Interpreting partial derivatives $\frac{\partial f}{\partial u}$ in differential geometry

Question

I'm reading Do Carmo's "Riemannian Geometry" and at some point he introduces the following notation:

Let $A \subset \mathbb{R}^2$ be an open region bounded by a piecewise differentiable curve and $f: A \to M$ be a differentiable mapping into a manifold $M$. Now define:

$$\frac{\partial f}{\partial u} (u,v) := T_{(u,v)}f \cdot \frac{\partial}{\partial u}$$

And similarily for $\frac{\partial f}{\partial v}$. Next in the book there's a remark saying both of these can be viewed as vector fields along $f$. What does that mean? Does it mean that $(p \mapsto T_{f^{-1}(p)}f \cdot \frac{\partial}{\partial u})$ is a smooth vector field defined on $f(A)$? What if $f^{-1}$ isn't differentiable?

Straight after he introduces the notation $\frac{D}{\partial u}$ for the covariant derivative along the curve $(u \mapsto f(u,v_0))$. Is it correct to translate that to $\nabla_{\frac{\partial f}{\partial u}}$ ?

What's really bothering me is that I'm finding the partial derivative notation very confusing. After a couple of pages the following expression appears:

$$\frac{\partial}{\partial u}<\frac{\partial f}{\partial v},\frac{\partial f}{\partial u}> = <\frac{D}{\partial u} \frac{\partial f}{\partial v},\frac{\partial f}{\partial u}> + <\frac{\partial f}{\partial v},\frac{D}{\partial u} \frac{\partial f}{\partial u}>$$ Which to me looks extremely confusingly simplistic. As i see it, $\frac{\partial}{\partial u}$ at the start of the line must really stand for $\frac{\partial f}{\partial u}$ and plugging it in it does make much more sense to me (and looks like the usual inner product rule for covariant derivatives).

Now, for the general question:

• How should I think of an object like $\frac{\partial f}{\partial u}$ in the context of differential geometry?

Should I always think of it $Tf \cdot \frac{\partial }{\partial u} : A \to TM$ ?

If $f$ is a diffeomorphism i can interpert it as a pushforward of the section $\frac{\partial }{\partial u} |_{(u,v)}$ by $f$ i.e.

$$\frac{\partial f}{\partial u} := f_* (\frac{\partial}{\partial u}) = Tf \circ \frac{\partial}{\partial u} \circ f^{-1} : f(A) \to TM$$

Is this the right interpretation?

I'm really lost here, thanks for the help.

Answer 1

This issue also confused me when I was reading Do Carmo so let me try to explain this in some detail.

Let $f \colon N \rightarrow M$ be a smooth map between manifolds and let $\pi_{TM} \colon TM \rightarrow M$. A vector field along $f$ is just a lift of $f$ to $TM$ - that is, a smooth map $V \colon N \rightarrow TM$ satisfying $\pi_{TM} \circ V = f$. In other words, $V$ is a smooth map that attaches to each point $p \in N$ a tangent vector $V(p) \in T_{f(p)}M$. For example, if $N = (0,1)$ and $\alpha \colon N \rightarrow M$ is a curve then the velocity $\dot{\alpha}(t)$ is a vector field along $\alpha$. If $f$ is not an embedding, it doesn't make sense to talk about vectors fields on $f(N)$ but it makes sense to talk about vector fields along $f$. Note that even if $f$ is an embedding, a vector field along $f$ is not the same as a vector field on $f(N)$ since $V(p)$ doesn't have to be tangent to $f(N)$ but can be any tangent vector in $T_{f(p)} M$. For example, if $N = S^2, M = \mathbb{R}^3$ and $f$ is the inclusion, the outward pointing unit normal to $N$ is a vector field along $f$.

An affine connection on $M$ allows us to covariantly differentiate vector fields on $M$ but in order to define geodesics, we need to be able to define the acceleration of a curve $\alpha$ in $M$. For that, we need to differentiate $\dot{\alpha}$ which is not a vector field on $M$ but only a vector field along $\alpha$. Luckily, using a connection on $M$ on can naturally differentiate any vector field along any smooth map $f \colon N \rightarrow M$. To be more precise, denote by $\Gamma(f^{*}(TM))$ the collection of all vector fields along $f$. This is a vector space and a $C^{\infty}(N)$ module. Given a vector field $Z$ on $M$, one can construct a vector field along $f$ by letting $V(p) = Z(f(p))$. Let us denote this $V$ by $f^{*}(Z)$. Then, there is a naturally induced and unique operator $D \colon \Gamma(TN) \times \Gamma(f^{*}(TM)) \rightarrow \Gamma(f^{*}(TM))$ satisfying:

1. $D_X(V + W) = D_X(V) + D_X(W)$ for all $X \in \Gamma(TN)$ and $V,W \in \Gamma(f^{*}(TM))$.

2. $D_{fX + gY}V = fD_X(V) + gD_Y(V)$ for all $f,g \in C^{\infty}(N)$ and $V \in \Gamma(f^{*}(TM))$.

