)\)-tensor \(F\) can be regarded as the same as the following linear map

\[\underbrace{\Omega^1(M)\times\cdots\times \Omega^1(M)}_\text{\small $r$ times}\times\underbrace{\mathfrak{X}(M)\times\cdots\times \mathfrak{X}(M)}_\text{\small $s$ times}\rightarrow C^\infty(M)\]

so \(\nabla_XF\) is uniquely determined by its values on \(\omega^1,\ldots,\omega^r\in\Omega^1(M)\) and \(Y_1,\ldots, Y_s\in\mathfrak{X}(M)\). For example, applying the Leibniz rule to a simple tensor \(F=X_1\otimes\cdots\otimes X_r\otimes\alpha^1\otimes\cdots\otimes\alpha^s\in\mathcal{T}^{r,s}(M)\) yields

\[\begin{aligned}\nabla_X(X_1\otimes\cdots\otimes X_r\otimes\alpha^1\otimes\cdots\otimes\alpha^s)&=(\nabla_XX_1)\otimes X_2\otimes\cdots\otimes X_r\otimes\alpha^1\otimes\cdots\alpha^s\\ &\phantom{=aa}+\cdots\\ &\phantom{=aaaa}+X_1\otimes X_2\otimes\cdots\otimes(\nabla_XX_r)\otimes\alpha^1\otimes\cdots\otimes\alpha^s\\ &\phantom{=aaaaaa}+X_1\otimes\cdots X_r\otimes(\nabla_X\alpha^1)\otimes\alpha^2\otimes\cdots\otimes\alpha^s\\ &\phantom{=aaaaaaaa}+\cdots\\ &\phantom{=aaaaaaaaaa}+X_1\otimes\cdots\otimes X_r\otimes\alpha^1\otimes\alpha^2\otimes\cdots\otimes(\nabla_X\alpha^s)\end{aligned}\]

so, computing the value of \(\nabla_XF\) on \(\omega^1,\ldots,\omega^r\in\Omega^1(M)\) and \(Y_1,\ldots, Y_s\in\mathfrak{X}(M)\), we obtain

\[\sum_{i=1}^r\omega^1(X_1)\omega^2(X_2)\cdots\omega^i(\nabla_XX_i)\cdots\omega^r(X_r)\alpha^1(Y_1)\cdots\alpha^s(Y_s)+\sum_{j=1}^s\omega^1(X_1)\cdots\omega^r(X_r)\left(X(\alpha^j(Y^j))-\alpha^j(\nabla_XY^j)\right)\]

and, replacing the first sum on the right-hand side using

\[\omega^i(\nabla_XX_i)=X(\omega^i(X_i))-(\nabla_X\omega^i)(X_i)\]

we obtain the formula

\[\begin{aligned}\nabla_XF(\omega_1,\ldots,\omega^r,Y_1,\ldots, Y_s)&=X(F(\omega^1,\ldots,\omega^r,Y_1,\ldots, Y_s))\\ &\phantom{=aa}-\sum_{i=1}^r F(\omega^1,\ldots,\nabla_X\omega^i,\ldots,\omega^r,Y_1,\ldots, Y_s)\\ &\phantom{=aaaa}-\sum_{j=1}^s F(\omega^1,\ldots, \omega^r,Y_1,\ldots, \nabla_XY_j,\ldots, Y_s)\end{aligned}\]

This formula holds for all simple \((r,s)\)-tensors, so it also holds for all \((r,s)\)-tensors. That this \(\nabla\) satisfies the desired conditions and is uniquely determined by the condition \(\nabla_Xf=Xf\) is now a straightforward computation.

Total Connection

Consider the connection \(\nabla\) on \(\mathcal{T}^{r,s}(M)\) defined above. For any \((r,s)\)-tensor \(F\), we can think of \(\nabla F\) as an \((r,s+1)\)-tensor taking \(r\) \(1\)-forms \(\omega^1,\ldots,\omega^r\) and \(s+1\) vector fields \(Y_1,\ldots, Y_s, X\) and yielding the \(C^\infty(M)\) function

\[(\nabla_XF)(\omega^1,\ldots,\omega^r,Y_1,\ldots, Y_s)\]

In particular, applying \(\nabla\) to a \((0,0)\)-tensor, i.e., a \(C^\infty\) function \(f\), yields a \((0,1)\)-tensor \(\nabla f\). By definition, the covector \(\nabla f\) is defined by the formula

\[X\mapsto \nabla_Xf=Xf\]

On the other hand, setting \(M=\mathbb{R}^m\) and, as in §Riemannian Metric, §§Musical Isomorphism, regarding the gradient vector \(\operatorname{grad} f\) of a function \(f\) as the covector defined by the formula

\[X\mapsto \langle X, \operatorname{grad} f\rangle\]

then by the property of the gradient vector, we know that \(\langle X,\operatorname{grad} f\rangle=Xf\). This explains why the gradient vector has traditionally also been denoted by \(\nabla f\). To treat this on a general manifold, a Riemannian metric is inevitably required, so we postpone the details to the next post.

Now the second covariant derivative \(\nabla_{X,Y}^2 F\) is defined as the \((r,s)\)-tensor \(\nabla_{X,Y}^2F\) satisfying the formula

\[\nabla_{X,Y}^2F(\ldots)=(\nabla^2 F)(\cdots, Y,X)\]

This \(\nabla_{X,Y}^2F\) is \(C^\infty(M)\)-linear in \(Y\), but \(\nabla_X\nabla_Y\) is not \(C^\infty\)-linear in \(Y\), so in general \(\nabla_{X,Y}^2\neq\nabla_X\nabla_Y\). However, the following holds.

Proposition 6 For any \((r,s)\)-tensor \(F\),

\[\nabla_{X,Y}^2F=\nabla_X(\nabla_YF)-\nabla_{\nabla_XY}F\]

holds.

Proof

One need only plug \((\omega^1,\ldots,\omega^r,Z_1,\ldots,Z_s)\) into the right-hand side.

In particular, applying this to a \((0,0)\)-tensor \(C^\infty(M)\) yields the covariant Hessian \(\nabla^2 u\).

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References

[Lee] John M. Lee. Introduction to Riemannian Manifolds, Graduate texts in mathematics, Springer, 2019

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