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Background: Let $f:x \mapsto x^2$. I first learned to think of the derivative of a function as giving the slope of line tangent to the graph of the function at the point.

For example, since $f':x \mapsto2x$, the tangent to the graph of $f$ at $x=1$ has slope $f'(1)=2(1)=2$.

However, consider now the function $F:(x,y)\mapsto y-x^2$. The graph of $f$ is just the level set $F^{-1}\{0\}$. To find the normal to the graph at $x=1$, we take the line spanned by the vector $$\left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}\right)=(-f',1).$$

The discrepancies in possible interpretations of differentiation don't stop with what was mentioned above. For example, consider now the graph of $f$, or equivalently $F^{-1}\{0\}$, to be the trace of the parametrized curve $\phi: t \mapsto (t,t^2)$. Then the tangent vectors are described by $$(\phi_1',\phi_2')=\left(\frac{d\phi_1}{dt},\frac{d\phi_2}{dt} \right) .$$ And yet simultaneously the following vector is normal to the trace of $\phi$: $$(\phi_1'',\phi_2'')=\left(\frac{d^2\phi_1}{d^2 t}, \frac{d^2 \phi_2}{d^2t} \right) .$$

Thus, despite the fact that tangents and normals are essentially opposite concepts, it seems like we can think of derivatives as characterizing either type of local behavior near a point.

Question: What is the "correct" intuition to think of differentiation? As producing tangents, as one might learn in high school, or as producing normals, as one might learn in college?

If this is an issue of duality, which type of duality? Does the duality require a choice of inner product (i.e. Riemannian metric), or is it independent of any such choice?

Further Discussion:
If we consider instead of just smooth curves arbitrary smooth manifolds, the ambiguity of "proper interpretation" remains. When applying the implicit function theorem, the partial derivatives span a vector space that is normal to the graph of the hypersurface, rather than tangent.

However, the total derivative, which defines the partial derivatives, also defines the directional derivatives, which are in one-to-one correspondence with the derivations of smooth functions which are the elements of the tangent space of the hypersurface.

I feel like this might be an issue of some type of duality which I was unaware of, with the kernel of the total derivative (considered as a linear transformation) corresponding to the tangent $(n-m)$-space and the image of the total derivative corresponding to the normal $m$-space.

The confusion might also be a result of failing to recognize implied identifications of covectors and vectors (like the identification of the outer product of two vectors in $\mathbb{R}^3$, which is a bivector, with a vector via Hodge duality and calling the resulting vector the cross product).

In either case, I have no idea how I could use either explanation to understand the discrepancies between derivatives as normals and tangents in the one-dimensional case (for smooth curves).

Is this related at all to curvature, e.g. the distinctions between Gaussian and principal curvatures, or between the curvature tensor and sectional curvatures? How much of this is linear algebra without an inner product, i.e. the distinction between vectors and covectors, between a vector space and its dual, and how much does it rely upon musical isomorphisms and Hodge duality and other canonical identifications between the tangent and cotangent bundle which implicitly assume the presence of a background Riemannian metric (inner product)?

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For $f(x)$, you're basing your interpretation on the graph $y=f(x)$ and the slope of its tangent line at a point. This picture is drawn in a two-dimensional $xy$ coordinate system.

In order to find a corresponding interpretation for $F(x,y)$, you must think about the graph $z=F(x,y)$ and the slope(s) of its tangent plane at a point, so you must have a three-dimensional $xyz$ coordinate system in mind.

When you start talking about level curves for $F(x,y)$, which are curves in the $xy$ plane, it's not surprising that the interpretations don't seem to agree!

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    $\begingroup$ Sorry, I didn't mean to be rude, I just found the whole question so very strange. The derivative of a function $f \colon M \to N$ (where $M$ and $N$ are manifolds) at a point $p \in M$ is a linear map between tangent spaces, $dF_p \colon T_p M \to T_{f(p)} N$. That's all there is, if you want to say what the derivative “really” is. But this map can tell you different things, depending on what questions you ask it. I just didn't see what could possibly be confusing about that if you already know all that other advanced material. $\endgroup$ – Hans Lundmark Nov 9 '16 at 13:12
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    $\begingroup$ But maybe what you are after is that the partial derivatives (if $N=\mathbb{R}$) form the components of a one-form, not a vector. And if you want to interpret this one-form as a vector normal to the level sets you need a metric (or something like that) on $M$? $\endgroup$ – Hans Lundmark Nov 9 '16 at 13:14
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    $\begingroup$ When you're talking of the implicit function theorem, you mention partial derivatives and graphs of the hypersurface, but you're not really clear about what functions you are referring to. Say $F(x,y)=C$ defines $y$ as a function of $x$. In this context, the derivatives of $F$ don't have the same interpretation as the derivative of $y(x)$, although they are related, of course: $y'(x)=-F_x(x,y(x))/F_y(x,y(x))$. Might it be that you are somehow confusing these interpretations, and that this is what's bothering you? $\endgroup$ – Hans Lundmark Nov 9 '16 at 13:20
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    $\begingroup$ For $F \colon M \to \mathbb{R}$, the differential is $dF_x \colon T_x M \to T_{F(x)} \mathbb{R}$, which can be viewed simply as $dF_x \colon T_x M \to \mathbb{R}$, that is, a linear map taking tangent vectors to numbers. (And it depends smoothly on $x$ as well, if $F$ is smooth.) And that's exactly what a differential one-form on $M$ is. $\endgroup$ – Hans Lundmark Nov 10 '16 at 12:41
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    $\begingroup$ This one-form tells you the slope of the tangent plane of the graph $z=F(x)$ in $M \times \mathbb{R}$, namely how fast the value of $F$ is growing as you move in various directions in $T_x M$. And its nullspace also tells you the directions in which the value of $F$ (instantaneously) doesn't change, and these are the directions which are tangent to the level set of $F$ at the point $x$. So I think the various interpretations just amount to reading off information differently from one single linear map. $\endgroup$ – Hans Lundmark Nov 10 '16 at 12:49
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Both interpretations are correct. For simplicity and clarity, let's think of hypersurfaces in $\Bbb R^n$.