3. $D_X(fV) = (Xf)V + fD_X(V)$ for all $f \in C^{\infty}(N)$ and $V \in \Gamma(f^{*}(TM))$.

4. $D_X(f^{*}(Z))|_p = \nabla_{df|_p(X(p))} Z$ for all $X \in \Gamma(TN)$ and $Z \in \Gamma(TM)$.

When $N$ is an open interval with a natural coordinate $t$, by $(2)$, one needs to only specify $D_{\frac{d}{dt}}$. This is what Do Carmo denotes by $\frac{D}{dt}$. When $N$ is an open subset of $\mathbb{R}^2$ with coordinates $u,v$, one needs only specify $D_{\frac{\partial}{\partial u}}$ and $D_{\frac{\partial}{\partial v}}$ which Do Carmo denotes by $\frac{D}{\partial u}$ and $\frac{D}{\partial v}$.

A more unified and elegant approach can be given using the language of connections on vector bundles. Let $f \colon N \rightarrow M$ be a smooth map between manifolds and let $\pi \colon E \rightarrow M$ be a vector bundle over $M$. A section of $E$ along $f$ is a smooth section of the pullback bundle $f^{*}(E)$ over $N$. Given a connection $\nabla$ on $E$, one has a naturally induced connection $f^{*}(\nabla)$ on the pullback bundle $f^{*}(E)$ over $N$ which allows one to covariantly differentiate sections of $E$ along $f$. This connection is characterized uniquely by the property that

$$ f^{*}(\nabla)_X (f^{*}(s))|_{p} = \nabla_{df|_p(X)} s $$

where $s \in \Gamma(E)$ is a section of $\pi \colon E \rightarrow M$, $X$ is a vector field on $N$ and $p \in N$. Using this language, a covariant differentiation along a curve or a parametric surface or whatever is just applying an induced connection on sections of a different bundle. In his book, Do Carmo avoids discussing general vector bundles and connections over them. Hence, for him a vector field along a curve $\alpha$ is a lift and not a section of the pullback bundle and the induced covariant differentiation is not treated as a particular instance of a connection on a vector bundle but as a different object denoted by $D$.

Finally, let us interpret all the objects in: $$\frac{\partial}{\partial u} \left< \frac{\partial f}{\partial v},\frac{\partial f}{\partial u} \right> = \left< \frac{D}{\partial u} \frac{\partial f}{\partial v},\frac{\partial f}{\partial u} \right> + \left< \frac{\partial f}{\partial v},\frac{D}{\partial u} \frac{\partial f}{\partial u} \right>$$

The $\frac{\partial f}{\partial v},\frac{\partial f}{\partial u}$ are vector fields along $f$. Like you wrote, $$\frac{\partial f}{\partial v} (u_0,v_0) = df|_{(u_0,v_0)} (\frac{\partial}{\partial v}|_{(u_0, v_0)}) = \frac{d}{dt} f(u_0, v_0 + t) |_{t=0} \in T_{f(u_0, v_0)} M $$ and similarly for $\frac{\partial f}{\partial v}$. The expression $\left< \frac{\partial f}{\partial v},\frac{\partial f}{\partial u} \right>$ is a smooth real valued function defined on $A$ and so one can take its partial derivatives. The expression $\frac{D}{\partial u} \frac{\partial f}{\partial v}$ is the (induced) covariant derivative of the vector field $\frac{\partial f}{\partial v}$ along $f$ in the direction $\frac{\partial}{\partial u}$.

Answer 2

If {$u$,$v$} are the usual coordinates in $\mathbb{R^2}$, take a look at the curve $\gamma_{v_0}$, $u \rightarrow f(u,v_0)$ whenever is a connected curve. Then in any point $(u_0,v_0)$, $\frac{df}{du}(u_0,v_0):=df(\frac{\partial}{\partial u}|_{(u_0,v_0)}) \in T_{f(u_0,v_0)=\gamma_{v_0}(u_0)}M$, so, this is a well defined vector field along the curve $\gamma_v$, when $v$ is moving in $A$. Analogously you can define $\frac{df}{dv}$ as a vector field along curves of the form $\gamma_{u_0}$, $v \rightarrow f(u_0,v)$.

Now, for the covariant derivative of a vector field $V$ along the curve $\gamma_{v_0}$, $u \rightarrow f(u,v_0)$, if $V$ is a vector field along this "surface" in $M$: $V$ when restricted to this curve, is a vector field along this curve, so it makes perfectly sense to define $\frac{DV}{\partial u}(u_0,v_0)=\nabla_{\gamma_{v_0}^{'}}V(u,v_0)$, the covariant derivative of its restriction to this curve. This is one way to see this, since you understand the derivative of a vector field along a curve, as a way to measure how it behaves in the direction of this curve.

As for your general question, if $f$ is a diffeomorphism, then your manifold $M$ is a connected $2$-manifold, (by the assumptions on $A$), so this is very restrictive. As you see, $f$ need not to be a diffeomorphism in order to understand this definitions, not even an immersion.

I hope this was useful to you.