3
$\begingroup$

I was doing exercise 8 from do Carmo's Riemannian geometry and I stumbled upon the definition of gradient given.

Let $M$ be a Riemannian manifold... $f \in \mathcal{D}(M)$ .. the gradient of $f$ as a vector field $\text{grad} \; f$ on $M$ defined by $$ \langle \text{grad} \; f, v \rangle = df_p(v) \;\; p \in M, v \in T_pM \;\;\;\;\; (1) $$

here $\langle \cdot , \cdot\rangle$ is the Riemannian metric on $M$ and $f$ is a differentiable function on $M$. No the Riemannian metric is a bilinear map $$\langle \cdot,\cdot \rangle : T_p M \times T_p M \to \mathbb{R}$$ but the differential $df_p$ is a map between tangent spaces, namely $$ df_p : T_p M \to T_{f(p)} \mathbb{R} \cong \mathbb{R} $$

So in a nutshell I'm confused about the equality in $(1)$ because the lhs is a scalar in the field while the rhs is vector, though isomorphic to the scalar field. This definition actually makes a bit tricky for me to understand how to do the exercises, because any of the computations I do give me equalities that don't really make sense.

Can you clarify how the gradient is actually defined? I also own Tu's Differential Geometry, but I don't see these definitions (I'm kind of reading the two in parallel).

$\endgroup$
18
  • $\begingroup$ I disagree, do Carmo defines $df_p$ as a differential which is a map between two tangent spaces. If I use differential forms notation however $df_p$ is a cotangent vector, and hence a functional and so your argument works. Unless he meant differential of $f$ at $p$ I don't see how the math can work. $\endgroup$ Jun 28, 2020 at 16:32
  • $\begingroup$ I see what you're getting at, I misunderstood you and my previous comment didn't address your point of confusion. I'll delete it. $\endgroup$ Jun 28, 2020 at 16:34
  • $\begingroup$ It's fine, as usual do Carmo's notation isn't the best despite the great content. $\endgroup$ Jun 28, 2020 at 16:35
  • 1
    $\begingroup$ I'm sure you noticed this but the equation in Euclidean space (or even in local coordinates) says exactly that $D_vf = \nabla f \cdot v$ where $D_v$ is the directional derivative in the direction of $v$. $\endgroup$ Jun 28, 2020 at 17:30
  • 1
    $\begingroup$ $df_p$ is a map from $T_pM \to \mathbb{R}$ (I know you wrote $T_{f(p)}\mathbb{R} \cong \mathbb{R}$ but not only are they isomorphic, they are canonically so, so you can consider $df_p$ as a map from $T_pM \to \mathbb{R}$ without any choices). Since $v \in T_pM$, $df_p(v)$ is just the evaluation sending $v$ to the element in $\mathbb{R}$ so it is a scalar. $\endgroup$ Jun 28, 2020 at 18:56

1 Answer 1

6
$\begingroup$

It's natural to have some confusion about these things. There are many similar things that come up in differential geometry and smooth manifold theory (and even much of other parts of math) where we take shortcuts or "make identifications" that make our lives easier once we understand their meaning, but can make the uninitiated's life needlessly difficult when it comes time to write proofs and ask if we really understand the shortcuts we take.

For any smooth map $f\colon M\to \mathbb R$ there is the global differential map, $df\colon TM\to T\mathbb R$ defined by $$ df(p,v) = (f(p),df_p(v)), $$ and the vector $df_p(v)$ acts on smooth functions $h$ on $\mathbb R$ by $df_p(v)(h) = v(h\circ f)$. For fixed $p\in M$, the map $df_p\colon T_pM\to T_{f(p)}\mathbb R$ is the differential of $\pmb f$ at $\pmb p$. For any point $q\in\mathbb R$, there is a canonical vector space isomorphism $L_q\colon \mathbb R\cong T_{q}\mathbb R$ defined by $$ L_q(v) = v\frac{d}{dt}\bigg|_q, $$ i.e., sending the number $v$ to the directional derivative with respect to the "vector" $v$ (which is of course merely multiplication of the number $v$ with the usual derivative operator for smooth functions on $\mathbb R$.) We can compose $L_{f(p)}$ with $df_p$ to get a linear map $$ \widetilde{df_p} \equiv L_{f(p)}\circ df_p\colon T_pM\to \mathbb R. $$ Local coordinates $(x^1,\dots,x^n)$ near $p$, give a basis $\partial_{x^1}|_p,\dots,\partial_{x^n}|_p$ for $T_pM$, with respect to which, the linear map $\widetilde{df_p}$ is simply the row vector $$ \begin{bmatrix} \displaystyle\frac{\partial f}{\partial x^1}(p) & \dotsb & \displaystyle\frac{\partial f}{\partial x^n}(p) \end{bmatrix}. $$ For $f\colon M\to\mathbb R$, we also have a well-defined covector field $df\colon M\to T^*M$. In local coordinates $(x^1,\dots,x^n)$ near $p$, we can express the covector field $df$ in terms of the local coframe $dx^1,\dots,dx^n$ (dual frame of $\partial_{x^1},\dots,\partial_{x^n}$) as $$ df = \sum_i\frac{\partial f}{\partial x^i}\,dx^i. $$ At each point $p$, we thus have a covector $df_p\colon T_pM\to \mathbb R$ expressed in terms of the basis $dx^1|_p,\dots,dx^n|_p$ by $$ df_p = \frac{\partial f}{\partial x^i}(p)\,dx^i|_p. $$ so with respect to the basis $dx^1|_p,\dots,dx^n|_p$, $df_p\in T_p^*M$ can be expressed as the row vector $$ \begin{bmatrix} \displaystyle\frac{\partial f}{\partial x^1}(p) & \dotsb & \displaystyle\frac{\partial f}{\partial x^n}(p) \end{bmatrix}. $$ So really, $df_p$ the differential and $df_p$ the covector are literally the same object up to the canonical isomorphism $L_{f(p)}$. I think that we remind ourselves of this isomorphism $L$ maybe the first few times we identify the differential $df_p$ and the covector $df_p$, but we will drop it entirely after we get used to it. With more experience, one comes to appreciate the "intent of the law" rather than strictly follow the "letter of the law," and the interpretations we make are ultimately dictated by the purposes we have in mind.

That said, if one wants to define $\mathrm{grad}f$ "right," without making identifications, then I'd say you need to be comfortable with covector fields, and the musical isomorphism $(\cdot)^\sharp\colon T^*M\cong TM$ that the metric $g$ gives us, so we can do things properly and say simply and without ambiguity that $\mathrm{grad} f = (df)^\sharp$.

$\endgroup$
1
  • $\begingroup$ I think my source of confusion lies in the fact that doCarmo defines the gradient as an operator acting on some function. Vectors and covector might be isomorphic but this only means you can associate to each vector a unique covector and viceversa linearly but they're not overall the same thing. The musical isomorphism is indeed a construction to provide a vector given a covector. I understand the point of being the same, but they're actually not. $\endgroup$ Aug 22, 2020 at 9:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .