4

When we apply $\nabla$ or $D$ to a function $f:\mathbb R^n\to \mathbb R$, then they in principle do the same operation.

However, in textbooks $\nabla$ is often written as a column vector $(\partial_1,...,\partial_n)^T$, whereas $D$ is written as a row vector $(\partial_1,...,\partial_n)$.

Is there a well defined difference between these two operators when they are applied to a single-valued real function?

user56834
  • 12,323
  • There is no functional difference, from what I know. Both represent the same object (the vector that represents the best linear approximation of $f$ near points close to some $x_0$) – rubikscube09 Apr 29 '18 at 05:11
  • IMO, the main confounding issue here is that the derivative came into practice before the distinction between vectors and covectors was widely recognized (in coordinates, those translate into column and row vectors respectively). Being covector valued is more natural, but being vector valued has a longer tradition, and IMO, actual usage of $\nabla$ is split, rather than everyone using $\nabla$ as being vector valued. –  Apr 29 '18 at 05:25

1 Answers1

5

A differential geometer will tell you that the differential $Df|_p$ at a point is the linear map $\mathbb R^n \to \mathbb R$ which sends a column vector $v \in \mathbb R^n$ to the inner product $(\partial_1|_p,...,\partial_n|_p) \cdot v$. The (total) differential $Df$ is the smooth map $\mathbb R^n \to (\mathbb R^n)^*$ (the dual as a vector space1) which sends $p$ to $Df|_p$. It makes sense to define the differential for a function $f : \mathbb R^n \to \mathbb R^m$ too; it will be a smooth map $\mathbb R^n \to \operatorname{Hom}(\mathbb R^n, \mathbb R^m)$.1

The gradient $\nabla f$ is a vector field (a tangent vector at each point) characterized by the property that $\langle (\nabla f)(p), v \rangle = Df|_p(v)$, where $\langle \cdot, \cdot \rangle$ is the inner product. That is, $(\nabla f)(p) = (\partial_1|_p,...,\partial_n|_p)^T$.

The differential $Df$ generalizes to any smooth map between differentiable manifold $M,N$ as a linear map between the tangent spaces at each point $p$ and its image $f(p)$. If $N = \mathbb R$, then $Df$ is a differential $1$-form. The gradient generalizes to Riemannian or pseudo-Riemannian manifolds, where we have a non-degenerate inner product.

The differential of a smooth real-valued function is a section of the cotangent bundle $\Omega^1(M)$. Its gradient is a section of the tangent bundle $TM$, which is the dual of $\Omega^1$.

1 The differential structure of $(\mathbb R^n)^*$ and $\operatorname{Hom}(\mathbb R^n, \mathbb R^m)$ are obtained by transporting the structure via a linear bijection with some $\mathbb R^N$, and it does not depend on the linear bijection. That is, you can take any basis and maps will be smooth iff their components in that basis are smooth.

Bart Michels
  • 26,985
  • 6
  • 59
  • 123
  • Thank you. I haven't thought of $Df|_p$ as a mapping before, but rather as the vector $(\partial_1|_p,...,\partial_n|_p)$ itself. Why is it defined differently, as a mapping from a vector to that dot product, rather than as the left vector itself. This seems simpler. – user56834 Apr 29 '18 at 08:36
  • So we can see $D(\cdot)|p$ as a mapping $(V\to \mathbb R)\to (V\to \mathbb R)$? It takes a function on a vector space, and returns another function on that vector space, (which is its directional derivative at that point) – user56834 Apr 29 '18 at 08:48
  • @Programmer2134 Good question. Because viewing a mapping as a vector requires the choice of a basis. (Here, the dual basis of the standard basis of $\mathbb R^n$.) For $\mathbb R^n$ there is a canonical choice for the basis, and the coordinate point of view is often more helpful. For general differentiable manifolds, there is no such canonical choice. – Bart Michels Apr 29 '18 at 08:49
  • Ah that is actually quite elegant! So do we retain a canonical bijection between dual vectors and vectors without a choice of basis for general differentiable manifolds? (i.e. the same way that there is a canonical bijection between the space of column vectors and row vectors in $\mathbb R^n$, by simply transposing them). – user56834 Apr 29 '18 at 08:53
  • For the second question: yes, where the first arrow $V \to \mathbb R$ means all smooth functions and the second arrow means linear mappings. (It isn't called directional deriviative though; the directional derivative is with respect to a vector.) – Bart Michels Apr 29 '18 at 08:54
  • Wait, shouldn't I correct myself on my second question? Shouldn't it be: $D(\cdot)|_p: (V\to \mathbb R)\to (T(V) \to \mathbb R)$, where $T(V)$ is the tangent space of $V$. – user56834 Apr 29 '18 at 08:56
  • Yes, I was assuming you meant $V = \mathbb R^n$, with the identification $T(V) = V$. – Bart Michels Apr 29 '18 at 08:57
  • For the third question: there is no such canonical isomorphism without a basis. The precise statement is that the duality functor is not canonically isomorphic to the identity functor on the category of finite vector spaces; see https://math.stackexchange.com/questions/622589 We do however have a canonical isomorphism if we fix a nondegenerate bilinear form; this is how one defines the gradient in the general setting of (pseudo-)Riemannian manifolds: apply the duality to the differential $Df$ to obtain a tangent vector. – Bart Michels Apr 29 '18 at 09:00
  • The choice of a nondegenerate bilinear form is in a sense "weaker" than that of a basis; since for every basis $(e_i)$ we can take the bilinear form with matrix $(\delta_{ij})$ in that basis. The isomorphism with the dual space via the dual basis is then the same as the one via the bilinear form. – Bart Michels Apr 29 '18 at 09:02
  • Correction, for a general differentiable manifold the $D(\cdot)|_p$ lands in $T_pV \to \mathbb R$, not $T(V) \to \mathbb R$. – Bart Michels Apr 29 '18 at 09:04
  • I'm having trouble rigorously deriving the nature of $D^2$. I suspect that we should have $D^2f|_p:T_pV\to T_pV$. Is this correct? I reasoned this intuitively though. Not sure how to find it rigorously by iterating over the definition of $D$. – user56834 Apr 29 '18 at 10:54
  • What do you mean by $D^2$? Feel free to post a new question. – Bart Michels Apr 29 '18 at 11:13
  • I have asked a new question here: https://math.stackexchange.com/questions/2758665/how-do-we-define-d2-formally-in-differential-geometry – user56834 Apr 29 '18 at 12:37