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lets say I have objects $f$ and $g$, for which one can define a derivative with the typical properties.

The product rule would be expected to be

$$ d(fg)=(df)g+f\,dg $$

But what if $f$ and $g$ are not commutative?

$$ [f,g]\neq 0\implies d(fg)=\text{?} $$

Does the product rule get modified?

Michael Hardy
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Anon21
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    You would have to define derivative in this space you are using – Anonmath101 Jun 27 '21 at 20:14
  • The rule you provided is generally correct even when $f$ and $g$ are not commutative. It would be wrong to say $d(fg) = f d(g) + g d(f)$, for example, because $g$ and $d(f)$ may not commute. To properly answer this question, I'd have to know what you mean by "typical properties" since usually, the product rule is such a typical property. – Mark Saving Jun 27 '21 at 20:14
  • @MarkSaving : Could I suggest writing $d(fg) = f,dg + g,df,$ with a thin space between $f$ and $dg$ and between $g$ and $df,$ just as one writes $f(x),dx$ rather than $f(x)dx$? – Michael Hardy Jun 27 '21 at 20:17
  • In at least some non-commutative contexts (e.g. matrices of functions of a single variable), we have things like $d(fg)=(\mathrm df)g+f,\mathrm dg$ (where the order of multiplication in things like $f,\mathrm dg$ matters). I would argue that this product rule is one of the "typical properties" (in fact, it's basically the definition of a derivation. Did you have a particular context in mind? – Mark S. Jun 27 '21 at 20:21
  • @MarkS. Yes when one creates a gauge covariant derivative, such as the ones used in physics. $D_\mu=\partial_\mu + iq A_\mu$. So I am thinking $D_\mu(fg)=?$, if $[f,g]\neq 0$. – Anon21 Jun 27 '21 at 20:29
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    That context would help refine your question greatly. Instead of "what happens with non-commutative stuff", it could be more like "what is the product rule for the noncommutative situation of a gauge covariant derivative?" – Mark S. Jun 27 '21 at 20:31

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