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Timeline for "Universal" differential identities

Current License: CC BY-SA 3.0

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May 17, 2017 at 14:05 history edited Michael Bächtold CC BY-SA 3.0
added comment to reflect the updated question
May 17, 2017 at 6:25 comment added Asaf Shachar The "Cofactor lemma" is a good non-trivial example of the kind of "combined operations" I want to discuss. For $d>2$ the cofactor of $df$ involves multiplication of partial derivatives. In fact I am also allowing for the $i$-th component of output to depend on partial derivatives of all components of $f$.
May 17, 2017 at 6:25 comment added Asaf Shachar $D_x \times D_y$ should take $f$ to $f_x \cdot f_y$, $D_x \circ D_y$ should take $f$ to $f_{xy} $, $(D_x \circ D_x) \times D_y$ should take $f$ to $f_{xx} f_y$ etc. (Here $f$ is a scalar function, to extend the operations to $\mathbb{R}^d$-valued maps, jut act on each component separately. I tried to elaborate more on this in my edit...
May 17, 2017 at 6:17 comment added Michael Bächtold I don't quite understand what you mean by multiplication vs composition. How does the multiplication of two partial derivatives act on a function?
May 16, 2017 at 17:18 comment added Asaf Shachar In fact, I am specifically interested whether the "cofactor identity" I mentioned in the question (divergence-free-rows) comes from the commutation of partial derivatives for $d>2$. (In $d>2$ multiplication is involved). So, we somehow need to consider all map $C^{\infty}(\mathbb{R}^n) \to C^{\infty}(\mathbb{R}^n)$ generated by the $D_i$ via composition, addition and multiplication... Can you think of the right algebraic setting for this? (Algebra? Module? this structure has $3$ operations...)
May 16, 2017 at 17:18 comment added Asaf Shachar Thanks, you are right. After thinking a bit more, it seems the "right" question that I meant to ask is a little different: Choosing the ring to be the one you specified (which was a just interpretation of my question) , I see that we missed something: multiplication. I want to allow relations of the form of $f_xf_y=f_yf_x$ (trivially true) or $f_{xx}=f_yf_{xy}$ (clearly false).
May 16, 2017 at 11:18 history answered Michael Bächtold CC BY-SA 3.0