4
$\begingroup$

Let $f, g \in \mathbb R^n \rightarrow \mathbb R$. Is there some way to make sense of the expression:

$$\frac{\partial f(x_1, x_2, \dots, x_n)}{\partial g(x_1, x_2, \dots, x_n)}$$

I want some way to measure "how much $f$ changes along $g$" --- I'm not sure what a reasonable definition of this. Here is one that came to mind.

Let us define $\frac{\partial f}{\partial g}\big(t \big)$: At each point $t \in \mathbb R^n$, we first compute $g'(t) \in \mathbb R^n$. Now, we compute the directional derivative of $f$ along $g'(t)$:

$$ \frac{\partial f(x_1, x_2, \dots x_n)}{g(x_1, x_2, \dots x_n)} (t) : \mathbb R^n \rightarrow \mathbb R\equiv (\nabla_{g'(t)} f)(t) = \lim_{h \rightarrow 0} \frac{f(t + hg'(t)) - f(t)}{h}$$

  • Is this a well-known idea? If so, what is it called?
  • My problem with this is that it returns a scalar at each point: what I am actually interested in is to find a new function which tells me "how to move $f$ infinitesimally so we can make it closer to $g$.

The given definition above clearly generalizes to any manifold: all I need is a directional derivative, which I do possess on a manifold: Can we say something interesting about this in a larger setting?

$\endgroup$
4
$\begingroup$

The construction you described doesn't actually work on manifolds - $g'(t)$ (or $dg$ as we would normally write) is not a tangent vector, but a cotangent vector; so the directional derivative $\nabla_{dg} f$ doesn't make sense without a metric (or some other additional structure) to identify $TM$ with $T^* M$.

It also doesn't seem to have the right behaviour to be casually called "$\partial f / \partial g$" - for example, $\nabla_{dg} f$ would double if you doubled $g$, which is the opposite scaling behaviour to what the notation would suggest.

Remember that partial derivatives are only defined in terms of a whole coordinate system - if you're given the two coordinate systems $(x,y)$ and $(x,z=y-x),$ the expression $\partial f/\partial x$ will have different values depending on whether you're holding $y$ or $z$ fixed! Thus I'm unsure how to interpret the intended spirit of "$\partial f/\partial g$".

Is this a well-known idea? If so, what is it called?

Not commonly enough to have a name; but it's a simple enough expression that you'll find it cropping up in equations here and there in vector calculus (as $\nabla g \cdot \nabla f$) and Riemannian geometry (as $\nabla_{\operatorname{grad}g}f$). The interpretation isn't really close to what you're looking for, however - it's just "the rate of change of $f$ in the direction $\nabla g$ (or vice versa).

what I am actually interested in is to find a new function which tells me "how to move f infinitesimally so we can make it closer to g".

I'm unsure how to interpret this - to me, the way to "move" a scalar function is to deform it by another scalar function (e.g. deform $f$ to $f + \epsilon \phi$ for a parameter $\epsilon$), in which case what you're describing would just be the scalar $g - f$? If you could formalize what you're asking for here (or at least give more of a geometric description) there might be something more useful to say.

$\endgroup$
  • $\begingroup$ You're right; I'll make my question more precise and ask again --- thanks a lot for pointing out that it doesn't really behave very well. If you could expand your answer on where on might see $\nabla_{grad~ g} f$, I'd be very interested! $\endgroup$ – Siddharth Bhat Feb 26 '20 at 9:51
0
$\begingroup$

I think $\frac{\partial f(x_1, x_2, \dots, x_n)}{\partial g(x_1, x_2, \dots, x_n)}$ should be able to predict small changes in $f$, given small changes in $g$. Now, if all we're given is a small change $dg$, then that's just $ \nabla g\cdot \vec{dr} $. Even knowing $\nabla{g}$ at the point, we can't invert this dot product to get back $\vec{dr}$. Without knowing $\vec{dr}$, we can't predict the change in $f$, which is $\nabla f\cdot \vec{dr}$. So in this sense, the derivative is not define-able.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.