If you are working with an implicit description $F(x_1, \dots, x_n) = 0$ then $\left( \frac {\partial F} {\partial x_1}, \dots, \frac {\partial F} {\partial x_n} \right)$ is a vector orthogonal to the surface. At the same time, the kernel of $\Bbb d F$ gives the tangent space. For the sphere $x^2 + y^2 + z^2 = 1$, notice that $(2x, 2y, 2z)$ is a normal vector.

If, you are working with a parametric description (I include here implicit descriptions as a particular case) $(u_1, \dots, u_n) \mapsto f(u_1, \dots, u_n) \in \Bbb R^n$, then the vectors $\frac {\partial f} {\partial u_i}$ are tangent to your surface (they form a basis at each point). This is what you do with the function $x \mapsto x^2$: rewrite it as $x \mapsto (x, x^2) \in \Bbb R^2$ and notice that this is a parametrization of the graph; the derivative $(1, 2x)$ is a vector tangent to the curve.

Your question is slightly mistaken in that you force an alternative ("either tangent or normal") where, in fact, the concept of differentiability underlies both.

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    $\begingroup$ @William: I understand you, but I maintain that the question is not addressable. Not all mathematical facts have intuition behind them, not even all the geometrical facts (which, by their nature, are somewhat more visual). Some of them must be accepted as they are. The idea is that differential calculus is such a powerful and generic tool, that you may express plenty of things with it, including concepts that are apparently contradictory to each other. $\endgroup$ – Alex M. Nov 9 '16 at 13:06
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A curve $\gamma:\mathbb R\to M$ in a manifold has a velocity $\gamma'=v$ which is a tangent vector.

A function $f:M\to\mathbb R$ on the manifold has a gradient $df=\phi$ which is a cotangent vector.

The dot product / dual pairing / contraction between $v$ and $\phi$ (at the point $\gamma(0)\in M$) is given by the derivative of the functions' composition:

$$f\circ\gamma:\mathbb R\to\mathbb R$$

$$\phi\cdot v=df\cdot\gamma'=(f\circ\gamma)'(0)$$

This is another example of duality: a tangent vector can be considered a linear differential operator on functions, and a cotangent vector can be considered a differential operator on curves.


(Now, for concreteness, I'll say $M$ is 3-dimensional.)

The vector $v$ can be visualized (we all know) as a little arrow, or as a linear approximation to the parametrized curve $\gamma$.

The covector $\phi$ can be visualized as a linear approximation of the level sets of the function $f$; a set of parallel planes equally spaced, with the same spacing as the actual level sets at a certain point. It measures a vector by the number of planes it crosses. If $v$ is parallel to these planes, $\phi$'s measure of $v$ is $0$.

(Also, covectors can be added and multiplied visually: the wedge product of two covectors is the planes' intersection lines, and their sum is the diagonal planes through those lines.)


This picture of a covector could easily be confused with a bivector. It could also be conflated with a vector or a bicovector; they're all in 3-dimensional spaces, and naturally isomorphic if there's extra structure.

Using index notation, and basis vectors $e_i$ and covectors $\epsilon^i$,

$$\phi=\phi_i\epsilon^i$$

A preferred trivector ("pseudoscalar") $V$, or a volume-form $\Omega$ (mutually definable by $\Omega\cdot V=1$), produces a bivector:

$$B=V^{ijk}\phi_ie_j\wedge e_k$$

(visualized as a parallelogram in one of $\phi$'s planes).

A metric tensor $g$ or musical isomorphism $\sharp$ produces a vector:

$$n=\phi^\sharp=(g^{-1})^{ij}\phi_ie_j$$

(this would be the normal vector).

A metric and a trivector induce a Hodge dual, to produce a bicovector:

$$\chi=\star\phi=g_{ij}g_{kl}V^{ikm}\phi_m\epsilon^j\wedge\epsilon^l$$

(visualized as a lattice of parallel lines perpendicular to $\phi$'s planes).


So a derivative is, by default, a tangent or cotangent vector, not a normal vector unless some kind of $\sharp$ is available.

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    $\begingroup$ "...a tangent vector can be considered a linear differential operator on functions, and a cotangent vector can be considered a differential operator on curves." There seems to be a lot of depth of insight in this answer, which will take me a long time to reflect on before being able to fully appreciate. Thank you for taking so much time to give such a thoughtful response. $\endgroup$ – Chill2Macht Sep 17 '18 at 16:20

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